Holographic Wilsonian RG and BH Sliding Membraneicts.ustc.edu.cn/chinese/seminar/transparencies/zhou yang/ZHOU YANG... · Holographic Wilsonian RG and BH Sliding Membrane Yang Zhou

Holographic Wilsonian RG and BH Sliding Membrane

Yang Zhoubased on 1102.4477 by SJS,YZ (JHEP05:030,2011)related references: 0809.3808 by N.Iqbal and H.Liu

1010.1264 by Heemskerk and J.Polchinski1010.4036 by Faulkner,Liu and Rangamani

July 7, 2011ICTS-USTC,Hefei

Yang Zhou based on 1102.4477 by SJS,YZ (JHEP05:030,2011) related references: 0809.3808 by N.Iqbal and H.Liu 1010.1264 by Heemskerk and J.Polchinski 1010.4036 by Faulkner,Liu and Rangamani ()Holographic Wilsonian RG and BH Sliding MembraneJuly 7, 2011 ICTS-USTC,Hefei 1 / 28

Outline

1 Introduction: Membrane paradigm

2 Holographic Wilsonian RG (HWRG)

3 Equivalence

4 Double Trace Flow


Introduction: Membrane paradigm

Motivations

Map the Hamilton-Jacobi flow and Exact Wilson RG of QuantumField Theory

Interpolating physical quantities evaluating at horizon and at theboundary



Motivations

Map the Hamilton-Jacobi flow and Exact Wilson RG of QuantumField Theory

Interpolating physical quantities evaluating at horizon and at theboundary



Formulations

Many approaches to Holographic RG after 1997 (AdS=CFT,Maldacena) which I apologize for not listing here.(E.T.Akhmedov,V.Balasubramanian eta, Susskind andWitten,J.d.Boer and Verlinde2)

Sliding membrane approach (2008, N.Iqbal and H.Liu)F

Wilsonian fluid/gravity (June.2010, Bredberg,Keeler,Lysov andStrominger.)

Holographic Wilsonian approach (Oct.2010,I.heemskerk andJ.Polchinski; Oct.2010, T.Faulkner,M.Rangamani and H.Liu)F



Formulations







Formulations







Formulations







Membrane paradigm

Black hole stretched horizon ⇐⇒ a fictitious membranewhich means, to an external observer, a black hole appears to behaveexactly like a dynamical fluid membrane.

Viscous response:

η =1

16πGN(Damour,1978) (1)

Electric response:

σ =1

e2(Znajek,1978; Damour,1978) (2)



Membrane paradigm


Viscous response:

η =1


Electric response:

σ =1




Membrane paradigm


Viscous response:

η =1


Electric response:

σ =1




The idea is:In-falling boundary conditions can be viewed asphysical properties of membrane located at the stretched horizon.

An action for membrane (Parikh and Wilczek 1998) is a surface termwhich cancel the boundary term action at the horizon.

For FFO to find a nonsingular field φ, near the horizon we have

φ(r , t, x) = φ(τ, x) (3)

where τ is the Eddington-Finklestein coordinate

dτ = dt +

√grr

gttdr . (4)

This implies in-falling condition ∂rφ =√

grr

gtt∂tφ at the horizon.






φ(r , t, x) = φ(τ, x) (3)


dτ = dt +

√grr

gttdr . (4)


grr







φ(r , t, x) = φ(τ, x) (3)


dτ = dt +

√grr

gttdr . (4)


grr




For a massless scalar probe with action

Sscalar = −1

2

∫r>rh

dd+1x√−g

1

q(r)(Oφ)2 (5)

the membrane action is

Ssurf =

∫Σ

ddx√−γ(

Π(rh, x)√−γ

)φ(rh, x) , Πmb ≡Π(rh, x)√−γ

= −√

g rr∂rφ

q(r)(6)

together with in-falling condition ∂rφ =√

grr

gtt∂tφ, give (orthonormal

basis)

ξmb ≡Πmb

∂tφ(rh)=

1

q(rh)=

1

16πGN(ShearViscosity !) (7)



For a Maxwell field probe:

SMaxwell = −∫

r>rh

dd+1x√−g

1

4g(r)2F 2 (8)

the membrane action is

Ssurf =

∫Σ

ddx√−γ(

jµ√−γ

)Aµ , Jµmb ≡

jµ√−γ

(9)

with in-falling condition Fri =√

grr

gttFti , give

σmb =1

g2(rh)(MembraneConductivity !) (10)



Sliding membrane

The idea is:View a cutoff surface at finite “rc” as a “sliding membrane”where the “disturb” and “response” can be defined.

For a massless scalar probe:

ξ(rc , kµ) =Π(rc , kµ)

iωφ(rc , kµ)(11)


σ(rc , kµ) =j i (rc , kµ)

Fit(rc , kµ)(12)



Sliding membrane




iωφ(rc , kµ)(11)



Fit(rc , kµ)(12)



Sliding membrane




iωφ(rc , kµ)(11)



Fit(rc , kµ)(12)



Flow involving EOM

Using the definition, one can represent EOM:

∂rσ = iω

√grr

gtt

[σ2

Σ(r)

(1− k2g ii

ω2g tt

)− Σ(r)

](13)

where

Σ(r) =

√−gg ii

√grrgtt

(14)

Regular condition of (13) at the horizon ⇐⇒ infalling boundarycondition.

Smooth interpolation between Horizon value and Boundary value

Even without any RG formulism, one does have flow equation forσ(rc)!

Yang Zhou based on 1102.4477 by SJS,YZ (JHEP05:030,2011) related references: 0809.3808 by N.Iqbal and H.Liu 1010.1264 by Heemskerk and J.Polchinski 1010.4036 by Faulkner,Liu and Rangamani ()Holographic Wilsonian RG and BH Sliding MembraneJuly 7, 2011 ICTS-USTC,Hefei 10 /

28


Flow involving EOM


∂rσ = iω

√grr

gtt

[σ2

Σ(r)

(1− k2g ii

ω2g tt

)− Σ(r)

](13)

where

Σ(r) =

√−gg ii

√grrgtt

(14)





28


Flow involving EOM


∂rσ = iω

√grr

gtt

[σ2

Σ(r)

(1− k2g ii

ω2g tt

)− Σ(r)

](13)

where

Σ(r) =

√−gg ii

√grrgtt

(14)





28


Flow involving EOM


∂rσ = iω

√grr

gtt

[σ2

Σ(r)

(1− k2g ii

ω2g tt

)− Σ(r)

](13)

where

Σ(r) =

√−gg ii

√grrgtt

(14)





28

Holographic Wilsonian RG (HWRG)

Split path integral in the bulk

Boundary Split:

Z =

∫boundary

DMkδ<1DMkδ>1e−S =

∫boundary

DMkδ<1e−S(δ)

:= 〈exp(−s(δ))〉 (15)

Bulk Split:

Z =

∫bulk

Dφz>εDφz=εDφz<εe−SIR(z>ε)−SUV (z<ε)

=

∫bulk

DφΨIR(ε, φ)ΨUV (ε, φ) (16)


28


Split path integral in the bulk

Boundary Split:

Z =

∫boundary

DMkδ<1DMkδ>1e−S =

∫boundary

DMkδ<1e−S(δ)

:= 〈exp(−s(δ))〉 (15)

Bulk Split:

Z =

∫bulk

Dφz>εDφz=εDφz<εe−SIR(z>ε)−SUV (z<ε)

=

∫bulk

DφΨIR(ε, φ)ΨUV (ε, φ) (16)


28


Map between Boundary and Bulk

Natural Map:

ΨIR(ε, φ) =

∫DM|kδ<1 exp

{−S0 +

1

κ2

∫ddx φi (x)Oi (x)

}(17)

s(δ) and SB :

e−s(δ) =

∫Dφe

∫ddxφ(x)O(x)e−SB (18)

withSB = − log ΨUV (ε, φz=ε) (19)


28


Map between Boundary and Bulk

Natural Map:

ΨIR(ε, φ) =

∫DM|kδ<1 exp

{−S0 +

1

κ2

∫ddx φi (x)Oi (x)

}(17)

s(δ) and SB :

e−s(δ) =

∫Dφe

∫ddxφ(x)O(x)e−SB (18)

withSB = − log ΨUV (ε, φz=ε) (19)


28


Holographic Wilsonian RGE

Cutoff independent of full amplitude:

0 =d

dεZ =

d

dε< e−s(δ) > (20)

In the bulk level, IR dynamics plus a boundary action SB should beuseful, since Wilson told us we need to integrate out the UV region.

Bulk HWRG equation then becomes:

∂ε

[∫ zH

εL + SB

]= 0 (21)

which gives∂εSB = −HIR (22)

.

What is the power of above HWRG?


28




0 =d

dεZ =

d

dε< e−s(δ) > (20)



∂ε

[∫ zH

εL + SB

]= 0 (21)


.



28




0 =d

dεZ =

d

dε< e−s(δ) > (20)



∂ε

[∫ zH

εL + SB

]= 0 (21)


.



28




0 =d

dεZ =

d

dε< e−s(δ) > (20)



∂ε

[∫ zH

εL + SB

]= 0 (21)


.



28


Holographic Wilsionian RG flow

One way to determine SB is taking use of

∂εSB [Aµ, ε] + HIR [Aµ,δSB

δAµ] = 0 (23)

For the Maxwell:

SB = −1

2

∫ddk

(2π)d

(−G 00(k, ε)A0(k)A0(−k) + G ii (k, ε)Ai (k)Ai (−k)

)(24)

Using flow equation, we have flow equations for coefficients:

∂εG00 = − (G 00)2√

−gg ttg zz+√−gg ttg iik2

∂εGii = − (G ii )2√

−gg iig zz−√−gg ttg iiω2 . (25)


28





δAµ] = 0 (23)

For the Maxwell:

SB = −1

2

∫ddk

(2π)d


)(24)


∂εG00 = − (G 00)2√


∂εGii = − (G ii )2√



28





δAµ] = 0 (23)

For the Maxwell:

SB = −1

2

∫ddk

(2π)d


)(24)


∂εG00 = − (G 00)2√


∂εGii = − (G ii )2√



28


RG Solutions

we want to find solutions for the RG equations (25). We start byassuming

1

G 00=

1

G 00(0)

+ k2 1

G 00(1)

+ · · · ,1

G ii=

1

G ii(0)

+ ω2 1

G ii(1)

+ · · · (26)

and

∂εG00(0) = −

(G 00(0))

2

√−gg ttg zz

, ∂εGii(0) = −

(G ii(0))

2

√−gg iig zz

. (27)

We have the following solutions by imposing simple vanishingboundary conditions for G 00

(1),Gii(1)

1

G 00(1)(z)

=

∫ z

0−√−gg ttg ii

G 00(0)

,1

G ii(1)(z)

=

∫ z

0

√−gg ttg ii

G ii(0)

. (28)

We see that k2 and ω2 correction are controlled by zero ordersolutions in the low frequency approximation.


28


Flow for transport coefficient

Define a transport coefficient:

σ ≡ J i

Ei=

J i

−∂0Ai + ∂i A0

. (29)

With the definition of current

Jµ ≡ δSB

δAµ(30)

From SB we have

− 1

G 00=

A0

J0,

1

G ii=

Ai

J i. (31)

Express σ by G1

σ=

i

ω

(ω2

G ii− k2

G 00

). (32)


28


Flow for transport coefficient

Define a transport coefficient:

σ ≡ J i

Ei=

J i

−∂0Ai + ∂i A0

. (29)

With the definition of current

Jµ ≡ δSB

δAµ(30)

From SB we have

− 1

G 00=

A0

J0,

1

G ii=

Ai

J i. (31)

Express σ by G1

σ=

i

ω

(ω2

G ii− k2

G 00

). (32)


28


Conductivity Flow

Using the flow equations for G coefficients we obtain

− ∂εσ

iω=

(σ2

(1√

−gg iig zz− k2

ω2

1√−gg ttg zz

)−√−gg ttg ii

),

(33)This is what we have in the sliding membrane paradigm!

Under coordinate transformation z = 1/r , one has

∂ε = −r2∂r ,√−g (z)g

ttg ii =√−g (r)g

ttg ii r2, (34)

1√−g (z)g

zzg ii=

1√−g (r)g

rrg iir2 . (35)


28


Conductivity Flow

Using the flow equations for G coefficients we obtain

− ∂εσ

iω=

(σ2

(1√

−gg iig zz− k2

ω2

1√−gg ttg zz

)−√−gg ttg ii

),

(33)This is what we have in the sliding membrane paradigm!

Under coordinate transformation z = 1/r , one has

∂ε = −r2∂r ,√−g (z)g

ttg ii =√−g (r)g

ttg ii r2, (34)

1√−g (z)g

zzg ii=

1√−g (r)g

rrg iir2 . (35)


28


0 10 20 30 40 50

1.0

1.5

2.0

2.5

r

ReH

ΣL

0 10 20 30 40 50

-6

-5

-4

-3

-2

-1

0

r

ImHΣ

L

Figure: r flow of AC conductivity, with d = 4 AdS-black hole background andω = 2, 1.5, 1, from up to down in the left Figure and inversely in the right one.For d > 4 AdS-black hole, the behavior of solutions are similar.


28

Equivalence

Equivalence between HWRG and Sliding Membrane

There are two RG flows. One is the flow given by the classicalequation of motion, and the other is the flow from integrating out thegeometry. Here we want to prove the equivalence of the two.

No split amplitude:

Z1[φ0] =

∫bulk

Dφe−S ' exp(−S [φc ]) (36)

Split amplitude:

Z2 =

∫bulk

DφΨIR(ε, φ)ΨUV(ε, φ) , (37)

Classical approximation:

ΨIR(ε, φ) =

∫bulk

Dφz>εe−S(z>ε) = e−SIR [φIR

c ] (38)

ΨUV(ε, φ) =

∫bulk

Dφz<εe−S(z<ε) = e−SUV [φUV

c ], (39)


28

Equivalence


There are two RG flows. One is the flow given by the classicalequation of motion, and the other is the flow from integrating out thegeometry. Here we want to prove the equivalence of the two.No split amplitude:

Z1[φ0] =

∫bulk

Dφe−S ' exp(−S [φc ]) (36)

Split amplitude:

Z2 =

∫bulk



ΨIR(ε, φ) =

∫bulk


c ] (38)

ΨUV(ε, φ) =

∫bulk


c ], (39)


28

Equivalence


There are two RG flows. One is the flow given by the classicalequation of motion, and the other is the flow from integrating out thegeometry. Here we want to prove the equivalence of the two.No split amplitude:

Z1[φ0] =

∫bulk

Dφe−S ' exp(−S [φc ]) (36)

Split amplitude:

Z2 =

∫bulk



ΨIR(ε, φ) =

∫bulk


c ] (38)

ΨUV(ε, φ) =

∫bulk


c ], (39)


28

Equivalence

Sufficient:

Explicitly we write

Sε[φH , φ, φ0] = SIR [φIRc [φH , φ]] + SUV [φUV

c [φ, φ0]]. (40)

Z2 =

∫Dφe−Sε[φ] = e−Sε[φH ,φ∗,φ0] (41)

where φ∗ is a solution for

δSε

δφ=

δSUV

δφ+

δSIR

δφ= 0, (42)

which is nothing but ΠIR = ΠUV . Momentum continuity isESSENTIAL for bulk classical calculation!In order for Z1 = Z2, it is SUFFICIENT to have

φ∗ = φc(z)∣∣z=ε

so that φ∗ = φc . (43)

We should also notice that it guarantees the ε independence of the Z2

and Sε, i.e, dSεdε = 0!


28

Equivalence

Sufficient:

Explicitly we write


c [φ, φ0]]. (40)

Z2 =

∫Dφe−Sε[φ] = e−Sε[φH ,φ∗,φ0] (41)


δSε

δφ=

δSUV

δφ+

δSIR

δφ= 0, (42)







28

Equivalence

Sufficient:

Explicitly we write


c [φ, φ0]]. (40)

Z2 =

∫Dφe−Sε[φ] = e−Sε[φH ,φ∗,φ0] (41)


δSε

δφ=

δSUV

δφ+

δSIR

δφ= 0, (42)

which is nothing but ΠIR = ΠUV . Momentum continuity isESSENTIAL for bulk classical calculation!

In order for Z1 = Z2, it is SUFFICIENT to have






28

Equivalence

Sufficient:

Explicitly we write


c [φ, φ0]]. (40)

Z2 =

∫Dφe−Sε[φ] = e−Sε[φH ,φ∗,φ0] (41)


δSε

δφ=

δSUV

δφ+

δSIR

δφ= 0, (42)







28

Equivalence

Necessary (on-shell)

What about the “necessary” part?

On-shell:

!"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]_abcdefghijklmnopqrstuvwxyz{|}~

ZH Z=0ε

ϕH ϕ0

~

ϕ

C

B

A

Figure: Flows with (A) and without (C) the momentum continuity. Theuniqueness of the smooth solution forbids a solution like B.


28

Equivalence

Necessary (on-shell)

What about the “necessary” part?On-shell:

!"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]_abcdefghijklmnopqrstuvwxyz{|}~

ZH Z=0ε

ϕH ϕ0

~

ϕ

C

B

A

Figure: Flows with (A) and without (C) the momentum continuity. Theuniqueness of the smooth solution forbids a solution like B.


28

Equivalence

Necessary (off-shell)

What about the off-shell case?

Look at the RG again:

∂εSB [φ] + HIR [φ,δSB [φ]

δφ] = 0. (44)

Taking the derivative of eq. (??) with respect to φ we get

∂εΠφ = −δHIR

δφ. (45)

Momentum Continuity gives the ε dependence of φ

Πφ =√−gg zz∂zφ(z)

∣∣ε=√−gg zz |ε∂εφ (46)

We recover the full EOM for φ! The equations (??) and (??)together with φ0, φH as the boundary condition of φ repeat theclassical solution in whole bulk.


28

Equivalence





δφ] = 0. (44)



δφ. (45)



∣∣ε=√−gg zz |ε∂εφ (46)



28

Equivalence





δφ] = 0. (44)



δφ. (45)



∣∣ε=√−gg zz |ε∂εφ (46)



28

Equivalence





δφ] = 0. (44)



δφ. (45)



∣∣ε=√−gg zz |ε∂εφ (46)



28

Equivalence





δφ] = 0. (44)



δφ. (45)



∣∣ε=√−gg zz |ε∂εφ (46)



28

Equivalence

So far, we conclude that, in the bulk classical level, HWRG andsliding membrane are exactly equivalent. Basically, all bulk classicalflows involve EOM.

Still, the important question: how does the flow look like in boundaryfield theory language?


28

Double Trace Flow

Double Trace Flow

We will obtain SB by computing on-shell UV action.Maxwell equations are written by

∂z

[√−g (∂z Aµ − ∂µ Az)

]+ ∂ν

[√−g (∂ν Aµ − ∂µ Aν)

]= 0 . (47)

We assume

Aµ(z , k) = A(0)µ (z) + k2A(1)

µ (z) + kµkνA(2)ν (z) + · · · (48)

and Az is independent on k. In the small momentum ∂∂ → 0 limit,the first term will dominate and the above equation can be solved by

A(0)µ (z)− A(0)

µ (z0) =

∫ z

z0

Cµ1√

−gg zzgµµdz . (49)

Cµ1 =

1∫ εz0

dz√−gg zzgµµ

(A(0)µ (ε)− A

(0)µ,z0) := fµ(A(0)

µ (ε)− A(0)µ,z0) (50)

1

fµ=

∫ ε

z0

1√−gg zzgµµ

dz (51)


28

Double Trace Flow

Now, we want to integrate out z in region z0 < z < ε. For the standardquadratic Maxwell action, we obtain the on-shell action as boundary termusing the equations of motion:

Son−shell[z0,ε]

= −1

2

∫ddx

√−gg zzgµµAc

µ∂z Acµ

∣∣∣∣εz0

= −1

2

∫ddx Cµ

1 Acµ

∣∣∣∣εz0

.

(52)

where we used ∂z A(0)µ =

Cµ1√

−gg zzgµµ . ϕ(xµ, z) =∫ zz0

Azdz , the gauge

invariant field A(0)µ = A

(0)µ − ∂µϕ.

Using solution (??), we finally obtain the zero order on shell action

Son−shell[z0,ε]

= −1

2

∫ddx

∑µ

fµ(Aµ(ε)− Aµ,z0)(Aµ(ε)− Aµ,z0). (53)

This is SB at low frequency limit! If we set z0 = 0.


28

Double Trace Flow

From the boundary point of view, we have deformations due to SB .

Naturally, the deformed theory should be defined at ε, and the sourceshould be Aµ(ε) and the current should comes from

Jµε ≡

δSB

δAµ,ε(54)

The Green function in linear level can be defined as

Gµµε = − Jµ

ε

Aµ,ε. (55)

The field theory effective action can be derived from

δSeff = − 1

GκδJ J = − 1

Gµµε

δJµε Jµ

ε . (56)

which is given by

Seff =

∫ddx

∑µ

[−1

2(Jµ)2

1

fµ− Jµ(Aµ,0 + ∂µϕ)

]. (57)

This is nothing but the Legendre transformation of SB !


28

Double Trace Flow



Jµε ≡

δSB

δAµ,ε(54)


Gµµε = − Jµ

ε

Aµ,ε. (55)


δSeff = − 1

GκδJ J = − 1

Gµµε

δJµε Jµ

ε . (56)

which is given by

Seff =

∫ddx

∑µ

[−1

2(Jµ)2

1


]. (57)



28

Double Trace Flow



Jµε ≡

δSB

δAµ,ε(54)


Gµµε = − Jµ

ε

Aµ,ε. (55)


δSeff = − 1

GκδJ J = − 1

Gµµε

δJµε Jµ

ε . (56)

which is given by

Seff =

∫ddx

∑µ

[−1

2(Jµ)2

1


]. (57)



28

Double Trace Flow



Jµε ≡

δSB

δAµ,ε(54)


Gµµε = − Jµ

ε

Aµ,ε. (55)


δSeff = − 1

GκδJ J = − 1

Gµµε

δJµε Jµ

ε . (56)

which is given by

Seff =

∫ddx

∑µ

[−1

2(Jµ)2

1


]. (57)



28

Double Trace Flow

Double Trace Coupling flow

Compare with Witten description: SB contains double tracedeformation!

For 4d AdS black hole

1

fi=

1

2ε2 ,

1

f0=

1

4ln

(1 + ε2

1− ε2

)(58)

Double trace deformed Green function in linear level:

Jµε = −Gµµ

ε Aµ,ε ,1

Gµµε

− 1

Gµµz0

=1

fµ+

∂µϕ

Jµ. (59)

where ϕ(xµ, z) =∫ zz0

Azdz .

when ε → 0, Gε → G0.


28

Double Trace Flow




1

fi=

1

2ε2 ,

1

f0=

1

4ln

(1 + ε2

1− ε2

)(58)


Jµε = −Gµµ

ε Aµ,ε ,1

Gµµε

− 1

Gµµz0

=1

fµ+

∂µϕ

Jµ. (59)


Azdz .



28

Double Trace Flow




1

fi=

1

2ε2 ,

1

f0=

1

4ln

(1 + ε2

1− ε2

)(58)


Jµε = −Gµµ

ε Aµ,ε ,1

Gµµε

− 1

Gµµz0

=1

fµ+

∂µϕ

Jµ. (59)


Azdz .



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Double Trace Flow

Conclusions and Future Questions

HWRG and Sliding membrane flow are equivalent.

Double trace flow comes from integrating out UV bulk geometry.

Our Green function flow from Membrane is equivalent to Legendretransformation of SB in 1010.1264 (I.heemskerk and J.Polchinski)

Open Question: what about the possible flow for non-local operator?Is it possible to establish the precise map between H-J flow and exactWilson RG in field theory?

Quick Question: Does this double trace flow mean anything inAdS/Nuclear or AdS/CMT? Running of everything we obtainedbefore? Dispersion relation? Transport Coefficients?


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Double Trace Flow








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Double Trace Flow








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Double Trace Flow








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Double Trace Flow








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Documents

Holographic Wilsonian RG and BH Sliding Membraneicts.ustc.edu.cn/chinese/seminar/transparencies/zhou yang/ZHOU YANG... · Holographic Wilsonian RG and BH Sliding Membrane Yang Zhou