and Yaron Oz a

We consider strings with large spin in AdS 3 × S 3 × M with NS-NS background. We construct the string configurations as solutions of SL(2, R) WZW theory. We compute the relation between the space-time energy and spin, and show that the anomalous correction is constant , and not logarithmic in the spin. This is in contrast to the S-dual background with R-R charge where the anomalous correction is logarithmic.


he anomalous
orrection is logarithmic.

Introduction

Examples where gauge theory perturbative computations can be extrapolated to strong coupling and compared to classical supergravity results are rare when the calculated quantities are not protected by supersymmetry.An important example has been studied by Gubser, Klebanov and Polyakov [1].They showed that the energy of a spinning string with spin S in AdS 5 in the limit that its size i adius is
E = S + √ λ π ln(S/ √ λ) .(1)
In global coordinates the energy is identified with the dimension of the dual operator.GKP proposed that this string configuration is dual to a twist 2 operator in N = 4 SYM such as
O (µ 1 • •µn) = Tr φ * D (µ 1 • • • D µn) φ .(2)
In perturbative gauge theory this operator has an anomalous dimension that grows as ln n, which is in agreement with the strong coupling result derived from supergravity.Quantum corrections to the energy of the spinning string were analyzed in [2,3] with no ln 2 S corrections found.One therefore expects that the leading term in the anomalous dimension to all orders in perturbation theory (and probably non-perturb c, i.e.
E = S + f (λ) ln(S/ √ λ) .(3)
Following the work of GKP many other string and membrane configurations were studied [4] - [13].For earlier works on semi-classical string solutions see [14].

In this paper we will study spinning strings in AdS 3 × S 3 × M with NS-NS 2-form background.This is of interest, since string theory on this background can be exactly solved in terms of the SL(2, R) WZW model [15,16].We will see that the "planetoid" spinning string solution of AdS 5 is not a solution in AdS 3 because of the NS-NS 2-form.In section 2 we will find the explicit form of the classical spinning string configuration, and calculate the energy-spin relation.We will see that the leading behavior is the same as in AdS 5 , but the anomalous correction is constant rather than logarithmic.This is in contrast with the S-dual background with R-R flux (for a review see [17]), where the anomalous correction is logarithmic.Indeed it confirms the fact that the near horizon limits of D1-D5 and F1-NS5 systems lead to very different theories.In section 3 we will analyze the conditions under which the classical string configuration is valid as a quantum state.In section 4 we consider more general string configurations that rotate on the S 3 part of the metric as well.In the last sec

on we discuss the results.


Classic
l spinning strings in AdS 3

String theory on AdS 3 × S 3 × M has many classical solutions.We will be interested in a class of classical solutions that correspond to spinning closed strings [18,19,20].Similar string configurations were also analyzed in [21] in th context of open strings.

The NS-NS AdS 3 background, in globa 2 1 ,(4)
with
B tφ = R 2 r 2 .
We will consider the following ansatz for a time dependent embed c 2 τ + φ (σ) , r = r (σ) .
c 1 and c 2 are constants, and c 1 is assumed to be positive to insure forward propagation in time.τ and σ denote the world-sheet coordinates and σ ≃ σ + 2π.The GKP ansatz is the one in which t = φ = 0 .At first sight this seems to be a solution of the classical equations of motion.The energy-spin relation for such a configuration will be as in [1].For strings larger than the AdS radius, it takes the form (3), with √ λ = R 2 /α ′ .However, we will see that while this configuration solves the constraints T ±± = 0, the SL(2, R) Kac-Moody currents associated with it are not holomorphic.Note, however, that for a fundamental spinning string in a R-R AdS 3 background, given by the near-horizon limit of the D1-D5 system, the simpler ans tz is perfectly valid.

The world-sheet action for a closed string embeddin 2 c 1 d φ dσ − c 2 d t dσ .
From (6) w )d φ dσ = c 1 r 2 − k 2 r 2 ,
where k 1 and k 2 are integration constants.The constraint on the energymome σσ + T τ τ ± 2T τ σ = 0 .(8)
In (8) we assumed that there are no other contributions to the energymomentum tensor from the internal CFT on S 3 × M. It is, however, easy to generalize this calculation to include contribu ions from the internal CFT.

In such a case the constraint is on the su dS ±± + T S 3 ×M ±± = 0 .(9)
We will discuss this generalization in section 4. From the requirement that
T στ = 0 we get that c 1 k 1 = c 2 k 2 .
To sim ,(10)β = c 2 1 + 2c 1 k 2 .
With these definitions the energy-mo k 2 2 β k 2 2 α − r 2 . (11)
The solution (α − β) sin √ 4αβ k 2 σ .(12)
For a closed string we need to impose r(2π) = r(0) which le n = √ 4αβ k 2 = integer . (13)
The angular coordinate can be tan n 2 σ − (α − β) √ 4αβ .(14)
Note that ( 14) is discontinuous whenever tan[ n 2 σ] diverges.In order to get a continuous function we should add π after each discontinuous point (nπ altogether).In order to ensure the periodicity φ(2π, τ ) = φ(0, τ ) mod 2π, we should have either (i) c 1 an integer and n even, or (ii) c 1 half integer and n odd.A few examples are plotted in figure 1
for τ = 0. c=1/2, n=3 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 angular coordinate is φ(σ).Note that these plots show the string at constant worldsheet time, which is not the same as target space time because of ( 5).

The time coordinate is given by
t(σ, τ ) = c 1 τ + c 2 σ + (15) − (c τ ) = t(0, τ ) is obeyed if k 2 satisfies (again note that arctan[...] gets shifted by nπ)
c 2 = (c 2 + k 1 ) √ αβ (k 2 2 + α)(k ions of α and β, and gives the allowed value for k 2
k 2 = 2c 3 1 n 2 − 4c 2 1 .(17)
Note ould be restricted to n > 2c 1 .At the point n = 2c 1 the size of the string diverges.We will analyze this limit below.

For n < 2c 1 we have the unphysical r sult r 2 (σ) < 0. Substituting k 2 back in equations ( 12), ( 14) and ( 15) we get
r 2 (σ) = 2 n 2 c 2 1 (n 2 + c 2 1 s in S of the system are given by
E = R 2 2πα ′ 2π 0 dσ δL δ ṫ = R 2 alculated explicitly, and we can express E and S in terms of c 1 and n as
E = R 2 (c 1 + k 2 ) α ′ = R 2 c is system and get the energy-spin relation
E(S, n) = R 2 α ′ n 2 + 2n 2 α ′ the limits where the string size is much smaller or much larger than the AdS 3 radius.These two limits will correspond to the 2c 1 ≪ n and the 2c 1 → n limits, respectively.When 2c 1 ≪ n the string is mostly concentrated near the origin of AdS 3 , as can be seen from taking the limit of (18)
r 2 (σ) → 2c 2 1 n 2 (1 + sin[nσ]) much less than the AdS 3 radius.In this limit we get the flat space relation between the energy of the string and its spin
S = α ′ E 2 R 2 n . (25)
On the oth n 2 − 4c 2 1 → 0 the asymptotic form of ( 18) is
r 2 (σ) → n 2 8ǫ (5 + 3 sin[nσ]) ing is much larger than the AdS 3 radius, and the leading relation between E and S is given by E = S .This linear behavior was also found for open rotating strings in [21].Using (22) we can actually calculate all the classical corrections to the leading linear behavior
E = S + 3 8 R 2 n α ′ − 10 256 R 4 .Note in comparison that in AdS 5 or AdS 3 with R-R background the leading correction is ln S.


Quantum spinning strings

The spe

rum of strings on AdS 3 w
th NS-NS background was analyzed in [16].In this section we will use a similar analysis in order to identify the SL(2, R) representation corresponding to the spinning string solution.We will investigate when is this state expected to be part of the physical spectrum of strings on AdS 3 .In order to identify the corresponding representation of SL(2, R), we switch to light-cone coordinates, and calculate the SL(2, R) currents associated with the spinning string solution.The solution of the previous section satisfies T AdS ±± = 0, which implies that the Casimir of SL(2, R), C 2 = −j(j − 1) = 0. We therefore should look for a representation with j = 1.

We define k = R 2 /α ′ .We also define new target space coordinates
u = 1 2 (t + φ) , v = 1 2 (t − φ) .(28) sheet are: x ± = τ ± σ .Using the conservation laws (7) we evaluate the right and left SL(2, R) currents.In the following we shall only need the explicit form of J 3
J 3 R = k ∂ + u + (1 + 2r 2 )∂ + v = k(c 1 etermined in the same way.

Note that had we used the ansatz with t = φ = 0 the currents would have been
J 3 R = k (c 1 + c 2 ) + (1 + 2r 2 (σ))( eft/right current holomorphic/antiholomorphic, but not both.In WZW theory the holomorphy of the currents is equivalent to the equations of motion, thus such a configuration does not solve the field equations of the SL(2, R) WZW model.

The zero modes of the currents (29) are 3 0 = 2π 0 dx + 2π J 3 R , J3 0 = 2π 0 3 0 and J3 0 which are given by
m = k n 2 c 1 − 2c 3 1 − nc 2 1 n 2 − 4c the solution has only a zero mode and no other higher modes, since J 3 ν ∼ J 3 L e iνx + dx + vanishes unless ν = 0.For this solution to be a valid quantum state it must be in a unitary representation of SL(2, R).The full quantum spectrum will be associated with the corresponding representation of the affine SL(2, R) subject to the physical state constraint.

Recall that the unitary representations f the SL(2, R) Lie algebra are:

• Discrete representations: D + j and D j for real j, for which m = j, j ±
1, j ± 2, • • •. • Continuous representations: C a j (0 ≤ < 1) for j = 1/2 + is (s is real); or for 1/2 < j < 1 and j − 1/2 < |a − 1/2|, for which m = a, a ± 1, a ± 2, • • • • Identity representation: j = 0 .
Our solution has j = 1 so it cannot corres ond to a unitary continuous representation.It can correspond to one of the discrete representations but only if m and m are integers.If they are not integers they can still belong to a non-unitary continuous representation, but we will not consider them as physical quantum states.It is clear that to have integer m and m, k must be rational.Let us examine the spectrum for k = 1 as an example.There do not seem to be solutions with integer m and m for odd n in this case.For even n there are solutions.The first one is n = 24, c 1 = 6.In this case the string state corresponds to the WZW state
|j = 1, m = 5 × | j = 1, m = 9 .
For higher values of k there will be other sol tions.It is not clear though whether all integers can be generated by (32) for a given value of k.

It was argued in [16] that using spectral flow ne can generate new classical solutions
t(τ, σ) → t(τ, σ) + ωτ ,(33)φ(τ, σ) → φ(τ, σ) + ωσ ,
where ω is an intege wn to be important in understanding the nature of long string states in AdS 3 .The SL(2, R) currents, and the energy-momentum tensor are modified by the spectral flow to
J 3 0 → J 3 0 + kω 2 ,(34)T ++ → T ++ − ωJ 3 0 − k 4 owed state is the same as the un-flowed state, since it is S = m − m.The energy, however, is shifted by kω.Since the energy tensor is modified by the spectral flow one must impose a new physical state condition
T (ω) ++ = T ++ − ωJ 3 0 − k 4 ω 2 = 0 ,(35)
and the same fo e original solutions, and J 3 0 = J3 0 , there cannot be a solution to the physical state condition other than ω = 0.In fact, this will also be true even if there are contributions from the CFT on S 3 , as long as T S 3 ++ = T S 3 −− .We now describe the standard way in which the quantum spectrum of strings in AdS 3 ×S 3 ×M is built around the classical spinning string solution we presented.We consider the state |ψ 0 = |j, j, m, m, h, h as the ground state, where (h, h) are the conformal weights of some state of the internal CFT on S 3 × M. In our case h = h = 0.The excited states are constructed This modifies (11) to be
dr dσ 2 = αβ k 2 2 r 2 r 2 − k 2 2 β k 2 2 α − r 2 − ω 2 (1 + r 2 ) .(44)
In order to get a closed string solution, we must assume that
∆ = (α − β) 2 − 0 this is automatically satisfied.However, for generic va check if (45) is obeyed.The solution for r(σ) is similar to the one found for ω = 0
r 2 (σ) = k 2 2 /2 αβ + k 2 2 ω 2 (α + β − ω 2 ) + √ ∆ sin 4αβ k 2 2 + 4ω 2 σ (46)
For the string to close r(σ) + 4ω 2 = integer .(47)
The s ay as for the ω = 0 case.
n particular, the periodic boundary conditions on the angular coordinate will lead to the same result as in the ω = 0 case with n replaced by n ω .The time coordinate is given by
t(τ, σ) = c 1 τ − c 2 σ + (c 2 + k 1 ) 4αβ + 4k 2 2 ω 2 n ω (k n[ nω 2 σ] + k 2 2 /2 αβ+k 2 2 ω 2 √ ∆ (k 2 2 +α Imposing t(τ, 0) = t(τ, 2π) gives k 2 = 2c 3 1 − 2c 1 ω 2 n 2 ω − 4c 2 1 .(49)
We now return to (45), and check if this assumption is correct.We first note that from th now bounded from below, n ω > 2ω.T c 1 ≥ ω.Combining all these re of k 2 in (45) that this is enough to ensure ∆ ≥ 0. When c 1 ning of this point.Using this result and ( 22) we can write down the energy and spin of the string
E = R 2 (c 1 + k 2 ) α ′ = R 2 c 1 α ′ n 2 ω − 2c 2 1 − 2ω 2 n 2 ω − 4c 2 1 ,(50)S = R 2 k 1 α ′ = R 2 α ′ n ω (c 2 1 − ω 2 ) n 2 ω − 4c 2 1 .
It is agai tant limits of our result can, however, be easily = 0 and retrieving the earlier result (22).The second one is by taking c 1 = ω.We get a string state with vanishing spin in AdS 3 .At this point the energy of the string depends only on ω in the expected way.
E = R 2 ω α ′ .(51)
The last interesting limit is that of "long" s Discussion

In this paper we analy NS-NS 2-form background.We saw that the "planetoid" spinning string solution of AdS 5 is not a solution in AdS 3 because of the NS-NS 2-form.According to the AdS/CFT duality string theory on AdS 3 × S 3 × M is dual to a superconformal field theory on a cylinder, which is the boundary of AdS 3 in global coordinates.In fact string theory on this background can be exactly solved in terms of the SL(2, R) WZW model and the dual (space-time) conformal field theory can be constructed.There is an exac al theory, and by J 3 the generator of U(1) ⊂ SU(2) subgroup of the (4, 4) superconformal algebra.These are not to be confused with L 0 and J 3 of the bulk WZW model.The relation to the bulk parameters E, S and ω is
E = L 0 + L0 , S = L 0 − L0 , J 3 = J 3 = ω . (53)
Using the known relation between the space-times energy in AdS 3 and the eigenvalues of J 3 0 and J3 0 from the WZW model we g and L0 = J3 0 .Thus, for a string state with a given (m, m) (32) we know the exact space-time energy E = m + m and spin rmation about the general semiclassical relation between the two, E

).We saw th
t the leading behavior of the relation is the same as in AdS 5 , but the anomalous correction is constant rather than logarithmic.

In comparison the S-dual background has R-R charge.The conformal theory in this case can be thought of the IR fixed point of a (1+1)-dimensional gauge theory on the D1-D5 system.A spinning fundamental string in this background is of the same, "planetoid", form as the GKP string, and is likely to be dual to a similar operator on the conformal theory side.In this case the relation between the energy and spin of the string exhibits the same ln S behavior in the "long" string limit.We saw that the conformal field theory dual to the AdS 3 NS-NS background is quite different in this respect.

Figure 1 :
1
Figure 1: Parametric polar plots depicting the string at τ = 0 for (i) n = 3, c 1 = 1/2.(ii) n = 4, c 1 = 1.The radial coordinate is r(σ) and the angular coordinate is φ(σ).Note that these plo ich is not the same as target space time because of (5).

Acknowledgments.We would like to thank A. Giveon and M. Karliner for a valuable discussion.The work of AL was supported in part by NSF grant-0071512.The work of YO is supported in part by the US-Israel Binational Science Foundation.by applying the operators J a −ν , with ν i = N and L −µ i with µ i = M (and the same for the anti-holomorphic sector)and imposing the physical state conditions[22](Land a similar constraint for the anti-holomorphic part.For the state to be invariant under arbit ary translation of the world-sheet coordinate σ, one must also impose a level matching conditionwhich can also be stated as4 Adding momentum on S 3In this section we would like to generalize the discussion of spinning strings in AdS 3 to include rotation in the S 3 part of the metric.The simplest ansatz is to assume that the string is point-like on S 3 .The S 3 background is given bywith the angular variables taken to be periodicConsider a string embedding in S 3 of the form θ = ωτ , and θ = x = 0.The contributions to T S 3 ++ equals that for T S 3 −− since the string embedding is σ indepen