OOpen Strings On The Rindler Horizon
Edward Witten
School of Natural Sciences, Institute for Advanced Study,Einstein Drive, Princeton, NJ 08540 USA
Abstract
It has been proposed that a certain Z N orbifold, analytically continued in N , can be used to describethe thermodynamics of Rindler space in string theory. In this paper, we attempt to implement this idea forthe open-string sector. The most interesting result is that, although the orbifold is tachyonic for positiveinteger N , the tachyon seems to disappear after analytic continuation to the region that is appropriate forcomputing Tr ρ N , where ρ is the density matrix of Rindler space and Re N >
1. Analytic continuation ofthe full orbifold conformal field theory remains a challenge, but we find some evidence that if such analyticcontinuation is possible, the resulting theory is a logarithmic conformal field theory, necessarily nonunitary. a r X i v : . [ h e p - t h ] N ov ontents Tr ρ N K Contribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.2 The K Contribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
A A Reformulation Of The Problem Of Analytic Continuation 23B An Alternative Formula For G ( z, τ ) In quantum field theory, the replica trick is often used to compute entropies. For example, to compute theentanglement entropy of quantum fields across a Rindler horizon in D -dimensional Minkowski spacetime, onefactors spacetime as R × R D − and replaces R by an N -fold cover, for some integer N , branched over theorigin in R . A formal argument shows that a path integral on this cover computes Tr ρ N , where ρ is thedensity matrix for an observer outside the horizon.To compute the Rindler entropy, one would like to analytically continue f ( N ) = Tr ρ N as a function of N tothe vicinity of N = 1, and then differentiate with respect to N to get the entropy − Tr ρ log ρ = − f (cid:48) (1). In thisparticular example, one can directly access non-integer N by considering a cone with a general opening angle.However, let us continue the discussion assuming that this option is not available, as that will be the case in our This technique may have originated in spin glass theory. See [1] for an introduction in that context. ρ is a density matrix, that is, a positive hermitian matrixsatisfying Tr ρ = 1, then f ( N ) = Tr ρ N is holomorphic in N for Re N ≥
1, and bounded in absolute value by1 in that half-plane. Therefore, we expect that the function f ( N ), initially defined for integer N , will have ananalytic continuation to the half-plane Re N ≥
1, bounded by 1. By Carlson’s theorem, such a continuation isunique if it exists, so once we know Tr ρ N for positive integer N , we can determine it throughout the half-planeRe N ≥ N = 1 will give theentropy.In local quantum field theory in a background spacetime, this procedure is not rigorously valid becauseobservations outside the horizon cannot really be described by a density matrix (basically because the algebraof observables outside the horizon is a von Neumann algebra of Type III, not of Type I). Still, in practice, thisprocedure frequently gives good results.In string theory, we cannot apply the replica trick in precisely the fashion stated, because there is no knownconformal field theory whose target space is an N -fold cover of R , branched over the origin. However, analternative has been suggested in the form of an R / Z N orbifold [3], where, for technical reasons, we willassume N to be odd; the even N case is slightly different. Now, formally, the orbifold computes Tr ρ /N . Again, one might hope to analytically continue from oddinteger values of N and eventually to compute the entropy by differentiating at N = 1. This program runsinto a variety of difficulties. Formally, Tr ρ /N should be given by the exponential of the sum of amplitudesfor connected string worldsheets, so the (unexponentiated) sum of connected worldsheets should computelog Tr ρ /N . The 1-loop approximation log Tr ρ /N (cid:12)(cid:12) would belog Tr ρ /N (cid:12)(cid:12)(cid:12) = Z ,N , (1.1)where Z ,N is the 1-loop vacuum amplitude of the orbifold.However, the closed-string spectrum of the orbifold is tachyonic, and as a result the integral in (1.1) isinfrared-divergent. This statement holds for both closed and open strings, though there is no tachyon in theopen-string spectrum. In an open-string 1-loop amplitude, the divergence results from closed-string exchangein the crossed channel.This problem in computing Tr ρ /N for N > ρ /N can be divergent if N >
1. But it does mean that we need a more refined starting point foranalytic continuation. For this, as in [3, 4], we use the standard representation of Z ,N as an integral over a The theorem says that a function f ( N ) that is holomorphic for Re N ≥
1, vanishes for positive integer N , and obeys certainbounds on its growth is identically zero. The required bounds are that f ( N ) is exponentially bounded throughout the half-planeRe N ≥
1, meaning that | f ( N ) | < C exp( λ | N | ) for some constants C, λ , and moreover that on the line Re N = 1, such a boundholds with λ < π . See [2], p. 153. The growth conditions exclude counterexamples such as sin πN , e − N N sin πN and e N sin πN .Carlson’s theorem has an obvious analog for a function that vanishes on the positive odd integers, which can be reduced to the caseof a function that vanishes on all positive integers by the mapping N → N − /
2. It is this version of the theorem that wewill actually use later, since we will consider a Z N orbifold for odd N . For a previous attempt to use this orbifold to compute the Rindler entropy, see [4]. For alternative approaches to entanglemententropy in string theory, see [5–8]. Z ,N = 12 (cid:90) ∞ d TT Z N ( T ) , (1.2)where T is a real proper time parameter and Z N ( T ) is the 1-loop partition function with modular parameter T . (To lighten notation, we write this as Z N ( T ) rather than Z ,N ( T ). Similarly, we will sometimes write, forexample, just Tr ρ /N , not Tr ρ /N (cid:12)(cid:12) .) This formula has a well-known analog for closed strings.Moreover, one can write an explicit formula for Z N ( T ) or its closed-string analog. The question thenarises of whether one can analytically continue these functions away from positive odd integer values of N in areasonable way. The most interesting version of the problem is for closed strings, since the graviton is describedby a closed string. But that version of the problem appears more difficult and in this paper we will consideropen strings. In fact, we will consider the simplest open-string problem, which is the case of a Dp-brane whoseworldvolume crosses the Rindler horizon. Thus in Type II superstring theory on R / Z N × R , we consider aDp-brane with worldvolume R / Z N × R p − .In this context, it is possible to explicitly describe the unique continuation of Z N ( T ) to the half-planeRe N ≥
1, with bounded growth, that is allowed by Carlsen’s theorem. As a bonus, it turns out that thisanalytically-continued function is actually holomorphic in a larger half-plane Re
N >
0. Note that Re
N > N > N = 1 /N . So in particular, we can analytically continue to Re N >
1. Thatis the region where we really hope for good behavior, since Tr ρ /N = Tr ρ N , and in quantum mechanics ingeneral, Tr ρ N is holomorphic for Re N > Z N ( T ) for small T . It turns out that to the extent that the closed-string sector can beprobed in this way, it is tachyon-free for Re N >
1. This means that there is no exponential divergence inthe integral that is supposed to compute log Tr ρ N . The integral may still have a power-law divergence for T → p is too large. Recall that 1-loop scatteringamplitudes of modes that propagate on the Dp-brane worldvolume are infrared-divergent for p >
6, becausethe Dp-brane acts as a source for massless bosons (the dilaton and graviton) that propagate in bulk, and thisproduces infrared-divergent effects if the number of dimensions transverse to the Dp-brane is too small. It turnsout, however, that the bound we need to make Tr ρ N infrared-finite is stronger than p ≤
6. For Re N >
1, Tr ρ N turns out to be infrared-finite for p ≤
4. After differentiating with respect to N and setting N = 1, one findsthat Tr ρ log ρ is infrared-finite for p ≤
2. The proper interpretation is far from clear. However, at a technicallevel, it appears that the unexpected behavior may be related to the fact that the orbifold theory, analyticallycontinued away from positive integer N , is a nonunitary logarithmic conformal field theory (see [9, 10] for earlywork, and [11] for a recent review). As we will explain, a logarithmic conformal field theory can potentiallylead to the sort of behavior that we will actually find.In section 2, we describe our strategy for analytic continuation. Section 3 consists of a derivation of standardformulas for the Dp-brane partition function. In section 4, we combine these ingredients to compute Tr ρ N and to investigate some of its properties. The paper also contains two appendices. Appendix A describes apossible starting point for analytic continuation. Appendix B describes an alternative formula for the partitionfunction. 3he results in this paper will probably only be useful if it turns out that the closed-string version of theproblem has a reasonable analytic continuation. If that is the case, and the analytically continued closed-stringtheory is indeed tachyon-free for Re N >
1, then presumably the one-loop closed-string contribution to Tr ρ N will be well-defined and finite. Even so, there would be a potential obstacle in going to higher orders: it willbe possible to get sensible answers for multiloop contributions to Tr ρ N only if dilaton tadpoles vanish. Notealso that we compute the annulus contribution to Tr ρ N without first discussing the disc contribution. Thereason for this is simply that a proper framework for the disc contribution is not clear. For a comment on thiscontribution, see the conclusion of section 4. The 1-loop amplitude for open strings is computed using a worldsheet that is an annulus I × S , where I isan interval and S a circle. The annulus partition function for strings in R can be written as a trace in theopen-string Hilbert space (but vanishes because of spacetime supersymmetry). The annulus partition functionfor strings on the orbifold is obtained in a standard way as a sum of traces. Let U be a generator of theorbifolding group Z N and, for integer k , let Z k,N be a trace in the open string Hilbert space with an insertionof U k (see section 3 for more detail). The orbifold partition function on the annulus is then Z N ( τ ) = 1 N N − (cid:88) k =0 Z k,N ( τ ) , (2.1)where τ is the modulus of the annulus. The factor of 1 /N reflects the fact that the projection operator onto Z N -invariant states is N (cid:80) N − k =0 U k .For analytic continuation, we will use the fact that there is a meromorphic function J ( z, τ ) with the followingproperties:1. The twisted partition function Z k,N ( τ ) can be expressed in terms of J by Z k,N ( τ ) = J ( k/N, τ ) . (2.2)2. J ( z, τ ) is a periodic and even function J ( z + 1 , τ ) = J ( z, τ ) = J ( − z, τ ) . (2.3)3. The residues at poles of J ( z, τ ) vanish exponentially for Im z → ± i ∞ . (More precisely, J and its residuesbehave near ± i ∞ in a way that permits the following computation.) Moreover, J (0 , τ ) = 0, because ofspacetime supersymmetry. A possible reason for such vanishing is as follows. A tadpole arises when a Riemann surface Σ degenerates to two componentsΣ (cid:96) and Σ r glued at a point p . Writing V for the dilaton vertex operator, the tadpole amplitude is proportional to a product (cid:104)V ( p ) (cid:105) Σ (cid:96) (cid:104)V ( p ) (cid:105) Σ r . At N = 1, both factors vanish due to supersymmetry. Differentiating with respect to N and then setting N = 1will potentially eliminate the vanishing of one of the two factors but not both. This suggests that a dilaton tadpole may notpresent a problem in multi-loop contributions to the entropy. However, the same reasoning suggests that there is such a difficulty inmulti-loop contributions to Tr ρ N with Re N >
1. The potential problem is an infrared divergence due to gravitational and dilatonback-reaction that is sourced by the tension of the Rindler horizon; this tension vanishes at tree-level assuming that the orbifoldCFT can be analytically continued in N , but it may be nonzero (for N (cid:54) = 1) in higher orders.
4. Poles of J ( z, τ ) are at Re z = 0 or 1 / Z . They are all simple poles except for a double pole at z = 1 / J is a restatement of standard facts, and will be explained in section 3.The orbifold partition function on the annulus is therefore Z N ( τ ) = 1 N N − (cid:88) k =1 J ( k/N, τ ) , (2.4)where we omit the term with k = 0 since J (0) = 0.Consider the function K ( z, N ) = N − (cid:88) k =1 π sin πz sin( πk/N ) sin π ( z − k/N ) . (2.5)It is a periodic function, K ( z + 1 , N ) = K ( z, N ), and bounded for Im z → ±∞ . The poles of K ( z, N ) in thestrip 0 ≤ Re z ≤ z = k/N , k = 1 , · · · , N − z ∼ = z +1 to define a cylinder, and view the function K ( z, N ) J ( z, τ )as a meromorphic function on the cylinder. We can view the orbifold partition function on the annulus as asum of residues of the function KJ at the poles of K on the cylinder: Z N ( τ ) = 1 N N − (cid:88) k =1 Res z = k/N ( K ( z, N ) J ( z, τ )) . (2.6)On the other hand, the sum over all residues of this function on the cylinder vanish. Here we need a conditionon the behavior of J ( z, τ ) for z → ± i ∞ ; the formulas in section 3 will make it clear that there is no problem.Therefore, we can get an alternative formula for Z N ( τ ) as minus the sum of residues at poles of J . (Since thepoles of J are at Re z = 0 , /
2, the set S of poles of J on the cylinder is disjoint from the set { /N, /N, · · · , ( N − /N } of poles of K .) Thus Z N ( τ ) = − N (cid:88) z ∈S Res z = z (cid:0) K ( z, N ) J ( z, τ ) (cid:1) . (2.7)If the poles of J ( z, τ ) were all simple poles, the formula would reduce to − N (cid:80) z ∈S K ( z , N )Res z = z J ( z, τ ).Actually, J ( z, τ ) has a double pole at z = 1 /
2, so setting S (cid:48) to be the set of simple poles of J ( z, τ ), the correctversion of the formula is Z N ( τ ) = − N (cid:88) z ∈S (cid:48) K ( z , N )Res z = z J ( z, τ ) − N ∂ z K ( z, N ) | z =1 / Res z =1 / (( z − / J ( z, τ )) . (2.8)(This treatment of the double pole at z = 1 / J ( z − /
2) = J ( − ( z − / z = 1 / J ( z, N ) behaves as c ( z − / + . . . , where c is a constant and the omitted terms areregular.) 5rom these formulas, it is clear that to analytically continue Z N ( τ ), all we need is an analytic continuationof K ( z, N ). By Carlson’s theorem, if there is an analytic continuation of K ( z, N ) that is holomorphic forRe N ≥ z , provided Re z is suitably constrained,the analytic continuation in N suggested by Carlson’s theorem does exist and can be written explicitly. Butone has to use different continuations for different values of Re z . There is not a holomorphic function of z and N that coincides with K ( z, N ) when N is an integer and satisfies the conditions of Carlson’s theorem asa function of N for fixed z .The poles of J ( z, τ ) are all at Re z = 0 or Re z = 1 /
2, so we only need to analytically continue K ( z, N )at those values of Re z . At Re z = 0, we proceed as follows. First, for integer N , an alternative formula for K ( z, N ) is K ( z, N ) = πN cot πN z − π cot πz. (2.9)For Re z = 0, K ( z, N ) is holomorphic in N for Re N ≥
1, and satisfies the appropriate exponential bounds. Itis the unique continuation of K ( z, N ) from positive integer values of N that has those properties. As a bonus,for Re z = 0, K ( z, N ) is actually holomorphic in a larger half-plane Re N > z = 1 /
2, we have to proceed more carefully and consider analytic continuation from odd positiveinteger values of N . (This is natural because we will study an orbifold defined for odd N ; an analogous orbifoldfor even N is slightly different [3].) For N an odd positive integer, an alternative formula for K ( z, N ) is K ( z, N ) = πN cot ( π ( N ( z − /
2) + 1 / − π cot πz. (2.10)For Re z = 1 / K ( z, N ) is the unique continuation of K ( z, N ) from positive odd integer values of N that isholomorphic in the half-plane Re N > z = 1 / K ( z, N ) is holomorphic in the larger half-plane Re N > N . If S is the set of polesof J ( z, τ ) with Re z = 0, S / is the set of poles with Re z = 1 /
2, and S (cid:48) / is the subset of S / with z (cid:54) = 1 / Z N ( τ ) that is holomorphic for Re N > Z N ( τ ) = − N (cid:88) z ∈S K ( z , N )Res z = z J ( z, τ ) − N (cid:88) z ∈S (cid:48) / K ( z , N )Res z = z J ( z, τ ) − N ∂ z K ( z, N ) | z =1 / Res z =1 / (( z − / J ( z, τ )) . (2.11)As is obvious in the formulas we have written, K ( z, N ) and K ( z, N ) can be continued away from the linesRe z = 0 and Re z = 1 / z and N . But those functions are different. In thissense, trying to analytically continue K ( z, N ) in the fashion suggested by Carlson’s theorem leads to differentresults depending on the assumed value of Re z .More generally, if one wants an analytic continuation of K ( z, N ) along the line Re z = a/b , where a, b arerelatively prime integers, one can restrict to N congruent to b mod 1 and use the function K [ a/b ] ( z, N ) = N cot ( π ( N ( z − a/b ) + a/b )) − cot πz. (2.12)6his has similar properties to K ( z, N ) and K ( z, N ).In section 4, we will learn that to the extent that the closed-string sector can be probed from a knowledgeof the open-string partition function, it is tachyon-free if K ( z, N ), after analytic continuation in z , is free ofpoles in the strip 0 < Re z <
1. From the formula in (2.10), we see that to get a pole in K ( z, N ), we need N ( z − /
2) = k − / k . Equivalently, with N = 1 /N , we need N = z − / k − / . (2.13)For k ∈ Z and 0 < Re z <
1, the right hand side has real part less than 1, so there is no pole in the strip ifRe
N ≥
1. This is why, as stated in the introduction, evaluation of the 1-loop open-string contribution to Tr ρ N for Re N ≥ J (0 , τ ) = 0, we would still get a correct formula for Z N ( τ ) if we add a multiple of cot πz to theoriginal function K ( z, N ); this would shift the coefficient of the cot πz terms in K ( z, N ) and K ( z, N ) by anequal amount. We have chosen this coefficient to try to make the formulas of section 4 as natural-looking aspossible. To be more precise, the coefficient was chosen so that an indication of non-diagonalizability of theclosed-string spectrum in the K contribution (which we tentatively identify with the Ramond-Ramond sectorof closed strings) only occurs for massive states. It is conceivable that this was not the best choice. In this section, we will study open strings on an annulus I × S , where I is the interval 0 ≤ σ ≤ π and S isthe circle σ ∼ = σ + 2 πT (fig. 1). For the metric of the annulus, we take d s = d σ + d σ . A path integralon the annulus has two useful interpretations. It describes an open string, of width π , parametrized by σ ,propagating in the σ direction for an imaginary proper time 2 πT . The initial and final states of the openstring are identified to make a trace. Alternatively, the same annulus describes a closed string, of circumference2 πT , parametrized by σ , propagating in the σ direction for an imaginary proper time π . In this description,there is no trace; the closed string is created from a “brane” at σ =0 and annihilated on the brane at σ = π .However, using conformal invariance, it is convenient to rescale lengths by a factor 1 /T so that the closed stringhas a standard circumference 2 π . Then σ , which is the proper time in the closed-string picture, has a range π/T . This is 2 π (cid:101) T where (cid:101) T is the closed-string proper time parameter (cid:101) T = 12 T . (3.1)The small T behavior of the annulus – which corresponds to the ultraviolet region in field theory – is bestunderstood in terms of the large (cid:101) T behavior of the closed-string description.In section 3.1, we construct the annulus partition function for the Dp-brane. In section 3.2, we computethe residues that are needed to evaluate the formula (2.11) for the analytically continued partition function. Insection 3.3, we discuss the closed-string sector. 7igure 1: An annulus in which σ parametrizes an interval and σ parametrizes a circle describes an open stringpropagating in the σ direction, or a closed string propagating in the σ direction, known as the crossed channel. Theclosed string states that contribute all have L = L , since the annulus is invariant under rotation of the circle parametrizedby σ . We will compute the annulus partition function for a Dp-brane in the orbifold using light-cone Green-Schwarzfermions. These are a set of eight world-sheet fermions that transform as a spinor of a copy of Spin(8) thatacts on, say, the first eight coordinates in R . We then construct an R / Z N × R orbifold in which Z N actson, say, the first two coordinates, which we will call X and X . Consider an element of Z N that rotates X = X + i X by a phase exp(2 π i r/N ), for some integer r . The Green-Schwarz fermions will then consist offour complex fermions that are rotated by a phase that is the square root of this. Here it is important whether N is odd or even. If N is odd, then a square root of exp(2 π i r/N ) is again an N th root of 1, which one canwrite as exp(2 π i r (1 − N ) / N ). We will focus on this case, both because it is slightly simpler and because weare interested in analytically continuing to N = 1, which is odd, so it is more natural to start with odd valuesof N . (See [3] for a description of the corresponding orbifolds with even N .)It is convenient to set k = r (1 − N ) /
2, which is an integer assuming that N is odd. Then a general elementof Z N acts on the four complex fermions by a phase exp(2 π i k/N ), and on X by a phase exp(4 π i k/N ). The Z N element with k = 1 mod N is a generator U of the Z N group.Let H be the Hilbert space of open strings on R , and H the corresponding Hamiltonian. The annuluspartition function on R in the Green-Schwarz description isTr H ( − F exp( − πHT ) . (3.2)This vanishes because of supersymmetry. The open-string Hilbert space H N of the orbifold is just the Z N -invariant subspace of H . The projector onto the Z N -invariant subspace is P = N (cid:80) N − k =0 U k , so the annulus What follows is a standard type of computation which we present for completeness; an equivalent set of formulas for thisparticular orbifold has been described and discussed in [3, 4]. Z N =Tr H N ( − F exp( − πHT ) = Tr H ( − F P exp( − πHT )= 1 N N − (cid:88) k =0 Tr H U k ( − F exp( − πHT ) (3.3)The k = 0 term in this sum coincides with (3.2), up to a factor of 1 /N , and so vanishes. Thus Z N = N − (cid:88) k =1 Z k,N , Z k,N = 1 N Tr H U k ( − F exp( − πHT ) . (3.4)We want to compute the annulus partition function for open strings that end on a Dp-brane whose world-volume is spanned by X , · · · , X p +1 . Except for bosonic zero-modes, the value of p is not important. One wayto see this is to observe that if one toroidally compactifies the dimensions X p +2 , · · · , X , then T -duality onsome of those coordinates will change the value of p without affecting the spectrum of worldsheet modes otherthan the bosonic zero-modes. For the bosonic zero-modes, there definitely is a dependence on p .As usual, modes of a free field on an interval of width π are conveniently compared to modes of a chiral freefield on a circle of circumference 2 π that is made by taking two copies of I glued together at the endpoints.This circle can be parametrized by an angular variable σ , 0 ≤ σ ≤ π . Together σ and σ parametrize atorus Σ with a modular parameter τ = i T and therefore q = exp(2 π i τ ) = exp( − πT ). In computing Z k,N from eqn. (3.4), all bosons and fermions obey periodic boundary conditions in the σ and σ directions, exceptfor the twist in the σ direction associated to the orbifolding.The twisted partition function Z k,N is a product of factors coming from bosons and fermions: Z k,N = Z Fk,N Z Bk,N . (3.5)First let us compute the fermionic partition function Z Fk,N . Let ψ, (cid:101) ψ be a conjugate pair of chiral fermions,twisted by angles ± πk/N . For every positive integer n , the mode expansion of either ψ or (cid:101) ψ contains a creationoperator that raises the energy by n units. The contribution of these modes to the partition function is a factor(1 − q n exp(2 π i k/N ))(1 − q n exp( − π i k/n )) . (3.6)More subtle is the role of the fermion zero-modes. Both ψ and (cid:101) ψ have a zero-mode, transforming respectivelyas exp( ± π i k/N ) under the twist. Quantizing these two modes gives a pair of states, of opposite statistics,transforming under the twist as exp( ± π i k/N ). The contribution to the partition function of these modes isthus a factor ± (exp( π i k/N ) − exp( − π i k/N )) = ±
2i sin( πk/N ) , (3.7)where the overall sign depends on which of the two states is bosonic and which is fermionic. Out of context,there is no natural choice of this sign, a fact that is related to the fact that exp( π i k/N ) is a square root of thetwisting phase exp(2 π i k/N ), but there is no natural way to choose the sign of the square root. This ambiguitywill disappear in a moment when we combine the contributions of the four complex fermions. Thus q is positive; in what follows, q α , for a real number α , will represent the positive number exp( − παT ). /
12. Altogether, the partition function of such a complex fermion is ±
2i sin( πk/N ) q / ∞ (cid:89) n =1 (1 − q n exp(2 π i k/N ))(1 − q n exp( − π i k/n )) . (3.8)Combining four such multiplets, we get the fermionic partition function Z Fk,N = (2 sin( πk/N )) q / ∞ (cid:89) n =1 (1 − q n exp(2 π i k/N )) (1 − q n exp( − π i k/n )) . (3.9)Here the sign is correct. This may be seen as follows. When we expand (2 sin( πk/N )) = (2i( e i πk/N − e − i πk/N )) in powers of exp(2 π i k/N ), only integer powers appear, so we do not need to choose a square root. The maximumpower that appears is exp(2 π i k/N ) = exp(4 πik/N ). This is the same as the phase by which the given Z N element acts on X = X + i X , so it corresponds to angular momentum 1 in rotation of that plane. Thecorresponding spin 1 mode is a gauge boson mode, which is bosonic, so it should appear with a positivecoefficient. We have chosen the sign in eqn. (3.9) to ensure this.We can conveniently write Z Fk,N = F ( z, τ ) , (3.10)with z = k/N (3.11)and F ( z, τ ) = ( w / − w − / ) q / ∞ (cid:89) n =1 (1 − q n w )(1 − q n w − ) , w = exp(2 π i z ) . (3.12)Now let us consider Z Bk,N . The nonzero modes of the bosons can be treated rather as before. First considerthe complex boson field X that is twisted by exp(4 π i k/N ). For every positive integer n , both X and X had amode of energy n . The partition function of these modes is11 − q n exp(4 π i k/N ) 11 − q n exp( − π i k/N ) . (3.13)This is just the inverse of the corresponding factor for fermions, except that the twist angle is twice as large.Similarly, the ground state energy of a complex boson (untwisted in the σ direction) is − /
12, which willgive a factor q − / . We also have to include in light cone gauge the nonzero modes of six untwisted bosons X , · · · , X . Including the ground state energy, these give a factor 1 /η ( τ ), where η is the Dedekind eta function η ( τ ) = q / (cid:81) ∞ n =1 (1 − q n ). Altogether, then the nonzero modes of the bosons give1 q / (cid:81) ∞ n =1 ((1 − q n exp(4 π i k/N ))(1 − q n exp( − π i k/N )) η ( τ ) . (3.14)So far everything is just the reciprocal of what we would get for fermions with the same twist angles.10owever, for the zero-modes the story will be different. For open strings ending on the Dp-brane, thefields X p +2 , · · · , X have no zero-modes. The fields X , · · · , X p +1 have untwisted zero-modes, as if we wereconsidering open strings in R . And the fields X , X have zero-modes in the σ direction that are twisted byan angle 4 πk/N in the σ direction; we call these modes twisted zero-modes.The action for free bosons X i on a Euclidean string worldsheet, in conformal gauge, is I = 14 πα (cid:48) (cid:90) d σ d σ (cid:88) i,α (cid:18) ∂X i ∂σ α (cid:19) . (3.15)The bosonic zero-modes are functions of the worldsheet “time” coordinate σ only. Integrating over σ , whichruns from 0 to π , and writing t for σ , we get an effective action for the zero-modes, I = 14 α (cid:48) (cid:90) d t (cid:88) i (cid:18) d X i d t (cid:19) . (3.16)The corresponding Hamiltonian is H = α (cid:48) (cid:80) i P i , where P i = − i ∂∂X i . The path integral with the action (3.16)on a line segment of length 2 πT (rather than a circle) with initial and final values (cid:126)X and (cid:126)X (cid:48) at the two endsof the interval, computes the heat kernel (cid:104) (cid:126)X (cid:48) | exp( − πT α (cid:48) (cid:126)P ) | (cid:126)X (cid:105) = 1(8 π α (cid:48) T ) n/ exp( −| (cid:126)X − (cid:126)X (cid:48) | / πα (cid:48) T ) , (3.17)where n is the number of components of (cid:126)X and (cid:126)X (cid:48) that are considered. For open strings ending on the Dp-brane, there are p − t direction, so for those modes, n = p − (cid:126)X (cid:48) = (cid:126)X and integrate over (cid:126)X . The integral gives V , the volume of that intersection, times 1 / (8 π α (cid:48) T ) ( p − / . The result is then a factor V (8 π α (cid:48) T ) ( p − / . (3.18)If we take the R / Z N × R model literally, then V is infinite. We could make V finite by toroidal compactificationof the coordinates X , · · · , X p +1 , without significantly altering the rest of the discussion. In practice, it is notnecessary to do this explicitly. At any rate, we are interested in identifying contributions (to the entropy, or tolog Tr ρ N ) that are proportional to V . If we were studying a D9-brane, V would be the volume of the Rindlerhorizon; for a more general Dp-brane, it is the volume of the brane worldvolume intersected with the Rindlerhorizon.Now let us consider the twisted zero-modes. In going around the circle, X and X are twisted by therotation matrix R = (cid:18) cos(4 πk/N ) sin(4 πk/N ) − sin(4 πk/N ) cos(4 πk/N ) (cid:19) . (3.19)The path integral on the circle is then given by setting X (cid:48) = RX and n = 2 in eqn. (3.17) and integrating over X , giving 18 π α (cid:48) T (cid:90) d (cid:126)X exp( −| (1 − R ) X (cid:48) | / πα (cid:48) T ) = 1 | det(1 − R ) | . (3.20) This is inevitable, because the path integral for the bosonic fields X i , in the absence of a worldsheet B -field, is manifestlypositive, while the integral over the fermion zero-modes gave a result that is not positive-definite. | det(1 − R ) | = 14 sin (2 πk/N ) . (3.21)This does not coincide with the inverse of the zero-mode contribution to a fermionic integral with the sametwist angle. From eqn. (3.7), but with k/N replaced by 2 k/N , we see that the zero modes of a complex fermionwith twist angle 4 πk/N would contribute a factor ±
2i sin(2 πk/N ) in the path integral. The zero modes of atwisted boson contribute the absolute value squared of the inverse of this. Thus there is an extra factor of1 / ( ∓
2i sin(2 πk/N )) = ± i / πk/N ) compared to what we would expect if the boson path integral were justthe inverse of a corresponding fermion path integral.Putting all these factors together, the bosonic partition function for k (cid:54) = 0 is Z Bk,N = V (8 π α (cid:48) T ) ( p − /
14 sin (2 πk/N ) q / (cid:81) ∞ n =1 (1 − q n exp(4 π i k/N )(1 − q n exp( − π i k/N )) 1 η ( τ ) . (3.22)This can be written more simply in terms of the function F ( z, τ ) introduced in eqn. (3.12): Z Bk,N = V (8 π α (cid:48) T ) ( p − / i2 sin 2 πz F (2 z, τ ) η ( τ ) . (3.23)The overall partition function Z k,N = Z Fk,N Z Bk,N is then Z k,N ( τ ) = V (8 π α (cid:48) T ) ( p − / i2 sin 2 πz F ( z, τ ) F (2 z, τ ) η ( τ ) , z = k/N. (3.24)The right hand side is the function J ( z, τ ) that we need in eqn. (2.11) for the analytically continued partitionfunction.This function has the properties that were claimed in section 2. To understand it better, let us look moreclosely at the function F ( z, τ ) defined in eqn. (3.12). The infinite product is absolutely convergent for | q | < F ( z, τ ) is holomorphic throughout that region. There is an obvious antiperiodicity F ( z + 1 , τ ) = − F ( z, τ )(which leads to periodicity of F ( z, τ ) /F (2 z, τ )), but actually there is a second periodicity under z → z + τ ,which corresponds to w → qw , with w = exp(2 π i z ). To prove this, we can freely rearrange terms in the productthat defines F ( z, τ ), since this product is absolutely convergent, and we find F ( z + τ, τ ) = F ( z, τ ) (1 − w − )( q / w / − q − / w − / )(1 − qw )( w / − w − / ) = F ( z, τ ) q − / w − . (3.25)It follows then that F (2 z + 2 τ, τ ) = F (2 z + τ, τ ) q − / ( qw ) − = F (2 z, τ ) q − w − . (3.26)So F ( z, τ ) /F (2 z, τ ) is invariant under z → z + τ . In other words, as a function of z , this function is doubly-periodic under the lattice shifts z → z + m + nτ .It is convenient to define G ( z, τ ) = F ( z, τ ) F (2 z, τ ) η ( τ ) . (3.27)12he function J ( z, τ ) is J ( z, τ ) = V (8 π α (cid:48) T ) ( p − / i2 sin 2 πz G ( z, τ ) . (3.28)The function G ( z, τ ) is doubly-periodic, meaning that for fixed τ , G ( z, τ ) is a meromorphic function on theelliptic curve (or genus 1 Riemann surface) E = C / ( Z + Z τ ). From the product representation (3.12), we seethat F ( z, τ ) vanishes if and only if q n w = 1 for some integer n , which means that z is an element of the lattice Z + Z τ . Similarly, F (2 z, τ ) vanishes if and only if q n w = 1 for some n , which means that 2 z is an elementof the lattice. For 2 z to be a lattice element means that z is congruent, modulo a lattice shift, to one of thehalf-lattice points 0 , / , τ / τ ) /
2. Combining these statements, G ( z, τ ) has simple poles at the non-zerohalf-lattice points z = 1 / , τ / , and (1 + τ ) /
2, coming from zeros of F (2 z, τ ), and it has a third order zero at z = 0. Up to a constant multiple, G ( z, τ ) is uniquely determined, for fixed q , by these poles and zeros, and inparticular the fact that the zero at z = 0 is of odd order implies that G ( z, τ ) is an odd function of z : G ( − z, τ ) = − G ( z, τ ) . (3.29)Analogous properties of the partition function Z ( z, τ ) follow immediately, taking into account the additionalfactor of 1 / sin 2 πz . In particular, J ( z, τ ) has precisely the same poles as G ( z, τ ) except that its poles at z = 1 / Z are double poles. Moreover J ( z, τ ) = J ( − z, τ ) = J ( z + 1 , τ ) . (3.30)It is possible to write G ( q, z ) in terms of theta functions, together with the Dedekind eta function [3, 4].This representation is related to the product formula that we have given by the Jacobi triple product formula.Because G ( q, z ) is a doubly-periodic meromorphic function, it is also possible to write it as a lattice sum; seeAppendix B. In view of eqn. (3.28), we can slightly rewrite the formula (2.11) for the analytically continued partition function Z N ( τ ) in terms of residues of G ( z, τ ) rather than J ( z, τ ):2(8 π α (cid:48) T ) ( p − / V Z N ( τ ) = − N (cid:88) z ∈S i K ( z , N )sin 2 πz Res z G ( z, τ ) − N (cid:88) z ∈S (cid:48) i K ( z , N )sin 2 πz Res z G ( z, τ ) − πN ∂ z K ( z, N ) | z =1 / Res z =1 / G ( z, τ ) . (3.31)To evaluate this, we need a formula for the residues of G ( z, τ ).We first consider the residue at z = τ / w = q / , where w =exp(2 π i z ). From the product formula for G ( z, τ ), one finds that G ( z, τ ) z → τ/ ∼ − − qw − · q − / (cid:81) ∞ n =1 (1 − q n − / ) η ( τ ) . (3.32)13ence the residue of G ( z, τ ) at z = τ / i4 π q − / (cid:81) ∞ n =1 (1 − q n − / ) η ( τ ) = i4 π η ( τ / η ( τ ) . (3.33)The other cases can be treated similarly. We note that z = (1 + τ ) / w = − q / . We have G ( z, τ ) z → (1+ τ ) / ∼ − qw − · q − / (cid:81) ∞ n =1 (1 + q n − / ) η ( τ ) , (3.34)so the residue of G at z = (1 + τ ) / − i4 π q − / (cid:81) ∞ n =1 (1 + q n − / ) η ( τ ) = − i4 π η ( τ ) η (2 τ ) η ( τ / . (3.35)Finally, let us look at the residue at z = 1 / w = −
1. We have G ( z, τ ) z → / ∼ − w − w − · q / (cid:81) ∞ n =1 (1 + q n ) η ( τ ) , (3.36)so the residue of G at z = 1 / π q / (cid:81) ∞ n =1 (1 + q n ) η ( τ ) = i4 π η (2 τ ) η ( τ ) . (3.37)As a check on these computations, the vanishing of the sum of the residues of the doubly-periodic differential G ( z, τ )d z at its poles at z = 1 / , τ /
2, and (1 + τ ) / q − / (cid:81) ∞ n =1 (1 − q n − / ) η ( τ ) − q − / (cid:81) ∞ n =1 (1 + q n − / ) η ( τ ) + 16 q / (cid:81) ∞ n =1 (1 + q n ) η ( τ ) = 0 . (3.38)This is the identity, due originally to Jacobi, that was used by Gliozzi, Olive, and Scherk [12] to demonstratesupersymmetry of the open-string spectrum using RNS fermions in light-cone gauge. The three terms are thecontributions to the partition function of the string from the three even spin structures on a torus. If we labela spin structure in the usual way as ±± , where the two signs correspond to periodic or antiperiodic boundaryconditions in the σ and σ directions, respectively, then the three terms in eqn. (3.38) correspond to the − +, −− , and + − spin structures.Of course, in our calculation, we started with Green-Schwarz fermions, not RNS fermions. But we introduceda Z N twist to construct the orbifold, and the evaluation of the partition functions in terms of residues involvedanalytically continuing with respect to the twist parameter and extracting residues that arise at the points atwhich the bosons are untwisted – leading to a pole in G ( z, τ ) from a bosonic zero-mode – while the fermionsare twisted by signs. The twisting of the fermions determines an even spin structure and the residue of G ( z, τ )at such a point is the RNS partition function for that spin structure. The residue is taken in the z variable, that is we want the residue of the one-form d z G ( z, τ ). .3 Closed-String Sector For closed strings, the R / Z N × R orbifold has NS-NS tachyons and massless Ramond-Ramond states in twistedsectors. This was shown in [3] using both the Green-Schwarz and RNS descriptions.A short explanation in the Green-Schwarz language is as follows. We parametrize a closed string by anangular variable σ , with σ ∼ = σ + 2 π . Consider closed strings that are twisted by an element U k ∈ Z N inthe sense that, for any worldsheet field Φ, Φ( σ + 2 π ) = U k Φ( σ ). Since the value of k only matters mod N , we can assume that − N/ < k < N/
2. For k >
0, the partition function that counts the right-movingoscillator modes in this twisted sector is − G (( k/N ) τ, τ ). (For k <
0, the formula is + G (( k/N ) τ, τ ), leading to asimilar analysis.) To construct the orbifold spectrum in the twisted sector, we have to combine the right-movingoscillator spectrum with a similar left-moving spectrum, and project onto the subspace of Z N invariants. Wehave − G (( k/N ) τ, τ ) = q − k/N − O ( q k/N ) . (3.39)The ground state, in RNS language, is an NS sector tachyon with L = − k/N , and the first excited level consistsof 4 Ramond sector states with L = 0. When we combine left- and right-movers and project onto the subspaceof Z N invariants, we are left with an NS-NS tachyon with L = L = − k/N and 4 = 16 Ramond-Ramondstates with L = L = 0. (The states with L = − k/N , L = 0, or vice-versa, are removed when one projectsonto Z N invariants.)In the annulus partition function, we expect to see closed-string states propagating in the crossed channel.Let us discuss the contribution that we should expect from the twisted RR states with L = L = 0. Weconsider a twisted sector closed-string state that is emitted on the left in fig. 1 of section 3, propagates fora proper time 2 π (cid:101) T , and is absorbed on the right. We have to compute the appropriate matrix element ofexp( − πH (cid:101) T ), where H = L + L is the Hamiltonian for the mode in question. Since we consider a modewhose oscillator content has L = L = 0, H receives a contribution only from the effective Hamiltonian ofthe untwisted bosonic zero-modes. This effective Hamiltonian can be worked out as in the discussion of eqn.(3.16), with the only change being α (cid:48) → α (cid:48) / π . Hence H = α (cid:48) (cid:80) i P i , and the analog of (3.17) is (cid:104) (cid:126)X (cid:48) | exp( − π (cid:101) T α (cid:48) (cid:126)P ) | (cid:126)X (cid:105) = 1(4 π α (cid:48) (cid:101) T ) exp( −| (cid:126)X − (cid:126)X (cid:48) | / πα (cid:48) (cid:101) T ) , (3.40)which gives the amplitude for a twisted sector closed-string state to be emitted at a point X on the intersectionof the Dp-brane worldvolume with the Rindler horizon and absorbed at a point X (cid:48) on that intersection. (Forclosed strings, we have set n = 8, since all 8 bosonic coordinates along the Rindler horizon have zero modes in theclosed-string sector.) To get the full amplitude, we have to integrate X and X (cid:48) over the intersection of the Dp-brane worldvolume with the Rindler horizon. The dimension of this intersection is p −
1, so that is the numberof components of X and X (cid:48) over which we integrate. The integral over X and X (cid:48) gives V / (4 π α (cid:48) (cid:101) T ) (9 − p ) / . Wealso have to integrate over the proper time (cid:101) T . The contribution of the massless modes to this integral behavesas V (4 π α (cid:48) ) (9 − p ) / (cid:90) ∞··· d (cid:101) T (cid:101) T (9 − p ) / , (3.41)and this will be the dominant behavior for large (cid:101) T if there are no tachyons. We will look for such behavior insection 4. 15 twisted sector state in which the oscillator modes have L = (cid:101) L = n would, by the same reasoning, givea contribution that would behave for large (cid:101) T as V (4 π α (cid:48) ) (9 − p ) / (cid:90) ∞··· d (cid:101) T (cid:101) T (9 − p ) / exp( − πn (cid:101) T ) . (3.42)We will find such contributions in section 4, but after analytic continuation, we will also find terms that cannotquite be put in this form. Tr ρ N We now have the information we need to make explicit the formula (3.31) for the analytically continued partitionfunction Z N ( τ ). We work out first the contributions proportional to K and then those proportional to K . K Contribution
The contribution to the right hand side of eqn. (3.31) that is proportional to K is a sum over points in thecylinder z ∼ = z + 1 that are congruent mod τ to τ /
2. Those points are at z = ( r + 1 / τ = ( r + 1 / T , r ∈ Z .A short calculation, using the residue formula (3.33), shows that the contribution of these points to the righthand side of eqn. (3.31) is − (cid:88) r ∈ Z (cid:18) coth πN ( r + 1 / T − N coth π ( r + 1 / T (cid:19) π ( r + 1 / T η (i T / η (i T ) . (4.1)Going back to eqns. (1.1) and (1.2), the corresponding contribution to log Tr ρ /N (cid:12)(cid:12) is − V π α (cid:48) ) ( p − / (cid:90) ∞ d TT ( p +1) / (cid:88) r ∈ Z (cid:18) coth πN ( r + 1 / T − N coth π ( r + 1 / T (cid:19) π ( r + 1 / T η (i T / η (i T ) . (4.2) T → ∞ is the infrared region in the open-string channel, the region in which one would expect to agree with a1-loop field theory computation using the massless states of the open string. The T → ∞ behavior is dominatedby r +1 / ± /
2. After taking account of the large T behavior of the eta function, η (i T ) ∼ exp( − πT / T as (cid:82) ∞··· d T /T ( p +1) / , and convergesfor T → ∞ if p >
1. This is the expected behavior in field theory. For p = 1, the integral is logarithmicallydivergent for T → ∞ . This infrared divergence is present in the field theory limit (even for integer N ) and wecannot expect to get rid of it by going to string theory. T → T → (cid:101) T = 1 / T , as we recalled in the introduction to section3. To determine the behavior for small T , as usual one uses the modular identity η (i T ) = 1 √ T η (i /T ) , (4.3)which implies η (i T / η (i T ) = (2 T ) η (2i /T ) η (i /T ) = (2 T ) η (4i (cid:101) T ) η (2i (cid:101) T ) . (4.4)So we can rewrite the integral (4.2) in terms of (cid:101) T : − V · (9 − p ) / (8 π α (cid:48) ) ( p − / (cid:90) ∞ d (cid:101) T (cid:101) T (11 − p ) / (cid:88) r ∈ Z (cid:18) coth( πN ( r + 1 / / (cid:101) T ) − N coth( π ( r + 1 / / (cid:101) T ) (cid:19) × π ( r + 1 / / (cid:101) T η (4i (cid:101) T ) η (2i (cid:101) T ) . (4.5)The sum over r is of the general form (cid:80) r ∈ Z W (( r + 1) / (cid:101) T ) with a certain function W that is holomorphicand bounded in a strip containing the real axis. For such a function, it is useful to write (cid:88) r ∈ Z W (( r + 1 / / (cid:101) T ) = (cid:88) m ∈ Z (cid:90) ∞−∞ d r exp(2 π i mr ) W (( r + 1 / / (cid:101) T ) = (cid:101) T (cid:88) m ∈ Z (cid:90) ∞−∞ d x exp(2 πim ( x (cid:101) T − / W ( x ) , (4.6)where in the last step we change variables from r to x = ( r + 1 / / (cid:101) T . This procedure is essentially Poisson re-summation. The terms with m (cid:54) = 0 vanish exponentially for (cid:101) T → ∞ , as one learns by deforming the integrationcontour above or below the real r axis. So modulo exponentially small terms, we reduce to (cid:101) T (cid:82) ∞−∞ d x W ( x ). Inthe present case, this is (cid:101) T Y ( N ) with Y ( N ) = (cid:90) ∞−∞ d x (cid:18)(cid:18) coth( πN x/ − N coth( πx/ (cid:19) πx (cid:19) . (4.7)There does not seem to be a convenient closed form for this integral, but some special cases relevant to theentropy are Y (1) = 0 , Y (cid:48) (1) = π − π . (4.8)The large (cid:101) T behavior of the integral (4.5) is − V · (9 − p ) / (8 π α (cid:48) ) ( p − / (cid:90) ∞··· d (cid:101) T (cid:101) T (9 − p ) / Y ( N ) . (4.9)As discussed in section 3.3, this is the expected behavior for tree level exchange of a massless twisted sectorclosed-string state. The integral converges for p ≤
6. Since 9 − p is the codimension of the Dp-brane worldvolumein spacetime, the divergence for p > (cid:101) T . First of all, tree level exchange of a closed-string state is expected to produce a contribution proportional to exp( − π (cid:101) T ( L + L )) (times the amplitudefor the state in question to be produced by the boundary on the left in fig. 1 of section 3 and annihilated onthe right). We note that 16 η (4i (cid:101) T ) η (2i (cid:101) T ) = 16 (cid:81) ∞ n =1 (1 + exp( − πn (cid:101) T )) − exp( − πn (cid:101) T ) (4.10)is the same as Tr exp( − πL ) in the Ramond sector of open strings, or equivalently in the right-moving Ramondsector of closed strings. This has a simple interpretation: a standard fact in D-brane physics is that the closed-string states that propagate in the crossed channel of the annulus have the same Fock space structure for left-and right-movers, so the enumeration of these states is the same as the enumeration of states in the right-movingsector, with the usual exp( − π ( L + L ) (cid:101) T ) replaced by exp( − πL (cid:101) T ), since these states have L = L . Thissuggests that the K contribution to the analytically continued partition function represents, for large (cid:101) T , thecontribution of Ramond-Ramond states in the crossed channel. (This interpretation may be oversimplified, asit does not take into account the sum over r in eqn. (4.5), which also depends on on (cid:101) T . If the interpretation isvalid, it depends on the precise choice we made of the K function with no pole at z = 0; see the conclusion ofsection 2.) The massless twisted sector modes responsible for the behavior in eqn. (4.9) are presumably simplythe massless twisted sector Ramond-Ramond states of the orbifold; see section 3.3.Some exponentially small terms come from the expansion of eqn. (4.10) in powers of exp( − π (cid:101) T ). Additionalexponentially small contributions come from terms with m (cid:54) = 0 in eqn. (4.6), which we have omitted so far.The integral for general m is (cid:90) ∞−∞ d x exp(2 π i m ( x (cid:101) T − / (cid:18)(cid:18) coth( πN x/ − N coth( πx/ (cid:19) πx (cid:19) . (4.11)Let us first suppose that N is an odd integer. The singularities of the integrand are then simple poles at x = 2i k/N , for nonzero integer k not divisible by N . The contribution of a simple pole at Im x = b is evaluatedby shifting the integration contour upward or downward in the complex x -plane depending on the sign of mb .This contribution is a constant multiple of exp( − π | mb | (cid:101) T ). With b = 2 k/N , this is exp( − π ( k (cid:48) /N ) (cid:101) T ), where k (cid:48) = | mk | . That is the expected behavior due to twisted sector states with L = L = k (cid:48) /N , which of courseare present in the orbifold theory at levels of positive mass.What happens if we analytically continue away from odd integer N ? The most important difference is thatnow there are double poles at x = 2i k for integer k , in addition to the simple poles. The contribution of sucha double pole is not a simple exponential. The residue at a double pole has an extra factor of (cid:101) T , leading to acontribution of the form (cid:101) T exp( − πk (cid:48) (cid:101) T ), again with k (cid:48) = | mk | . What is the meaning of the extra factor of (cid:101) T ?It is not possible to get such an extra factor in a unitary quantum field theory, in which L and (cid:101) L can bediagonalized. However, such contributions are possible in a logarithmic conformal field theory [9–11]. In suchtheories, which are necessarily nonunitary, L and L are not diagonalizable. Suppose that in a two-dimensionalsubspace, L = k (cid:48) I + (cid:18) (cid:19) , (4.12)18here I is the 2 × − π (cid:101) T L ) = exp( − πk (cid:48) (cid:101) T ) (cid:18) − π (cid:101) T (cid:19) . (4.13)If the boundary state associated to the Dp-brane, when projected to this two-dimensional subspace, is a genericlinear combination of the two states, then the contribution of these states in the closed-string channel of theannulus will have a term proportional to (cid:101) T exp( − πk (cid:48) (cid:101) T ).Thus there is an indication that analytic continuation away from integer N , if it can be carried out sys-tematically, gives a logarithmic conformal field theory. For the K contribution, which we tentatively interpretas the Ramond-Ramond contribution in the closed-string channel, we have found such an indication only formassive modes. For the K contribution, we will find such behavior also for massless states. K Contribution
The contribution to the right hand side of eqn. (3.31) that is proportional to K is a sum over points in thecylinder z ∼ = z + 1 that are congruent mod τ to either (1 + τ ) / /
2. Let us first consider the contributionsat z = 1 / r + 1 / τ , with integer r . Using the residue formula (3.35), one finds that the contribution ofthese points to the right hand side of eqn. (3.31) is − (cid:88) r ∈ Z (cid:18) tanh πN ( r + 1 / T − N tanh π ( r + 1 / T (cid:19) π ( r + 1 / T η (i T ) η (i T / η (2i T ) . (4.14)The contributions at z = 1 / rτ are similar; we just replace r + 1 / r and use the residue formula (3.37),to get 14 (cid:88) r ∈ Z (cid:18) tanh πN rT − N tanh πrT (cid:19) πrT η (2i T ) η (i T ) . (4.15)(The r = 0 term here actually comes from the last term in eqn. (3.31), which originates in the double pole ofthe partition function at z = 1 / T → ∞ in the open-string channel comes from the terms in eqn. (4.14) with r + 1 / ± / r = 0 contribution in eqn. (4.15). Other contributions vanish exponentially.Let us look at the opposite limit T →
0. We have with the help of eqn. (4.3) η (i T ) η (i T / η (2i T ) = T η (2i (cid:101) T ) η (i (cid:101) T ) η (4i (cid:101) T ) , η (2i T ) η (i T ) = T η (i (cid:101) T ) η (2i (cid:101) T ) . (4.16)The analog of eqn. (4.5) is − V · (9 − p ) / (8 π α (cid:48) ) ( p − / (cid:90) ∞ d (cid:101) T (cid:101) T (11 − p ) / (cid:32) A ( (cid:101) T ) η (2i (cid:101) T ) η (i (cid:101) T ) η (4i (cid:101) T ) + B ( (cid:101) T ) η (i (cid:101) T ) η (2i (cid:101) T ) (cid:33) , (4.17)19ith A ( (cid:101) T ) = − (cid:88) r ∈ Z (cid:18) tanh πN ( r + 1 / / (cid:101) T − N tanh π ( r + 1 / / (cid:101) T (cid:19) π ( r + 1 / / (cid:101) TB ( (cid:101) T ) = 14 (cid:88) r ∈ Z (cid:18) tanh πN r/ (cid:101) T − N tanh πr/ (cid:101) T (cid:19) πr/ (cid:101) T . (4.18)The functions η (2i (cid:101) T ) η (i (cid:101) T ) η (4i (cid:101) T ) = exp(2 π (cid:101) T ) (cid:81) ∞ n =1 (1 + exp( − π (2 n − (cid:101) T )) (1 − exp(4 πn (cid:101) T )) = exp(2 π (cid:101) T ) + 8 + O (exp( − π (cid:101) T )) η (i (cid:101) T ) η (2i (cid:101) T ) = exp(2 π (cid:101) T ) (cid:81) ∞ n =1 (1 − exp( − π (2 n − (cid:101) T )) (1 − exp(4 πn (cid:101) T )) = exp(2 π (cid:101) T ) − O (exp( − π (cid:101) T )) (4.19)can be interpreted, respectively, as Tr exp( − πL T ) and as Tr ( − F exp( − πL T ), in the right-moving Neveu-Schwarz sector of a closed string. This suggests that the K contribution to the partition function represents,for large (cid:101) T , the contribution of NS - NS states in the crossed channel. (Like the analogous comment in section4.1, this interpretation may be oversimplified.)The contributions that can potentially grow exponentially for large (cid:101) T come from the exp(2 π (cid:101) T ) terms onthe right hand side of eqn. (4.19). The corresponding contributions to the A and B terms in eqn. (4.17), aftersetting r = s/
2, can be conveniently combined together and written14 exp(2 π (cid:101) T ) (cid:88) s ∈ Z ( − s (cid:18) tanh πN s/ (cid:101) T − N tanh πs/ (cid:101) T (cid:19) πs/ (cid:101) T . (4.20)We will return to this sum in a moment, but first let us look at the contributions that come from theconstant terms on the right hand side of eqn. (4.19). One might expect that they correspond to the exchangeof massless closed-string states. Now the A and B terms combine to − (cid:88) s ∈ Z (cid:18) tanh πN s/ (cid:101) T − N tanh πs/ (cid:101) T (cid:19) πs/ (cid:101) T . (4.21)The large (cid:101) T behavior of this sum can be analyzed by a similar method to what we used for the sum that weencountered in eqn. (4.5). Modulo exponentially small terms, the sum can be replaced by an integral − (cid:101) T (cid:90) ∞−∞ d x (cid:18) tanh πN x/ − N tanh πx/ (cid:19) πx/ , (4.22)leading again to a contribution to the (cid:101) T integral in eqn. (4.17) of the form (cid:82) ∞··· d (cid:101) T / (cid:101) T (9 − p ) / . There are alsoexponentially small corrections, controlled by poles of the function V ( x ) = (cid:18) tanh πN x/ − N tanh πx/ (cid:19) πx/ . (4.23)20or odd integer N , this function has simple poles, corresponding to contributions of closed-string states with L >
0. Upon continuation away from odd integer N , V ( x ) has double poles, suggesting a nondiagonalizabilityof the spectrum, again at L >
0, as we found in section 4.1.Going back to the potentially tachyonic sum in eqn. (4.20), Poisson resummation converts it to14 exp(2 π (cid:101) T ) (cid:88) m ∈ Z (cid:90) ∞−∞ d s exp(i π (2 m + 1) s ) (cid:18) tanh πN s/ (cid:101) T − N tanh πs/ (cid:101) T (cid:19) πs/ (cid:101) T = (cid:101) T π (cid:101) T ) (cid:88) m ∈ Z (cid:90) ∞−∞ d x exp(2i π (2 m + 1) x (cid:101) T ) (cid:18) tanh πN x/ − N tanh πx/ (cid:19) πx , (4.24)where x = s/ (cid:101) T . All contributions to the integral are exponentially small for large (cid:101) T . The dominant contribu-tions come from 2 m + 1 = ± P ( x ) = (cid:18) tanh πN x/ − N tanh πx/ (cid:19) πx . (4.25)Let us first discuss the situation for odd integer N . All poles of P ( x ) are simple poles. The poles closestto the real axis are at x = ± i /N . But we can get a somewhat clearer picture by considering all of the polesfor | Im x | ≤
1. As we will see in a moment, poles at | Im x | < | Im x | = 1 will make a contribution with a constantor polynomial dependence on (cid:101) T . For odd integer N , the poles with | Im x | ≤ x = ± i (1 − k/N ), k = 1 , · · · , ( N − / N . After shifting the contour in the direction of positive or negative Im x ,we find that the contribution of such a pole to the integral in eqn. (4.24) is a constant times exp( − π | x | (cid:101) T ) =exp( − π (cid:101) T (1 − k/N )). Including the prefactor exp(2 π (cid:101) T ) in eqn. (4.24), the contribution of such a pole tothe analytically continued partition function is proportional to exp(4 π ( k/N ) (cid:101) T ). This is exp( − π (cid:101) T L ), with L = − k/N . That is the appropriate value for the tachyonic twisted sector closed-string modes, as reviewed insection 3.3.What happens when we analytically continue away from odd integer N ? There is then a double poleat x = ± i, so the closed-string spectrum apparently becomes undiagonalizable. The double pole makes acontribution (cid:101) T exp( − π (cid:101) T ) to the integral. Combining this with the explicit factor (cid:101) T exp(2 π (cid:101) T ) that is alreadypresent in eqn. (4.24), we find that the double pole contributes a term (cid:101) T to the expression that appears inparentheses in the formula (4.17) for the K contribution to the analytically continued partition function. Thisleads to an integral over (cid:101) T whose large (cid:101) T behavior is (cid:82) ∞··· d (cid:101) T / (cid:101) T (7 − p ) / , not the familiar (cid:82) ∞··· d (cid:101) T / (cid:101) T (9 − p ) / .As long as Re 1 /N <
1, this term is subleading relative to exponentially growing contributions from poleswith | Im x | <
1. However, as soon as Re 1 /N >
1, there are no poles for | Im x | < x = ± i. (This reflects the fact that for Re 1 /N >
1, the function K ( z, N ) has no pole with 0 < Re z <
1. See eqn. (2.13).) Thus, as stated in the introduction, when continuedto Re 1 /N > N > N = 1 /N , the theory appears to be nontachyonic. In thisrange of N , the dominant contribution for large (cid:101) T comes from the double pole at x = ± i and has an extraand difficult to interpret factor of (cid:101) T relative to what one would get from conventional massless closed-stringexchange. 21s in eqn. (4.12), the double pole may correspond to a two-dimensional space of closed-string states thatappears when N is not an odd integer, and in which L = (cid:18) (cid:19) . (4.26)These are states that propagate only on the Rindler horizon (analogous to twisted sector states of the orbifold)since their contribution is proportional to the volume V of the intersection of the Dp-brane worldvolume withthe Rindler horizon, and not to the full Dp-brane worldvolume. If it is true that the K contribution correspondsin the closed-string channel to the NS-NS sector, then the two states in question are NS-NS states.To compute the 1-loop contribution to the entropy, we have to differentiate the 1-loop partition functionwith respect to N at N = 1. Differentiating the integral in (4.24) with respect to N and setting N = 1, we get (cid:90) ∞−∞ d x exp(2i π (2 m + 1) x (cid:101) T ) (cid:18) πx/ πx/ πx/ (cid:19) πx . (4.27)Now the integrand has triple poles at x = ± i. The residue of a triple pole contributes an extra factor of (cid:101) T , sothat the integral in (4.27) behaves as (cid:101) T exp( − π (cid:101) T ) for large (cid:101) T . This means that the integral controlling theentropy behaves for large (cid:101) T as (cid:82) ∞··· d (cid:101) T / (cid:101) T (5 − p ) / . For the K contribution discussed in section 4.1, differentiatingwith respect to N at N = 1 gives the expected behavior, with a coefficient that is determined by eqn. (4.8).Concretely, the integrand in eqn. (4.24) has a double pole at x = i and a simple pole at x = i /N . For N near 1 and x near i, the integrand in (4.24) behaves as1( x − i) − x − i)( x − i /N ) = i(1 − /N )( x − i) ( x − i /N ) . (4.28)For generic N near 1, there is a double pole at x = i and a simple pole at x = i /N . At N = 1, both of thesepoles disappear, but if one differentiates with respect to N before setting N = 1, one is left with a triple poleat x = i. To describe this situation requires a three-dimensional space of states, in which L can look like − /N . (4.29)This will reproduce the large (cid:101) T behavior we have found, in all regimes.The extra factors of (cid:101) T and (cid:101) T that we have obtained from double and triple poles at x = ± i are difficultto interpret. Taking what we have found at face value, it appears that the 1-loop contribution to log Tr ρ N ,Re N > p ≤
4, while the corresponding 1-loopcontribution to the entropy is finite only for p ≤
2. One would have expected the bound p ≤ / | x | − p where | x | is the distance from the Dp-brane.22he integral of this over 9 − p transverse directions behaves as (cid:82) ∞··· d − p x | x | − (7 − p ) and is infrared divergent.Thus the Dp-brane corrects the classical Bekenstein-Hawking contribution to the Rindler entropy – the volumeof the horizon – by an infrared-divergent amount. In string theory, this might appear as a disc contribution,though the proper machinery for that computation is not currently known. If the disc contribution to theopen-string entropy is infrared divergent, perhaps one should not be too surprised to have such a divergence inthe annulus contribution. Still, the rationale for the precise power-law behavior that occurs at large (cid:101) T is notclear. Acknowledgements
I thank V. Rosenhaus, P. Sarnak, and especially A. Dabholkar for discussions concerningthis topic. Research partially supported by NSF Grant PHY-1606531.
A A Reformulation Of The Problem Of Analytic Continuation
Let us first recall Buscher’s path integral derivation of T -duality [13]. We start with some two-dimensionaltheory with a U(1) global symmetry and an action that we will schematically call I ( X ), with bosonic fields X , some of which are charged under U(1) (there may also be fermion fields). The first step is to gauge thesymmetry, adding a U(1) gauge field A , with field strength F = d A , and to promote I ( X ) to a gauge-invariantaction I ( X, A ). Then we add a circle-valued field B , with period 2 π , and consider the extended action on atwo-manifold Σ: (cid:98) I ( X, A, B ) = i (cid:90) Σ BF π + I ( X, A ) . (A.1)The extended theory with this action is precisely equivalent to the original theory. To see this, one first performsthe path integral over B . If B were a real-valued field, the path integral over B would give a delta functionsetting F = 0, and nothing more, meaning that the connection A would be flat, but not necessarily trivial.Because B is circle-valued, one also has to sum over winding modes of B . This gives a further delta functionforcing the global holonomies of A to be trivial. This statement depends on the precise normalization of the BF term in the action, and will change when this coefficient is changed (as we do below). With the chosencoefficient, after summing over winding modes, we learn that A is gauge-equivalent to 0. So one can go to thegauge A = 0, and one then gets back the original theory with action I ( X ).On the other hand, given that the fields X include scalar fields charged under U(1), one can go to a “unitarygauge” (at least in a suitable portion of field space) in which the phase of some charged field is set to 0, leavinga reduced set of fields X (cid:48) . In such a unitary gauge (assuming the original action I ( X ) is quadratic in firstderivatives) A has a nondegenerate kinetic energy without derivatives and can be integrated out in a localfashion. This gives a new action (cid:101) I ( X (cid:48) , B ), which describes the T -dual of the original theory.A fairly well-known variant of this arises if we modify the coefficient of the BF term in the action and For a complete gauge fixing of this type to be possible, the action of U(1) on the target space of the original theory should befaithful, meaning in the case of a linear model that the U(1) charges of the original fields should be relatively prime. Then someproduct of those fields has charge 1, and the unitary gauge is defined by requiring this product to be positive. (cid:98) I N ( X, A, B ) = i N (cid:90) Σ BF π + I ( X, A ) . (A.2)Here, since B has period 2 π , and the flux of F is an integer multiple of 2 π , the coefficient N must be an integerin order that the action is well-defined mod 2 π i, as is necessary in order for the Feynman path integral of thetheory to make sense. Repeating the derivation, the integral over small fluctuations of B still gives a deltafunction setting F = 0, but the sum over winding modes of B now gives a delta function telling us that theholonomies of N A are trivial, not necessarily the holonomies of A . This means that A reduces to a Z N gaugefield and the theory is equivalent to a Z N gauge theory, with only a Z N subgroup of the original U(1) globalsymmetry being gauged. This Z N gauge theory is alternatively called a Z N orbifold of the original theory. Thepath integral of the Z N gauge theory consists of summing over Z N twists around all cycles in Σ and dividingby an overall factor of N (this last step is the Z N version of Faddeev-Popov gauge fixing). That is the usualrecipe for a Z N orbifold [14].So a restatement of the problem of analytically continuing the orbifold with respect to N is to analyticallycontinue the theory with action (cid:98) I N ( X, A, B ) with respect to N . Unfortunately, it is not clear how to proceed.The problem is somewhat similar to the problem of analytically continuing three-dimensional Chern-Simonsgauge theory with respect to its integer coupling constant k . That problem actually has an answer that isuseful for some purposes [15], but what is known in Chern-Simons theory does not have an obviously usefulanalog for our present problem.Going back to eqn. (A.1), one can ask what would be an orbifold in the dual description. Suppose it weretrue that the extended action (cid:98) I ( X, A, B ) had a dual Z N symmetry B → B + 2 π/N . This is not really the case,since that operation actually shifts the action by the topological invariant (2 π i /N ) (cid:82) F/ π , which is an integermultiple of 2 π i /N and not of 2 π i. Still, suspending disbelief, after orbifolding by B → B + 2 π/N , the periodof the field B would be reduced to 2 π/N . To get back to a scalar field with the standard period 2 π , we couldintroduce C = N B . In terms of C the action would be (cid:98) I /N ( X, A, C ) = i N (cid:90) Σ CF π + I ( X, A ) . (A.3)This does not lead by any standard method to a well-defined quantum theory, since the coefficient of the CF term is not properly quantized. Still, we see that, formally speaking, a Z N orbifold of the dual variablewould give the same thing we might hope to get by analytic continuation from N to 1 /N . As discussed in theintroduction, replacing N by 1 /N is similar to considering an N -fold cover of the target space of the originaltheory, rather than an N -fold quotient. In this paper, we have found some evidence that, if I ( X ) is a standardworldsheet action in string theory, then the theory associated to such an N -fold cover, if it exists, is a nonunitarylogarithmic conformal field theory. B An Alternative Formula For G ( z, τ ) The function G ( z, τ ) that we encountered in section 3 is a doubly-periodic meromorphic function, so, for fixed τ , it is can be uniquely characterized, up to a constant multiple, by its zeros and poles. These consist of triple This paragraph is a response to a comment by J. Maldacena. Z + τ Z , and simple poles at the non-zero half-lattice points.Let us compare this to the behavior of some doubly-periodic functions. The Weierstrass P -function isdefined by P ( z, τ ) = (cid:88) n,m ∈ Z z − n − mτ ) . (B.1)The sum requires some regularization, but does define a doubly-periodic function whose only poles are doublepoles at lattice points. It is convenient to denote P ( z, τ ) as x . The derivative of the P -function is y = ∂P ( z, τ ) ∂z = − (cid:88) n,m ∈ Z z − n − mτ ) . (B.2)Clearly x is an even function of z and y is an odd function. The definition of y makes it obvious that y has a triple pole at lattice points and no other poles; because y is an odd function of z , it vanishes at thenonzero half-lattice points 1 / , τ / , and (1 + τ ) /
2, as well as translates of these points by lattice vectors.Since y has only three poles (or rather one triple pole) in each fundamental cell of the lattice, it can have noadditional zeros. Thus the function 1 /y has the same zeros and poles as G ( q, w ) and for fixed τ , the two mustcoincide, up to a constant multiple. To fix the multiple, we just observe that near z = 0, 1 /y ∼ − z /
2, while G ( z, τ ) ∼ (2 π i z ) / (4 π i z ) = − i(2 π ) z/
2. So G ( z, τ ) = − i(2 π ) y . (B.3)A standard result is that the functions x = P ( z ) and y = P (cid:48) ( z ) are related by y = 4 x − g x − g , (B.4)which is the usual description of a Riemann surface of genus 1 as an algebraic curve. Here g and g dependon τ only, and are modular forms of weights 4 and 6, respectively : g ( τ ) = 60 π (cid:88) m,n ∈ Z (cid:54) =0 m + nτ ) , g ( τ ) = 140 π (cid:88) m,n ∈ Z (cid:54) =0 m + nτ ) . (B.5)Eqn. (B.4) can be proved by comparing the expansions of the left and right hand sides near z = 0. Thestatement y = P (cid:48) = d x/ d z is equivalent to d z = d xy . (B.6) References [1] M. Mezard, G. Parisi, and M. Virasoro,
Spin Glass Theory And Beyond (World Scientific, 1987).[2] R. P. Boas, Jr.,
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