Introduction q A q - City University of New Yorkwebsupport1.citytech.cuny.edu/Faculty/ccalinescu/A_2n_2... · 2016. 5. 18. · cipal subspace WT L ˆV T L and describe certain ideals

VERTEX ALGEBRAIC STRUCTURE OF PRINCIPAL

SUBSPACES OF BASIC A(2)2n -MODULES

CORINA CALINESCU, ANTUN MILAS AND MICHAEL PENN

Abstract. We obtain a presentation of the principal subspace of the basic

A(2)2n –module, n ≥ 1. We also show that its full character is given by the

Nahm sum of the tadpole Dynkin diagram Tn = A2n/Z2. An evidence formodularity of specialized characters was presented based on the twisted affine

Lie algebra of type A(2)4 .

1. Introduction

An important problem in number theory is to understand a precise connectionbetween multifold q-hypergeometric series and modular forms. For certain r-foldhypergeometric series this is also related to Nahm’s conjecture [N, Za]. It is widelybelieved that q-hypergeometric series (also known as Nahm sums):∑

n=(n1,...,nr)∈Zr≥0

qntAn

(q)n1· · · (q)nr

where A is the Gram matrix (or its inverse) of ADET type are modular, in the sensethat adding an appropriate power of q makes it modular. These and related sumsare also expected to be characters of modules of rational conformal field theories.For more about this subject see [Ki, N, Za, VZ].

Multi q-hypergometric series naturally appear from considerations of charactersof principal subspaces of highest weight integrable modules of affine Kac-Moodyalgebras. This approach was initiated in an influential work of B. Feigin and Stoy-anovsky [FS1]. As explained in [FS1], principal subspaces, also known as FS princi-pal subspaces, and their characters can be also studied by using the geometry of theinfinite-dimensional flag variety and principal variety. This approach would providea different bosonic (vs. hypergeometric or fermonic) form of characters of principalsubspaces. Although this is expected to give a better insight into aforementionedmodularity of characters, this has not materialized except in a few low examples. Inaddition, there is another approach to bosonic formulas based on quantum groups,although even in the case of g = sl3 such formulas are rather complicated [FFJMM].

There are other algebraic structures that lead to Nahm sums and q-series iden-tities such as double affine Hecke algebras [CF]. And of course, independent ofall this, there are purely non-representation theoretic methods based primarily onmodular forms and symmetric functions (see [WZ, BCFK], etc.).

C.C was partially supported by the Simons Foundation Collaboration Grant for Mathemati-

cians, and by PSC-CUNY Research Awards.A.M. was partially supported by the Simons Foundation Collaboration Grant for

Mathematicians.

1

2 CORINA CALINESCU, ANTUN MILAS AND MICHAEL PENN

In this paper, motivated by a long line of work [FS1, CalLM1, CalLM2, CalLM3,CalLM4, Je, MPe, P, S, B], some by the authors, we are exploring FS principalsubspaces associated to basic modules of twisted affine Kac-Moody algebras of type

A(2)2n , n ≥ 1. Our main results can be viewed as a generalization of the A

(2)2 case by

Lepowsky and the first two authors [CalLM4].Let us briefly outline the content of the paper. In Section 2, we recall a con-

struction of the Kac-Moody Lie algebra A(2)2n and its level one standard module V TL ,

viewed as a twisted module for the lattice vertex operator algebra VL, where L isthe root lattice A2n. Results in this part are standard and mostly taken from [L1]applied to our special case (see also [CalLM4]). In Section 3, we introduce the prin-cipal subspace WT

L ⊂ V TL and describe certain ideals J and IΛ inside the universalenveloping algebra U(n[ν]). Then in Section 4, which is the most technical part ofthe paper, we define certain maps acting on WT

L and on U(n[ν]). These results areneeded to determine the annihilator of the highest weight vector in W (cf. Lemma4.1). We should say that related maps were also considered in [CalLM3, CalLM4].Finally, in Section 5 we completely described the annihilating ideal of the highestweight vector inside WT

L (see Theorem 5.1). This is then used in Theorem 5.2 tosetup certain exact sequences. This leads to a system of q-difference equations forthe character of WT

L (see Corollary 5.1). The main result is then the statement ofCorollary 5.2 giving the character of WT

L as the Nahm sum of the tadpole Dynkindiagram:

• • • • •��

It is still unknown whether or not the Nahm sum associated to the tadpole diagram

is modular. We present a proof in the case of A(2)4 . We also compare our findings

with Zagier’s work [Za] (see Section 6).Acknowledgements: The second named author would like to thank to S.

Zwegers for a helpful discussion pertaining to Section 6.

2. The setting

Let g = sl(2n + 1,C) and let h ⊂ g be the standard Cartan subalgebra of g ofdiagonal matrices. Denote by ∆ the corresponding root system

(2.1) ∆ = {±(εi − εj) | 1 ≤ i < j ≤ 2n+ 1}.

Fix simple roots to be

(2.2) α1 = ε1 − ε2, . . . , α2n = ε2n − ε2n+1.

Let 〈a, b〉 = tr(ab), where a, b ∈ sl(2n+1,C), be symmetric nondegenerate invariantbilinear form.

Consider the root lattice of sl(2n+ 1,C)

L = Z∆ = Zα1 ⊕ · · · ⊕ Zα2n ⊂ h∗.

We also consider the isometry ν : L→ L given by

ν(αi) = α2n−i+1

for 1 ≤ i ≤ 2n. This is of course the isometry that is equivalent to the Dynkindiagram folding symmetry. Observe that ν2 = 1 yet, in the spirit of [CalLM4], we

PRINCIPAL SUBSPACES OF BASIC A(2)2n -MODULES 3

take k = 4 so that we have

(2.3)⟨νk/2α, α

⟩∈ 2Z

and

(2.4)

⟨k−1∑j=0

νjα, α

⟩∈ 2Z

for all α ∈ L.Now we fix η = i to be our primitive fourth root of unity with η0 = (−1)4η = η

as in section 2 of [CalLM4]. We now extend ν linearly to an automorphism of

(2.5) h = C⊗Z L,

our Cartan subalgebra.We have two central extensions of L by 〈i〉: L and Lν with commutator maps

C0 and C respectively. Choose normalized sections e of L and Lν sending α ∈ Lto eα ∈ L (resp. Lν). We also have normalized cocycles εC and εC0

such that

(2.6) eαieαj = εC0(αi, αj)eαi+αj

and

(2.7)εC0

(αi, αj)

εC0(αj , αi)= (−1)〈αi,αj〉

thus,

(2.8) eαieαj = (−1)〈αi,αj〉eαjeαi in L.

Explicitly, we may take εC0: L× L→ 〈i〉 to be

(2.9)εC0(αi, αj) = 1 if i ≤ j,

εC0(αi, αj) = (−1)〈αi,αj〉 if i ≥ j.

In particular, we have

(2.10)

εC0(αi, αi+1) = 1 for 1 ≤ i ≤ 2n− 1,

εC0(αi+1, αi) = −1 for 1 ≤ i ≤ 2n− 1,

εC0(αi, αj) = 1 for 1 ≤ i, j ≤ 2n

and j /∈ {i− 1, i+ 1}.

Notice that Equations 2.17-2.19 of [CalLM4] hold as needed.Observe that εC0

interacts with ν via the relationship given by

(2.11) εC0(α, β) = εC0

(νβ, να) for all α, β ∈ L.

Explicitly we can check for α = r1α1 + · · ·+ r2nα2n and β = s1α1 + · · ·+ s2nα2n

(2.12)εC0

(α, β) = (−1)r2s1+r3s2+···+r2ns2n−1

= εC0(νβ, να).

Our next goal is to lift our isometry ν to an automorphism ν of L and Lν thatsatisfies

(2.13) νa = a if νa = a.


In order to explicitly write down this automorphism we will make use of the fol-lowing notation to write the roots of sl(2n+ 1,C). Define

(2.14) α(j)i = αi + · · ·+ αi+j−1 =

j−1∑k=0

αi+k = εi − εi+j

(recall (2.2)). This notation is chosen because α(j)i is the sum of j consecutive

simple roots beginning with αi. We have

(2.15) ∆ = {±α(j)i |1 ≤ i ≤ 2n, 1 ≤ j ≤ 2n− i+ 1}

(recall (2.1)). Furthermore we have

(2.16) ν(α

(j)i

)= α

(j)2n−i−j+2.

We use the notation

(2.17) α′

i = α(2n−2i+2)i

for 1 ≤ i ≤ n. Note that ν(α′

i) = α′

i.We now introduce the following definition to simplify notation in the future.

Definition 2.1. Let i, j, k, l such that 1 ≤ i, k ≤ 2n, 1 ≤ j ≤ 2n − i + 1 and1 ≤ l ≤ 2n− k + 1. We say that:

(1) α(j)i and α

(l)k are adjacent if 〈α(j)

i , α(l)k 〉 = −1.

(2) α(j)i and α

(l)k are conjugate-adjacent if ν

(α

(j)i

)and α

(l)k are adjacent,

i.e. 〈ν(α

(j)i

), α

(l)k 〉 = −1.

(3) α(j)i and α

(l)k are disconnected if they are neither adjacent nor conjugate-

adjacent, i.e. 〈α(j)i , α

(l)k 〉 ≥ 0 and 〈ν

(α

(j)i

), α

(l)k 〉 ≥ 0.

Now we can choose ν such that

ν (eαi) = −eα2n−i+1if i /∈ {n, n+ 1}(2.18)

ν (eαn) = ieαn+1(2.19)

ν(eαn+1

)= ieαn(2.20)

and then extend this to an automorphism of the whole space.

We say that α(j)i contains αm if i ≤ m ≤ i+ j−1, thus it appears as a summand

when α(j)i is written as the sum of positive roots.

Proposition 2.1. This definition of ν is completely determined by the following

ν(eα

(j)i

)= −e

α(j)2n−i−j+2

if α(j)i does not contain αn or αn+1(2.21)

ν(eα

(j)i

)= ie

α(j)2n−i−j+2

if α(j)i contains exactly one of αn or αn+1(2.22)

ν(eα

(j)i

)= e

α(j)2n−i−j+2

if α(j)i contains both αn and αn+1.(2.23)

and analogous formulas for ν(e−α(j)i

).


Observe that Proposition 2.1 implies that ν(eα′i

)= eα′i

for all 1 ≤ i ≤ n (recall

(2.17)). In other words, our lifting satisfies (2.13), as desired. It is clear from thisconstruction that ν2eα = ±eα for all α ∈ L and thus ν4 = 1.

Now we form the vertex operator algebra VL (after [LL]) and extend the auto-

morphism ν of L to an automorphism of VL, which we will also call ν. Considerthe induced L-module

C{L} = C[L]⊗C[〈i〉] C,

which is isomorphic linearly with C[L]. Set

ι(a) = a⊗ 1 ∈ C{L}

for a ∈ L. Recall that we have a linear isomorphism,

(2.24) VL ∼= S(h−)⊗ C[L],

where

(2.25) h− = h⊗ t−1C[t−1].

Thus the automorphism lifted to VL acts as ν ⊗ ν.Now we will look at the decomposition of h in the spirit of [CalLM4]. For n ∈ Z,

set

(2.26) h(n) = {x ∈ h|ν(x) = inx} ⊂ h.

So we have

(2.27) h =∐

n∈Z/4Z

h(n),

where we have identified h(n mod 4) with h(n). In view of our isometry ν : L → Lwe have

(2.28)h(0) = span{αi + α2n−i+1|1 ≤ i ≤ n}h(2) = span{αi − α2n−i+1|1 ≤ i ≤ n}

with h(1) = h(3) = 0. Define Pn mod 4 : h → h(n) to be the canonical projectionsand set h(n) = Pn mod 4(h). For 1 ≤ i ≤ 2n we have

(2.29)(αi)(0) =

1

2(αi + α2n−i+1) ∈ h(0)

(αi)(2) =1

2(αi − α2n−i+1) ∈ h(2).

Now we can form the ν-twisted affine Lie algebra associated to the abelian Liealgebra h:

(2.30) h[ν] =∐m∈Z

h(m) ⊗ tm/4 ⊕ Ck =∐

m∈ 14 Z

h(4m) ⊗ tm ⊕ Ck

where k is a central element and

(2.31) [α⊗ tr, β ⊗ ts] = 〈α, β〉 rδr+s,0k

for r, s ∈ 14Z and α ∈ h(4r), β ∈ h(4s). By the above decomposition, we have

(2.32) h[ν] = h(0) ⊗ C[t, t−1]⊕ h(2) ⊗ t1/2C[t, t−1]⊕ Ck.


There is also a natural 12Z-grading, with

(2.33)wt(α⊗ tm) = −m

wt(k) = 0,

with m ∈ 12Z and α ∈ h(4m).

Consider the Heisenberg subalgebra of h[ν] given by

(2.34) h[ν] 14 Z =

∐m∈ 1

4 Zm6=0

h(4m) ⊗ tm ⊕ Ck

and its subalgebras

(2.35) h[ν]± =∐

m∈ 14 Z

±m>0

h(4m) ⊗ tm.

Now consider the induced module

(2.36) S[ν] = U(h[ν]

)⊗U(

∐m≥0 h(4m)⊗tm⊕Ck) C ∼= S

(h[ν]−

).

This is an irreducible h[ν] 14 Z-module, which is Q-graded such that

(2.37) wt(1) =n

16

(cf. (2.62) in [CalLM4]).Recall the spaces N , M , and R defined in [L1] (see also [CalLM4]). In our case

we have

(2.38) N = {α ∈ L|〈α, h(0)〉 = 0} = spanZ{αi − α2n−i+1|1 ≤ i ≤ n}

and

(2.39) M = (1− ν)L = spanZ{αi − α2n−i+1|1 ≤ i ≤ n}.

For α, β ∈ L we have the commutator map

(2.40)

CN (α, β) = i∑3j=0〈jνjα,β〉

= i4〈να,β〉+2〈α,β〉

= (−1)〈α,β〉.

One can easily check that 〈α, β〉 ∈ 2Z for all α, β ∈ N and thus CN (α, β) = 1,which implies that

(2.41) R = {α ∈ N |CN (α,N) = 1} = N.

Therefore, we have

(2.42) N = M = R,

and thus,

(2.43) N = M = R.

By Proposition 6.1 in [L1] we have a unique homomorphism

(2.44) τ : M = N → C×


such that

(2.45)τ(i) = i

τ(aνa−1) = i−12

∑3j=0〈νja,a〉

Let Cτ be the one dimensional N -module C with character τ , write T = Cτ , andconsider the induced Lν-module

(2.46) UT = C[Lν ]⊗C[N ] T∼= C[L/N ],

which is graded by weights and on which Lν , h(0), and xh for h ∈ h(0) all naturallyact. Set

(2.47) V TL = S[ν]⊗ UT ∼= S(h[ν]−

)⊗ C[L/N ],

which is naturally acted upon by Lν , h[ν] 14 Z, h(0), and xh for h ∈ h.

For α ∈ h and m ∈ 14Z, define operators (on V TL )

(2.48) α(km) ⊗ tm 7→ αν(m)

and set

(2.49) αν(x) =∑m∈ 1

4 Z

αν(m)x−m−1.

Observe that since α(1) = α(3) = 0 for α ∈ {α1, . . . , α2n}, we have

(2.50) ανj (x) =∑m∈ 1

2 Z

ανj (m)x−m−1.

for j ∈ {1, 2, . . . , 2n}.Consider the ν-twisted vertex operator acting on V TL for eα ∈ L

(2.51) Y ν(ι(eα), x) = 4−〈α,α〉

2 σ(α)E−(−α, x)E+(−α, x)eαxα(0)+

〈α(0),α(0)〉2 − 〈α,α〉2

as defined in [CalLM4], where

(2.52) E±(−α, x) = exp

∑n∈±Z+

−α(n)

nx−n

,

and σ is the normalizing factor given by

(2.53) σ(α) = (1 + i)〈να,α〉2〈α,α〉

2

for α ∈ h. Define the component operators xνα (m) for m ∈ 14Z and α ∈ L by

(2.54) Y ν(ι(eα), x) =∑m∈ 1

4 Z

xνα (m)x−m−〈α,α〉

2 .

Following [CalLM4] (cf. [L1]), we define the nonassociative algebra (g, [·, ·]) overC:

(2.55) g = h⊕∐α∈∆

Cxα

with h = C⊗Z L and {xα}α∈∆ such that h is commutative,

(2.56) [h, xα] = 〈h, α〉xα = −[xα, h], for h ∈ h,


and,

(2.57) [xα, xβ ] =

εC0(α,−α)α if α+ β = 0

εC0(α, β)xα+β if 〈α, β〉 = −1

0 if 〈α, β〉 ≥ 0

for h ∈ h and α, β ∈ ∆.Observe that g is a Lie algebra isomorphic to sl(2n + 1,C). We can extend the

bilinear form 〈·, ·〉 to g by

(2.58) 〈h, xα〉 = 〈xα, h〉 = 0

and

(2.59) 〈xα, xβ〉 =

{εC0

(α,−α) if α+ β = 00 if α+ β 6= 0

Using our choice of εC0 (see (2.9) and (2.10)) we have(2.60)

[xα, xβ ] =

α if α+ β = 0 and α ∈ ∆

xα+β if α = α(j)i and β = α

(k)i+j with 1 ≤ i ≤ 2n, 1 ≤ j ≤ 2n− i+ 1, 1 ≤ k ≤ 2n− i− j + 1

0 if 〈α, β〉 ≥ 0

with

(2.61) 〈xα, xβ〉 =

{1 if α+ β = 0 and α ∈ ∆0 if α+ β 6= 0

As in [CalLM4] (cf. [L1]) it will be useful to define the map

(2.62) ψ : (Z/4Z)× L→ 〈i〉

by the condition

(2.63) νp (ι(eα)) = ψ(p, α)ι (eνpα)

with our choice of ν (extended to C{L}). In our set up we have

(2.64) ψ(0, α) = 1, ψ(1, α) = −1, ψ(2, α) = 1, ψ(3, α) = −1

for α = α(j)i , where α

(j)i does not contain αn or αn+1,

(2.65) ψ(0, α) = 1, ψ(1, α) = i, ψ(2, α) = −1, ψ(3, α) = −i


(j)i contains exactly one of αn or αn+1, and

(2.66) ψ(0, α) = 1, ψ(1, α) = 1, ψ(2, α) = 1, ψ(3, α) = 1


(j)i contains both of αn or αn+1.

Now we extend the automorphism ν of h to an automorphism of g, denoted byν, as follows:

(2.67) νxα = ψ(1, α)xνα, for α ∈ ∆,

and thus,

(2.68) νpxα = ψ(p, α)xνpα, for 0 ≤ p ≤ 3


and ν4 = 1 on g. Similar to our calculations in Proposition 2.1, for 1 ≤ k ≤ n wehave

ν(xα

(l)k

)= −x

α(l)2n−k−l+2

if α(l)k does not contain αn or αn+1(2.69)

ν(xα

(l)k

)= ix

α(l)2n−k−l+2

if α(l)k contains exactly one of αn or αn+1(2.70)

ν(xα

(l)k

)= x

α(l)2n−k−l+2

if α(l)k contains both αn and αn+1,(2.71)

and analogous formulas for ν(x−α(l)

k

).

It follows that for α′

k defined in (2.17) we have

(2.72) ν(xα′k

)= xα′k

.

For m ∈ Z set

(2.73) g(m) = {x ∈ g|ν(x) = imx}.Form the ν-twisted affine Lie algebra associated to g and ν:

(2.74) g[ν] =∐m∈Z

g(m) ⊗ tm/4 ⊕ Ck =∐

m∈ 14 Z

g(4m) ⊗ tm ⊕ Ck

with

(2.75) [x⊗ tr, y ⊗ ts] = [x, y]⊗ tr+s + 〈x, y〉 rδr+s,0kand

(2.76) [k, g[ν]] = 0,

for r, s ∈ 14Z, x ∈ g(4r), and y ∈ g(4s) This is isomorphic to A

(2)2n and is 1

4Z-graded.Set

(2.77) g[ν] = g[ν]⊕ Cd,

with

(2.78) [d, x⊗ tm] = mx⊗ tm,for x ∈ g(4m), m ∈ 1

4Z and [d, k] = 0,

We now recall the following result that gives V TL the structure of a g[ν]-module.

Theorem 2.1. (Theorem 9.1 [L1], Theorem 3 [FLM2]) The representation of h[ν]on V TL extends uniquely to a Lie algebra representation of g[ν] on V TL such that

(xα)(4m) ⊗ tm 7→ xνα (m)

for all m ∈ 14Z and α ∈ L. Moreover V TL is irreducible as a g[ν]-module.

As in Section 2 of [CalLM4] (also Section 6 of [L1]) we have a tensor productgrading on V TL given by the action of Lν(0) with

(2.79)

Lν(0)1 =1

64

3∑j=1

j(4− j)dim h(j)1

=1

64

(3dim h(1) + 4dim h(2) + 3dim h(3)

)1

=n

161,


so we will write wt(1) = n16 . For a homogeneous element v ∈ V TL we have

(2.80) Lν(0)v =(

wt(v) +n

16

)v

We also have

(2.81) [Lν(0), xνα (m)] =

(−m− 1 +

1

2〈α, α〉

)xνα (m)

and thus

(2.82) wt(xνα (m)) = −m− 1 +1

2〈α, α〉 .

Let λ1, . . . , λ2n be the fundamental weights of g. For i = 1, . . . , n consider theelements λi + λ2n−i+1 , which are fixed under the automorphism ν. Note that

(2.83) λi + λ2n−i+1 = α′

1 + · · ·+ α′

i

for 1 ≤ i ≤ n. The space V TL also has n-charge gradings each given by the eigenvalueof (λi + λ2n−i+1)(0). We can arrange these into an n-tuple so that

(2.84) ch(xνα (m)

)= (〈λ1 + λ2n, α〉 , 〈λ2 + λ2n−1, α〉 , . . . , 〈λn + λn+1, α〉) ,

for m ∈ 14Z. Observe that for 1 ≤ i, j,≤ n we have

(2.85) 〈λj + λ2n−j+1, αi〉 = 〈λj + λ2n−j+1, α2n−i+1〉 = δi,j .

It follows that

(2.86)⟨λj + λ2n−j+1, α

′

i

⟩= 2

when i ≤ j and

(2.87)⟨λj + λ2n−j+1, α

′

i

⟩= 0

when i > j.Recall from Section 2 of [CalLM4] the following property of the ν-twisted oper-

ator:

(2.88) Y ν(νv, x) = limx1/k→η−1x1/k

Y ν(v, x),

which generalizes to

(2.89) Y ν(νrv, x) = limx1/k→η−rx1/k

Y ν(v, x)

for any integer r. These formulas specialized to our setting (k = 4 and η = i) are

(2.90) Y ν(νv, x) = limx1/4→(−i)x1/4

Y ν(v, x)

and

(2.91) Y ν(νrv, x) = limx1/4→(−i)rx1/4

Y ν(v, x),

and they lead to the following two results.

Lemma 2.1. For α ∈ ∆+, and integers i and j such that 1 ≤ i ≤ 2n and 1 ≤ j ≤2n− i+ 1 we have

Y ν(ι(e

α(j)i

), x)

=∑m∈ 1

2 Z

xνα

(j)i

(m)x−m−1, if α(j)i does not contain αn or αn+1,


Y ν(ι(e

α(j)i

), x)

=∑

m∈ 14 + 1

2 Z

xνα

(j)i

(m)x−m−1, if α(j)i contains exactly one of αn or αn+1,

and

Y ν(ι(e

α(j)i

), x)

=∑m∈ 1

2 Z

xνα

(j)i

(m)x−m−1, if α(j)i contains both αn or αn+1.

Moreover, for 1 ≤ i ≤ n we have

Y ν(ι(eα′i

), x)

=∑m∈Z

xνα′i

(m)x−m−1.

Lemma 2.2. We have

(1) xνα

(j)2n−i−j+2

(m) = −xνα

(j)i

(m) if m ∈ Z and α(j)i does not contain αn or αn+1

(2) xνα

(j)2n−i−j+2

(m) = xνα

(j)i

(m) if m ∈ 12 + Z and α

(j)i does not contain αn or

αn+1

(3) xνα

(j)2n−i−j+2

(m) = xνα

(j)i

(m) if m ∈ 14 +Z and α

(j)i contains exactly one of αn

or αn+1

(4) xνα

(j)2n−i−j+2

(m) = −xνα

(j)i

(m) if m ∈ 34 + Z and α

(j)i contains exactly one of

αn or αn+1

(5) xνα

(j)2n−i−j+2

(m) = xνα

(j)i

(m) if m ∈ Z and α(j)i contains both αn and αn+1

(6) xνα

(j)2n−i−j+2

(m) = −xνα

(j)i

(m) if m ∈ 12 + Z and α

(j)i contains both αn and

αn+1

The commutator formula among twisted vertex operators gives

[Y ν(u, x1), Y ν(v, x2)] =1

4x−1

2 Resx0

∑j∈Z/4Z

δ

(ij

(x1 − x0)1/4

x1/42

)Y ν(Y (νju, x0)v, x2

) ,

and this implies the following result:

Lemma 2.3.

(1) For α(j)i and α

(l)k disconnected, we have [xν

α(j)i

(r), xνα

(l)k

(s)] = 0.


(l)k adjacent but not conjugate adjacent, we have

(2.92) [xνα

(j)i

(r), xνα

(l)k

(s)] =1

2xνα

(j)i +α

(l)k

(r + s),


(l)k conjugate adjacent but not adjacent we have:

(a)

(2.93) [xνα

(j)i

(r), xνα

(l)k

(s)] = − (−1)2r

2xνν(α

(j)i )+α

(l)k

(r + s)

if α(j)i does not contain αn or αn+1.

(b)

(2.94) [xνα

(j)i

(r), xνα

(l)k

(s)] = − i4

(i−4r − (−i)−4r

)xνν(α

(j)i )+α

(l)k

(r + s)

if α(j)i contains exactly one of αn or αn+1.


(c)

(2.95) [xνα

(j)i

(r), xνα

(l)k

(s)] =(−1)2r

2xνν(α

(j)i )+α

(l)k

(r + s)

if α(j)i contains both αn or αn+1.


(l)k conjugate adjacent and adjacent we have

(2.96) [xνα

(j)i

(r), xνα

(l)k

(s)] =1

2xνα

(j)i +α

(l)k

(r + s) +(−1)2r

2xνν(α

(j)i )+α

(l)k

(r + s),

where the only sub case is that α(j)i contains both αn and αn+1.

The following special cases of Lemma 2.3 will be useful in Section 3:

Corollary 2.1. We have

(1) [xναi (r) , xναi (s)] = 0 for r, s ∈ 12Z and 1 ≤ i ≤ n− 1

(2) [xναn (r) , xναn (s)] = − i4

(i−4r − (−i)−4r

)xνα′n

(r + s) for r, s ∈ 14 + 1

2Z

(3) [xνα (r) , xνα′n

(s)] = 0 for r ∈ 14Z, s ∈ Z and α ∈ {α1, . . . , αn−2, αn, α

′

n}.

Now we can decompose g into its in-eigenspaces with respect to ν for n ∈ Z/4Z.

(2.97)

g(0) =

n∐i=1

(Cxα′i ⊕ C (αi + α2n−i+1)⊕ x−α′i

)⊕n−1∐k=1

n∐j=k+1

C(x±α(2n−j−k+2)k

+ x±α(2n−j−k+2)j

)

g(1) =

n∐j=1

(C(x

α(n+1−j)j

+ xα

(n+1−j)n+1

)⊕ C(x−α(n+1−j)j

+ x−α(n+1−j)n+1

))

g(2) =

n∐i=1

C (αi − α2n−i+1)⊕n−1∐k=1

n−k∐j=1

C(x±α(k)j± x±α(k)

2n−j−k+2

)

⊕n−1∐k=1

n∐j=k+1

C(x±α(2n−j−k+2)k

− x±α(2n−j−k+2)j

)

g(3) =

n∐j=1

(C(x

α(n+1−j)j

− xα

(n+1−j)n+1

)⊕ C(x−α(n+1−j)j

− x−α(n+1−j)n+1

).)

As this decomposition is quite daunting we will now give a few examples of typicalelements from each component

(2.98)

α′

1, xα′1, xα

(3)n−1

+ xα

(3)n

= xαn−1+αn+αn+1+ xαn+αn+1+αn+2

∈ g(0)

xα

(2)n−1

+ xα

(2)n+1

= xαn−1+αn + xαn+1+αn+2∈ g(1)

α′′

i , xα(2)n−3

+ xα

(2)n+3

= xαn−3+αn−2+ xαn+3+αn+4

∈ g(2)

xα

(2)n−1− x

α(2)n+1

= xαn−1+αn − xαn+1+αn+2∈ g(1)


Observe that we have

(2.99)

dim g(0) = n2 + 2n

dim g(1) = 2n

dim g(2) = 3n2 − 2n

dim g(3) = 2n

as well as the following decomposition(2.100)

g[ν] = g(0) ⊗ C[t, t−1]⊕ g(1) ⊗ t1/4C[t, t−1]⊕ g(2) ⊗ t1/2C[t, t−1]⊕ g(3) ⊗ t3/4C[t, t−1]⊕ Ck.

Consider the following ν-stable Lie subalgebra of g:

(2.101) n =∐α∈∆+

Cxα.

and its ν-twisted affinization

(2.102) n[ν] =∐r∈Z

n(r) ⊗ tr/4 ⊕ Ck =∐r∈ 1

4 Z

n(4r) ⊗ tr ⊕ Ck,

where n(r) is the ir-eigenspace of ν in n. As in previous work [CalLM4], we take

(2.103)

n[ν] =∐r∈ 1

4 Z

n(4r) ⊗ tr

n[ν]+ =∐

r∈ 14 Z≥0

n(4r) ⊗ tr

n[ν]− =∐

r∈ 14 Z<0

n(4r) ⊗ tr

3. Principal subspaces of standard A(2)2n -modules

In the spirit of [CalLM4] and other works involving principal subspaces (after[FS1]) we recall the following definition.

Definition 3.1. For any standard g[ν]-module V we define its principal subspaceW to be

W = U (n[ν]) · v,where v ∈ V is a highest weight vector.

Denote by WTL the principal subspace of the standard g[ν]-module V TL ,

(3.1) WTL = U (n[ν]) · vΛ,

where Λ ∈(h(0) ⊕ Ck ⊕ Cd

)∗is the fundamental weight of g[ν] defined by 〈Λ, k〉 =

1,⟨Λ, h(0)

⟩= 0, and 〈Λ, d〉 = 0, and vΛ is a highest weight vector of V TL . So we

have

(3.2) WTL = U (n[ν]−) · vΛ.

We take vΛ = 1 = 1⊗ 1 ∈ V TL (∼= S[ν]⊗ UT ).Consider the surjection

(3.3)FΛ : U (g[ν])→ V TL

a 7→ a · vΛ


and its restriction

(3.4)fΛ : U (n[ν])→WT

L

a 7→ a · vΛ

As in previous work ([CalLM1]-[CalLM4], [P]), our main goal is to describe thekernel of fΛ, giving us a presentation of WT

L .Following [CalLM4](cf. [CalLM1]-[CalLM3]) we set

(3.5) NTL = U (g[ν])⊗U(g[ν])≥0

CvNΛ ,

i.e. the generalized Verma module where,

(3.6) g[ν]≥0 =∐n≥0

g(n) ⊗ tn/4 ⊕ Ck.

We also use the principal subspaces of the generalized Verma module

(3.7) WT,NL = U (n[ν]) · vNΛ ⊂ NT

L .

Now we define the following related surjections

(3.8)

FNΛ : U (g[ν])→ NTL

a 7→ a · vNΛfNΛ : U (n[ν])→WT,N

L

a 7→ a · vNΛas well as

(3.9)

Π : NTL → V TL

a · vNΛ 7→ a · vΛ

π : WT,NL →WT

L

a · vNΛ 7→ a · vΛ.

We now work towards constructing relations which be used to describe WTL as a

quotient of U (n[ν]). Using

(3.10) E+(α, x1)E−(β, x2) = E−(β, x2)E+(α, x1)∏

p∈Z/4Z

(1− ipx

1/42

x1/41

)〈νpα,β〉we obtain the following result:

Theorem 3.1. On V TL we have

(1) Y ν(ι(eα

(j)i

), x)Y ν(ι(eα

(l)k

), x) = 0 if α(j)i and α

(l)k are disconnected.

(2) limx1/42 →ix1/4

1

(x1/21 − x1/2

2 )Y ν(ι(eα

(j)i

), x1)Y ν(ι(eα

(l)k

), x2) = 0 if α(j)i and α

(l)k

are adjacent, but not conjugate-adjacent.

(3) limx1/42 →x1/4

1

(x1/21 +x

1/22 )Y ν(ι(e

α(j)i

), x1)Y ν(ι(eα

(l)k


(l)k are

conjugate-adjacent, but not adjacent.

(4) limx1/42 →x1/4

1

(x1 − x2)Y ν(ι(eα

(j)i

), x1)Y ν(ι(eα

(l)k


(l)k are

adjacent and conjugate-adjacent.


By extracting the appropriate coefficients from the operators in Theorem 3.1 weproduce the following expressions, all of which take any v ∈ V TL to zero via leftaction. For t ∈ 1

4Z,

(3.11) R(α

(j)i , α

(l)k |t)

=∑

m1+m2=−txνα

(j)i

(m1)xνα

(l)k

(m2) ,

where α(j)i and α

(l)k are disconnected,

(3.12)

R(α

(j)i , α

(l)k |t)

=∑

m1+m2+ 12 =−t

(xνα

(j)i

(m1 +

1

2

)xνα

(l)k

(m2)− xνα

(j)i

(m1)xνα

(l)k

(m2 +

1

2

)),

where α(j)i and α

(l)k are adjacent, but not conjugate-adjacent,

(3.13)

R(α

(j)i , α

(l)k |t)

=∑

m1+m2+ 12 =−t

(xνα

(j)i

(m1 +

1

2

)xνα

(l)k

(m2) + xνα

(j)i

(m1)xνα

(l)k

(m2 +

1

2

)),

where α(j)i and α

(l)k are conjugate-adjacent, but not adjacent, and

(3.14)

R(α

(j)i , α

(l)k |t)

=∑

m1+m2+1=−t

(xνα

(j)i

(m1 + 1)xνα

(l)k

(m2)− xνα

(j)i

(m1)xνα

(l)k

(m2 + 1)),

where α(j)i and α

(l)k are adjacent and conjugate-adjacent. In the above expressions,

we take m1 and m2 from the appropriate subsets of 14Z as described in Lemma 2.1.

Observe that Lemma 2.2 implies that the above expressions are constant multi-ples of

(3.15) R(α

(j)i , α

(l)k |t),

where 1 ≤ i, k ≤ n, 1 ≤ j ≤ 2n− 2i+ 2, and 1 ≤ l ≤ 2n− 2k + 2.

Remark 3.1. Of particular interest will be expressions of the form

(3.16) R (αi, αi|t) =∑

m1,m2∈ 12 Z

m1+m2=−t

xναi (m1)xναi (m2) , for 1 ≤ i ≤ n− 1,

(3.17)

R (αn, αn|t) =∑

m1,m2∈ 14 + 1

2 Zm1+m2+ 1

2 =−t

(xναn

(m1 +

1

2

)xναn (m2) + xναn (m1)xναn

(m2 +

1

2

)),

(3.18) R(α′

n, α′

n|t)

=∑

m1,m2∈Z,m1+m1=−t

xνα′n

(m1)xνα′n

(m2) ,

and

(3.19) R(α′

n, αn|t)

=∑

m1∈Z, m2∈ 14 + 1

2 Zm1+m1=−t

xνα′n

(m1)xναn (m2) ,


Lemma 3.1. The above expressions (3.16) – (3.19) can be written in the form

R(α, β|t) = R0(α, β|t) + a,

where R0(α, β|t) is a related finite sum and a ∈ U (n[ν]) n[ν]+.

Proof. We will focus on the expression R(αn, αn|t), the rest will will follow similarly.We can write

R(αn, αn|t) =∑

m1,m2∈ 14 + 1

2 Z<0

m1+m2+ 12 =−t

(xναn

(m1 +

1

2


(m2 +

1

2

))

+∑

m1,m2∈ 14 + 1

2 Zm2≥ 1

4

m1+m2+ 12 =−t

(xναn

(m1 +

1

2


(m2 +

1

2

))

+∑

m1,m2∈ 14 + 1

2 Zm1≥ 1

4

m1+m2+ 12 =−t

(xναn

(m1 +

1

2


(m2 +

1

2

))

Observe that the first summand has finitely many components and the second is inU (n[ν]) n[ν]+ so we will focus on the third. Using Corollary 2.1 we may write

xναn

(m1 +

1

2

)xναn (m2) = xναn (m2)xναn

(m1 +

1

2

)+i

4

(i−4m1 − (−i)−4m1

)xνα′n

(m1 +m2 +

1

2

)and

xναn (m1)xναn

(m2 +

1

2

)= xναn

(m2 +

1

2

)xναn (m1)

− i

4

(i−4m1 − (−i)−4m1

)xνα′n

(m1 +m2 +

1

2

)which implies

xναn

(m1 +

1

2


(m2 +

1

2

)= xναn (m2)xναn

(m1 +

1

2

)+ xναn

(m2 +

1

2

)xναn (m1) .

Which shows that the third summand is a member of U (n[ν]) n[ν]+.�

Remark 3.2. Indeed, all of the expressions (3.11) – (3.14) can be written in theform described in Lemma 3.1. As it turns out (3.16) – (3.19) will play a specialrole, and thus we focus on them to the exclusion of the other similar relations.


Now we consider the following finite sums

R0(αi, αi|t) =∑

m1,m2∈ 12 Z<0

m1+m2=−t

xναi (m1)xναi (m2) for 1 ≤ i ≤ n− 1

(3.20)

R0(αn, αn|t) =∑

m1,m2∈ 14 + 1

2 Z<0

m1+m2+ 12 =−t

(xναn

(m1 +

1

2


(m2 +

1

2

))(3.21)

R0(α′

n, α′

n|t) =∑

m1,m2∈Z<0m1+m2=−t

xνα′n

(m1)xνα′n

(m2)

(3.22)

R0(αn, α′

n|t) =∑

m1∈ 14 + 1

2 Z<0

m2∈Z<0m1+m2=−t

xναn (m1)xνα′n

(m2)

(3.23)

Observe that for w ∈ WTL we have R0(α, β|t)w = 0 for all t ∈ 1

2Z where we view

R0(α, β|t) ∈ U (n[ν])Define the following left ideals of U (n[ν]):

(3.24)

J =U (n[ν])

n∑i=1

∑t≥ 1

2

R0(αi, αi|t)

+ U (n[ν])∑t≥2

R0(α′

n, α′

n|t)

+ U (n[ν])∑t≥ 5

4

R0(αn, α′

n|t)

and

(3.25) IΛ = J + U (n[ν]) n[ν]+.

Note that the above elements which generate the ideal J are homogeneous withweight t,

(3.26) wt(R0(α, β|t)

)= t,

and charge given by

(3.27)

ch(R0(αi, αi|t)

)= 2εi, 1 ≤ i ≤ n,

ch(R0(α

′

nα′

n|t))

= 4εn,

ch(R0(αn, α

′

n|t))

= 3εn,

where εi = (0, . . . , 0, 1, 0, . . . , 0) - the ith-standard basis vector of Rn.


4. Important maps

Define

(4.1) γj =1

2(λj + λ2n−j+1) (∈ h(0)),

for 1 ≤ j ≤ n, where λ1, . . . , λ2n are the fundamental weights of g. Following[CalLM4], we consider the family of characters of the lattice L

(4.2) θj : L→ C×

given by

(4.3)

θj(αj) = −iθj(α2n−j+1) = i

θj(αk) = 1,

for 1 ≤ j ≤ n and k /∈ {j, 2n−j+1}. We also consider the Lie algebra automorphism

(4.4)τγj ,θj : n[ν]→ n[ν]

xνα (m) 7→ θj(α)xνα(m+

⟨α(0), γj

⟩).

This extends to an automorphism of U (n[ν]) which we will also denote by τγj ,θjand has the following action

τγj ,θj

(xνβ′1

(m1) · · ·xνβ′k

(mk))

= θj(β′

1 + · · ·+ β′

k)xνβ′1

(m1 +

⟨(β′

1)(0), γj

⟩)· · ·xν

β′k

(mk +

⟨(β′

k)(0), γj

⟩),

where β′

j ∈ ∆+ and mj ∈ 14Z. Notice that

(4.5) τ−1γj ,θj

= τ−γj ,θ−1j.

For 1 ≤ i, j ≤ n we have

τγi,θi

(xναj (m)

)=θi(αj)x

ναj

(m+

1

2δi,j

),(4.6)

τγi,θi

(xνν(αj)

(m))

=θi(ν(αj))xνν(αj)

(m+

1

2δi,j

),(4.7)

τγi,θi

(xνα′j

(m))

=θi(α′j)x

να′j

(m+ δi,j) if i ≤ j.(4.8)

More generally, for any β ∈ h and 1 ≤ j ≤ n, we have the Lie algebra automor-phisms,

(4.9)τβ,θj : n[ν]→ n[ν]

xνα (m) 7→ θj(α)xνα(m+

⟨α(0), β(0)

⟩),

and their natural extensions to U(n[ν]).

Lemma 4.1. We have

(4.10) τγi,θi

(IΛ + U (n[ν])xναi

(−1

2

))= IΛ,

for 1 ≤ i ≤ n− 1, and

(4.11) τγn,θn

(IΛ + U (n[ν])xναn

(−1

4

))= IΛ.


Proof. Fix 1 ≤ i ≤ n. We will begin by showing

(4.12) τγi,θi(IΛ + U (n[ν])xναi (−k)

)⊂ IΛ,

where k ∈{

14 ,

12

}as appropriate. Observe that

IΛ+U (n[ν])xναi (−k)

= J + U (n[ν]) n[ν]+ + U (n[ν])xναi (−k)

and its is clear that

(4.13) τγi,θi(U (n[ν]) n[ν]+ + U (n[ν])xναi (−k)

)⊂ U (n[ν]) n[ν]+.

Next we will show that for 1 ≤ i ≤ n we have

(4.14) τγi,θi (J) ⊂ IΛ.First assume 1 ≤ i ≤ n− 1. By using Corollary 2.1 we have

τγi,θi(R0 (αi, αi|t)

)= θ(αi)

2R0(αi, αi|t− 1) + 2θ(αi)2xναi (−t+ 1)xναi (0)

∈ J + U (n[ν]) n[ν]+ = IΛ,

for t ≥ 12 . Since the other generators of the ideal J are fixed by τγi,θi , then (4.14)

holds for 1 ≤ i ≤ n− 1.Now assume i = n. Using again Corollary 2.1 we obtain

τγn,θn(R0(αn, αn|t)

)=

θ(αn)2R0(αn, αn|t− 1) + 2θ(αn)2

(xναn

(3

4− t)xναn

(1

4

)+ xναn

(1

4− t)xναn

(3

4

))∈ IΛ

for t ≥ 12 ,

τγn,θn(R0(α′n, α

′n|t)

= θn(α′n)2R0(α′n, α′n|t− 2) + 2θn(α′n)2xνα′n(t+ 2)xνα′n(0) ∈ IΛ

for t ≥ 2, and

τγn,θn(R0(αn, α

′n|t))

=

θn(αn)θn(α′n)R0

(αn, α

′n|t−

3

2

)+ θn(αn)θn(α′n)xνα′n

(5

4− t)xναn

(1

4

)+ θn(αn)θn(α′n)xναn

(3

2− t)xνα′n(0) ∈ IΛ

for t ≥ 5/4, and thus (4.14) is true for i = n. Then (4.12) holds for any 1 ≤ i ≤ n.Now we show

(4.15) τ−1γi,θi

(IΛ) ⊂ IΛ + U (n[ν])xναi (−k) + U (n[ν])xνα′i

(−1) ,

where k ∈{

14 ,

12

}as needed, and 1 ≤ i ≤ n. At the end we will show that

xνα′i(−1) ∈ IΛ + U (n[ν])xναi (−k) .

It is clear that

τ−1γi,θi

(U (n[ν]) n[ν]+) ⊂ U (n[ν]) n[ν]+ + U (n[ν])xναi (−k) + U (n[ν])xνα′i

(−1) ,

and via the above calculations we see that

τ−1γi,θi

(J) ⊂ IΛ + U (n[ν])xναi (−k) + U (n[ν])xνα′i

(−1) ,

thus showing (4.15).


Using Lemma 2.3 we can find a, b, c ∈ C such that

xνα′n

(−1) =aR0(αn, αn|1) + bxναn

(−5

4

)xναn

(1

4

)+ cxναn

(−3

4

)xναn

(−1

4

)∈ IΛ + U (n[ν])xναn

(−1

4

).

Now assume 1 ≤ i ≤ n − 1. By Lemmas 2.2 and 2.3, we can find a, b, c ∈ C suchthat(4.16)

xνα′i

(−1) =aR0(αi, αi|1)xνα′i+1

(0) +

(bxναi

(−1

2

)xνα′i+1

(0) + cxνα′i+1+αi

(−1

2

))xναi

(−1

2

)∈ IΛ + U (n[ν])xναi

(−1

2

).

This finishes the proof of our lemma.�

For 1 ≤ i ≤ n− 1 define the maps

(4.17)

ψαi,θi : U (n[ν])→ U (n[ν])

a 7→ τ−αi,θ−1i

(a)xναi

(−1

2

),

and

(4.18)

ψαn,θn : U (n[ν])→ U (n[ν])

a 7→ τ−αn,θ−1n

(a)xναn

(−1

4

).

Let 1 ≤ j ≤ n. Notice that for 1 ≤ i ≤ n− 1 we have

(4.19) ψαi,θiτγi,θixναj (m) =

xναj

(m− 1

2

)xναi

(− 1

2

)if i = j

xναj(m+ 1

2

)xναi

(− 1

2

)if j = i± 1

xναj (m)xναi(− 1

2

)otherwise,

and

(4.20) ψαn,θnτγn,θnxναj (m) = xναj

(m+

1

2δj,n−1

)xναn

(−1

4

).

Lemma 4.2. For 1 ≤ i ≤ n− 1, we have

ψαi,θiτγi,θi

(IΛ + U (n[ν])xναi

(−1

2

))⊂ IΛ,

and

ψαi,θiτγn,θn

(IΛ + U (n[ν])xναn

(−1

4

))⊂ IΛ.

Proof. Let us begin by focusing on the case when i 6= n. Recall that

IΛ = J + U (n[ν]) n[ν]+.

We will first show that ψαi,θiτγi,θi (U (n[ν]) n[ν]+) ⊂ IΛ. It suffices to focus onthe terms of the form xναj (0), where j ∈ {i− 1, i, i+ 1} \ {n}. Suppose that j = iand observe that

ψαi,θiτγi,θixναi (0) = xναi

(−1

2

)xναi

(−1

2

)= R0(αi, αi|1) ∈ J ⊂ IΛ.


Now suppose that j = i± 1 (j 6= n) and observe that

ψαi,θiτγi,θixναj (0) = xναj

(1

2

)xναi

(−1

2

)= xναi

(−1

2

)xναj

(1

2

)+

1

2xναi+αj (0)

∈ U (n[ν]) n[ν]+ ⊂ IΛ.

Now we will show that ψαi,θiτγi,θi(J) ⊂ IΛ. It suffices to the check that this holdsfor the generators of J which do not commute with xναi

(− 1

2

). Assume j = i ± 1

(j 6= n). Then by Lemma 2.3 we have

(4.21)

ψαi,θiτγi,θi(R0(αj , αj |t)

)=

∑m1,m2≤− 1

2m1+m2=−t

xναj

(m1 +

1

2

)xναj

(m2 +

1

2

)xναi

(−1

2

)

= xναi

(−1

2

)R0(αj , αj |t− 1) +

1

2

∑m1,m2≤− 1

2m1+m2=−t

xναi+αj (m1)xναj

(m2 +

1

2

)

+1

2

∑m1,m2≤− 1

2m1+m2=−t

xναj

(m1 +

1

2

)xναi+αj (m2) .

Notice that the first term is in IΛ so we will concentrate our efforts on the secondtwo terms on the right hand side. Observe that

(4.22)

∑m1,m2≤− 1

2m1+m2=−t

xναi+αj (m1)xναj

(m2 +

1

2

)= xναi+αj

(−t+

1

2

)xναj (0)

+∑

m1,m2≤− 12

m1+m2=−t+ 12

xναi+αj (m1)xναj (m2)

and similarly,

(4.23)

∑m1,m2≤− 1

2m1+m2=−t

xναj

(m1 +

1

2

)xναi+αj (m2) = xναj (0)xναi+αj

(−t+

1

2

)

+∑

m1,m2≤− 12

m1+m2=−t+ 12

xναj (m1)xναi+αj (m2) .

Notice that the first terms in the right hand sides of (4.22) and (4.23) are mem-bers of IΛ, so we will focus on

∑m1,m2≤− 1

2

m1+m2=−t+ 12

xναi+αj (m1)xναj (m2) +∑

m1,m2≤− 12

m1+m2=−t+ 12

xναj (m1)xναi+αj (m2) .


We also observe that on one hand we have,(4.24)[R0

(αj , αj |t−

1

2

), xναi (0)

]= R0

(αj , αj |t−

1

2

)xναi (0)− xναi (0)R0

(αj , αj |t−

1

2

)∈ J + U (n[ν]) n[ν]+ = IΛ,

and on the other hand we have

(4.25)

[R0

(αj , αj |t−

1

2

), xναi (0)

]=

1

2

∑m1,m2≤− 1

2

m1+m2=−t+ 12

xναi+αj (m1)xναj (m2)

+1

2

∑m1,m2≤− 1

2

m1+m2=−t+ 12

xναj (m1)xναi+αj (m2) ,

which together with our previous calculations imply

ψαi,θiτγi,θi(R0(αj , αj |t)

)⊂ IΛ for j = i± 1, j 6= n, i 6= n.

Now we shift our attention to ψαn−1,θn−1τγn−1,θn−1(R0(αn, αn|t)). Using Lemma2.3 again we obtain

ψαn−1,θn−1τγn−1,θn−1

(R0(αn, αn|t)

)=

∑m1,m2∈ 1

4 + 12 Z<0

m1+m2+ 12 =−t

(xναn (m1 + 1)xναn

(m2 +

1

2

)+ xναn

(m1 +

1

2

)xναn (m2 + 1)

)xναn−1

(−1

2

)

= xναn−1

(−1

2

)R0(αn, αn|t− 1)

+1

2

∑m1,m2∈ 1

4 + 12 Z<0

m1+m2+ 12 =−t

(xναn (m1 + 1)xναn−1+αn (m2) + xναn

(m1 +

1

2

)xναn−1+αn

(m2 +

1

2

))

+1

2

∑m1,m2∈ 1

4 + 12 Z<0

m1+m2+ 12 =−t

(xναn−1+αn

(m1 +

1

2

)xναn

(m2 +

1

2

)+ xναn−1+αn (m1)xναn (m2 + 1)

)

= xναn−1

(−1

2

)R0(αn, αn|t− 1) +

[R0(αn, αn|t), xναn−1

(1

2

)]∈ IΛ.

We now compute(4.26)

ψαn−1,θn−1τγn−1,θn−1

(R0(α′n, α

′n|t))

=∑

m1,m2≤−1m1+m2=−t

xνα′n(m1 + 1)xνα′n(m2 + 1)xναn−1

(−1

2

)

= xναn−1

(1

2

)R0 (α′n, α

′n|t− 2) +

[R0 (α′n, α

′n|t) , ανn−1

(1

2

)]+ b ∈ IΛ,

where b ∈ U (n[ν]) n[ν]+. Similarly one can show that

(4.27) ψαn−1,θn−1τγn−1,θn−1

(R0(αn, α

′n|t))∈ IΛ.


Observe also that for 1 ≤ i ≤ n− 1

ψαi,θiτγi,θi

(xναi

(−1

2

))= xναi (−1)xναi

(−1

2

)=

1

2R0

(αi, αi|

3

2

)and

Now we will consider the map ψαn,θnτγn,θn . It is easy to see that ψαn,θnτγn,θn (U (n[ν]) n[ν]+) ⊂IΛ. In order to prove that ψαn,θnτγn,θn (J) ⊂ IΛ we will again consider only thegenerators of J which do not commute with xναn(−1/4). Notice that we have

ψαn,θnτγn,θn(R0(αn−1, αn−1|t)

)=

∑m1,m2≤− 1

2m1+m2=−t

xναn−1

(m1 +

1

2

)xναn−1

(m2 +

1

2

)xναn

(−1

4

)

=xναn

(−1

4

)R0(αn−1, αn−1|t− 1)

+1

2

∑m1,m2≤− 1

2m1+m2=−t

xναn−1

(m1 +

1

2

)xναn−1+αn

(m2 +

1

4

)

+1

2

∑m1,m2≤− 1

2m1+m2=−t

xναn−1+αn

(m1 +

1

4

)xναn−1

(m2 +

1

2

).

The first term on the right hand side of the above equation is a member of J so wewill focus on the second two, the second of which may be rewritten as follows(4.28)∑

m1,m2≤− 12

m1+m2=−t

xναn−1

(m1 +

1

2

)xναn−1+αn

(m2 +

1

4

)= xναn−1

(0)xναn−1+αn

(−t+

3

4

)

+∑

m1,m2≤− 12

m1+m2=−t+ 12

xναn−1(m1)xναn−1+αn

(m2 +

1

4

),

and similarly the third(4.29)∑

m1,m2≤− 12

m1+m2=−t

xναn−1+αn

(m1 +

1

4

)xναn−1

(m2 +

1

2

)= xναn−1+αn

(−t+

3

4

)xναn−1

(0)

+∑

m1,m2≤− 12

m1+m2=−t+ 12

xναn−1+αn

(m1 +

1

4

)xναn−1

(m2) ,

Observe that the first two terms in the right hand sides of (4.28) and (4.29) aremembers of IΛ so we consider∑m1,m2≤− 1

2

m1+m2=−t+ 12

xναn−1(m1)xναn−1+αn

(m2 +

1

4

)+

∑m1,m2≤− 1

2

m1+m2=−t+ 12

xναn−1+αn

(m1 +

1

4

)xναn−1

(m2) .

Using a calculation similar to that carried out in (4.24) and (4.25), except thistime with


[R0

(αn−1, αn−1|t−

1

2

), xναn

(1

4

)]we finish this case.

We also have

ψαn,θnτγn,θn(R0(αn, αn|t)

)= xναn

(−1

4

)R0(αn, αn|t) + aR0

(α′

n, αn|t+1

4

)+ b,

where a ∈ C is determined by Lemma 2.3 and b ∈ U (n[ν]) n[ν]+. We also have

ψαn,θnτγn,θn

(R0(α

′

n, αn|t))

= xναn

(−1

4

)(R0(α

′

n, αn|t) + cR0

(α′

n, α′n|t+

1

4

)and

ψαn,θnτγn,θn

(R0(α

′

n, α′

n|t))

= xναn

(−1

4

)R0(α

′

n, α′

n|t),

where the nonzero constant c ∈ C is determined by Lemma 2.3.We finish the proof by observing that

ψαn,θnτγn,θn

(xναn

(−1

4

))= xναn

(−1

4

)xναn

(−1

4

)= R0

(αn, αn|

1

2

).

This finishes the proof of our lemma. �

For 1 ≤ i ≤ n consider the maps

(4.30) eαi : V TL → V TL

and their restrictions to the principal subspace WTL ⊂ V TL . Observe that we have

(4.31) eαixνα (m) = C(α, αi)x

να

(m−

⟨α(0), αi

⟩)eαi

and

(4.32) eαi · 1 =4

σ(αi)xναi (−k) · 1,

where k ∈{

14 ,

12

}as appropriate. For i < n, we may write

(4.33) eαixναj (m) =

C(αj , αi)x

ναj (m− 1) eαi if i = j

C(αj , αi)xναj

(m+ 1

2

)eαi if j = i± 1

C(αj , αi)xναj (m) eαi otherwise

and

(4.34) eαnxναn (m) = C(αn, αn)xναn

(m− 1

2

)eαn .

Thus

(4.35) eαi(a · 1) = Aσ(),θi()C(·,·) ψαi,θi(a) · 1,

where a ∈ U (n[ν]) and Aσ(),θi()C(·,·) ψγi,θi is a non zero constant depending on the three

indexed maps: (2.53), (4.2), and the commutator map.We will now introduce another useful map, ∆T . An untwisted version of this

map was introduced in [Li2] and a twisted version was used in [CalLM4]. Set

(4.36) ∆T (λi,−x) = i(4n+2)(λi)(0)x(λi)(0)E+(−λi, x),


where

(4.37) E+(−λi, x) = exp

∑n∈ 1

4 Z+

−λi(n)

nx−n

∈ (EndV TL)

[[x−1/4]]

and thus

(4.38) ∆T (λi,−x) ∈(EndV TL

)[[x1/4, x−1/4]].

Observe that we have the following

E+(−λi, x)E−(−αj , xj) =

(1−

x1/2j

x1/2

)δi,jE−(−αj , xj)E+(−λi, x)(4.39)

E+(−λi, x)E−(−α′

j , xj) =(

1− xjx

)δi,jE−(−α

′

j , xj)E+(−λi, x),(4.40)

With this construction we have

∆Tc (λi,−x) :WT

L →WTL(4.41)

a · 1 7→ τγi,θi(a) · 1(4.42)

5. The Main Results

Theorem 5.1. We have

(5.1) Ker fΛ = IΛ,

or equivalently

(5.2) Ker πΛ = IΛ · vNΛ .

Proof. As has become standard in these results involving principal subspaces wefollow an argument similar to that in [CalLM1]-[CalLM4] and [P]. First we observethat IΛ · vNΛ ⊂ Ker πΛ. Assume by way of contradiction that there is an elementa ∈ U (n[ν]) such that a ·vNΛ ∈ Ker πΛ\IΛ ·vNΛ . In addition we may assume that a ishomogeneous with respect to all gradings. We assume that a is of minimal chargewith respect to the partial order, and of smallest possible weight for this minimalcharge.

We will first show that we have

(5.3) a ∈ IΛ + U (n[ν])xναn

(−1

4

).

Suppose that this is not true. Then by Lemma 4.1 we see that

(5.4) τγn,θn(a) · vNΛ /∈ IΛ · vNΛ .

On the other hand, since a · vNΛ ∈ Ker πΛ we have a · 1 = 0. By using (4.41) we get

(5.5) τγn,θn(a) · 1 = 0,

and thus

(5.6) τγn,θn(a) · vNΛ ∈ Ker πΛ.

The elements τγn,θn(a) and a have the same charge, and

(5.7) wt (τγn,θn(a)) < wt(a),


which contradicts the minimality of a and proves (5.3). There exist homogeneouselements b ∈ IΛ and c ∈ U (n[ν]) such that

(5.8) a = b+ cxναn

(−1

4

).

Observe that wt(a) = wt(b) and wt(c) < wt(a), a and b have the same charge andthe charge of c is less than the charge of a.

Next we will show that

(5.9) cxναn

(−1

4

)∈ IΛ.

Since

0 = (a− b) · 1 = cxναn

(−1

4

)· 1 = eαn(ταn,θn(c) · 1),

where the last equality holds up to a nonzero constant, and eαn is injective, thenwe have ταn,θn(c · 1) = 0, and thus

τan,θn(c) · vNΛ ∈ Ker πΛ.

Note that the charge of ταn,θn(c) is less than the charge of a, and so we have

ταn,θn(c) · vNΛ ∈ IΛ · vNΛ .

By Lemma 4.1 we have

τ−γn,θ−1n

(ταn,θn(c)) ∈ IΛ + U (n[ν])xναn

(−1

4

),

and by Lemma 4.2, we have

cxναn

(−1

4

)= ψαn,θnτγn,θnτ−γn,θ−1

n(ταn,θn(c)) ∈ IΛ,

which proves (5.9).Now (5.8) and (5.9) imply that

a = b+ cxναn

(−1

4

)∈ IΛ,

and thus a · vNΛ ∈ IΛ · vNΛ , which contradicts our original assumption and finishesthe proof.

�

Theorem 5.2. We have the following family of short exact sequences

(5.10) 0→WTL

eαi−−→WTL

∆Tc (λi,−x)−−−−−−−→WT

L → 0

for 1 ≤ i ≤ n.

Proof. We begin by observing that eαi is clearly injective while ∆Tc (λi,−x) is clearly

surjective.Suppose that a · 1 ∈ ker

(∆Tc (λi,−x)

). So we have

0 = ∆Tc (λi,−x)(a · 1) = τγi,θi(a) · 1,

which implies that τγi,θi(a) ∈ IΛ and thus by Lemma 4.1 we see that

a ∈ IΛ + U (n[ν])xναi (−k) ,


where k = 12 for 1 ≤ i ≤ n−1 and k = 1

4 if i = n. So we have the equivalence givenby

(5.11) a · 1 ∈ ker(∆Tc (λi,−x)

)if and only if a ∈ IΛ + U (n[ν])xναi(−k).

Now we switch gears and suppose that a · 1 ∈ im (eαi). So there is b ∈ U (n[ν])such that a · 1 = bxναi (−k) · 1 and thus

a ∈ IΛ + U (n[ν])xναi (−k) .

So we have the equivalence given by

(5.12) a · 1 ∈ im (eαi) if and only if a ∈ IΛ + U (n[ν])xναi (−k) .

By (5.11) and (5.12) we have

a · 1 ∈ im (eαi) if and only if a · 1 ∈ ker(∆Tc (λi,−x)

),

which finishes the proof. �

We now recall the multi-grading on V TL as described in Section 2. In particular,the weight grading given by the eigenvalue of the Lν (Virasoro) operator and thecharge given by an n − tuple each entry being the eigenvalue of λi + λ2n−i+1 =(λi+λ2n−i+1)(0) as previously defined. We now restrict these gradings to WT

L andinstead use the weight grading given by 2Lν(0) and define the graded dimension ofWTL :

(5.13) χ(x; q) = tr|WTLxλ1+λ2n

1 · · ·xλn+λn+1n q2Lν ∈ q n2 C[[x, q]],

where x1,. . . , xn and q are commuting formal variables with x = (x1, . . . , xn). Inthe spirit of [CalLM4] ([CalLM3], [P]) we use the following modification:

(5.14) χ′(x; q) = q−

n2 χ(x; q) ∈ C[[x, q]].

We denote(WTL

)(k,l)

to be the homogeneous subspace of WTL given by elements

with charge k = (k1, . . . , kn) and weight l. Equipped with the homogeneous gradedcomponents and the graded dimension we can state a Corollary to Theorem 5.2.

Corollary 5.1. For 1 ≤ i ≤ n− 1, the following sequences are exact(5.15)

0→(WTL

)(k−εi,l+ki−1−2ki+ki+1+1)

eαi−−→(WTL

)(k,l)

∆Tc (λi,−x)−−−−−−−→

(WTL

)(k,l−ki)

→ 0,

and(5.16)

0→(WTL

)(k−εn,l+kn−1−kn+ 1

2 )

eαn−−→(WTL

)(k,l)

∆Tc (λn,−x)−−−−−−−→

(WTL

)(k,l−kn)

→ 0,

where εi = (0, . . . , 1, . . . , 0) with the 1 in the ith-component. Moreover, we have(5.17)

χ′(x; q) = χ

′(x1, . . . , xiq, . . . , xn; q)+xiqχ

′(x1, . . . , xi−1q

−1, xiq2, xi+1q

−1 . . . , xn; q),

for 1 ≤ i ≤ n− 1 and

(5.18) χ′(x; q) = χ

′(x1, . . . , xn−1, xnq; q) + xnq

12χ′(x1, . . . , xn−1q

−1, xnq; q),

Finally we can solve the above recursion (cf. [A]) giving the following.


Corollary 5.2. We have

(5.19) χ′(x; q) =

∑k∈(Z≥0)

n

qktMk

2

(q)k1 · · · (q)knxk11 · · ·xknn ,

where for 1 ≤ i, j ≤ n, Mi,j = 2⟨(αi)(0), (αj)(0)

⟩. In other words

M =

2 −1 0 · · · 0 0 0−1 2 −1 · · · 0 0 00 −1 2 · · · 0 0 0...

......

. . ....

......

0 0 0 · · · 2 −1 00 0 0 · · · −1 2 −10 0 0 · · · 0 −1 1

Although there is a unique level one standard A

(2)2n -module (as well as a unique

twisted VL-module), in parallel with [CalLM3], we can also consider certain shiftedprincipal subspaces and their characters. We do not go into details of constructionof these shifted subspaces, so instead we only consider their normalized characters.For 0 ≤ i ≤ n, we denote

χ′i(x; q) := χ′(x1, ..qxi.., xn; q); 1 ≤ i ≤ n− 1, and χ′n(x; q) := χ

′(x1, ..., q

12xn; q)

where χ′

0(x1, ..., xn; q) := χ′(x1, ...., xn; q) is given by (5.19).

6. Towards modularity of χ′

i(1; q)

In this section we discuss ”modularity” of specialization of χ′

i(x; q) in connectionto Nahm’s conjecture [Za]. Because the Nahm sum involves the Cartan matrix ofthe tadpole diagram Tn we expect

Conjecture 1. After the inclusion of q−an/24 for some an ∈ Q (”central charge”),

the sum q−an/24χ′

i(1; q) is a weight zero modular form, where 1 = (1, ..., 1).

Here we prove the conjecture in the case of n = 2, that is for A(2)4 , For n = 1,

the above conjecture trivially follows from a pair of Euler’s identities

(−q1/2; q)∞ =

∞∑n=0

qn2

2

(q; q)n,

(−q; q)∞ =

∞∑n=0

qn2+n

2

(q; q)n.

For A(2)4 we have


Theorem 6.1.

∞∑n,m=0

qm2+n2

2 −mn

(q)m(q)n=

(−q1/2;q)2∞(q;q)∞

∑n∈Z

(−1)nq2n2−n2 .(6.1)

∞∑n,m=0

qm2+m+n2

2 −mn

(q)m(q)n=

(−q1/2;q)2∞(q;q)∞

∑n∈Z

(−1)nq2n2− 3n2(6.2)

∞∑n,m=0

qm2+n2+n

2 −mn

(q)m(q)n=

(−q;q)2∞(q;q)∞

∑n∈Z

(−1)nq2n2

(6.3)

Proof. We start from Gauss’ identity

(−xq1/2; q)∞ =

∞∑n=0

qn2/2xn

(q; q)n.

This leads to

∞∑n,m=0

qm2+n2

2 −mn

(q)m(q)n=

∞∑m=0

qm2

(−q1/2−m; q)∞(q; q)m

(−q1/2; q)∞

∞∑m=0

qm2/2(−q1/2; q)m

(q; q)m,

where we used (−q1/2−m; q)∞ = q−m2/2(−q1/2; q)∞(−q1/2; q)m. The remaining

sum is essentially the sum side appearing in Ramanujan-Slater/Gollnitz-Gordontheorem (see formula (2.8.4) in [LSZ]). After we rewrite the product side in (2.8.4)loc.cit. we get

∞∑m=0

qm2

(−q; q2)m(q2; q2)m

=(−q; q2)∞(q2; q2)∞

∑n∈Z

(−1)nq4n2−n.

This proves the first identity.For other two identities, similar easy manipulation with q-series reduces ev-

erything to known Ramanujan-Slater type identities (see formulas (2.8.1)-(2.8.3)loc.cit.). We omit the details. �

As it is well-known, Weber’s function f(τ) = q−1/48(−q1/2; q)∞ is modular, as

is η(τ) = q1/24(q; q)∞ and θ(τ) =∑n∈Z(−1)nq2(n− 1

8 )2 . Therefore

q−596χ

′(1; q) =

f(τ)2

η(τ)θ(τ),

also modular with a2 = 54 . Similarly for two other cases.

Remark 6.1. All identities in Theorem 6.1 were already predicted in [Za]. Moreprecisely, based on extensive numerical searches, Zagier lists possible candidates forrank 2 triples (A,B,C), with ”central charge” c := L(ξA)/L(1) ∈ Q (here L(·) isRoger’s dilogarithm function [Ki, Za]) for which

fA,B,C(τ) =∑

n=(n1,n2)∈Z2≥0

qntAn+ntB+C

(q)n1(q)n2


is (or is expected) to be modular (see Table 2 loc.cit.). In our case

A =

[2 −1−1 1

]The corresponding Q-system is 1 − Q1 = Q2

1Q−12 , 1 − Q2 = Q−1

1 Q2, which has a

unique solution inside (0, 1)2: Q1 =√

2− 1 and Q2 = 1−√

22 . Therefore L(ξA) :=

L(√

2− 1) + L(1−√

22 ) = 5

24π2 and c = 5

4 .In addition to our three identities, Zagier’s list predicts two additional identities

corresponding to vectors B = (2,−3/2) and B = (1,−1/2). For B = (1,−1/2) weget a closely related identity

(6.4)

∞∑n,m=0

qm2+n2−n

2 +m−mn

(q)m(q)n= 2

(−q; q)2∞

(q; q)∞

∑n∈Z

(−1)nq2n2+n,

which is clearly modular after an inclusion of a power of q. For B = (2,−3/2) weget

(6.5)

∞∑n,m=0

qm2+2m+n2−3n

2 −mn

(q)m(q)n= 2q−1 (−q; q)2

∞(q; q)∞

∑n∈Z

(−1)nq2n2

,

which is also modular after multiplication with q25/24. Presumably (see p.14 in[VZ]), our five examples give a complete list of modular q-hypergometric serieswith A as above.

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Heidelberg-New York (2006), 3-65.


Department of Mathematics, The Graduate Center and New York City Col-

lege of Technology, City University of New York, New York, NY 10016

E–mail address: [email protected]

Department of Mathematics and Statistics, University at Albany (SUNY),

Albany, NY 12222


Department of Mathematics and Computer Science, Colorado College, Col-

orado Springs, CO 80903


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Introduction q A q - City University of New Yorkwebsupport1.citytech.cuny.edu/Faculty/ccalinescu/A_2n_2... · 2016. 5. 18. · cipal subspace WT L ˆV T L and describe certain ideals