UvA-DARE ( Digital Academic Repository ) Deforming the D 1 D 5 CFT away from the orbifold point

The D1D5 brane bound state is believed to have an ‘orbifold point’ in its moduli space which is the analogue of the free Yang Mills theory for the D3 brane bound state. The supergravity geometry generated by D1 and D5 branes is described by a different point in moduli space, and in moving towards this point we have to deform the CFT by a marginal operator: the ‘twist’ which links together two copies of the CFT. In this paper we find the effect of this deformation operator on the simplest physical state of the CFT — the Ramond vacuum. The twist deformation leads to a final state that is populated by pairs of excitations like those in a squeezed state. We find the coefficients characterizing the distribution of these particle pairs (for both bosons and fermions) and thus write this final state in closed form.


and thus wri
e this final state in closed form.

Introduction

The D1D5 bound state is perhaps the best system to tackle the physics of black holes.This system gives a nonzero entropy at extremality, both for the 2-charge D1D5 bound state and its excitation, the 3-charge D1D5P bound state [1].Non-extremal excitations of this state collide and exit the bound state at exactly the rate at which Hawking radiation is produced from the corresponding black hole [2].The microstate structure of 2-charge states and a large number of 3-charge states has been found, and the 'fuzzball' nature of these states resolves the well known Hawking information paradox since these micr states do not have a traditional horizon [3].

AdS/CFT duality [4] relates the D1D5 CFT to the gravity solution produced by the bound state.But the CFT has been mostly studied at its 'free' point, which is believed to be an 'orbifold CFT' [5].The CFT at this point is dual to a gravitational solution which is very singular, and not in a domain interesting for gravitational physics like the formation of black holes.To get a description of interesting gravitational phenomena we will have to move in the moduli space of the CFT to a point away from the 'orbifold' point.In particular we need to turn on the 'blow up' mode of the orbifold, which is given by a 'twist operator' in the orbifold CFT.Such a deformation has been discussed in various contexts in earlier work [6].

Our goal in this paper will be to study the effect of this deformation on excitations of the CFT.The orbifold CFT is given by a 1+1 dimensional sigma model with target space (M 4 ) N 1 N 5 /S N 1 N 5 , the symmetric product of N 1 N 5 copies of a 4-manifold M.Here M can be T 4 or K3; we will take it to be T 4 .We will take the spatial circle σ of the CFT to be compact.Each copy of T 4 gives rise to a free c = 6 CFT.Fig. 1 shows the effect of the twist operator: it takes two copies of the c = 6 CFT and links them together to make one copy of the CFT living on a doubly wound circle.

In this paper we do the simplest computation involving the deformation operator.We start with the vacuum state for each of the two initial copies of the c = 6 CFT.In the physical problem of the black hole the vacuum states are in the Ramond sector, and we will let both copies have the 'spin down' Ramond ground state.We then act with the deformation operator at a location w 0 = τ 0 + iσ 0 on the cylinder describing the 1+1 dimensional CFT.For τ > τ 0 we have one copy of the c = 6 CFT living on a doubly wound circle, with a set of bosonic and fermionic excitations that are created by the effect of the deformation operator.Our goal is to find the state at τ > τ 0 .We argue that the excitations in this state have the structure of excitations in a squeezed state (schematically ∼ e γa † a † for the bosons, and similarly for the fermions).Thus we find the state by finding the coefficients γ mn for the bosonic and fermionic excitations.The deformation operator also has a supersymmetry current G applied to this squeezed state, and after taking that into account we write down the full final state obtained by the action of the deformation operato on our chosen vacuum state.

To use this result for the physics of the D1D5 system we should integrate over the location w 0 , and also allow different possible excitations in the initial state.We will carry out those steps elsewhere, and restrict ourselves to finding the basi squeezed state here.(2.1)

Wrap N 1 D1 branes on S 1 , and N 5 D5 branes on S 1 × T 4 .The bound state of these branes is described by a field theory.We think of the S 1 as being large compared to the T 4 , so that at low energies we look for excitations only in the direction S 1 .This low energy limit gives a conformal field theory (CFT) on the circle S 1 .We can vary the moduli of string theory (the string coupling g, the shape and size of the torus, the values of flat connections for gauge fields etc.).These changes move us to different points in the moduli space of the CFT.It has been conjectured that we can move to a point called the 'orbifold point' where the CFT is particularly simple [5].At this orbifold point the CFT is a 1+1 dimensional sigma model.The 1+1 dimensional base space is e 0 ≤ y < 2πR (2.2) is a coordinate along the S 1 , and t is the time of the 10-d string theory.For our CFT computations, we rotate time to Euclidean time, and also use scaled coordinates (σ, τ ) where the space direction of the ontinue back to Lorentzian signature at the end.
τ = t R σ = y R . (2
The target space of the sigma model is the 'symmetrized product' of
N 1 N 5 copies of T 4 , (T 4 ) N 1 N 5 /S N 1 N 5 ,(2.4)
with each copy of T4 giving 4 bosonic excitations X 1 , X 2 , X 3 , X 4 .It also gives 4 fermionic excitations, which we call ψ 1 , ψ 2 , ψ 3 , ψ 4 for the left movers, and ψ1 , ψ2 , ψ3 , ψ4 for the right movers.

The fermions can be antiperiodic or periodic around the σ circle.If they are antiperiodic on the S 1 we are in the Neveu-Schwarz (NS) sector, and if they are periodic on the S 1 we are in the Ramond (R) sector 4 .The is c = 6 (2.5)
The total central charge of the entire system is t

s 6N 1 N 5 .


Symmetr
es of the CFT

The D1D5 CFT has (4,4) supersymmetry, which means that we have N = 4 supersymmetry in both the left and right moving sectors.This leads to a superconformal N = 4 symmetry in both the left and right sectors, generated by operators L n , G ± r , J a n for the left movers and Ln , Ḡ± r , Ja n for the right movers.The algebra generators and their OPEs and commutators are given in ppendix A.

Each N = 4 algebra has an internal R symmetry group SU(2),5 so there is a global symmetry group SU (2) L × SU (2) R .We denote the quantum numbers in these two SU (2) groups as
SU (2) L : (j, m); SU (2) R : (, m). (2.6)
In the geometrical setting of the CFT, this symmetry arises from the rotational symmetry in the 4 space directions of M 4,1 in Equat R . (2.7)
Here the subscript E stands for 'external', which denotes that these rotations are in the noncompact directions.These quantum numbers therefore give the angular momentum of quanta in the gravity description.We have another SO(4) symmetry in the four directions of the T 4 .This symmetry we call SO( 4) I (where I stands for 'internal').This symmetry is broken by the compactification of the torus, but at the orbifold point it still provides a useful organizing principle.We . (2.8)
We use spinor indices α, α for SU (2) L and SU (2) R respectively.We use spinor indices A, Ȧ for SU (2) 1 and SU (2) 2 respectively.The 4 real fermions of the left sector can be grouped into complex fermions ψ αA with the reality constraint (ψ † ) αA = −ǫ αβ ǫ AB ψ β (2.9)

The right fermions have indices ψ α Ȧ with a similar reality constraint.The bosons X i are a vector in the T 4 .They have no charge under SU (2) L or SU (2) R and are 4 are the three Pauli matrices and the identity.(The notations described here are explained in full detail in Appendix A.)
[X] ȦA = 1 √ 2 X i (σ i )

. (2

States and ope
ators

Since we orbifold by the symmetric group S N 1 N 5 , we generate 'twist sectors', which can be obtained by acting with 'twist operators' σ n on an untwisted state.Suppose we insert a twist operator at a point w 0 .As we circle the point w 0 , different copies of T 4 get mapped into each other.Let us denote the copy number by a subscript a = 1, 2, . . .n.The twist operator is labeled by the permutation it generates.For instance, every time one circles the twist operator 2.11) the fields X (a) i get mapped as
σ (123...n) ,( i , (2.12) and the other copies of X (a) i are unchanged.We have a similar action on the fermionic fields.Each set of linked copies of the CFT is called one 'component string'.

T e simplest states of the CFT are in the 'untwisted sector' where no copy of the c = 6 CFT is linked to any other copy; i.e. all component strings have winding number unity.Consider one component string, and consider the theory defined on the cylinder.The fermions on this string can be either periodic around the σ circle of the cylinder (Ramond sector R) or antiperiodic (Neveu-Schwarz sector NS).Consider one copy of the c = 6 CFT.The simplest state of this theory is the NS sector vacuum
|0 N S : h = 0, m = 0 (2.13)
But the CFT arising from the D1D5 brane bound state is in the Ramond (R) sector.One can understand this because the periodicities of the fermions around the S 1 are inherited from the behavior of fermionic supergravity fields around the S 1 in (2.1).These supergravity fields must be taken to be periodic, since otherwise we would generate a nonzero vacuum energy in our spacetime and the metric far from the branes would not be flat.We can relate the state (2.13) to a Ramond ground state using spectral flow [7].Spectral flow maps amplitudes in the CFT to amplitudes in another CFT; under this map dimensions and charges change as (we write only the left sector)
h ′ = h + + cα 12 (2.14)
We have c 6.Setting α = −1 gives
|0 − R : h = of the Ramond ground states of the c = 6 CFT for a component string with winding number unity.Other Ramond ground states are obtained by acting with fermion zero modes, so that we have four states in all
|0 − R , ψ ++ possibilities for the right moving sector).

The deformation operator involves the twist σ 2 .As we go around a point of insertion of this twist, the fermions in the first copy change to fermions in the second copy, and after another circle return to their original value.Creating such a twist automatically brings in a 'spin field' at the insertion point, which has h = 1 2 , j = 1 2 [8].Thus there are two possible insertions of such a twist, with m = 1 2 and with m = − 1 2 .We write these as σ + 2 and σ − 2 respectively.The operator σ + 2 is a chiral primary and σ − 2 is an anti-chiral primary.


The deformation

perator

Let us describe
he deformation operator in some detail.


The structure o

the operator

The deformation
perator is a singlet under SU (2) L × SU (2) R .To obtain such a singlet we apply modes of
G ∓ Ȧ to σ ± 2 . A singlet under SU (2) L × SU (2)
R can be made as (writing both left and right moving sectors) [6]
Ô Ȧ Ḃ ∝ 1 4 ǫ αβ ǫ α β 1 2πi ndix B we show that
1 2πi w can write the deformation operator as (we chose its normalization at this stage)
Ô Ȧ Ḃ ( normalization of σ ++ 2 will be specified below.The indices Ȧ, Ḃ indices can be contracted to rewrite the above four operators as a singlet and a triplet of SU (2) 1 .6


N

depicted in figure .2.Let us
write down all the states and operators in this amplitude.


The initial state

We have two component strings.Si

e each is in the R
mond sector, we have to choose one of the Ramond ground states (2.16).Let us take the state
|Ψ i = |0 −− R (1) ⊗ |0 −− R(2)
(3.1)


The final state

int w 0 on the c
linder.Thus the final state that we want to find is given by
|Ψ f = Ô Ȧ Ḃ (w 0 )|Ψ i = 1 2πi w 0 dwG − Ȧ (w) 1 2πi ring with winding number 2, since the deformation operator contains the twist σ 2 .

From this stage on, we will write only the left moving part of the state, and join it up with the right moving part at the end.Thus we write
|Ψ f = |ψ | ψ (3.3)
and work with |ψ in what follows.


Outline of the computation

Let us outline our

teps for computing |ψ .

(a
The essence of the computation lies in the ature of the deformation operator.This operator is given by a supercharge acting on the twist operator σ + 2 .This supercharge is given by a contour integral of G − Ȧ around the twist insertion.We first deform this contour to a pair of contours: one above and one below the insertion.These contours give zero modes of the supercurrent on the states before and after the twist insertion.We handle these zero modes at the end, and focus first on the state produced by just the twist insertion σ + 2 ; we call this state |χ .

(b) Let us now look at the nature of the twist opera or for bosonic fields.As we circle the twist, the two copies of the boson go into each other.The twist operator is defined by cutting a small hole around its insertion w 0 , and taking boundary condition at the edge of this hole given by filling the hole in the covering space with a disc; i.e. there are no operator insertions in this covering space and we have just the vacuum state [10].To use this structure of the twist operator, we first map the cylinder to the plane via z = e w , and then pass to the covering space t by the map z = z 0 + t 2 (here z 0 = e w 0 is the location of the twist).The small hole cut out on the cylinder around w 0 becomes a small hole around t = 0. Since the boundary condition on the edge of this hole is generated by filling this hole by a disc, we get just the vacuum state at the origin in the t plane.This observation takes into account the entire effect of the twist on the bosons.

(c) On the cylinder we can specify the initial state of t e system on the two circles at τ → −∞ corresponding to the two copies of the c = 6 CFT.On the t plane these circles map to punctures at t = ±z 1 2 0 ≡ ±ia.Since we have taken no bosonic excitations in our initial state, the bosonic part of the states at these punctures is just the vacuum, and we can close these punctures smoothly, just like the hole at t = 0. Thus we have no insertions anywhere in the t plane.

(d) Our goal is to find the state at a circle τ → ∞ on the ylinder.But this circle maps to the circle |t| = ∞ on the t plane.Thus what we need is the state in the t plane at infinity.But since there are no insertions anywhere on the t plane, this state is just the t plane vacuum |0 t .One might think that this means there are no excitations in the final state, but this is not the case: the vacuum on the t plane is killed by positive energy modes defined with respect to the t coordinate, and these will map to a linear combination of positive and negative energy modes in the original cylinder coordinate w.Thus all we have to do is express the state |0 t in terms of the modes on the cylinder, and we would have obtained the bosonic part of the state arising from the twist insertion.

(e) Let us now ask if we can guess the nature of this state in t rms of the modes on the cylinder.In the treatment of quantum fields on curved space we often come across a situation where we have to express the vacuum with respect to one set of field modes in terms of operators defined with respect to another set of field modes.The state with respect to the latter modes has the form of an exponential of a quadratic, i.e. of the form e γmna † m a † n |0 .The essential reason for getting this form for the state can be traced to the fact that free fields have a quadratic action, and if there are no operator insertions anywhere then in all descriptions of the state we can only observe exponentials of a quadratic form.

For our problem, we make the ansatz that the state |χ has a similar exponential form.We find the γ mn by computing the inner product of the state with a state containing a pair of operator modes.In Appendix C we prove that this exponential ansatz is indeed correct to all orders.Such a form for the state is termed a squeezed state in atomic physics.

(f) Let us now ask if similar arguments can be applied to the fermioni field.The initial state on the cylinder has Ramond vacua for the two copies of the CFT.If we map to the t plane these would give nontrivial states at t = ±ia.Thus we first perform a spectral flow on the cylinder, which maps the problem to one where these Ramond vacua map to NS vacua at τ → −∞ on the cylinder.These NS vacua will map to NS vacua at t = ±ia, so there will be no operator insertions in the t plane at these points.We should also check the effect of this spectral flow on the twist σ + 2 (w 0 ).From (2.23) we find that σ + 2 (w 0 ) will change by just a multiplicative constant.This constant will not matter at the end since we know the normalization of our final state by (2.30).

We can now pass to the covering space t.We must now ask for the state at the dge of the hole around t = 0.One finds that the fermions in the t plane are antiperiodic around t = 0 [8].Thus the state given by the operator σ + 2 corresponds to having the positive spin Ramond vacuum |0 + R t .As it stands this implies that we have a nontrivial state at t = 0, but we perform another spectral flow, this time in the t plane.Under this second spectral flow the Ramond vacuum |0 + R t maps to the NS vacuum of the t plane |0 t .We take the normalization t 0|0 t = 1 (3.4) for this NS vacuum.At this stage we have indeed no insertions anywhere in the t plane, and the state at t = ∞ is just the t plane vacuum for the fermions.Since all coordinate maps and spectral flows were linear in the operator modes, we again expect the state to be given by the exponential of a bilinear in fermion modes.We write such an ansatz, and find the coefficients in the exponential.

(g) We can summarize the above discussion in the following general relation
0 R,− | Ô1 , Ô2 , . . . Ôn σ + 2 (w 0 )|0 − R (1) ⊗ |0 − R (2) = t 0| Ô′ 1 , Ô′ 2 , . .
ertion (we will need to insert boson and fermion modes in finding the coefficients γ mn ).On the RHS, the operators Ô′ i are obtained by mapping the operators Ôi through all coordinate changes and spectral flows till we reach the t plane with the NS vacuum at t = 0.The normalizations (2.29) and (3.4) tell us that there is no extra constant relating the LHS to the RHS of (3.5).

(h) After obtaining the state |χ generated by the action of the twist σ 2) , we act with the

ero mode of the supercurrent to obt
in the final state |ψ obtained by the action of the full deformation operator on
+ 2 on |0 − R (1) ⊗ |0 − R(|0

the twist, an
we have simply cut and rejoined the copies into one long loop.Thus the state |χ can be considered as built up by adding excitations (with total charge zero) to the Ramond vacuum |0 − R of the doubly twisted theory (we assume that this vacuum is normalized to unit norm)
|χ = {α C Ċ,m i } {ψ β Ḋ,n j }|0 − R (4.19)
As discussed in section (3.3), the state |χ should have the form of an exponential in the boson and fermion creation operators.Let us write down the ansatz and then explain some of its points.In Appendix C we show that for the bosonic case this ansatz is correct to all orders in the number of excitations; the fermionic case can be done in a similar way.We will write
|χ = e α +−,−n ] e m≥0,n≥1 γ F mn [d ++ −m d −− −n −d +− −m d −+ −n ] |0 − R (4.20)
Below we will define more precisely the modes α A Ȧ,n , d αA n on the double circle.For now, Let us note some points about the above expression:

(a) From eq. ( 2.30) we see that there will not be any additional multiplicative constant on the RHS; the coefficient of first term obtained by expan ymmetry group (2.7) in the torus directions.The operator σ + 2 is a singlet under this group also.Thus the state |χ will have to be a singlet und r this group.Thus we have written the ansatz in a way that the A, Ȧ indices of α A Ȧ, ψ αA are grouped to make singlets under SO(4) I .

(c) We have
α A Ȧ,0 |0 − R = 0 (4.21)
since there is no momentum in the st te.Thus the mode summations for the bosons start with m, n = 1 and not with m, n = 0.

(d) We have
d −A 0 |0 − R = 0 (4.22)
Thus the sum over fermion modes starts with n = 1 for d −− , d −+ and with m = 0 for d ++ , d +− .By writing modes this way we remove a normal ordering term that can arise from the anticommutation of zero modes; such a c ntribution would then have to be reab orbed in an overall normalization constant in front of the exponential.We will find later that the value m = 0 does not occu r that case; in fact we will find that the γ B mn , γ F mn are nonzero only for m, n odd.


The first spectral flow

Let us continue to work with the state |χ , and return to |ψ at the end.

First we perform a spectral flow (2.14) with parameter α = 1.Let the resulting state be called |χ α=1 .This spectral flow has the following effects: (2)  (4.23)(b) The operator σ + 2 changes by a phase which depends on its charge q; this charge is q = 1 2 .So we get
(a) The Ramond ground states |0 − R (i) , i = 1, 2 cha

onsider the circle on the cyli
der at τ → −∞; this is the location of the initi and then to t = ±ia on the t plane.There were two copies of the CFT at τ → −∞ on the cylinder, and the initial state of one copy (copy (1)) will map to the point t = ia and the state of the other copy (copy (2)) will map to t = −ia.

Note that in our present problem the states of these copies, were |0 − (i) to start with, which became NS vacua |0 (i) after the first spectral flow.Now when we map them to the t plane we find that there is just the t plane vacuum at the points ±ia, so we may smoothly close the punctures at these points with no insertions at the puncture.


R

Figure 4: The z plane is mapped to the covering space -the t plane -by the map z = z 0 + t 2 .The point z = 0 corresponds to τ → −∞ on the cylinder, and the two copies of the CFT there correspond to the points t = ±ia.The location of the twist operator maps to t = 0.The top the cylinder τ → ∞ maps to t → ∞.After all maps and spectral flows, we have the NS vacuum at t = 0, ±ia, and so we can smoothly close all these punctures.The state |χ is thus just the t plane vacuum; we must write this in terms of the original cylinder modes and apply the supercharge to get the final state |ψ .

(b) The location of the twist insertion σ + 2 maps to t = 0.At this location we have the state |0 + R t , the spin up Ramond vacuum of the t plane.

(c) The operator modes become before the twist
α (1) A Ȧ,n → 1 2π t=ia ∂ t X A Ȧ(t)(z 0 + t 2 ) n dt (4.39) α (2) A Ȧ,n → 1 2π t=−ia ∂ t X A Ȧ(t)(z 0 + t 2 ) n dt (4.40) d (1)+A n → 1 2πi t=ia ψ +A (t)(z 0 + t 2 ) n−1 (2t) 1 2 dt (4.41) d (2)+A n → 1 2πi t=−ia ψ +A (t)(z 0 t 2 ) n−1 (2t) 1 2 dt (4.42) d (1)−A n → 1 2πi t=ia ψ −A (t)(z 0 + t 2 ) n (2t) 1 2 dt (4.43) d (2)−A n → 1 2πi t=−ia ψ −A (t)(z 0 + t 2 ) n (2t) 1 dt (4.44)
After the twist we have
α A Ȧ,n → 1 .This will have the foll
nential in |χ , generating negative index mod s.We would like to write |ψ with only negative index modes acting d −A −n α A ,n − i 2 ∞ n>0 d −A n α A Ȧ,−n (6.3)
Recall that γ B mn , γ F mn a
spectrum of the corresponding microstate geometr

s, and radi
tion from specific black hole microstates agrees exactly with the expected radiation from the corresponding CFT state [3,9,11].But to proceed further with the study of microstates we need to study the CFT away from the orbifold point, since the gravity solutions relevant to black hole physics do not correspond to the orbifold point.We can move away from the orbifold point by adding the deformation operator to the Lagrangian of the theory.In this paper we have computed the effect of this deformation operator on the simplest state of the CFT: two 'singly wound' copies of the c = 6 CFT, each in the Ramond ground state |0 − R .We argued that the effect of the twist operator should be given by an expression that is of the form of a squeezed state, and we found the coefficients in the exponential that describe this state.We also have a supercharge applied to this twist operator, and after taking this into account we found the final state given in (6.10).

To apply this expression to a physical problem we have to integrate the location of the deformation operator over the τ, σ space of the 1+1 dimensional CFT, and also consider more general initial states.We will discuss these steps elsewhere, but from the expression (6.10) we can already see some aspects of the general physics: the deformation operator can generate an arbitrary number fermion and boson pairs (besides the fermion and boson supplied by the supercharge).The falloff with energy of th se excitations are given by the expressions for γ B mn , γ F mn ; this falloff is a power law rather than an exponential.Thus we can have a long 'tail' in the distribution with an ever larger number of particle pairs.


B Showing
G − σ + 2 ∼ G + σ − 2
Here we prove that the two operators
1 2πi w 0 dwG − Ȧ (w)σ + 2 (w 0 ) (B.1) and 1 2πi w 0 dwG + Ȧ (w)σ − 2 (w 0 ) (B.2)
are proportional to each other, so we do not need to add over both these possibilities in the deformation operator (2.17).

Under the map z = e w (B.1) gives
1 2πi w 0 dwG − Ȧ (w)σ + 2 (w 0 ) = 1 2πi z 0 dzG − Ȧ (z)z 1 2 σ + 2 (z 0 ) (B.3) Under the map z = z 0 + t 2 we get 1 2πi z 0 dzG − Ȧ (z)z 1 2 σ + 2 (z 0 ) = 1 2πi t=0 dtG − Ȧ (t)(2t) − 1 2

form in the t plane is gi
en by eq.( 4.56).We can expand the latter form in terms of natural modes on the t plane (4.59) which agrees with (C.12).Thus we have verified the ansatz to order four in the bosonic field operators.Proceeding in this way we can verify the complete exponential ansatz.
α A Ȧ,n → 1 2π t=∞ ∂ t X A Ȧ(t)(z 0 + t 2 ) n 2 dt = 1 2π t=∞ ∂ t dtX A Ȧ(t) k≥0 n 2 C k z k t n−2k = k≥0 n 2 C k z k 0 αA Ȧ,
The and we have the NS vacuum at t = 0. On this t plane the modes must appear in pairs to allow the amplitude to be nonvanishing, thus we must have an even number of modes in each term in the ansatz.Our ansatz allows all modes that are nonvanishing on the chosen vacuum state; thus the situation is similar to the bosonic case where we allowed all negative index bosonic oper e bosonic part.

Figure 1 :
1
Figure 1: The effect of the twist contained in the deformation operator: two circles at earlier times get joined into one circle after the insertion of the twist.


Figure 3 :
3
Figure 3: (a)  The supercharge in the deformation operator is given by integrating G − Ȧ around the insertion at w 0 .(b) We can stretch this contour as shown, so that we get a part above the insertion and a part below, joined by vertical segments where the contributions cancel.(c) The part above the insertion gives the zero mode of the supercharge on the doubly wound circle, while the parts below give the sum of this zero mode for each of the two initial copies of the CFT.


( 1 )( 2 )
12
Further, the set {α ++,m , α −−,n } decouples from the set {α +−,m , α −+,n }.Thus in what follows we write only the modes {α ++,m , α −−,n }.We have from the basic relation (3.5)0 R,− | α ++,n 1 α −−,n 2 . . .σ + 2 (w 0 )|0 − R ⊗ |0 − R = t 0| α ′ ++,n 1 α 2 can contract with the operators from either of the two γ B factors.The RHS of (C.1) gives 1 ,p 2 ,p 3 ,p 4B n 1 ,p 1 B n 2 ,p 2 B n 3 ,p 3 B n 4 ,p 4 α++,p 1 α−−,p 2 α++,p 3 α−−,p 4 |0 t = p 1 >0 p 1 B n 1 ,p 1 B n 2 ,−p 1 p 3 >0 p 3 B n 3 ,p 3 B n 4 ,−p 3 + p 1 >0 p 1 B n 1 ,p 1 B n 4 ,−p 1 p 3 >0 p 3 B n 3 ,p 3 B n 2 ,−p (C.13)Onusing(C.9) this givesn 1 n 2 n 3 n 4 γ B n 2 n 1 γ B n 4 n 3 + γ B n 4 n 1 γ B n 2 n 3
This gives1 2!(−1) 2n−2k(C.3)
t 0| p (C.14)

The periodicities flip when we map the cylinder to the complex plane because of a Jacobian factor.
In fact, the full R symmetry group of the N = 4 algebra is SO(4); however, the other ve a cu rent associated with it within the algebra.
Since we can write the deformation operator in terms of G − σ + 2 or in terms of G + σ − 2 , it provides a good check on the results that we get the same fin l state each way.
AcknowledgementsWe thank Justin David for several helpful discussions.We also thank Sumit Das, Antal Jevicki, Yuri Kovchegov, Oleg Lunin and Emil Martinec for many helpful comments.The work of SGA and SDM is supported in part by DOE grant DE-FG02-91ER-40690.The work of BDC was supported fermions ψ 1 , ψ 2 , ψ 3 , ψ 4 which we group into doublets ψ αAHere α = (+, −) is an index of the ubgroup SU (2) L of rotations on S 3 and A = (+, −) is an index of the subgroup SU (2) 1 from rotations in T 4 .We have the reality conditionsThe 2-point functions arewhere we have defined the ǫ symbol asThere are 4 real left moving bosons X 1 , X 2 , X 3 , X 4 which can be grouped into a matrix.6)where σ i = σ a , iI.This givesand the reality conditionThe 2-point functions areThe chiral algebra is generated by the operatorsThese operators generate the algebraThe above OPE algebra gives the commutation relationsAll we need to know is that this is a linear relationwith some constant coefficients B np .Since the relation (C.4) is linear in the field operators, we will have as many operators inserted between the brackets () on the RHS of (C.1) as on the LHS.But on the RHS