A Note on Large Gauge Transformations in Double Field Theory

8202019 A Note on Large Gauge Transformations in Double Field Theory

httpslidepdfcomreaderfulla-note-on-large-gauge-transformations-in-double-field-theory 134

MIT-CTP4665

April 2015

Prepared for submission to JHEP

A note on large gauge transformations in double

field theory

Usman Naseer

Center for Theoretical Physics

Massachusetts Institute of Technology

Cambridge MA 02139 USA

E-mail unaseermitedu

Abstract We give a detailed proof of the conjecture by Hohm and Zwiebach in double

field theory This result implies that their proposal for large gauge transformations in terms of

the Jacobian matrix for coordinate transformations is as required equivalent to the standard

exponential map associated with the generalized Lie derivative along a suitable parameter



Contents

1 Introduction 1

2 Preliminaries 4

21 Review 4

22 Decomposition of G (ξ (ξ )) 8

3 Obtaining differential equations 11

31 Differential equation for F (λξ ) 11

32 Differential Equation for G (ξ (λξ )) 13

4 Solving for the parameter ξ (ξ ) 14

41 Comparison and an iterative method 1442 Order by Order checks 17

421 First and Second Order 17

422 Cubic Order 18

423 Quartic Order 19

424 Quintic Order 20

5 Determining ξ to all orders 22

6 Conclusions and Outlook 26

1 Introduction

Double field theory was developed to make manifest the O(D D) T-duality symmetry in

the low energy effective field theory limit of string theory In addition to the usual spacetime

coordinates lsquowindingrsquo coordinates are introduced The metric and the Kalb-Ramond two-form

are combined into a lsquogeneralized metricrsquo This generalized metric transforms linearly under

global O(D D) transformations Gauge transformations of the fields can also be written

in an O(D D) covariant form and they can be interpreted as the lsquogeneralized coordinate

transformationsrsquo in the doubled spacetime as discussed in [1] The action of double field

theory written in terms of the generalized metric is then manifestly invariant under these

transformations Double field theory is a restricted theory The so-called strong constraint

restricts the theory to live on a D-dimensional subspace of the doubled spacetime Different

ndash 1 ndash



solutions of the strong constraint are then related by T-duality Double field theory was

developed in [2ndash7] and earlier ideas can be found in [8ndash11] Further developments of double

field theory are discussed in [12ndash43] For recent reviews on this subject see [44] and [45]

The important issue of large gauge transformations in double field theory was discussed

in [1] where a formula for finite gauge transformations of fields has been proposed Thesetransformations are induced by generalized coordinate transformations Under the transfor-

mation from cordinates X M to coordinates X M fields are claimed to transform by the matrix

F given by

F N M =

1

2

partX P

partX M

partX P

partX N +

partX M

partX P

partX N

partX P

(11)

A generalized vector V M transforms as

V M (X ) = F N M V N (X ) (12)

or in index-free notation

V (X ) = F V (X ) (13)

and fields with more than one O(D D) index transform tensorially where each index is trans-

formed by F N M If the finite transformation of the coordinates is generated by a parameter

ξ M (X ) ie

X M

= eminusξP part P X M (14)

then F can be expressed in terms of this new parameter ie F = F (ξ )By considering an infinitesimal coordinate transformation X

M = X M minus ξ M [1] the

transformation rule (12) leads to the infinitesimal transformation of a generalized vector

which is given by the generalized Lie derivative Lξ as follows

V M (X ) = V M (X ) + ξ K part K V M (X ) +

part M ξ K minus part K ξ M

V K (X ) = V M (X ) +

LξV (X )

M

(15)

Generalized Lie derivatives define a Lie algebra [7] but the closure of the Lie algebra requires

the strong constraint

part P part P (middot middot middot) = 0 (16)

where lsquomiddot middot middot rsquo indicate arbitrary fields parameters or their products (so that part P part P A = 0 and

part P A part P B = 0 for any fields or parameters A and B) We can also realize large gauge

transformations by exponentiating the generalized Lie derivative Then as discussed in [1]

ndash 2 ndash



these transformations will form a group For a vector they are given by

V (X ) = eminusξP part P eLξ V (X ) equiv G (ξ ) V (X ) (17)

where we make a useful definition

G (ξ ) equiv eminusξP part P eLξ (18)

From (17) and (13) we would expect that

F (ξ ) = G (ξ ) (19)

but this is not true and this fact does not lead to any inconsistency The important question

is whether there exists a generalized coordinate transformation for which the corresponding

field transformation given by (11) equals G (ξ ) This is equivalent to asking whether it is

possible to find a parameter ξ (ξ ) such that F (ξ ) = G (ξ (ξ )) This is precisely the question

we address in this paper and the answer is positive Consistency requirements lead to the

condition that ξ and ξ can only differ by a quasi-trivial parameter

ξ P (ξ ) = ξ P +

i

ρi(ξ )part P ηi(ξ ) (110)

where we have defined a quasi-trivial parameter to have the form

i ρi(ξ )part P ηi(ξ ) where the

free index is carried by a derivative We will also refer to the parameters of the form part M f as

lsquotrivialrsquo parameters Note that every trivial parameter is also quasi-trivial It is also easy to

see using the strong constraint that Lξ is zero for a trivial parameter and it is not zero in

general for a quasi-trivial parameter

These insights were used in [1] to determine such ξ (ξ ) to cubic order in ξ We also note

a small difference in notation and approach as compared to [1] where parameters ξ and Θ (ξ )

are related by G (ξ ) = F (Θ (ξ )) Since Θ (ξ ) is just a sum of ξ and higher order terms in ξ

this relation can be inverted to obtain ξ (Θ) With this in mind the conjecture by Hohm and

Zwiebach (HZ conjecture) can be stated as

HZ conjecture For every parameter ξ (X ) there exists a parameter ξ (ξ ) such that ξ M (ξ ) =

ξ M + δ M (ξ ) where δ M (ξ ) is quasi-trivial and

G

ξ (ξ )

= F (ξ ) (111)

The main aim of this paper is to prove this conjecture by giving a procedure which can

be used to obtain ξ (ξ ) to all orders in ξ We will do this by finding a parameter ξ (ξ ) such

ndash 3 ndash



that G (ξ (λξ )) and F (λξ ) satisfy the same first order differential equation in λ and the same

initial condition at λ = 0 By the uniqueness of the solution we can then deduce that

F (λξ ) = G (ξ (λξ )) (112)

and by setting λ = 1 we get (111)

The HZ conjecture was examined explicitly in [46] see also [47] and [48] for other

relevant discussion In [46] Berman Perry and Cederwall considered this issue in the context

of equivalence classes of transformations These equivalence classes were defined modulo the

so-called lsquonon-translatingrsquo transformations which are the transformations generated by the

quasi-trivial parameters These transformations differ from the identity by nilpotent matrices

Important progress was made by showing that F (ξ ) and G (ξ ) belong to the same equivalence

class but the issue of the existence of ξ (ξ ) such that (111) holds was not addressed completely

Our results in section (22) play a key role in extending the results of [ 46] to prove HZ

conjecture Our construction here is more straightforward and one can in fact show without

finding the explicit form of ξ (ξ ) that there exists a parameter ξ (ξ ) such that (111) holds

This paper is organized as follows In section 21 we review the basics of gauge trans-

formations in double field theory We also review how the consistency requirements for the

transformations of the scalar field constrain the form of ξ (ξ ) In section 22 we show that

G (ξ (ξ )) can be written as the product G (ξ ) G (χ (ξ )) where χ (ξ ) is a quasi-trivial parameter

Further we show how χ (ξ ) can be used to find ξ (ξ ) Differential equations for F (λξ ) and

G (ξ (λξ )) are derived in sections 31 and 32 respectively By comparing the two differential

equations we obtain an iterative relation (46) In section 41 we give a procedure to use this

relation systematically to obtain ξ (ξ ) to all orders in ξ proving the HZ conjecture We useour procedure to compute ξ (ξ ) up to quintic order in section 42 Up to quartic order the

results are found to be in agreement with [1] For the quintic order the result is checked by

explicit computation using Mathematica In section 5 we give an explicit formula for χ (ξ ) to

all orders in ξ We also discuss the connection between our results and those in [46] in section

5 In section 6 we discuss the implications of our results on the composition of coordinate

transformations in double field theory Finally we comment on finite gauge transformations

in exceptional field theory

2 Preliminaries

21 Review

In this section we review the basics of gauge transformations in DFT and introduce

notation and conventions required to derive the relevant differential equations Most of these

relations were first derived in [1] and we present them here without proof

ndash 4 ndash



Consider a finite coordinate transformation which is generated by a parameter ξ M ie

X M = X M minus ζ M (ξ ) equiv

eminusξP part P

X M (21)

or in a more transparent index free notation

X = X minus ζ (ξ ) = eminusξP part P X (22)

Using the definition of F as in (11) we can express it in terms of the generating parameter

ξ M as follows

F (ξ ) = 1

2(eminusξeξ+aeminusξeξminusat + eminusξeξminusateminusξeξ+a) (23)

Here a N M equiv part M ξ N and ξ appearing on the RHS is to be understood as the differential

operator ξ middot

f equiv

ξ

P

part P f This expression forF

may look like a differential operator but it isa matrix function as demonstrated in section 41 of [1] The proof essentially follows from the

fact that eminusξ eξ+a is not a differential operator and it can be written in the following form

which makes its matrix nature manifest

eminusξeξ+a =1 middot eminus

larrminusξ +a

(24)

where we have introduced the notation

β larrminusξ Λ equiv

ξ P part P β

Λ (25)

Note that we can also write G (ξ ) in a similar manner as

G (ξ ) = eminusξe Lξ = eminusξeξ+aminusat =

1 middot eminus

larrminusξ +aminusat

(26)

We make the useful definition

A N M (ξ ) equiv part M ζ N (ξ ) = minus

infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)

where ζ N (ξ ) is read from equation (21) Further we make the definition

B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1

1 minus A(ξ ) = eminusξeξ+a =

1 middot eminus

larrminusξ +a

(29)

The equalities in the last line follow from equation (22) and the definition of A as in equation(27) Details can be found in section 41 of [1] Using these definitions F (ξ ) can be written

in terms of the matrix B(ξ ) We suppress the explicit ξ dependence here and write

F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)

Now we turn our attention to the finite transformations realized as the exponentiation

of the generalized Lie derivative As mentioned previously the infinitesimal transformation of

a vector field V (X ) is given by the action of the generalized Lie derivative ie

V (X ) = V (X ) + LξV (X ) (211)

with

LξV M (X ) = ξ K part K V M (X ) +


V K (X ) (212)

or in the index free notation

LξV (X ) = ξ middot V (X ) +

a minus at

V (X ) with a N

M = part M ξ N (213)

Generalized Lie derivatives define a Lie algebra under commutation We have Lξ1 Lξ2

= L[ξ1 ξ2]C

(214)

where [middot middot]C is the C-Bracket defined as follows

[ξ 1 ξ 2]M C = ξ P

1 part M ξ P 2 minus ξ P

2 part P ξ M 1 minus

1

2

ξ P 1 part M ξ 2P minus ξ P

2 part M ξ 1P

(215)

The Jacobi identity also holds for the commutation relation (214) because the Jacobiator of

three parameters ξ 1 ξ 2 ξ 3 is a trivial parameter and the generalized Lie derivative of a trivial

parameter vanishes [6 7] This allows us to realize finite transformations which form a groupby exponentiation of the generalized Lie derivative

V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf

(X ) = f (X ) [1] we can write

V (X ) = eξV (X ) = eξ F V (X ) (217)

If we naiumlvely compare (216) and (217) we get the equality in (19) However it can be

seen easily that the equality holds only for a very special class of parameters the quasi-trivial

parameters For an arbitrary parameter it holds only up to the second order in the parameter

However this does not lead to any inconsistency as argued in [ 1] Due to the strong constraint

double field theory allows some extra freedom This freedom can be exploited to change the

parameter so that the transformation works out for the vector field Transformation properties

of the scalar field put some restrictions on the nature of the modified parameter By definition

under any coordinate transformation X rarr X a scalar field φ(X ) transforms as

φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)

We now see how the finite transformation of the scalar field is realized by exponentiating

the generalized Lie derivative Then we compare it with the transformation rule (219) The

action of the generalized Lie derivative on a scalar field φ(X ) is given by

Lξφ(X ) = ξ P part P φ(X ) (220)

Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)

which is consistent with (219) We want the modified parameter ξ (ξ ) to preserve this con-

sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)

This condition can be satisfied if the following holds

ξ P part P (φ(X )) = ξ P part P (φ(X )) (223)

It is possible to satisfy the above requirement if ξ (ξ ) and ξ differ by a quasi-trivial parameter

ndash 7 ndash



as defined earlier and we aim to develop a procedure which determines such ξ (ξ ) to any

desired order in ξ

22 Decomposition of G (ξ (ξ ))

In this section we will prove important results which will help us in deriving a differentialequation for G (ξ (λξ )) in section 32 We will demonstrate how G (ξ (ξ )) can be written as a

product of G (ξ ) and G (χ(ξ )) where χ (ξ ) is a quasi-trivial parameter We will also derive an

expression for ξ (ξ ) in terms of ξ and χ (ξ )

For a quasi-trivial parameter δ we can use the strong constraint to obtain

G (δ ) = eminusδe Lδ = e

Lδ (224)

Note that due to the strong constraint we have

Lδ = δ P part P + b minus bt = b minus bt where

b N M equiv part M δ N Both indices of b are carried by derivatives therefore b bt and b minus bt are all

nilpotent matrices and we can simplify equation (224) to get

G (δ ) = ebminusbt = 1 + b minus bt (225)

The relations (224) and (225) for a quasi-trivial parameter will be used extensively in the rest

of this paper Let us consider a collection of quasi-trivial parameters δ 1 δ 2 middot middot middot and define

(bi) N M equiv part M δ N

i Using the strong constraint it is easy to see that the following identities

hold

bib j = bibt j = bt

ib j = btibt

j = 0 forall i j (226)

Now we consider the product G (δ 1) G (δ 2) middot middot middot By using equation (225) this product can bewritten as follows

G (δ 1) G (δ 2) middot middot middot =

1 + b1 minus bt1

1 + b2 minus bt

2

middot middot middot (227)

= 1 + b1 minus bt1 + b2 minus b2 + middot middot middot (228)

= eb1+b2+middotmiddotmiddotminus(bt1+bt2+middotmiddotmiddot) (229)

Since all δ irsquos are quasi-trivial the relation (224) implies that

G (δ 1) G (δ 2) middot middot middot = G (δ 1 + δ 2 + middot middot middot) (230)

The equation (230) will be useful in establishing our main result for this section

Theorem 21 We have ξ = ξ + δ with δ quasi-trivial if and only if

G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash



with χ also quasi-trivial Moreover

δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C

+ middot middot middot (232)

up to the addition of trivial parameters Here [ ]C is the C-Bracket and lsquo middot middot middot

rsquo contains nested C-brackets between ξ and χ given by the BCH formula

Proof The proof consists of two parts In the first part we assume that ξ and ξ differ by

a quasi-trivial parameter and show that this leads to the decomposition (231) Using the

definition (18) we can write

G

ξ

= eminusξ e Lξ = eminusξminusδ e

Lξ+δ = eminusξ e Lξ + Lδ (233)

where the last equality follows due to the strong constraint We make use of the lsquoZassenhausrsquo

formula [49] which can be written as follows

eX +Y = eX eY ige2

eC i(XY[middotmiddot]) (234)

where C i (XY [middot middot]) is the ith order Lie polynomial in X and Y The precise form of

C i (XY [middot middot]) is not essential for our arguments here However the fact that it only involves

commutators between X and Y is crucial C i (XY [middot middot]) can be determined as explained in

[49] For example C 2 (XY [middot middot]) = minus 12 [X Y ] and a higher order C i (XY [middot middot]) involves i

nested commutators between X and Y So we have

G ξ = eminusξ e Lξ e Lδige2

eC i( Lξ Lδ[middotmiddot]) (235)

From the Lie algebra of the generalized Lie derivatives we know that Lξ Lδ

= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L

C i(ξδ[middotmiddot]c) (237)

so we can write G (ξ ) as follows

G

ξ

= eminusξ e Lξ e

Lδ ige2

e LC i(ξδ[middotmiddot]c) (238)

ndash 9 ndash



It was shown in section 51 of [1] that the C-bracket involving a quasi-trivial parameter is also

quasi-trivial This means that δ as well as all of the C i (ξδ [middot middot]c) are quasi-trivial and we can

use the equation (224) to write

G ξ = G (ξ ) G (δ )

ige2G (C i (ξδ [middot middot]c)) (239)

Now the result in equation (230) implies that

G

ξ

= G (ξ ) G

δ +ige2

C i (ξδ [middot middot]c)

(240)

equiv G (ξ ) G (χ) (241)

In the last step we have defined χ equiv δ +

ige2 C i (ξδ [middot middot]c) which is clearly quasi-trivial

This completes the first part of the proof

Now we turn to the second part of the proof We assume that the composition in (231)holds and show that this leads to an expression for ξ which differs from ξ by a quasi-trivial

parameter ie we show that δ is quasi-trivial From the definition (18) the product G (ξ )G (χ)

can be written as

G (ξ )G (χ) = eminusξe Lξeminusχe

Lχ = eminusξe Lξe

Lχ (242)

where the second equality follows from the strong constraint Now we use the well known

Baker-Campbell-Hausdorff formula to write

e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ


where lsquomiddot middot middot rsquo contains further nested commutators between Lξ and Lχ From the defining

relation of the Lie algebra of the generalized Lie derivatives (214) we deduce that

ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c


up to addition of trivial parameters Here lsquomiddot middot middot rsquo contains further nested C-brackets between ξ

and χ We also note that in the above expression ξ and ξ differ by a quasi-trivial parameter

because χ and C-brackets involving χ are all quasi-trivial Using this result we can write

G (ξ )G (χ) = eminusξe Lξ (246)

ndash 10 ndash



Due to the strong constraint ξ P part P = ξ P part P and we have

G (ξ )G (χ) = eminusξe Lξ = G

ξ

(247)

with ξ = ξ + δ δ can be read off from (245) and matches with the expression in equation

(232) up to addition of trivial parameters This completes the proof of theorem ( 22)

We note that in the context of the conjecture (111) both δ and χ should be understood

as functions of ξ however results of this section hold in general We conclude this section by

simplifying G (ξ ) further Since χ is a quasi-trivial parameter we can use relation (225) to

write

G (χ) = 1 + ∆ minus ∆t with ∆ N M equiv part M χ

N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)

We mentioned in the introduction that the conjecture (111) will be proven by finding

the parameter ξ (ξ ) Equation (249) implies that our problem of finding ξ (ξ ) has now reduced

to finding ∆(ξ ) or χ(ξ ) If we can find a quasi-trivial χ (ξ ) then the theorem proven in this

section would guarantee that the HZ conjecture (111) holds

3 Obtaining differential equations

31 Differential equation for F (λξ )

In this section we will derive a differential equation for F To do this we introduce a λ

dependence in the coordinate transformations by letting ξ rarr λξ so that we have

X M

= X M minus ζ M (λξ ) equiv eminusλξP part P X M (31)

where the last equality defines ζ M as a function of λ Consequently the matrices A and B

defined earlier also become functions of λ and we write

A N

M (λ) equiv part M ζ N

(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e

λminuslarrminusξ +a

(33)

ndash 11 ndash



Here and in the rest of the paper the matrix a is to be understood as a N M = part M ξ N ie

without the λ dependence because it has already been taken into account The matrix F also

becomes λ-dependent and we can express it as follows

F (λξ ) =

1

2 B(λ) Bt

(λ)minus1 + B(λ)tminus1 B(λ) (34)

Now we are in a position to derive a first order differential equation for F (λξ ) in λ For

brevity explicit λ dependence is suppressed in this section From the definition of B it is

easy to see that B(λ) and

Btminus1

(λ) satisfy the differential equations

dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1

minuslarrminusξ minus at

(35)

Straightforward applications of the chain rule then yields

ddλB Btminus1 = B Btminus1 minuslarrminusξ + B a Btminus1 + B Btminus1 (a minus at) minus B Btminus1 a (36)

d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)

Now we make use of the following identities which are a consequence of the strong constraint

atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)

Using these we can simplify the above differential equations and combine them to get a

differential equation for F (λ)

d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus

larrminusξ + a minus at

+

1

2

B a

Btminus1

minus Ba + Bat minus at

(310)

We can also write Btminus1

=

1

1 minus A

tminus1

= 1 minus At (311)

such that

Ba Btminus1 = Ba(1 minus At) = Ba minus BaAt (312)

Bat minus at = B(at minus Bminus1at) = B(at minus (1 minus A)at) = B Aat (313)

ndash 12 ndash



Plugging these relations into (310) we finally arrive at a differential equation for F (λ)

d

dλF = F (minus

larrminusξ + a minus at) +

1

2 B

minusaAt + Aat

(314)

This equation will play an important role in our analysis in rest of the paper We will laterobtain a differential equation for G which has a very similar form We also re-write this

differential equation in a more visually appealing form by using the strong constraint By the

strong constraint Btminus1A = A and

Btminus1

a = a This allows us to write

F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2

B(Aat minus aAt) + (B +

Btminus1

minus 1)(Aat minus aAt)

(316)

= 1

2 B(Aat minus aAt) + Aat minus Aat + B(Aat minus aAt) minus Aat + aAt

(317)

= B Aat minus aAt (318)

Using this result in (314) we get

d

dλF = F

minus

larrminusξ + a minus at +

1

2Aat minus

1

2aAt

(319)

There appears to be no systematic way of solving the differential equation (319) to

obtain a closed form expression for F (λξ )

32 Differential Equation for G (ξ (λξ ))

In this section we derive a first order differential equation for G (ξ (λ)) by a procedure

which is similar to the one used in section (31) We introduce a parameter dependence by

ξ rarr λξ Doing this makes ξ (ξ ) χ(ξ ) ∆(ξ ) and hence G (ξ (ξ )) λ-dependent We use the

theorem (21) and write the relations (231) and (248) with a λ-dependence

G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)

where we also made a slight change of notation by defining G(λ) Now we are in a position

to obtain a differential equation for G(λ) We will suppress the explicit λ dependence from

here on Using the strong constraint it is easy to show that the following identities hold

at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash



Since these identities are a consequence of the strong constraint and the index structure of the

matrices ∆ and ∆t they also hold for their derivatives d∆dλ and d∆t

dλ Now we differentiateG (λ) with respect to λ to get

dG

dλ = dG

dλ 1 + ∆ minus ∆t+ G ddλ ∆ minus ∆t (323)

By letting ξ rarr λξ in equation(26) we see that

d

dλG = G

minus


(324)

Using this and the strong constraint identities recorded above we obtain

d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)

We want to put equation(325) in a form similar to (314) To do that we need to computeG

minus


After a straightforward but somewhat tedious computation one finds that

G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)

Using the last result in equation (325) we get the desired differential equation

d

G

dλ

= G minuslarrminusξ + a minus at+ Bξ + a +

d

dλ∆ minus ∆t+ ∆ minus ∆t at (327)

Note that this equation has very similar form as the differential equation for F (λ) (314)

Now our goal is to find ∆ such that the two equations are exactly the same This is the

subject of the next section

4 Solving for the parameter ξ (ξ )

41 Comparison and an iterative method

It is easy to see that F (λξ ) and G (ξ (λξ )) satisfy the same initial condition at λ = 0

ie F (0) = G (ξ (0)) = 1 For F (λξ ) to be equal to G (ξ (λξ )) they must satisfy the same

differential equation We compare the two differential equations (314) and (327) to getξ + a +

d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash



Without loss of generality we take ∆(λ) to be of the following form

∆(λ) =infin

n=1

λn∆n (42)

where ∆n is a λ-independent term which is nth order in ξ Also it is obvious that we canrsquot

have a zeroth order term in ∆ Following the same notation we write

A(λ) equiv

infinn=1

λnAn (43)

where An can be read off from (31) and (32) It takes the following form

(An) N M = part M

ζ N (n)

(44)

Here ζ N (n) is the nth order term in ξ We can read off ζ N (n) from equation (27) and

ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)

Now from (41) we get

(n + 1)

∆n+1 minus ∆tn+1

+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)

The last expression is an iterative equation for ∆n minus ∆tn There is a systematic procedure

which can be used to obtain ξ to all orders Before describing our procedure we want toshow that in general ∆n minus ∆t

n obtained from equation (46) will be of the form

∆n minus ∆t

n

N

M = part M f part N g minus part M gpart N f (47)

where f and g are ξ -dependent quantities and they are of the form f sim ξ k middot ξ P and g sim

ξ nminuskminus2 middot ξ P where the powers are chosen so that the whole term is n th order in ξ

To show this first note that from equation (46) it follows trivially that ∆1 minus ∆t1 =

∆2minus∆t2 = 0 which can certainly be written in the form (47) Now we use inductive argument

and suppose that ∆n minus ∆tn is of the form (47) ie

∆n minus ∆tn sim part M

ξ k middot ξ P

part N

ξ nminuskminus2 middot ξ P

minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash



then a short calculation yields

(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)

sim part M ξ k+1 middot ξ P

part N


minus part M ξ nminuskminus2 middot ξ P

part N

ξ k+1 middot ξ P

(410)

+ part M ξ k middot ξ P part N ξ nminuskminus1 middot ξ P minus part M ξ nminuskminus1 middot ξ P part N ξ k middot ξ P (411)

sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)

which is of the form (47) We can also read off the RHS of equation (46) and see that

Anat minus aAtn sim part M

ξ nminus1 middot ξ P

part N ξ P minus part M ξ P part N


(413)

This is also of the form (47) Since all the terms in (46) are of the form (47) we deduce

that ∆n+1 minus ∆tn+1 will also be of the form (47) and this completes our inductive argument

In the following we outline our procedure to obtain ξ (ξ )

bull Using the fact that ∆0 minus ∆t0 = 0 the equation (46) can be solved iteratively to find all

∆n minus ∆tn which will be of the form (47)

bull Only ∆n minus ∆tn can be found uniquely from equation ( 46) and there are many different

choices for ∆n which differ from one another by symmetric matrices We can choose

(∆n) N M = part M

f part N g

(414)

bull

Using (42) one can obtain ∆(λξ ) to all orders in ξ and by setting λ = 1 we get ∆(ξ )

bull Since all ∆nrsquos are total derivatives ∆(ξ ) will also be a total derivative therefore

∆ N M = part M χ

N (415)

can be solved to obtain an expression for χ(ξ ) Note that χ (ξ ) is determined only up

to the addition of trivial terms

bull By using (232) one can find δ (ξ ) to all orders in ξ Since ξ (ξ ) = ξ + δ (ξ ) we can

finally obtain ξ (ξ ) to all orders in ξ and this was our aim

We also note that this procedure does not give a unique ξ and it is determined only up to the

addition of trivial parameters This non uniqueness arises precisely because of the fact that

the generalized Lie derivative with respect to a trivial parameter is zero

ndash 16 ndash



42 Order by Order checks

In this section we demonstrate how the procedure outlined above can be used system-

atically to obtain ξ (ξ ) to quintic order in ξ Recall that ξ (ξ ) = ξ + δ (ξ ) and we only need

to find out δ (ξ ) We will show that our results here are in agreement with the computation of

δ (ξ ) done previously in [1] to quartic order The quintic order result is also verified explicitly

Before proceeding further we would introduce some useful notations here

bull We will represent matrices of the form lsquof M gN rsquo in an index free manner as follows

(B) N M = f M g

N equiv f g (416)

ie in an expression the terms carrying free matrix indices are represented with an

overhead arrow

bull

If the free vector index of a parameter is carried by a partial derivative ie V M

=A part M B we will denote it as follows

V = A part B (417)

bull We will often use the following short hand notation

A minus At equiv

A minus (middot middot middot )t

(418)

bull We will use χn to denote the O(ξ n) term appearing in χ(ξ ) ie

χ(ξ ) equiv χ1 + χ2 + χ3 + middot middot middot (419)

It is then trivial to see that (∆n) N M = part M χ

N n

bull We will compute δ (ξ ) upto the quintic order In the analysis to follow we will also find

that χ1 = χ2 = 0 Using this and the equation (232) we can write ξ (ξ ) up to quintic

order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)

421 First and Second Order

First and second order checks are trivial By expanding G (ξ ) and F (ξ ) to second order it

can bee seen that they are equal We will obtain the same conclusion by using our procedure

ndash 17 ndash



Using the fact that A does not have a zeroth order term ie A0 vanishes it is trivial to

see that ∆1minus∆t1 = 0 from (46) Without loss of generality we can choose ∆1 = 0 rArr χ1 = 0

this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)

For the second order case we see that for n = 1 equation (46) becomes

2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)

From (44) we see that A1 = a this implies that ∆2 minus ∆t2 = 0 Without loss of generality we

have ∆2 = 0 rArr χ2 = 0 which implies

δ M (ξ ) = 0 + O(ξ 3) (423)

This shows that F (ξ ) and G (ξ ) are equal up to quadratic order in ξ in agreement with [1]

422 Cubic Order

First and quadratic order computations were trivial in the sense that we found δ (ξ ) =

0 + O(ξ 3) Things get more interesting at the cubic order as we will get non-trivial results for

δ (ξ ) and hence for ξ (ξ )

bull For n = 2 from (46) we get

3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)

A2 can be read off from equation (44) After a short computation we obtain the follow-

ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P

part N ξ P minus part M ξ P part N (ξ middot ξ P )

(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)

where in the last step we added a symmetric part to the both terms so that they can

written as total derivatives Now we use the notation introduced at the beginning of

this subsection to write

A2at minus aAt2 = minus

1

2

part

(ξ middot ξ P ) partξ P

minus (middot middot middot )t

(427)

ndash 18 ndash



Using this in equation (424) we get

∆3 minus ∆t3 = minus

1

12

part



(428)

bull Now we can choose

∆3 = minus 1

12 part


rArr χ3 = minus

1

12(ξ middot ξ P ) partξ P (429)

bull From equation (420) we can write down δ (ξ ) to cubic order in ξ as follows

δ M (ξ ) = minus 1

12(ξ middot ξ P )part M ξ P + O(ξ 4) (430)

So we deduce that

F (ξ ) = G ξ minus 1

12

(ξ middot ξ P ) partξ P + O(ξ 4) (431)

This relationship can be inverted easily to get

F

ξ +

1

12(ξ middot ξ P ) partξ P

= G (ξ ) + O(ξ 4) (432)

This is in complete agreement with results of section(4) of [1]

423 Quartic Order

Now we turn to compute δ (ξ ) to quartic order in ξ I n [1] it was shown by explicit

computation that there is no quartic order term in δ (ξ ) We will see that our procedure

reproduces this result through very non trivial cancellation between quartic order terms

bull For n = 3 the equation (46) becomes

4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)

Let us compute different parts of the above equation now All the computations are

fairly straightforward and the final results are presented here

A3at minus aAt3 =

1

6 part (ξ 2 middot ξ P ) partξ P minus (middot middot middot )t (434)

(ξ + a) ∆3 = part

ξ 2 middot ξ K partξ K + part

ξ middot ξ K part (ξ middot ξ K ) minus part

ξ middot ξ K partξ P part P ξ K (435)

(ξ + a) ∆t3 = part (ξ middot ξ K ) part

ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K

minus partξ K partξ P part P

ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K

partξ P part P ξ K minus partξ K

partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash



We can determine ∆4 minus ∆t4 by using these results in equation (433)

∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K

minus (middot middot middot)t

(438)

bull Again without loss of generality we can choose

∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)

Note that we have a non trivial quartic order term

bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)

the C-bracket [ξ χ3]C can be computed easily by using the value of χ3 given in equation

(429) and the definition of the C-bracket (215) After a straightforward computation

and some index relabeling it is easy to show that

1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P

partξ P = minusχ4 (441)

Hence we see that the quartic order contributions from χ4 and 12 [ξ χ3]c cancel each

other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)

which agrees with the quartic order result of [1]

424 Quintic Order

Now we turn to the quintic order computation for ξ (ξ ) In contrast with the previous

orders this computation has not been done before Here we will determine ξ (ξ ) to fifth order

in ξ and then check our result by explicitly expanding F (ξ ) and G (ξ (ξ )) to the relevant order


5 ∆5 minus ∆t5+ (ξ + a) ∆4 minus ∆t

4+ ∆4 minus ∆t4 at =

1

2 A4at minus aAt4 (443)

This equation can be solved for ∆5 minus ∆t5 by similar kind of computation as was done

ndash 20 ndash



for the quartic order Key results are summarized below

(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P

part ξ Q part Q ξ P

(444)

minus (ξ + a) ∆t4 = minus

1

24 part ξ middot ξ P part ξ 2 middot ξ P + part ξ P part ξ 3 middot ξ P +

1

24 minus

part ξ P part ξ Q part Q

ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q

part P ξ Q part ξ P minus part ξ Q part P

ξ 2 middot ξ Q

part ξ P

(446)

A4 at minus a At4 = minus

1

24

part

ξ 3 middot ξ P

part ξ P minus part ξ P part

ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P

part (ξ middot ξ P ) minus (middot middot middot)t

(448)

bull Now we follow the familiar procedure and make a choice of ∆5 which is consistent with

the result found for ∆5 minus ∆t

5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P

part (ξ middot ξ P ) minus

ξ middot ξ P

part

ξ 2 middot ξ P

(450)

bull From ∆5 we can read off

χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)

bull We use the fact that 12 [ξ χ3]c = minusχ4 in equation (420) to get

δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)

After a straightforward computation one finds

[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)

Notice that all three terms appearing in [ξ χ4]c are also present in χ5 These terms add

1 A straightforward choice would have been to take

∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ

P part (ξ middot ξP )

(449)

This choice is perfectly fine but leads to an expression for δ (ξ) which is rather cumbersome

ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P

part M (ξ middot ξ P )

minus ξ middot ξ P part

M ξ 2

middot ξ P + O ξ 6 (454)

It can be verified explicitly that this value of δ (ξ ) indeed satisfies F (ξ ) = G (ξ (ξ )) up to quintic

order It was done using symbolic manipulations in Mathematica Let ξ M (n) denote the nth

order term in ξ M Then the fifth order contribution in G (ξ ) due to ξ M (5) is simply given by

part M ξ N (5) minus part N ξ (5)M equiv ∆ N

M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

To show that F (ξ ) = G ξ + ξ (3) + ξ (5) + O(ξ 6) we first expanded F (ξ ) and G ξ + ξ (3)to the quintic order Then we computed the difference F (ξ ) minus G ξ + ξ (3) up to quintic order

and noticed that this difference equals ∆

5 Determining ξ to all orders

From our results in the last section we have a systematic way of finding ξ (ξ ) to any

desired order in ξ This really involves two key steps first we find χ (ξ ) using the iterative

equation (46) and then we use the BCH formula (232) to obtain ξ (ξ ) In this section we

complete the first step of this computation to all orders in ξ ie we solve for χ (ξ ) to all orders

From equation (46) only ∆n minus ∆t

n can be obtained uniquely Here we obtain a formula for∆n minus ∆t

n for any integer n Then ∆ (ξ ) and χ (ξ ) can be obtained using their definitions

As argued earlier χ (ξ ) is determined only up to the addition of trivial parameters This

ambiguity in χ (ξ ) is not important because of the fact that the generalized Lie derivative

with respect to a trivial parameter is zero

From our previous analysis we expect ∆n to be an nth order matrix function of ξ whose

free indices are carried by the partial derivatives Then it is easy to see the most general form

for ∆n minus ∆tn is the following

∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P

part ξ nminuskminus2

middot ξ P minus (middot middot middot)t (51)

where we have used the notation A minus At =

A minus (middot middot middot)t

and αnk is some n and k dependent

coefficient which we want to find so that the ansatz ( 51) satisfies equation (46) We record

ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)

Now let us compute different components of the above equation for the ansatz ( 51) Bystraightforward computations one obtains the following

(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)

Now we add the last two results along with the term (n + 1)


to obtain the

LHS of equation (52) After a straightforward relabeling of the summation index k we obtain

LHS =

k=nminus1k=1

((n + 1) αn+1k + αnk + αnkminus1) part ξ k middot ξ P part ξ nminus

kminus1 middot ξ P minus (middot middot middot)

t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)

The RHS of equation (52) can be written by using the definition of An

RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)

Let us write αnk as product of two factors

αnk = 12

(minus1)n

n times β nk (57)

where β nk is some other n and k dependent function For LHS to be equal to the RHS we

ndash 23 ndash



get following conditions on β nk

minusβ n+10 + β n0 + β nnminus1 = 1 (58)

minusβ n+1k + β nk + β nkminus1 = 0middot (59)

Let us focus on the second condition first and write it as β n+1k = β nk + β nkminus1 this is just

the lsquoPascalrsquos rulersquo in combinatorics if we choose β nk =

nminus1k

Now for the first condition

minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)

so it is also satisfied In summary we deduce that

∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)

satisfy equation (52) for any n One can check that this formula agrees with the results

obtained in the last section up to quintic order From this formula we can make a choice of

∆n and hence ∆ =

n∆n can be determined From ∆ one can then find χ (ξ ) Using the

procedure we described in the last section one particular choice of χ (ξ ) is

χM (ξ ) =k=nminus2

k=0

αnk

ξ k middot ξ P

part M


middot (512)

Once we have χ (ξ ) we can use (232) to obtain ξ (ξ ) to any order in ξ

In [46] Berman et al considered the HZ conjecture in the context of equivalence classes

of finite transformations Equivalence classes were defined modulo the lsquonon-translatingrsquo trans-

formations These are the transformations which differ from the identity by a nilpotent matrix

Let F (ξ ) be a representative of this equivalence class then it can be formally defined as follows

[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)

Then it was shown that lsquoG minus1(ξ ) F (ξ )rsquo belongs to the same equivalence class as the identity

An expression for lsquoG minus1(ξ ) F (ξ )rsquo was computed explicitly which can be written in the following

form

G minus1(ξ ) F (ξ ) =infin

n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash



where κnk is an n and k dependent coefficient and M nk is a nil-potent matrix given by

κnk = 1

2

(minus1)n (n minus 2k minus 1)

(n + 1) (k + 1) (n minus k) and M nk = part

ξ k middot ξ P

part


(515)

We want to compare the result (514) with our analysis here Since we have shownthat F (ξ ) = G (ξ (ξ )) = G (ξ ) G (χ (ξ )) we expect the RHS of (514) to agree with our

expression for G (χ (ξ )) In the following we will show that this is indeed the case

We use the definition of G (χ (ξ )) = 1 + ∆ minus ∆t and our result in equation (511) We

also recall that ∆n minus ∆tn = 0 for n le 2 Using these facts we can write

G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)

The two terms in the sum can be combined by replacing the dummy index k by n minus k minus 2

G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

(αnk minus αnnminuskminus2)

part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0

(αnk minus αnnminuskminus2) M nminus1k (518)

where we have used the definition of M nk from equation (515) We now replace the dummy

index n by n + 1 to get

G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0

(αn+1k minus αn+1nminuskminus1) M nk (519)

Since the M nk are nil-potent matrices we can write

G (χ (ξ )) =infin

n=2

nminus1k=0

1 + (αn+1k minus αn+1nminuskminus1) M nk

(520)

ndash 25 ndash



From equation (511) we use the expression for αnk to compute

αn+1k minus αn+1nminuskminus1 = 1

2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1

k (n minus k minus 1) 1

n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)

Using this in equation (520) we see that

G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)

which is precisely the RHS of equation (514) and we deduce that our result for G (χ (ξ )) is

in agreement with [46] However the relationship between χ and ξ was not discussed in [46]

We made explicit the connection between χ(ξ ) and ξ (ξ ) due to our results in the section (22)

which play a crucial rule in proving Hohm and Zwiebachrsquos conjecture

6 Conclusions and Outlook

We conclude this paper by summarizing our results and discussing the important issue

of composition of finite transformations in double field theory We will also comment on the

finite transformations in exceptional field theoryWe have shown that the formula for finite gauge transformations in double field theory

is equivalent to the transformations obtained by exponentiating the generalized Lie derivative

In particular we proved that for every parameter ξ M we can find a parameter ξ M (ξ ) such

that

F (ξ ) = G

ξ (ξ )

(61)

and we showed that G (ξ (ξ )) can be decomposed as G (ξ ) G (χ (ξ )) with χ (ξ ) being a quasi-

trivial parameter We have also given an iterative procedure which can be used to determine

ξ M (ξ ) to any order in ξ We used this procedure to obtain results up to quintic order and

verified them explicitly In the last section we also gave an explicit formula for χ (ξ ) to allorders in ξ We also showed that ξ (ξ ) obtained using this procedure is not unique This

non-uniqueness can be understood in terms of the inherent non uniqueness of the generalized

Lie derivative because the generalized Lie derivative with respect to a trivial parameter is

zero

ndash 26 ndash



We note that the composition rules for G (ξ ) are well understood After proving (111)

we are now in a better position to understand the composition of F (ξ ) Composition of G (ξ )

was studied in section 52 of [1] and it was shown that

G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1

2 [ξ 2 ξ 1]C + middot middot middot (63)

where lsquomiddot middot middot rsquo contains further nested C-brackets between ξ 2 and ξ 1 given by the BCH formula

The issue of composition was also discussed in [1] where generalized coordinate transforma-

tions generated by a parameter Θ (ξ ) were considered such that F (Θ (ξ )) = G (ξ ) In our

language this means that ξ (Θ (ξ )) = ξ So under the coordinate transformations

X = eminusΘ(ξ1) X X = eminusΘ(ξ2) X (64)

the composition of F is given by

F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)

which implies that the composition of coordinates is given by

X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)

We want to understand this composition more directly ie in terms of generalized coordinate

transformations given by

X = eminusξ1X X = eminusξ2X (69)

The corresponding F matrices are given by

F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)

where δ (ξ i) and χ (ξ i) are determined using equations (232) and (512) It is now easy to see

ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)

This composition rule then implies that the composition of the generalized coordinate trans-

formations is given by

X = eminusΘ(ξC (ξ2+δ(ξ2)ξ1+δ(ξ1))) X (615)

We conclude the discussion about the composition by noting that the composition rule (615)

is very unconventional and it has important consequences regarding associativity as discussed

in section 6 of [1]

In the context of exceptional field theory (see refs [50ndash66]) the issue of finite trans-

formations still needs to be understood In particular there is no analogue for the matrix F

which gives finite transformations of the fields We conclude this paper with some remarks

about this issue We will follow the notation of [66] closely In exceptional field theory the

generalized Lie derivative takes the following form

Lξ = ξ + a minus aY (616)

where

aY N

M equiv Y

QN M P a P

Q a N M = part M ξ N and Y

QN MP is the E n(n) invariant tensor The

notion of C-bracket is replaced by a generalized Lie bracket called lsquoE-bracketrsquo defined as

follows

[ξ 1 ξ 2]M E = ξ P

1 part P ξ M 2 minus ξ P

2 part P ξ M 1 minus 1

2Y MN

P Qξ

Q1 part N ξ P

2 minus ξ Q2 part N ξ P

1 (617)

Similar to the case of double field theory generalized Lie derivatives form a Lie algebra given

by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)

The closure of this Lie algebra puts some constraints on Y M N P Q These constraints include an

analogue of the strong constraint in double field theory (see section 64 of [45] for a complete

list of constraints)

Y QN

MP part Qpart N (middot middot middot) = 0 (619)

where lsquomiddot middot middot rsquo is any product of fields and parameters Now we modify the definition of the

quasi-trivial parameter and for the exceptional case we define a parameter of the following

form to be quasi-trivial

V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash



It is easy to see using equation (619) that V P part P gives zero when acting on product of fields

and parameters The analogue of the matrix G (ξ ) is defined as

G E (ξ ) = eminusξeξ+aminusaY (621)

The relevant similarities between double field theory and exceptional field theory end here

Now note that we do not have analogue of equation (225) This follows due to the fact that

for a quasi-trivial parameter ξ M the matrix a minus aY is not nilpotent in general This in turn

follows from the fact that the E-bracket(as opposed to the C-bracket) of two quasi-trivial

parameters is not zero in general Also we do not have the analogue F E of the matrix F

which implements the finite transformations of the fields and can be written in terms of the

generalized coordinate transformations (ie in terms of partX partX and partX

partX )

While finding the matrix F E remains a challenging open question we argue that regard-

less of the final form of F E we should be able to express it in the form F E (ξ ) = G E (ξ + γ (ξ ))

where γ (ξ ) is quasi-trivial parameter and X = eminusξX This conclusion follows by considering

the transformation of a scalar field and using similar kind of arguments as in section 21 Our

analysis in section 22 can also be generalized2 to the case of exceptional field theory and we

can deduce that G E (ξ + γ (ξ )) = G E (ξ ) G E (Γ (ξ )) where γ and Γ satisfy a similar relation as

in equation (232) (with δ χ and C-brackets replaced by γ Γ and E-brackets respectively)

So we deduce that F E (ξ ) = G E (ξ ) G E (Γ (ξ )) Now the problem of finding F E can be stated

as Is there an expression for Γ (ξ ) such that the RHS of the last equation can be written in

terms of partX partX and partX

partX where X = eminusξX We stress that this reformulation of the problem

does not make it any easier as it is not immediately obvious how one should approach this

new problem However it does provide a different way to think about the problem of findingF E

Acknowledgments

I would like to thank Barton Zwiebach for numerous discussions very useful comments

and kind supervision throughout the course of this work This work is supported by the US

Department of Energy under grant Contract Number DE-SC00012567

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is preserved

ndash 29 ndash



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ndash 30 ndash



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ndash 32 ndash



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[64] O Hohm and Y Wang Tensor Hierarchy and Generalized Cartan Calculus in SL(3) times SL(2)

Exceptional Field Theory arXiv150101600 [hep-th]

[65] A Abzalov I Bakhmatov E T Musaev Exceptional Field Theory SO(5 5)


[66] J Edlund A tensor formalism for exceptional geometry Chalmers University of Technology

CPL ID 195735



Contents

1 Introduction 1

2 Preliminaries 4

21 Review 4

22 Decomposition of G (ξ (ξ )) 8

3 Obtaining differential equations 11

31 Differential equation for F (λξ ) 11

32 Differential Equation for G (ξ (λξ )) 13

4 Solving for the parameter ξ (ξ ) 14

41 Comparison and an iterative method 1442 Order by Order checks 17

421 First and Second Order 17

422 Cubic Order 18

423 Quartic Order 19

424 Quintic Order 20

5 Determining ξ to all orders 22

6 Conclusions and Outlook 26

1 Introduction

Double field theory was developed to make manifest the O(D D) T-duality symmetry in

the low energy effective field theory limit of string theory In addition to the usual spacetime

coordinates lsquowindingrsquo coordinates are introduced The metric and the Kalb-Ramond two-form

are combined into a lsquogeneralized metricrsquo This generalized metric transforms linearly under

global O(D D) transformations Gauge transformations of the fields can also be written

in an O(D D) covariant form and they can be interpreted as the lsquogeneralized coordinate

transformationsrsquo in the doubled spacetime as discussed in [1] The action of double field

theory written in terms of the generalized metric is then manifestly invariant under these

transformations Double field theory is a restricted theory The so-called strong constraint

restricts the theory to live on a D-dimensional subspace of the doubled spacetime Different

ndash 1 ndash









F given by

F N M =

1

2

partX P

partX M

partX P

partX N +

partX M

partX P

partX N

partX P

(11)


V M (X ) = F N M V N (X ) (12)


V (X ) = F V (X ) (13)



ξ M (X ) ie

X M








V K (X ) = V M (X ) +

LξV (X )

M

(15)







ndash 2 ndash








F (ξ ) = G (ξ ) (19)







ξ P (ξ ) = ξ P +

i















G

ξ (ξ )

= F (ξ ) (111)



ndash 3 ndash





F (λξ ) = G (ξ (λξ )) (112)


























2 Preliminaries

21 Review




ndash 4 ndash





eminusξP part P

X M (21)




ξ M as follows

F (ξ ) = 1



operator ξ middot

f equiv

ξ

P






larrminusξ +a

(24)



ξ P part P β

Λ (25)



1 middot eminus


(26)



infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)


B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735









F given by

F N M =

1

2

partX P

partX M

partX P

partX N +

partX M

partX P

partX N

partX P

(11)


V M (X ) = F N M V N (X ) (12)


V (X ) = F V (X ) (13)



ξ M (X ) ie

X M








V K (X ) = V M (X ) +

LξV (X )

M

(15)







ndash 2 ndash








F (ξ ) = G (ξ ) (19)







ξ P (ξ ) = ξ P +

i















G

ξ (ξ )

= F (ξ ) (111)



ndash 3 ndash





F (λξ ) = G (ξ (λξ )) (112)


























2 Preliminaries

21 Review




ndash 4 ndash





eminusξP part P

X M (21)




ξ M as follows

F (ξ ) = 1



operator ξ middot

f equiv

ξ

P






larrminusξ +a

(24)



ξ P part P β

Λ (25)



1 middot eminus


(26)



infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)


B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735








F (ξ ) = G (ξ ) (19)







ξ P (ξ ) = ξ P +

i















G

ξ (ξ )

= F (ξ ) (111)



ndash 3 ndash





F (λξ ) = G (ξ (λξ )) (112)


























2 Preliminaries

21 Review




ndash 4 ndash





eminusξP part P

X M (21)




ξ M as follows

F (ξ ) = 1



operator ξ middot

f equiv

ξ

P






larrminusξ +a

(24)



ξ P part P β

Λ (25)



1 middot eminus


(26)



infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)


B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





F (λξ ) = G (ξ (λξ )) (112)


























2 Preliminaries

21 Review




ndash 4 ndash





eminusξP part P

X M (21)




ξ M as follows

F (ξ ) = 1



operator ξ middot

f equiv

ξ

P






larrminusξ +a

(24)



ξ P part P β

Λ (25)



1 middot eminus


(26)



infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)


B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





eminusξP part P

X M (21)




ξ M as follows

F (ξ ) = 1



operator ξ middot

f equiv

ξ

P






larrminusξ +a

(24)



ξ P part P β

Λ (25)



1 middot eminus


(26)



infinn=1

(minus1)n

n part M

ξ P part P

nminus1ξ N

(27)


B N M (ξ ) equiv

partX N

partX M (28)

ndash 5 ndash



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735



which implies that

B(ξ ) = 1


1 middot eminus

larrminusξ +a

(29)



F = 1

2

B

Bminus1t

+

Bminus1t

B

(210)




V (X ) = V (X ) + LξV (X ) (211)

with



V K (X ) (212)



a minus at

V (X ) with a N



= L[ξ1 ξ2]C

(214)


[ξ 1 ξ 2]M C = ξ P



1

2


2 part M ξ 1P

(215)




V (X ) = eLξ V (X ) (216)

ndash 6 ndash



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735



By using the fact

eξf


V (X ) = eξV (X ) = eξ F V (X ) (217)









φ X = φ (X ) (218)

which implies

φ (X ) = eξ φ (X ) (219)





Therefore

φ(X ) = eLξφ(X ) = eξφ(X ) (221)


sistency ie

eξ(ξ)φ (X ) = eξφ (X ) (222)




ndash 7 ndash




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




desired order in ξ







Lδ (224)










hold

bib j = bibt j = bt

ib j = btibt




1 + b1 minus bt1

1 + b2 minus bt

2








G

ξ

= G (ξ )G (χ) (231)

ndash 8 ndash




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




δ = χ + 1

2 [ξ χ]C +

1

12

[ξ [ξ χ]C ]C + [χ [χ ξ ]C ]C







G

ξ





eX +Y = eX eY ige2










= L[ξδ]c

(236)

Therefore

C i

Lξ

Lδ [middot middot]

=

L



G

ξ

= eminusξ e Lξ e

Lδ ige2


ndash 9 ndash






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735






G ξ = G (ξ ) G (δ )



G

ξ

= G (ξ ) G

δ +ige2


(240)

equiv G (ξ ) G (χ) (241)






can be written as



Lχ (242)



e Lξe

Lχ = e Lξ (243)

where

Lξ = Lξ + Lχ + 1

2

Lξ Lχ

+

1

12

Lξ Lξ Lχ

+ Lχ

Lχ Lξ




ξ = ξ + χ + 1

2 [ξ χ]c +

1

12

[ξ [ξ χ]c]c + [χ [χ ξ ]c]c






ndash 10 ndash





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





ξ

(247)






write


N (248)

so that

G

ξ

= G (ξ ) + G (ξ )

∆ minus ∆t

(249)









X M




A N


(λξ ) (32)

and

B(λ) equiv partX

partX =

1

1 minus A(λ) =

1 middot e


(33)

ndash 11 ndash






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735






F (λξ ) =

1

2 B(λ) Bt





Btminus1


dB

dλ = B

minus

larrminusξ + a

and

d

Btminus1

dλ =

Btminus1


(35)



d

dλ

Btminus1

B

=

Btminus1

B

minuslarrminusξ

+

Btminus1

B

a minus at

+

Btminus1

Bat minus

Btminus1

atB(37)


atB = at

Btminus1

B = B +

Btminus1

minus 1

Btminus1

a = amiddot (38)



d

dλF =

d

dλ

1

2

B

Btminus1

+

Btminus1

B

(39)

= F

minus


+

1

2

B a

Btminus1


(310)


=

1

1 minus A

tminus1

= 1 minus At (311)

such that



ndash 12 ndash




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




d

dλF = F (minus


1

2 B

minusaAt + Aat

(314)




Btminus1


F

Aat minus aAt

= 1

2

B

Btminus1

(Aat minus aAt) +

Btminus1

B(Aat minus aAt)

(315)

= 1

2


Btminus1


(316)

= 1


(317)



d

dλF = F

minus


1

2Aat minus

1

2aAt

(319)








G (λ) equiv G

ξ (λξ )

= G (λξ ) G (χ (λξ ) ) (320)

G (χ (λξ )) = 1 + ∆(λξ ) minus ∆t(λξ ) (321)




at∆ = at∆t = ∆a = ∆ta = 0 G ∆ = B∆ G ∆t = B∆t (322)

ndash 13 ndash






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735






dG

dλ = dG



d

dλG = G

minus


(324)


d

G

dλ = G

minus


1 + ∆ minus ∆t

+ B

d

dλ

∆ minus ∆t

(325)


minus



G

minus


= G

minus


1 + ∆ minus ∆t

minus B (ξ + a)

∆ minus ∆t

minusB

∆ minus ∆t

at (326)


d

G

dλ


d










d

dλ

∆ minus ∆t

+

∆ minus ∆t

at = 1

2

Aat minus aAt

(41)

ndash 14 ndash




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




∆(λ) =infin

n=1

λn∆n (42)



A(λ) equiv

infinn=1

λnAn (43)


(An) N M = part M

ζ N (n)

(44)


ζ N (n) =

(minus1)n+1

n

ξ P part P

nminus1ξ N

(45)


(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(46)




∆n minus ∆t

n

N








ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(48)

ndash 15 ndash




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




(ξ + a)

∆n minus ∆tn

+

∆n minus ∆tn

at (49)


part N



part N

ξ k+1 middot ξ P

(410)


sim part M

ξ k middot ξ P

part N


minus part M


part N

ξ k middot ξ P

(412)






(413)








(∆n) N M = part M

f part N g

(414)

bull



∆ N M = part M χ

N (415)








ndash 16 ndash










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735










(B) N M = f M g

N equiv f g (416)


overhead arrow

bull



V = A part B (417)


A minus At equiv


(418)




N n



order as follows

δ (ξ ) = χ3 +

χ4 +

1

2 [ξ χ3]c

+

χ5 +

1

2 [ξ χ4]c +

1

12 [ξ [ξ χ3]c]c

+ O(ξ 6) (420)




ndash 17 ndash





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





this implies then

δ M

(ξ ) = 0 + O(ξ 2

) (421)


2

∆2 minus ∆t2

=

1

2

A1at minus aAt

1

(422)



δ M (ξ ) = 0 + O(ξ 3) (423)


422 Cubic Order





3 ∆3 minus ∆t

3 =

1

2 A2a

t

minus aA

t

2 (424)


ing

A2at minus aAt

2

N

M = minus

1

2

part M

ξ middot ξ P


(425)

= minus1

2

part M

ξ middot ξ P

part N ξ P

minus part N

ξ middot ξ P

part M ξ P

(426)





1

2

part



(427)

ndash 18 ndash





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





1

12

part



(428)


∆3 = minus 1

12 part


rArr χ3 = minus

1





So we deduce that


12



F

ξ +

1


= G (ξ ) + O(ξ 4) (432)


423 Quartic Order





4

∆4 minus ∆t4

+ (ξ + a)

∆3 minus ∆t

3

+

∆3 minus ∆t3

at =

1

2

A3at minus aAt

3

(433)



A3at minus aAt3 =

1


(ξ + a) ∆3 = part





ξ middot ξ K

+ partξ K

part

ξ 2 middot ξ K


ξ middot ξ K

(436)

∆3 minus ∆t3

at = part

ξ middot ξ K


partξ P part P

ξ middot ξ K

(437)

ndash 19 ndash




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




∆4 minus ∆t4 =

1

24

part

ξ 2 middot ξ K

partξ K


(438)


∆4 = 1

24 part

ξ 2 middot ξ K

partξ K

rArr χ4 =

1

24

ξ 2 middot ξ P

partξ P (439)


bull Now

δ (ξ ) = χ3 + χ4 + 1

2 [ξ χ3]C + O(ξ 5) (440)




1

2 [ξ χ3]C = minus

1

24

ξ 2 middot ξ P



other and we get


12

ξ middot ξ P

part M ξ P + O(ξ 5) (442)


424 Quintic Order







1



ndash 20 ndash




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




(ξ + a) ∆4 = 1

24

part

ξ 3 middot ξ P

partξ P + part

ξ 2 middot ξ P

partξ P minus part

ξ 2 middot ξ P


(444)


1


1

24 minus


ξ 2 middot ξ P

(445)

∆4 minus ∆t4

at =

1

24

part

ξ 2 middot ξ Q


ξ 2 middot ξ Q

part ξ P

(446)


1

24

part

ξ 3 middot ξ P


ξ 3 middot ξ P

(447)

∆5 minus ∆t5 =

minus1

240

3 part

ξ 3 middot ξ P

part ξ P + 2 part

ξ 2 middot ξ P


(448)



5

We choose 1

∆5 = minus1

240 part

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(450)


χ5 = minus1

240

3

ξ 3 middot ξ P

part ξ P +

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(451)


δ (ξ ) = χ3 + χ5 + 13

[ξ χ4]c + O(ξ 6) (452)


[ξ χ4]c = 1

24

ξ 3 middot ξ P

partξ P +

1

48

ξ 2 middot ξ P


ξ middot ξ P

part

ξ 2 middot ξ P

(453)



∆5 = minus1

240 part

3ξ3 middot ξ

P partξP + 2

ξ2 middot ξ


(449)


ndash 21 ndash



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735



up nicely to give


12

ξ middot ξ P

part M ξ P +

1

720

ξ 3 middot ξ P

part M ξ P +

ξ 2 middot ξ P



M ξ 2






M ie

G (ξ + ξ (3) + ξ (5)) = G (ξ + ξ (3)) + ∆ + O(ξ 6) (455)

















∆n minus ∆t

n =

k=nminus2

k=0

αnk part ξ

k

middot ξ P







ndash 22 ndash



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735



equation (46) here

(n + 1)


+ (ξ + a)

∆n minus ∆t

n

+

∆n minus ∆tn

at =

1

2

Anat minus aAt

n

(52)


(ξ + a)

∆n minus ∆tn

=

k=nminus2k=0

αnk

part

ξ k+1 middot ξ P

part



+ αnk

part

ξ k+1 middot ξ P

part



minus αnk

part

ξ k middot ξ P

partξ Qpart Q


+ αnk

part


partξ Qpart Q

ξ k middot ξ P

(53)

and

∆n minus ∆t

n

at =

k=nminus2k=0

αnk part

ξ k middot ξ P

partξ Qpart Q


minus αnk

part


partξ Qpart Q

ξ k middot ξ P

middot (54)



to obtain the


LHS =

k=nminus1k=1



t+ ((n + 1) αn+10 + αn0 + αnnminus1)

part

ξ P

part



middot (55)


RHS = 1

2

(minus1)n

n

part

ξ P

part



middot (56)


αnk = 12

(minus1)n

n times β nk (57)


ndash 23 ndash








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735








nminus1k


minus

n

0

+

n minus 1

0

+

n minus 1

n minus 1

= minus1 + 1 + 1 = 1 (510)


∆nminus∆tn =

k=nminus2

k=0

αnk

part

ξ k middot ξ P

part



with αnk =

1

2

(minus1)n

n

n minus 1

k

middot

(511)







k=0

αnk

ξ k middot ξ P

part M


middot (512)






[F (ξ )] =

F (ξ ) (1 + Q) where Q2 = 0

(513)



form


n=2

k=nminus1k=0

( 1 + κnk M nk) (514)

ndash 24 ndash




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




κnk = 1

2



ξ k middot ξ P

part


(515)





G (χ (ξ )) = 1 +infin

n=3

nminus2k=0

αnk

part

ξ k middot ξ P

part



(516)


G (χ (ξ )) = 1 +infin

n=3

nminus2k=0


part

ξ k middot ξ P

part


(517)

= 1 +infin

n=3

nminus2k=0




G (χ (ξ )) = 1 +

infin

n=2

nminus1

k=0



G (χ (ξ )) =infin

n=2

nminus1k=0


(520)

ndash 25 ndash





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735





2

(minus1)n+1

(n + 1)

n

k

minus

n

n minus k minus 1

(521)

= 1

2

(minus1)n+1

(n + 1)

1


n minus k minus 1

k + 1 (522)

= 1

2

(minus1)n

(n + 1)

n minus 2k minus 1

(k + 1) (n minus k) (523)

= κnk (524)


G (χ (ξ )) =infin

n=2

nminus1k=0

1 + κnk M nk

(525)











that

F (ξ ) = G

ξ (ξ )

(61)







zero

ndash 26 ndash






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735






G (ξ 2) G (ξ 1) = G ξ C

(ξ 2 ξ 1) (62)

with

ξ C (ξ 2 ξ 1) = ξ 2 + ξ 1 + 1








F (Θ (ξ 1)) F (Θ (ξ 2)) = G (ξ 2) G (ξ 1) (65)

= G

ξ C (ξ 2 ξ 1)

(66)

= F

Θ

ξ C (ξ 2 ξ 1)

(67)


X

= eminusΘ

(ξC (ξ2ξ1)

) X (68)





F (ξ 1) = G (ξ 1 + δ (ξ 1)) = G (ξ 1) G (χ (ξ 1)) (610)

F (ξ 2) = G (ξ 2 + δ (ξ 2)) = G (ξ 2) G (χ (ξ 2)) (611)


ndash 27 ndash



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735



that

F (ξ 2) F (ξ 1) = G (ξ 2 + δ (ξ 2)) G (ξ 1 + δ (ξ 1)) (612)

= G

ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1))

(613)

= F Θ ξ C (ξ 2 + δ (ξ 2) ξ 1 + δ (ξ 1)) (614)






in section 6 of [1]







where

aY N

M equiv Y

QN M P a P




follows

[ξ 1 ξ 2]M E = ξ P



2Y MN

P Qξ

Q1 part N ξ P


1 (617)


by Lξ1 Lξ2

= L[ξ1ξ2]E

(618)




Y QN





V N =

i

Y QN

MP ρM i part QηP

i (620)

ndash 28 ndash













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735













partX )














Acknowledgments




References




is preserved

ndash 29 ndash




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




arXivhep-th9305073





JHEP 1007 (2010) 016 arXiv10035027 [hep-th]


0909 090 (2009) arXiv09081792 [hep-th]




B 242 163 (1990)


Phys B 350 395 (1991)




44 085404 (2011) arXiv10114101 [hep-th]


(2010) arXiv10082746 [hep-th]




096 (2011) arXiv11032136 [hep-th]


Lett 107 171603 (2011) arXiv11065452 [hep-th]


(2011) arXiv11070008 [hep-th]



[hep-th]


126 (2012) arXiv11125296 [hep-th]

ndash 30 ndash

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735

















D84 044022 (2011) arXiv11056294 [hep-th]


025 (2011) arXiv11092035 [hep-th]










125 (2011) arXiv11064036 [hep-th]


JHEP 1109 025 (2011) arXiv11070876 [hep-th]


JHEP 1111 052 (2011) arXiv11090290 [hep-th]




ndash 31 ndash









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735









arXiv13051907






(2013) 098 arXiv13045946 [hep-th]






231601 (2013) arXiv13081673v2 [hep-th]


arXivhep-th0701203




JHEP 0809 123 (2008) arXiv08041362 [hep-th]


(2009) arXiv09011581 [hep-th]



[hep-th]





ndash 32 ndash




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735




1307 028 (2013) arXiv13026736 [hep-th]











CPL ID 195735

Documents

A Note on Large Gauge Transformations in Double Field Theory