NUPHB114679114679S0550-3213(19)30165-810.1016/j.nuclphysb.2019.114679The AuthorHigh Energy Physics – TheorySuperconformal indices on S1×(S5/Zp)AndreasGustavssonagbrev@gmail.comDepartment of Physics and Astronomy, Uppsala University, Box 516, SE-75120 Uppsala, SwedenDepartment of Physics and AstronomyUppsala UniversityBox 516UppsalaSE-75120SwedenDepartment of Physics and Astronomy, Uppsala University, Box 516, SE-75120 Uppsala, SwedenEditor: Leonardo RastelliAbstractWe obtain generating functions associated to the abelian superconformal indices for 6d (1,0) tensor and hypermultiplets on S1×(S5/Zp). We extract the superconformal indices and their high and low temperature behaviors. We consider round and generically squashed S5 in turn. We show that the unsquashed limit of the superconformal indices is smooth. We examine S-duality in the large p limit that acts by exchanging the Hopf circle with the temporal circle.1IntroductionThe partition function for 5d SYM was first computed perturbatively (i.e. by suppressing instanton contributions) on a round S5 in [4], [5]. Later this result was extended to the instanton sector in [7], [8] by regularizing with generic squashing parameters a,b,c subject to the relation a+b+c=0. For abelian gauge group, the 5d partition function including the instanton sector, was matched with the 6d superconformal index on S1×S5 in [8], providing supporting evidence for the M5/D4 correspondence [2], [3].It remains difficult to generalize the abelian tests of the M5/D4 correspondence to nonabelian gauge groups (except for the case of infinite rank gauge group where we may use the AdS/CFT correspondence) since that requires a definition of the nonabelian M5 brane theory itself.11However, some recent tests have been made in this direction. The half-BPS index has been computed for Ak 5d SYM by localization in [10] and corresponding expressions for the Dk and Ek gauge groups have been conjectured in [11]. These half-BPS indices have been recently deconstructed for the Ak and the Dk series from corresponding 4d quiver theories in [16], [17]. I thank A. Bourget for bringing these references to my attention. But we can also try to generalize the test to other geometries while keeping the gauge group abelian. As a first step in that direction, in this paper we will generalize the computation of the abelian 6d superconformal index to S1×(S5/Zp). Keeping (2,0) supersymmetry of the M5 brane leads to a very restricted class of possible 6d geometries. But these lens space geometries belong to that class. The lens spaces have non-trivial topology that is detected by the Ray-Singer torsion. In [15] we found a mismatch (a factor that is related to the Ray-Singer torsion) between the 6d index and the 5d partition function, for the maximally topologically twisted 5d maximally supersymmetric Yang-Mills. This mismatch was traced to the nontrivial circle reduction of the selfdual two-form from 6d to 5d. It is then natural to ask whether one can find the same type of mismatch also between the 6d (1,0) theories on R×(S5/Zp) and 5d SYM on S5/Zp.The abelian 6d superconformal index on S1×(S5/Zp) has been computed previously in [10]. We reproduce that result in eq. (5.5). We then use this result to extract its high temperature expansion, which is what we should get when we compute the partition function for 5d SYM theory on S5/Zp. On the other hand, in [10] the low temperature expansion of this index was found to match with the partition function of 5d SYM theory on R×CP2.The abelian M5 brane superconformal index on S1×S5 was first obtained in [1] by using radial quantization. We reproduced and generalized this in [13] by using Hamiltonian quantization. Two types of squashings were considered, squashing of the fiber and of the base manifold respectively. These squashings were also independently found in [12]. Three squashing parameters A,B,C are introduced associated to the three Cartan rotations of SO(6), which is the isometry group of S5. The trace parameterh=13(A+B+C) squashes the Hopf fiber of S5. The three traceless squashing parametersa=A−hb=B−hc=C−h which are subject to the constraint a+b+c=0, squashes the base manifold CP2. We consider a mass deformed M5 brane theory that we will call (2,0)⁎ theory. Mass deformation breaks supersymmetry by half and the (2,0) tensor multiplet splits into one (1,0) tensor multiplet and one (1,0) hypermultiplet with a mass parameter mH. The R-symmetry is SU(2)R and the flavor symmetry is SU(2)F. We have the following commuting charges: three Cartans ji of SO(6), one Cartan R1+R2 of SU(2)R, one Cartan R1−R2 of SU(2)F. We can associate one chemical potential to each of these Cartan generators while preserving one supercharge Qj1j2j3R1R2=Q−−−−− with the SO(6) Cartan charges ji=−1/2, R charges R1=R2=−1/2 and scaling dimension Δ=1/2. The following operators commute with Q−−−−−,O1=Δ+12(R1+R2)O2=R1−R2O3=j1−j2O4=j2−j3O5=j1+j2+j3−2R1−R2 In [1], [8], [7] the charge O5 that is associated with squashing of the Hopf fiber was not included. The existence of O5 was first noted in [6], [10], although there again only four independent charges were again included in the 6d superconformal index. But we can have five independent mutually commuting charges that commute with Q−−−−−. In particular the generator O5 is crucial in this paper when we consider the theory on a lens space. We assign chemical potentials to each of these charges and define the superconformal index asI(β,ω,a,b,c,mH)=tr(−1)FωO5e−β(O1+mHO2+aj1+bj2+cj3) Generating functions associated to this index are [12], [13](1.1)ftensor(β,ω,a,b,c)=e−3βω3−e−2βω2(eβa+eβb+eβc)(1−ωe−β(1+a))(1−ωe−β(1+b))(1−ωe−β(1+c))(1.2)fhyper(β,ω,a,b,c,mH)=e−32β(ωeβmH+ω2e−βmH)(1−ωe−β(1+a))(1−ωe−β(1+b))(1−ωe−β(1+c)) for the tensor and hypermultiplets respectively. Here mH is a parameter that determines the hypermultiplet mass and we will put the radius of S5 to be r=1.We will obtain the abelian indices on R×(S5/Zp) by Fourier transforming with respect to the chemical potential ω. We do this first for the special case a=b=c=0 in section 4, and later in full generality in section 5 and obtain the results presented in eqs. (5.9) and (5.11). We dualize the corresponding indices and obtain their high temperature expansions by using either one of three different dualization methods: zeta function regularization and the Abel-Plana formula method following [9] for the unsquashed case a=b=c=0, and the plethystic dualization method following [8] for the squashed case with generic a,b,c. We then show that the unsquashed limit a,b,c→0 is both well-defined (independent of how we take the limit) and smooth. In the limit, we recover the previously obtained results for the unsquashed case with a=b=c=0. We summarize these three dualization methods in three appendices. Finally, in section 6 we test an asymptotic S-duality conjecture [14] in the limit of large p.2Supersymmetry enhancementWe define the M5 brane generating functionfM5=ftensor+fhyper But for generic mass parameter mH this is a mass deformed version of the M5 brane. It has the same field content as the M5 brane, but not enough supersymmetry to make this a theory of a single (2,0) tensor multiplet. We call this a (2,0)⁎ theory. The preserved supercharges can be easily counted for each new chemical potential we insert into the index. Supercharges that preserve O1 have charges such that R1+R2=−1, hence they are Qj1j2j3−−. There are 8 such supercharges. This is the amount of supersymmetry that we have in the (2,0)⁎ theory for generic mH since these supercharges also commute with O2 whose chemical potential is mH. But let us now consider the linear combination O1+mHO2 that appears in the index. The condition for this to commute with the supercharges, is that12+(12+mH)R1+(12−mH)R2=0 For generic values of mH the only solution is R1=R2=−1/2. But for mH=1/2 we only need R1=−1/2 while R2 can be either of ±1/2. This means we have enhancement of supersymmetry from 8 to 16 supercharges at mH=1/2. A similar enhancement of supersymmetry happens at mH=−1/2 where instead R1 can be either of ±1/2 and R2=−1/2. Let us now turn on O5. Then for mH=1/2 we find the four preserved supercharges, Q−−−−−,Q+−−−+,Q−+−−+,Q−+−−+ and for mH=−1/2 we find the four preserved supercharges Q−−−−−,Q++−+−,Q+−++−,Q−+++−. Had we instead defined O5 the symmetric way asO5sym=j1+j2+j3−32(R1+R2) we would only get one preserved supercharge Q−−−−− when we turn on the chemical potential for O5sym. We would like to preserve as much supersymmetry as possible for each new chemical potential that we turn on, and therefore we will not define O5 in the symmetric way.If we pick mH=1/2 and a=b=c=0, then we find a simplification also when we turn on the chemical potential ω (corresponding to O5). The generating function becomesfM5(β,ω)=ωe−β1−ωe−β This simplification would not occur had we instead used O5sym, but also we do not get this simplification if we take mH=−1/2 despite then again we have four supercharges. Instead, the simplification at mH=−1/2 occurs if we replace O5 with O5∨=j1+j2+j3−R1−2R2. We normalized O5 such that e2πiO5=1 on bosonic states. This gives the quantization condition O5∈Z. By lensing, we will identify the rotation by the angle 2π/p with the identity operator, and this gives the quantization condition O5∈pZ. To project out all the integer modes except multiples of p, we put η=e2πi/p and introduce the projection operator1p∑ℓ=0p−1ηℓ=∑ℓ∈Zδn,ℓp Thus we get the generating function on R×(S5/Zp) asfM5(β,p)=1p∑ℓ=0p−1fM5(β,ηℓ) Lensing by using the generator O5 leads to the lens space L(p;1,1). For the lens space S5/Zp=L(p;q1,q2) we shall replace O5 with the operatorOq1,q2=q1j1+q2j2+j3−(q1+q2)R1−R2 Then for mH=1/2 we have generically only two conserved supercharges Q−−−−−,Q−−+−−, and for mH=−1/2 we preserve Q−−−−−,Q++−+−. It is easy to compute the generating function on R×L(p;1,1) for mH=1/2. The result isfM5(β,p)=e−pβ1−e−pβ Lensing for the (1,0) tensor multiplet does not depend on the choice of mH nor on the detailed definition of O5, whether we use O5,O5∨,O5sym or some other combination. This is simply because all fields in the (1,0) tensor multiplet are neutral with respect to the R-charges R1,R2. But for the hypermultiplet, lensing will depend on the choice of O5. Does this mean that lensing leads to an ambigous result for the hypermultiplet? We will now explain that the answer is either yes or no depending on the amount of supersymmetry that we like to preserve. When we put a theory on a curved manifold, we always need to specify the amount of supersymmetry that we like to preserve. At the point where a=b=c=0 and mH=1/2, the preserved supercharges may tell us how we shall choose O5 so that there is no ambiguity in the lensing procedure. In this paper we will consider the (2,0)⁎ theory that has four preserved supercharges Q−−−−−,Q+−−−+,Q−+−−+,Q−−+−+ at the point specified by a=b=c=0 and mH=1/2. This specification uniquely tells us that we shall use the operator O5 to define the superconformal index to squash the Hopf fiber, rather than O5∨ or some other combination. If we would like to preserve less than four supercharges at this particular point, then we will find a certain ambiguity in how to pick O5. But this is a common situation. When we lower the amount of supersymmetry we always find more freedom in how to set up a corresponding theory that preserves that amount of supersymmetry.3From generating function to superconformal indexIf we have a spectrum of discrete energy levels, then this may be encoded in a generating functionf(β)=∑ndne−βEn where En are the energy levels with degeneracies dn and β is the inverse temperature. (To get the generating function of an index, we count degeneracies of fermionic fields with a minus sign.) The index is an infinite productI(β)=∏ℓ∈Z∏n(aℓ2+En2)−dn/2 whereaℓ=2πℓβ The product is divergent and needs to be regularized. We may use zeta function regularization and define [9](3.1)ζ(s)=μ2s∑n∑ℓ∈Zdn(aℓ2+En2)−s where μ is an energy scale that we need to insert in order for ζ(s) to be dimensionless. Then the index will be given by(3.2)I(β)=e−βF(β)=e12ζ′(0) In [9] it was shown by a direct computation that (3.2) can be rewritten as a low temperature expansion(3.3)I(β)=e−βE∏n(1−e−βEn)−dn where E is the Casimir energy. The nice feature with the derivation in [9], is that the Casimir energy factor appears automatically and does not need to be multiplied by hand. If we take the logarithm, we getlogI(β)=−βE−∑ndnlog(1−e−βEn)=−βE+∑n∑k=1∞1kdne−kβEn If we exchange the sums, the result can be expressed as the plethystic sum of the generating functionlogI(β)=−βE+∑k=1∞1kf(kβ) The plethystic sum may involve a divergent sum in certain applications. Then, even if this divergent sum is canceled out in the dualization process that takes us from the low temperature expansion (3.3) to a high temperature expansion, we may nevertheless miss out some logarithmic terms (or prefactors of the superconformal index). Another drawback is that the plethystic method can only be applied to generating functions f(β) that are antisymmetric, f(β)=−f(−β). The advantage with the plethystic method is that it leads to simpler computations and in particular it is easy to extract the Stefan-Boltzmann terms from the generating function, which is much more involved to do by using the other methods.4Turning off squashing parametersIn this section we put the squashing parameters to zero, a=b=c=0, and obtain the generating functions on S1×(S5/Zp) by Fourier transforming with respect to the chemical potential ω. We also obtain the high temperature expansions of the corresponding indices by using zeta function dualization method, the Abel-Plana dualization method, and the plethystic dualization method. We outline each of these dualization methods in three appendices.4.1The (1,0) tensor multiplet indexThe generating function for the (1,0) tensor multiplet on R×S5 is given byftensor(β,ω)=ω3e−3β−3ω2e−2β(1−ωe−β)3 where we just turn on one chemical potential ω=e−βh that corresponds to a rotation along the Hopf fiber (and h was introduced in the Introduction). By noting that1(1−e−β)3=∑n=0∞(n+1)(n+2)2e−βn we can write this asftensor(β,ω)=∑n=1∞(1−n2)ωne−βn To obtain the generating function on the lens space S5/Zp=L(p;1,1), we sum over all ω that are taken as the p different p-th roots of unity. This sum amounts to projecting onto the lens space where we are identifyingψ∼ψ+2πp where ψ is the 2π ranged Hopf fiber coordinate on S5. We then use the identity1p∑ℓ=0p−1e2πiℓn/p=∑q∈Zδn,pq and getftensor(β,p)=∑n=1∞(1−n2p2)e−βpn which we evaluate to(4.1)ftensor(β,p)=e−pβ1−e−pβ−p24coshpβ2(sinhpβ2)3 The small-β′ expansion is(4.2)ftensor(β′,p)=−2pβ′3+1pβ′−12+p3+10p120β′+O(β′2) The low temperature (or large β) behavior of the full index is governed by the Casimir energy E,Itensor(β,p)∼e−βE The Casimir energy can be read off from the small-β′ expansion of the generating function, by interpreting β′ as a small regulator, and after subtracting the divergent terms proportional to β′−3 and β′−1, thus defining a renormalized fren. The Casimir energy is then computed asE=−12ddβ′ftensor,ren(β′,p)|β′=0 We getEtensor=−p3+10p240 If we use zeta function regularization, we will compute the Casimir energy as follows. We begin by definingE(s)=12∑n=1∞(1−p2n2)(pn)1−s=12p1−2s(ζ(s−1)−p2ζ(s−3)) and thenE(0)=12p(ζ(−1)−p2ζ(−3))=−p3+10p240 is the Casimir energy. The advantage with this latter method is that we do not need to worry about how to remove the singular terms proportional to β′−3 and β′−1.The zeta function that is associated to the generating function ftensor(β,p) is(4.3)ζ(s)=μ2s∑ℓ∈Z∑n=1∞(1−n2p2)(n2p2+aℓ2)−s We decompose ζ(s)=ζℓ=0(s)+∑ℓ≠0ζℓ≠0(s) and rewriteζℓ=0(s)=μ2s(p−2sζ(2s)−p2−2sζ(2s−2))ζℓ≠0(s)=μ2s∑n=1∞((1+aℓ2)(n2p2+aℓ2)−s−(n2p2+aℓ2)1−s) Then by using ζ(0)=−12, ζ(−2)=0 and ζ′(0)=−12log(2π), ζ′(−2)=−ζ(3)4π2 we getζℓ=0′(0)=−logμp−log(2π)+p2ζ(3)2π2 For ℓ≠0 we get22We have a sum over n=1,2,... that we wish to dualize. To this end, we add the term that we would get by taking n=0. We also add the terms with n replaced by −n, which is possible since only n2 appears in the summand. The details are in appendix A.ζℓ≠0,n=0′(0)=−2π445pβ3+π23pβ+log(μβ) By also including the sectors with n≠0, we get the result(4.4)−βFtensor=−π445pβ3+π26pβ+12logpβ2π+p2ζ(3)4π2+∑ℓ=1∞∑n=1∞p2n3(12π2+2ℓnpβ+n2p2+(2πℓnpβ)2)e−4π2ℓnpβ In particular, we see that the dependence on the scale μ cancels out. We computed the contribution from terms with n≠0 by using the zeta function dualization method that we outline in appendix A, following [9].4.2The (1,0) hypermultiplet indexIf we pick the hypermultiplet mass parameter mH=1/2, then we have the following generating functionfhyper(β)=e−2β+e−β(1−e−β)3 We may use1(1−e−β)3=∑n=0∞12(n+1)(n+2)e−βn to bring this into the formfhyper(β)=∑n=1∞n2e−βn Now let us take mH=1/2+ϵ. Then the generating function isfhyper(β)=12∑n=0∞(n−1)ne−β(n+ϵ)+12∑n=0∞(n+1)ne−β(n−ϵ) We will now use the Abel-Plana method to obtain the high temperature expansion. We outline this method in the appendix. Associated to this is the functionf(z)=12(z−1)zlog(1−e−β(z+ϵ))+12(z+1)zlog(1−e−β(z−ϵ)) Now we Taylor expand this function in ϵf(z)=z2log(1−e−βz)−βϵze−βz1−e−βz−12β2ϵ2z2e−βz(1−e−β)2+O(ϵ3) We have∫0∞dxf(x)=−π445β3−ϵπ26β−ϵ2π26β+O(ϵ3) which is consistent withπ445β3+12mH2−312π26β by taking mH=1/2+ϵ and expanding in ϵ. A rather curious fact is that the series terminates at order ϵ2. But the integrals at order ϵn for n>2 all diverges, although by summing them all up, we shall of course expect a finite result, and actually the contribution from all those divergent integrals should sum up to zero.It is easy to get the Casimir energy from the generating function. We getf(β′,mH)=2β′3+4mH2−14β′+1960(17−120mH2+80mH4)β′+O(β′2) Then we see that the Casimir energy isE=1240+ϵ124−ϵ3112−ϵ4124 as an exact expression in ϵ.4.2.1RefinementIf we take mH=1/2+ϵ, then we havefhyper=12∑n=0∞(n+1)(n+2)(e−β(n+2+ϵ)ωn+2+e−β(n+1−ϵ)ωn+1) Now we can lens this expression, and get(4.5)fhyper=12∑q=0∞(pq−1)pqe−β(pq+ϵ)+12∑q=0∞(pq+1)pqe−β(pq−ϵ)=p2cosh(βϵ)coshpβ2+psinh(βϵ)sinhpβ24(sinhpβ2)3 From this we can extract the Casimir energyEhyper=p3240+ϵp24−ϵ3112p−ϵ4124p For the high temperature expansion, we definef(z,ϵ)=12((pz−1)pzlog(1−e−β(pz+ϵ))+(pz+1)pzlog(1−e−β(pz−ϵ))) that we Taylor expand in ϵ and then we are able to compute its integral to the first few orders, with the result∫0∞dxf(x,ϵ)=−π445pβ3−ϵπ26pβ−ϵ2π26pβ+O(ϵ3) Let us move on to the other term in the Abel-Plana formulai∫0∞dxf(ix)−f(−ix)e2πx−1 We will content ourselves to compute up to linear order in ϵ, and thus we only need to work with f expanded to first orderf(z,ϵ)=p2z2log(1−e−pβz)+ϵpβze−pβz1−e−pβz To this order, we getf(ix,ϵ)−f(−ix,ϵ)=p2x2log(−eipβx)+ipxβϵ=p2x2i(pβx−2πn−π)+ipxβϵ where∫0∞dx=∑n=0∞∫2πn2π(n+1)dx Thus we like to compute (and here we changed the sign of f(z))∑n=0∞∫2πn/(pβ)2π(n+1)/(pβ)dxp2x2(pβx−2πn−π)+pxβϵe2πx−1 We report the following partial results,∫0∞dxx3e2πx−1=1240∫0∞dxx2e2πx−1=ζ(3)4π3∫0∞dxxe2πx−1=124 We may expand the denominator ase−2πx1−e−2πx=∑k=1∞e−2πkx and consider the integrals∫dxx2e−2πkx=−14π3k3(1+2πkx+2π2k2x2)e−2πkx We getp3+10ϵp240β−p2ζ(3)4π2 plus the double sump2π2∑k,n=1∞nk3[(1+2πkx+2π2k2x2)e−2πkx]2πn/(pβ)2π(n+1)/(pβ)=−∑k,n=1∞p2k3(12π2+2knpβ+(2πknpβ)2)e−4π2knpβ Upon adding the Casimir energy term −βEhyper we find a cancellation. Summarizing, we have got−βFhyper=π445pβ3+ϵπ26pβ+ϵ2π26pβ−p2ζ(3)4π2−∑k,n=1∞p2k3(12π2+2knpβ+(2πknpβ)2)e−4π2knpβIf we add the contributions of the tensor and the hypermultiplets, we get−β(Ftensor+Fhyper)=π2(1+ϵ+ϵ2)6pβ+12logpβ2π+∑n,k=1∞1ne−4π2knpβ If we put ϵ=0, we may express this result asI(β)=pβ2πeπ26pβexp(∑n=1∞1ne−4π2npβ1−e−4π2npβ) This result was also obtained in [8] by using the known modular property of the Dedekind eta function.We can also easily perform the sum over n. For the hypermultiplet and if we put ϵ=0, we then get−βFhyper=π445pβ3−∑k=1∞[p24π2k3cosh2π2kpβsinh2π2kpβ+p2k2β1(sinh2π2kpβ)2+π2kβ2cosh2π2kpβ(sinh2π2kpβ)3]4.2.2The same result from the plethystic exponentLet us return to the generating function (4.5) and take ϵ=0 for simplicity,(4.6)fhyper(β,p)=p24coshpβ2(sinhpβ2)3=2pβ3−p3120β+O(β2) We can use this to compute the index by taking the plethystic exponent,logI˜hyper(β,p)=∑n=1∞1nfhyper(nβ,p)=∑n=−∞∞∫ϵ∞dssfhyper(sβ,p)e2πins The correct index is given by Ihyper=e−βEI˜hyper. For the second identity to hold, we need to pick up the points s=n for n=1,2,... from∑n∈Ze2πins=∑n∈Zδ(s−n) To this end, we shall take lower integration bound such that 0<ϵ<β.Dualization is now performed as follows. First we regularize as followsf(s)=fsing(s)+freg(s) where freg(s)=f(s)−fsing(s) and fsing(s) involves terms of the form s−n for n>0 such that freg(0) is finite. Next we can compute the contribution from fsing(s). We have the small-s expansionfhyper(sβ,p)=2pβ3s3−p3βs120+O(s3) The first term in this expansion is the singular term, which gives rise to the Stefan-Boltzmann term∑n=−∞∞∫ϵ∞ds2pβ3s4e2πins=2pβ3ζ(4)=π445pβ3 The remaining piece is∑n=1∞∫−∞∞dssfreg(βs)e2πins+12∫−∞∞dssfreg(βs)−12freg(s)s|s=0 There are triple poles ats=2πikpβ,k∈Z and by encircling those triple poles that lie in the upper half plane and picking up the residues, we get(4.7)∑n=1∞∫−∞∞dssfreg(βs)e2πins=−∑n=1∞∑k=1∞p2k3(12π2+2knpβ+(2πknpβ)2)e−4π2knpβ(4.8)12∫−∞∞dssfreg(βs)=−p24π2ζ(3) By a general argument that we prove in the appendix, we have that−12freg(s)s|s=0 is canceled by the Casimir energy.By comparing the two computations, and by looking at the term in (4.8), we see that what was referred to as the topological subleading term in [9] corresponds to what was referred to as the perturbative contribution in [8].5Turning on squashing parametersWe will now consider a more complicated situation with generic squashing parameters turned on. For this case, the plethystic method is easy to use, whereas the other two methods, the generalized zeta function and the Abel-Plana formula, become difficult to use. The generating functions for general squashing parameters a,b,c and generic lensing parameter p, are quite complicated. However, already from the small β expansion of these generating functions, we can extract the high temperature and low temperature asymptotic behavior of the corresponding indices, so that is where we will start.We begin by extracting the Stefan-Boltzmann terms with squashing. Given a generating function f(β), the Stefan-Boltzmann terms are obtained by computing the following quantity,∑n∈Z∫ϵ∞dsse2πinsβfsing(s) For the explicit computations, we will only need the following results,∑n∈Z∫ϵ∞dsse2πinsβ1sn=ζ(n+1)βn and in this paper we will encounterζ(4)=π490ζ(2)=π265.1The tensor multipletThe refined generating function for the tensor multiplet is given by (1.1). We begin by writing a series expansion for(5.1)1(1−ωe−β(1+a))(1−ωe−β(1+b))(1−ωe−β(1+c))=−∑n=0∞ωne−βnfn(a,b,c) wherefn(a,b,c)=e−βa(n+1)(eβb−eβc)+e−βb(n+1)(eβc−eβa)+e−βc(n+1)(eβa−eβb)(1−e−β(a−b))(1−e−β(b−c))(1−e−β(c−a)) We now need to multiply this by the numeratorω3e−3β−ω2e−2β(eβa+eβb+eβc) We then get two terms, ωn+3(...)+ωn+2(...). The trick is to shift the sum for the first term to bring both terms into the same form ωn+2(...+...), and then replace n+2 by mp where m=1,2,... for p>1. (If p=1, then we shall take m=2,3,....) For p>1 we getftensor(p,β,a,b,c)=−∑m=1∞e−βpm(fpm−3(β,a,b,c)−(eβa+eβb+eβc)fpm−2(β,a,b,c)) This is a geometric sum, which we evaluate to(5.2)ftensor(p,β,a,b,c)=−1+eβ(b−c)(1−e−β(a−b))(1−e−β(c−a))11−epβ(1+a)+cycl We note thatftensor(p,β,a,b,c)+ftensor(p,−β,a,b,c)=−1 This enables us to write this in the manifestly antisymmetric form by adding −1/2,ftensor(p,β,a,b,c)=(coshβ2(b−c)4sinhβ2(a−b)sinhβ2(a−c)coshpβ2(1+a)sinhpβ2(1+a)+cycl)−12 The small β expansion readsftensor(p,β,a,b,c)=−2Npβ3+6+5(ab+bc+ca)6Npβ−12+p3+10p120β−a4+b4+c4−36abc240Npβ+O(β2) where N:=(1+a)(1+b)(1+c) and is a generalization of (4.2) to the squashed case. From the singular terms we extract the Stefan-Boltzmann termsβFtensor=π445Npβ3−(6+5(ab+bc+ca))π236Npβ and from the linear term we extract the Casimir energyEtensor=−p3+10p240+a4+b4+c4−36abc480Np5.2The hypermultipletWe pick mH=1/2+ϵ and put t=eβϵ where we have the following refined generating function,fhyper(β,ω,t)=e−2βω2t−1+e−βωt(1−ωe−β(1+a))(1−ωe−β(1+b))(1−ωe−β(1+c)) We again use the series expansion (5.1) for the denominator and writefhyper(β,ω)=−∑n=0∞ωne−βn(e−2βω2t−1+e−βωt)fn(a,b,c)=−∑n=−1∞ωn+2e−β(n+2)(fnt+fn+1t−1) Summing over ω picks out n+2=mp for m=1,2,.... We getfhyper(β,p,a,b,c,t)=∑m=1∞e−βmp(tfmp−1+t−1fmp−2) The sum can be evaluated with the result(5.3)fhyper(β,p,a,b,c,t)=te−βc+t−1eβb(1−e−β(a−b))(1−e−β(c−a))11−epβ(1+a)+cycl By noting thatfhyper(p,β,a,b,c,t)+fhyper(p,−β,a,b,c,1/t)=0 we can write this asfhyper(p,β,a,b,c,t)=coshβ2(b+c+2ϵ)4sinhβ2(a−b)sinhβ2(a−c)coshpβ2(1+a)sinhpβ2(1+a)+cycl The small β expansion reads(5.4)fhyper=2Npβ3+ab+bc+ca+6(ϵ+ϵ2)6Npβ+Mβ120Np+O(ϵ2,β2) whereM=12(a4+b4+c4)+2abc−Np4+(20(ab+bc+ca)+10abc−10Np2)ϵ+10(ab+bc+ca)ϵ2+20ϵ3+10ϵ4 which is a generalization of (4.6) to the squashed case. From the singular terms we extract the Stefan-Boltzmann terms−βFhyper=π445Npβ3+(ab+bc+ca+6(ϵ+ϵ2))π236Npβ and from the linear term we extract the Casimir energyEhyper=p3240+pϵ24−ϵ312Np−ϵ424Np−a4+b4+c4+4abc480Np−2(ab+bc+ca)+abc24Npϵ−ab+bc+ca24Npϵ25.3Summing the contributionsIf we sum the contributions from tensor and hypermultiplets,fM5(β,p)=ftensor(β,p)+fhyper(β,p) we get the result(5.5)fM5=sinhβ(ϵ−c)2sinhβ(ϵ−b)2sinhβ(a−b)2sinhβ(c−a)211−epβ(1+a)+cycl This result agrees precisely with eq. (4.5) in [10]. At ϵ=0, the small β expansion readsfM5=1+ab+ac+bcNpβ−12+112pβ+abc6Npβ+O(β2) From the divergent term we obtain the Stefan-Boltzmann termlogISB=1+ab+bc+caNpπ26β and from the linear term we obtain the Casimir energy,EM5=−p24−abc12Np5.4More on the exact resultsThe exact expressions for generating functions that we have obtained in (5.2) and (5.3) have not yet been written in the fully reduced form, by which we mean the following. If we write the sum of the three cyclic permutations on a common denominator (for i={tensor,hyper} respectively)fi=Pi(1−eβ(b−a))(1−eβ(c−b))(1−eβ(a−c))(1−epβ(1+a))(1−epβ(1+b))(1−epβ(1+c)) then this can be always further reduced to the formfi=Pi,reduced(1−epβ(1+a))(1−epβ(1+b))(1−epβ(1+c)) by which we mean that the poles associated to the vanishing of 1−eβ(a−b) and any of its cyclic permutations are all removable poles. We have tested this up to large values of p and seen that this cancellation of poles always happens so we conjecture this always happens for all values of p, but we have no proof. We will here obtain a general formula for Pi,reduced, which again will be a conjecture. For notational simplicity, we putu=eβav=eβbw=eβc which are subject to the constraint uvw=1, and we putx=eβ In this notation, we havefhyper(p)=1/w+v(1−v/u)(1−u/w)(1−xpup)+cyclftensor(p)=−1+v/w(1−v/u)(1−u/w)(1−xpup)+cycl We rewrite these in the formfi=gi1−xpup+cycl whereghyper=−u+1(u−v)(u−w)gtensor=u(v+w)(u−v)(u−w) By adding the three cyclic terms, we getfi=ai−xpbi+x2pci(1−xpup)(1−xpvp)(1−xpwp) whereai=gi+cyclbi=(vp+wp)gi+cyclci=1upgi+cycl Explicity we findahyper=0atensor=−1 For the other two terms, we have a rather complicated dependence on p. For the first few values of p we find thatbhyper(1)=1bhyper(2)=1+u+v+wbhyper(3)=u+v+w+u2+v2+w2+uv+uw+vwbtensor(1)=−u−v−wbtensor(2)=−u2−v2−w2−uv−uw−vwbtensor(3)=−u3−v3−w3−u2v−uv2−u2w−v2w−uw2−vw2−2uvwchyper(1)=−1chyper(2)=−1−1u−1v−1wchyper(3)=−1u−1v−1w−1u2−1v2−1w2−1uv−1vw−1wuctensor(1)=0ctensor(2)=u+v+wctensor(3)=2+uv+vu+uw+wu+vw+wv We have the relationschyper(p,u,v,w)=−bhyper(p,1/u,1/v,1/w)ctensor(p,u,v,w)=−btensor(p,1/u,1/v,1/w)−1up−1vp−1wp These relations determine ci once we know bi. We also have the relationbtensor(p)+bhyper(p+1)−bhyper(p)+bhyper(p−1)=0 that we can use to determine btensor once we know bhyper. Our task has been reduced to determine bhyper, before we have used the relation uvw=1. Let us now switch to a short notation. If bhyper(2)=1+u+v+w, then we will write this as p=2:(100),(010),(001),(000) where (100) represents the term u and so on. We will also suppress all terms that are obtained by trivial permutations, so instead of writing out (100),(010),(001), we will just write (100). This way we get for the first few values of p the following results,33We carried out this computation up to p=6 by using Mathematica.p=1:[(000)]p=2:[(100)],[(000)]p=3:[(200),(110)],[(100)]p=4:[(300),(210),(111)],[(200),(110)]p=5:[(400),(310),(220),(211)],[(300),(210),(111)]p=6:[(500),(410),(320),(221),(311)],[(400),(310),(220),(211)] where we have grouped the elements into two classes. From this, we see the following patternp=p:[(p−1,0,0),(p−2,1,0)...],[(p−2),(p−3,1,0)...] where the first class of elements are all those elements whose entries sum up to p−1, and the second class are all those elements whose entries sum up to p−2. We have now in principle completed the computation, although the result has not been presented in an explicit way. This situation can be improved by restricting to (u,v,w)=(u,1/u,1) where we getbhyper(p,u,1/u,1)=(p−1)(u+u−1)+(p−2)(u2+u−2)+⋯+(up−1+u−(p−1))+pchyper(p,u,1/u,1)=−bhyper(p,u,1/u,1)btensor(p,u,1/u,1)=−(p−1)(u+u−1)−(p−2)(u2+u−2)−⋯−(up−1+u−(p−1))−p−(up+u−p)ctensor(p,u,1/u,1)=(p−1)(u+u−1)+(p−2)(u2+u−2)+⋯+(up−1+u−(p−1))+p−1 For the sum we have a closed form,(5.6)bhyper(p,u,1/u,1)=(up/2−u−p/2u1/2−u−1/2)2 Using this result, we getfhyper(p,u,1/u,1)=−(up/2−u−p/2u1/2−u−1/2)2xp+x2p(1−xpup)(1−xpvp)(1−xpwp)|v=1/u,w=1ftensor(p,u,1/u,1)=(up/2−u−p/2u1/2−u−1/2)2xp+x2p(1−xpup)(1−xpvp)(1−xpwp)|v=1/u,w=1+−1+(up+u−p)xp−x2p(1−xpup)(1−xpvp)(1−xpwp)|v=1/u,w=1 The second term in the second line can be simplified by noting that(1−xpup)(1−xpu−p)=1−xp(up+u−p)+x2p Then, by adding the two contributions, we get the result,fM5(p,u,1/u,1)=1−1+xp We notice that surprisingly this M5 brane generating function does not depend on the squashing parameter a.To get a nontrivial dependence on squashing parameters for the M5 brane generating function at mH=1/2, we shall consider generic a,b,c. Then we shall return to our result above. To streamline the notation, we definesq(u,v,w)=∑r+s+t=qurvswt andQq1,q2,...(u,v,w)=sq1(u,v,w)+sq2(u,v,w)+... Then we have(5.7)bhyper(p,u,v,w)=Qp−1,p−2(u,v,w) Let us further define the setQq1,q2,...={r,s,t|r+s+t=q1}∪{r,s,t|r+s+t=q2}∪... We then conjecture the following general expression,(5.8)fhyper(p,β,a,b,c)=∑r+s+t∈Qp−1,p−2cosh[pβ2+β(ar+bs+ct)]4sinhpβ2(1+a)sinhpβ2(1+b)sinhpβ2(1+c) As consistency checks, we note that if we take the limit u,v,w→1, then the sum (5.7) reduces tobhyper(p,1,1,1)=p(p+1)2+(p−1)p2=p2 and if we put (u,v,w)=(u,1/u,1) in (5.8), then it reduces to (5.6).Let us move on to the tensor multiplet. We clearly seem to havebtensor(p,u,v,w)=−Qp,p−3(u,v,w) As checks, we see that in special cases, this reduces to the previous results,btensor(p,1,1,1)=−p2−2btensor(p,u,1/u,1)=−(up/2−u−p/2u1/2−u−1/2)2−(up+u−p) We are then ready to conjecture the general result for the generating function,ftensor(p,x,u,v,w)=−1+xpQp,p−3+x2p(Qp,p−3∨−u−p−v−p)(1−xpup)(1−xpvp)(1−xpwp) where we defineQp,p−3∨(β)=Qp,p−3(−β) We can take out a simple term from this and write the rest in a manifestly antisymmetric form,ftensor=−11−xpwp+xp2(Qp,p−3∨−u−p−v−p)+x−p2(Qp,p−3−up−vp)((xu)p2−(xu)−p2)((xv)p2−(xv)−p2)((xw)p2−(xw)−p2) which can also be written as(5.9)ftensor=e−pβ(1+c)1−e−pβ(1+c)+coshpβ(1−2a)2+coshpβ(1−2b)2−∑r,s,t∈Qp,p−3coshpβ(1−2p(ra+sb+tc))24sinhpβ(1+a)2sinhpβ(1+b)2sinhpβ(1+c)2 It is easy to see that we reproduce the previously known result if we put p=1. For p=1 we have∑r,s,t∈Qp,p−3coshpβ2(1−2p(ra+sb+tc))|p=1=coshβ(1−2a)2+coshβ(1−2b)2+coshβ(1−2c)2 and so we get(5.10)ftensor=e−β(1+c)1−e−β(1+c)−coshβ(1−2c)24sinhβ(1+a)2sinhβ(1+b)2sinhβ(1+c)2Let us now return to the hypermultiplet. To understand how to generalize to general mH, it is sufficient to just look at say the case with p=2 for which we getfhyper(2,x,u,v,w,t)=(t−1+t(u+v+w))x−1+(t+t−1(u−1+v−1+w−1))x(ux−u−1x−1)(vx−v−1x−1)(wx−w−1x−1) If we express this in the formfhyper(2,x,u,v,w,t)=∑r,s,t∈Q0t−1urvswt+tu−rv−sw−t(ux−u−1x−1)(vx−v−1x−1)(wx−w−1x−1)+∑r,s,t∈Q1t−1urvswt+tu−rv−sw−t(ux−u−1x−1)(vx−v−1x−1)(wx−w−1x−1) then it seems clear that this should generalize as(5.11)fhyper(p,β,a,b,c,ϵ)=∑r,s,t∈Qp−2cosh[pβ2−β(ar+bs+ct)−βϵ]4sinhpβ2(1+a)sinhpβ2(1+b)sinhpβ2(1+c)+∑r,s,t∈Qp−1cosh[pβ2−β(ar+bs+ct)+βϵ]4sinhpβ2(1+a)sinhpβ2(1+b)sinhpβ2(1+c) As two consistency checks for our conjectured formula (5.11), we first notice that the small β expansion agrees with (5.4) for arbitrary p, and second, if we put a=b=c=0 and keep p arbitrary, then we reproduce (4.5) to all orders in β.5.5Dualization using the plethystic methodHaving obtained the generating functions in their fully reduced forms, we are now ready to dualize these generating functions using the plethystic method as we outline in the appendix following [8]. We have already extracted the Stefan-Boltzmann terms. Let us move on to compute the integrals12∫−∞∞dssfi,reg(s) for i running over hyper and tensor multiplets. If there are only simple poles, the instanton contribution is computed in a similar way,∑k=1∞∫−∞∞dssfi,reg(s)e2πiks/β and if simple poles are located at (5.12), then the sum over k becomes a geometric series,∑k=1∞e−4π2knpβ(1+a)=e−4π2npβ(1+a)1−e−4π2npβ(1+a)5.5.1The hypermultipletWe note that this integral does not depend on β so it should be part of the perturbative contribution from the 5d viewpoint. The advantage with turning on generic a,b,c parameters is that there now will appear only contributions from simple poles located at(5.12)s=2πinp(1+a)s=2πinp(1+b)s=2πinp(1+c) and by closing the contour in the upper halfplane, we will pick up contributions only from those poles with n=1,2,3,.... We expand around a pole,sinhps(1+a)2=(−1)np(1+a)2(s−2πinp(1+a))+... and get the corresponding residueHn(a,b,c)=−∑Qp−212ncosπn1+a(2+a−2p(ar+bs+ct)−2ϵp)sinπn(b−a)1+asinπn(c−a)1+a−∑Qp−112ncosπn1+a(2+a−2p(ar+bs+ct)+2ϵp)sinπn(b−a)1+asinπn(c−a)1+a after some computation. The contribution from the other two poles can be obtained by cyclic permutations of a,b,c. Thus we have obtained12∫−∞∞dssfhyper,reg(s)=∑n=1∞(Hn(a,b,c)+Hn(b,c,a)+Hn(c,a,b)) If we take p=1 we get only the contribution from r=s=t=0 from the sum in the second line,Hn(a,b,c)=−12ncosπn1+a(2+a+2ϵ)sinπn(b−a)1+asinπn(c−a)1+a which then leads to a result that is in a good agreement with eq. (2.67) in [8].To take the unsquashed limit, we first assume that ϵ=0 and defineH˜n(λ,a,b,c)=Hn(λa,λb,λc)+Hn(λb,λc,λa)+Hn(λc,λa,λb) and then Taylor expandH˜n(λ,a,b,c)=−12n3π2+2abc15nπ2λ3+O(λ4) and we get12∑n=1∞(Hn(a,b,c)+Hn(b,c,a)+Hn(c,a,b))=−p24π2ζ(3)+p215abcζ(−1)+O(|a,b,c|4) where the factor p2 comes from the sum ∑Qp−1,p−21=p2. This result shows that the unsquashed limit is smooth and we reproduce our previous result (4.8), if we interpret the sum by means of zeta function regularization.If we keep ϵ nonzero, then there will be a correction to this result on the formH˜n(λ,a,b,c)=−12n3π2+O(ϵ3,λ4) which is still consistent with our previous result that we computed up to order O(ϵ2).Let us move on to the instanton contribution. We then consider the sum∑n=1∞∑k=1∞(Hn,k(a,b,c)+Hn,k(b,c,a)+Hn,k(c,a,b)) where we defineHn,k(a,b,c)=Hn(a,b,c)e−4π2nkpβ(1+a) To take the unsquashed limit, we defineK˜n,k(λ;a,b,c)=Kn,k(λa,λb,λc)+Kn,k(λb,λc,λa)+Kn,k(λc,λa,λb) and expand K˜ up zeroth order in λ,K˜n,k(λ;a,b,c)=(2kβn2p+12π3n3+4π2k2p2β2n)e−4π2knpβ+O(λ) We notice that no singular terms appear in this expansion, and that the finite term is independent of a,b,c, which means that the unsquashed limit is well-defined and does not depend on how we let a,b,c approach to zero as long as a+b+c=0. This result then leads to a perfect agreement with our previous result in eq. (4.7). Hence also for the instanton contribution, the unsquashed limit is smooth.5.5.2The tensor multipletFor the tensor multiplet, the cyclic symmetry in a,b,c is hidden once we separate out the first term in (5.9). We will now dualize the terms in the second line in (5.9) which are antisymmetric under β→−β so that we can apply the plethystic dualization method on these terms alone.44Later we will take back this statement, due to regularization issues. We get the following residues at a given n when we compute the integral ∫dsftensor,reg(s)/s from the first two terms on the second line of eq. (5.9),(5.13)−cosπn3a1+a+cosπn(a+2b)1+a2nsinπn(b−a)1+asinπn(c−a)1+a−cosπn(b+2a)1+b+cosπn3b1+b2nsinπn(a−b)1+bsinπn(c−b)1+b−cosπn(c+2a)1+c+cosπn(c+2b)1+c2nsinπn(a−c)1+csinπn(b−c)1+c These are cyclic permutations up to the following terms−cosπn3b1+b−cosπn(b+2c)1+b2nsinπn(a−b)1+bsinπn(c−b)1+b=1n and−cosπn3a1+a−cosπn(a+2c)1+a2nsinπn(b−a)1+asinπn(c−a)1+a=1n where we have used the trigonometric identity 2sinAsinB=cos(A−B)−cos(A+B). The last term on the second line in eq. (5.9) contributes something that has already cyclic permutation symmetry,∑cosπn(a+2(ra+sb+tc)/p)1+a2nsinπn(b−a)1+asinπn(c−a)1+a+cycl Summing all the contributions coming from the second line in eq. (5.9) we get the result12∫−∞∞dssftensor,reg,2ndline(s)=∑n=1∞12n(tn(a,b,c)+tn(b,c,a)+tn(c,a,b)+2) wheretn(a,b,c)=−cosπn(b−c)1+asinπn(b−a)1+asinπn(c−a)1+a+∑r,s,t∈Qp,p−3cosπn(a+2(ra+sb+tc)/p)1+a2sinπn(b−a)1+asinπn(c−a)1+a For p=1 this reduces totn(a,b,c)=12Tn(a,b,c)−1 whereTn(a,b,c)=cosπn(b−c)1+a2sinπn(b−a)1+asinπn(c−a)1+a and we get(5.14)12∫−∞∞dssftensor,reg(s)=∑n=1∞12n(Tn(a,b,c)+Tn(b,c,a)+Tn(c,a,b)−1) This is in good agreement with eq. (2.66) in [8]. Let us expand around a=b=c=0 up to cubic order. For simplicity let us take p=1. Thus we defineT˜n(λ,a,b,c)=Tn(λa,λb,λc)+Tn(λb,λc,λa)+Tn(λc,λa,λb) The small λ-expansion readsT˜n(λ,a,b,c)=14π31n3+12n−3π25abcnλ3+O(λ4) The problematic term 1/(2n) cancels against −1/(2n). Then after carrying out the summation over n using zeta function regularization, we get12∫−∞∞dssftensor,reg(s)=14π3ζ(3)−3π25abcζ(−1)+O(|a,b,c|4) Up to cubic order, this expression is completely symmetric in a,b,c and there is no reason to expect this symmetry will be broken at higher orders. But there is obviously a problem here since if we dualize the first simple term in (5.10), then we get a term proportional to(5.15)12log(β(1+c)) which breaks the permutation symmetry among a,b,c. To restore it by adding the contribution that we get from dualizing the second term, the second term can not lead to a result that is completely symmetric in a,b,c [8]. We may not be allowed to dualize the second term by the plethystic method, since we can not dualize the first simple term in (5.10) by the plethystic method. We demonstrate this fact at the end of appendix C. To use the plethystic method to dualize the first term, we need to regularize a divergent plethystic sum ∑1/n which can not be regularized using the zeta function. We may for instance multiply the generating function by a gaussian regulator e−ϵβ2 (which is symmetric in β) and then at the end take ϵ→0. This effectively places a cutoff at n∼Nϵ∼1/ϵ for the sum over n. The same regularization should then be used throughout, hence to the whole expression in (5.10). Now, if we apply this regularization to the second term as well, it will amount to a multiplication of each of the three terms at the three lines in eq. (5.13) by their corresponding regulator factor e−ϵs2 where s is evaluated at the three poles in (5.12) for each line respectively. The correction is proportional to ϵ for the zeroth order term in λ, and so it goes to zero as we take ϵ to zero. But interesting regularization effects can show up at cubic order where we are regularizing a divergent sum ∑n. This will then contain a divergent piece ∑n=1Nϵn∼Nϵ2∼1/ϵ, which is canceled against the order ϵ correction to (5.13), which is asymmetric in a,b,c. A detailed such computation would involve the error function and so it would be quite involved. Let us here content ourselves with noting that in the limit a,b,c→0 we reproduce the correct result, which agrees with our previous results that we obtained by rigorous methods. When a,b,c is away from zero, we may not have got the entirely correct result by the plethystic dualization method, but the error should be well-confined and small as long as λ is small. The error we made is only concerning the perturbative part. No instanton sum is affected by this issue, since the instanton sums are convergent.Let us move on to the instanton sum. We defineTn,k(a,b,c)=Tn(a,b,c)e−4π2nkβ(1+a) andT˜n,k(λ,a,b,c)=Tn,k(λa,λb,λc)+Tn,k(λb,λc,λa)+Tn,k(λc,λa,λb) This has the small λ-expansionT˜n,k(λ,a,b,c)=(kβn2+12n+14π2n3+2π2kβ2n)e−4π2knβ+O(λ2) By finally adding the contribution coming from the first term in (5.9), which gives the contributionπ26β+12logβ2π we reduce to our previous result (4.4) in the unsquashed limit.55For the comparison with (4.4) we should remember to multiply T˜k by 2 since we defined this out of T that is in the perturbative part where we have the factor of 1/2 multiplying the integral ∫dsfreg/s, while there is no such factor 1/2 for the corresponding integral for the instanton contributions.6Asymptotic S-dualityIn [14] it was argued that for the geometry Sβ1×(Sr5/Zp) there appears an emergent rectangular T2 spanned by Sβ1 and the Hopf fiber of S5, in the limit when p becomes very large. It was then argued that there would be an S-duality associated with this emergent T2. The radius of the temporal Sβ1 is β, while the radius of the Hopf fiber is 2πr/p where r is the radius of S5. In order to exchange these two circles, it is convenient to follow [14] and put β=2πr/k for some integer k. The S-dual geometry will then correspond to SβD1×(Sr5/Zk) with βD=2πr/p. For a T2 to emerge on both sides of the duality, we need to assume that both p and k are very large integer numbers. For the duality to relate high and low temperature behaviors, we need to in addition assume that k<<p. Then βD<<β which means that the S-dual geometry corresponds to the high temperature side of the duality.We will now test whether asymptotic S-duality holds, which we can do since we know both the low temperature and the high temperature behaviors of the logarithm our indices (free energies). The high temperature behavior of the free energy is governed by the Stefan-Boltzmann terms, in which we shall put β=2πr/k. The low temperature behavior is governed by the Casimir energy computed on S5/Zk multiplied by βD=2πr/p.6.1Accidental asymptotic S-duality for the indexWe begin with listing our results for the Stefan-Boltzmann termsβFtensor=π445Npβ3−(6+5(ab+bc+ca))π236NpββFhyper=−π445Npβ3−(ab+bc+ca+6(ϵ+ϵ2))π236Npβ and for the Casimir energiesEtensor=−p3240−p24+a4+b4+c4−36abc480NpEhyper=p3240+pϵ24−ϵ312Np−ϵ424Np−a4+b4+c4+4abc480Np−2(ab+bc+ca)+abc24Npϵ−ab+bc+ca24Npϵ2 We are now ready to test asymptotic S-duality. In the Stefan-Boltzmann terms we put β=2πr/k and getβFtensor=πk3360p−(6+5(ab+bc+ca)πk72NpβFhyper=−πk3360p−(ab+bc+ca)πk72Np and for the Casimir energies we replace p by k and multiply by βD=2πr/p to getβDEtensor=−πk3120p−πk12p+π(a4+b4+c4−36abc)240NkpβDEhyper=πk3120p+πkϵ12p−πϵ36Nkp−πϵ412Nkp−π(a4+b4+c4+4abc)240Nkp−π(2(ab+bc+ca)+abc)12Nkpϵ−π(ab+bc+ca)12Nkpϵ2 S-duality for (1,0) supermultiplets would hold if we had βFi=βDEi for i={tensor,hyper}. Clearly we do not have such an S-duality. Things improve if we consider (2,0) theory for which we put ϵ=0. Then we have for the sumβFM5=−πk12Np−(ab+bc+ca)πk12Np=−πk12p+abcπk12Np where in the second step we used N=1+ab+bc+ca+abc. We also haveβDEM5=−πk12p−abc6Nkp Thus we have βFM5=βDEM5 if and only if abc=0. Moreover, if we put a=b=c=0 and keep ϵ arbitrary, then we haveβFM5=−πk12p−ϵπk12p−ϵ2πk12pβDEM5=−πk12p+ϵπk12p−ϵ3π6kp−ϵ4π12kp Hence only when ϵ=0 and abc=0 can we have asymptotic S-duality. We believe that this asymptotic S-duality that we see here is rather accidental, and a result of two competing effects. On the one hand we have an increased amount of supersymmetry at ϵ=0 and a=b=c=0. On the other hand we have with increased amount of supersymmetry also further cancellation of leading powers that lowers the leading power from T3 down to T in the Stefan-Boltzmann terms in the large T limit, where T=1/β is the temperature. The cancellation of leading power appears to make asymptotic S-duality less likely to hold, but then increased supersymmetry apparently compensates for that so that we can see asymptotic S-duality nevertheless.But asymptotic S-duality was expected to hold by a very general argument in [14], and thus we would not expect to only see this duality by some accident. We have found that in general there is no such asymptotic S-duality for indices, other than for a rather accidental choice of parameters. We could then ask ourselves why this is so. We believe that the answer is due to the fact that in 5d the generic leading term in the Stefan-Boltzmann law should generically grow like T5 for large T and asymptotic S-duality is expected only for this leading term. But for the supersymmetric indices, we have no such high power leading term as T5 due to supersymmetric cancellation. The asymptotic S-duality only holds in the very large p and k limits and thus is expected to be seen only for the T5 term in the Stefan-Boltzmann law, which grows like k5 when we put β∼1/k. To see those terms, we may instead consider the contribution to the index coming from each individual field before the cancellation has taken place. Or we may consider the partition function rather than the index. Indeed, here we will see asymptotic S-duality that seems to be generic, rather than accidental.6.2Generic asymptotic S-duality for individual fieldsWe will now demonstrate asymptotic S duality at leading order T5 for each individual field in the (1,0) tensor multiplet.For the fields in the tensor multiplet, the scalar field (S), the tensor gauge field (T) and the Weyl fermions (F), we have on a round S5 the following refined degeneracies [13]dnS(ω)=∑m=0ndm,n−mω2m−ndnT(ω)=∑m=0n−1(dm,n−m−1ω2m−n−2+dm,n−mω2m−n+dm,n−m+1ω2m−n+2)dnF(ω)=∑m=0n(dm,n−mω2m−n−3/2+dm,n−m+1ω2m−n+1/2) wheredp,q=12(p+1)(q+1)(p+q+2) Although we have the relation 4dnS=dn−1F+dnF for the unrefined degeneracies, this relation does not extend to the refined case.We defineDnS(ω)=dn−2S(ω)DnT(ω)=dn−2T(ω)DnF(ω)=dn−3F(ω)ω−3/2+dn−2F(ω)ω3/2 and then we have the refined generating functionsfi(β,ω)=∑n=0∞Dni(ω)e−βn for i=S,T,F and where one may check that we can extend the sum all the way down to n=0 since for n=0,1 there is no nonzero contribution to the sum. Although there are no nice and simple explicit expressions for these refined dimensions, we are able to repackage these refined dimensions into manageable closed form expressions for the refined generating functions,fS(β,ω)=e−2β−e−4β(1−ωe−β)3(1−ω−1e−β)3fT(β,ω)=e−3β(ω−3+3ω−1+6ω)−3e−4β(ω−2+ω2+3)+3e−5β(ω−1+ω)−e−6β(1−ωe−β)3(1−ω−1e−β)3fF(β,ω)=e−2β(1+3ω2)+e−3β(ω−3+3ω−1−3ω−ω3)−e−4β(1+3ω−2)(1−ωe−β)3(1−ω−1e−β)3 We get a simplification when we compute the refined tensor multiplet generating function,ftensor=fS+fT−fF=e−3βω3−3e−2βω2(1−ωe−β)3 but we will not consider this object here, but rather the contributions from the individual fields. To obtain the corresponding generating functions on lens space L(p;1,1), we expand the denominator in an infinite seriesfi(β,ω)=∑λωλfλi(β)(1−ωe−β)3(1−ω−1e−β)3=∑λfλi(β)∑n,m=0∞14(n+1)(n+2)(m+1)(m+2)e−β(n+m)ωn−m+λ and then we sum over ω running over all the p distinct p-th roots of unity that will put n−m+λ=pq. Let us assume that p is sufficiently large, such thatλ<p Let us furthermore restrict ourselves to the case that λ≥0. Since n=m+pq−λ≥0, we see that for q≥1 there will be no further restriction on m coming from requiring that m≥λ−pq since by our assumptions we will have λ−pq<0. Hence part of our sum will consist of∑q=1∞∑m=0∞f(m+pq−λ,m) Let us next assume that q≤−1. We then bring this into q≥1 by first exchanging m and n assuming that the summand has this exchange symmetry, and next replacing λ by λ′=−λ which is negative. We then need to analyze the case when −p<λ′≤0. Here we find no restrictions at all, so we have the contribution∑q=1∞∑m=0∞f(m+pq+λ,m) Now only remains the case when q=0. Then n=m−λ. If λ≥0, then we have the contribution∑n=0∞f(n,n+λ)=∑m=0∞f(m+λ,m) If λ is negative, we have the contribution∑m=0∞f(m−λ,m) Summing all contributions, we get a quantity that we call Sλ,Sλ=∑q=1∞∑m=0∞(f(m+pq−λ,m)+f(m+pq+λ,m))+∑m=0∞f(m+|λ|,m) where we shall takef(n,m)=14(n+1)(n+2)(m+1)(m+2)e−β(n+m) We then get the lensed indices as follows,fS=(e−2β−e−4β)S0fT=e−3β(S3+9S1)−e−4β3(2S2+3S0)+e−5β6S1−e−6βS0fF=(e−2β−e−4β)(S0+3S2) Unlensed indices are reproduced by taking p=1 and arefS=e−2β+e−3β(1−e−β)5fT=e−5β−5e−4β+10e−3β(1−e−β)5fF=4e−2β+e−3β(1−e−β)5 and in totalfS+fT−fF=e−3β−3e−2β(1−e−β)3 Lensing givesfS+fT−fF=e−3pβ−(2+p2)e−2pβ+(1−p2)e−pβ(1−e−pβ)3 but the expressions for the individual contributions are quite lengthy. Let us therefore only present their small β expansions,fS=2pβ5−16pβ3+−160+168p2+21p4+2p630240pβfT=6pβ5−52pβ3+1pβ−12+−832+1848p2−63p4+2p610080pβfF=8pβ5−23pβ3+−664+798p2−105p4+2p67560pβ The leading order Stefan-Boltzmann terms are associated withζ(6)=π6945 and are given byβFS=−2945π6r5pβ5βFT=−2315π6r5pβ5βFF=−8945π6r5pβ5 The Casimir energy contributions in the large p limit areES=−p530240rET=−p510080rEF=−p57560r In the Stefan-Boltzmann behavior in the high temperature limit, we put β=2πr/k, to getβFS=−2πk530240pβFT=−2πk510080pβFF=−2πk57560p We would now like to reproduce this from the low temperature and the Casimir energy. We then take βD=2πr/p and consider L(k;1,1) and getβDES=−2πk530240pβDET=−2πk510080pβDEF=−2πk57560p We thus have got a complete agreementβFi=βDEi for all the individual fields in the tensor multiplet and so we have an asymptotic S-duality at order T5 that appears to be generic. We also can see that at subleading orders, we have no such agreement, just as was to be expected.7DiscussionWe have found that the unsquashing limit a,b,c→0 is smooth. In the 5d localization computation, the unsquashed case is difficult to analyze since the instanton particles will spread out as we take a=b=c=0 [8]. As long as the squashing parameters are not exactly zero, then no matter how small they are, once we fix their values and then take the localization limit, these instanton particles become localized at three fixed points on CP2, in the localization limit. But we may worry that the unsquashed limit is singular, or that the limit is discontinuous and that we get a different result when we put a=b=c=0 compared to what we get by taking the limit a,b,c→0. Indeed such a discontinuity is natural to expect because we need a different computation when a=b=c=0 in the 5d theory. It is therefore a nontrivial result to have found that the limit a,b,c→0 is actually smooth. It would be interesting to see if this smooth behavior extends to nonabelian gauge groups.AcknowledgementsI would like to thank Seok Kim and Maxim Zabzine for discussions. This work was supported by the grant Geometry and Physics from Knut and Alice Wallenberg foundation.Appendix AThe zeta function dualization methodWe follow [9] and decompose the zeta function (3.1) into two parts,ζ(s)=ζℓ=0+2ζℓ>0(s) where we notice that ζℓ<0=ζℓ>0, which explains the factor 2. By a rewriting of the zeta function, we encounter coefficientsCλ(aℓ2)=∑σCλ,naℓ2σ that are polynomials in aℓ2, whereaℓ=2πℓβ Let us illustrate how such coefficient polynomials arise by a simple example. We consider the tensor multiplet zeta function (4.3) that (for p=1) involves terms of the form(n2−1)(n2+aℓ2)−s=(n2+aℓ2−aℓ2−1)(n2+aℓ2)−s=(n2+aℓ2)1−s−(aℓ2+1)(n2+aℓ2)−s In this example, we have the coefficient polynomialsC0(aℓ2)=−1−aℓ2C1(aℓ2)=1 In a more general situation, these coefficient polynomials are defined through the following expansions of the zeta function components aboveζℓ=0(s)=(μr)2s∑λ∑n=1∞Cλ,0(np)2λ−2s=(μr)2s∑λCλ,0p2λ−2sζ(2s−2λ) andζℓ>0(s)=(μr)2s∑λ∑ℓ=1∞∑n=1∞Cλ(aℓ2)(aℓ2+(np)2)λ−s respectively. We then apply the Mellin transform that puts aℓ2+(np)2 in the exponent,(aℓ2+(np)2)λ−s=1Γ(s−λ)∫0∞dttts−λe−(aℓ2+n2p2)t We now wish to dualize the sum with respect to n. To this end, we extend the sum over n=1,2,... to also include n=0,−1,−2,..., which we can do since n2 is even. But then we have to remember to subtract the term with n=0 again, and also divide the result by 2. Once we have got a sum over n∈Z, we can apply the Poisson resummation formula to that sum,∑n∈Ze−tn2=πt∑nD∈Ze−π2nD2t For the terms with nD≠0 we then use the following integral formula for the modified Bessel function∫0∞dtttνe−at−bt=2(ab)ν/2Kν(2ab) while the integral we get for nD=0 is gives a Gamma function. Then we must, as we have said, also subtract the term with n=0 (and then divide the whole thing by 2). This way, we end up with the following result,ζℓ>0(s)=ζℓ>0,n=0(s)+ζℓ>0,n≠0(s) whereζℓ>0,n=0(s)=π2pμ2s∑λ,σCλ,σΓ(s−λ−1/2)Γ(s−λ)(2πβ)1+2λ+2σ−2sζ(2s−2σ−2λ−1)−12μ2s∑λ,σCλ,σ(2πβ)2λ+2n−2σζ(2s−2σ−2λ) andζℓ>0,n≠0(s)=πpμ2s∑λ∑ℓ=1∞∑n=1∞CλΓ(s−λ)(nβ2pℓ)s−λ−1/2Ks−λ−1/2(4π2nℓpβ) Explicit forms of the Bessel functions that we will encounter areK±1/2(x)=π2xe−xK±3/2(x)=π2x(1+1x)e−x To get the index, we need to compute the derivative of the zeta function at s=0. We haveζℓ=0′(0)=∑λCλ,0p2λ2ζ′(−2λ) For ζℓ>0′(0) we first bring up an overall factor of s by using 1/Γ(s)=s/Γ(s+1) and 1/Γ(s−1)=s(s−1)/Γ(s+1). We then obtain the zeta function on the form ζ(s)=sζ˜(s) and the derivative is then simply given by ζ′(0)=ζ˜(0) since in all our examples ζ˜(s) will be regular at s=0. Thus to compute the derivative, we never need to actually compute any derivative.It would be more natural to apply Poisson resummation with respect to the sum over ℓ that is already over Z. But then we would get the low temperature expansion (3.3) as was shown in [9]. That computation is very elegant. In particular the Casimir energy factor drops out automatically without any need to consider normal ordering and zero point energies.Appendix BThe Abel-Plana dualization methodIf we write the log of the index as a low temperature expansionlogI=−βE+∑ndnlog(1−e−βEn) then we may compute the sum over n using the Abel-Plana integral formula. The analytic function that we need to consider in this application is given byf(z)=d(z)log(1−e−βE(z)) where d(n)=dn and E(n)=En for n=0,1,2,..., and analytically continued away from these integer values. If we compute the sum over n by the Abel-Plana integral formula, then we automatically turn this sum into a high temperature expansion.Let us now present the Abel-Plana integration formula. A sum over n may (under certain conditions) be computed by a contour integral,∑n=0∞f(n)=∮Cdzf(−z)e−2πiz−1 Here C is a counter clockwise contour surrounding the positive real axis, including the origin. Next we assume that f is analytic in the positive halfplane and behave nicely at infinity, to rotate the contour to the imaginary axis. Then the right-hand side becomes12f(0)+i∫−∞∞dxf(ix)e2πx−1 where we add f(0)/2 because the integral contour that goes through the pole at z=0 picks up the other half of that same residue, to make up f(0)/2+f(0)/2=f(0) in total. We separate the integral domain into two pieces [−∞,∞]=[−∞,0]∪[0,∞] and rewrite the former integrali∫−∞0dxf(ix)e2πx−1=i∫0∞dxf(−ix)e−2πx−1=−i∫0∞dxf(−ix)e2πx−1e2πx=−i∫0∞dxf(−ix)−i∫0∞dxf(−ix)e2πx−1 The first integral is along the imaginary axis, but by assumption our function f is well-behaved at infinity and analytic in the positive real halfplane, and the integral can be Wick rotated to the positive real axis,−∫0∞dxf(x)+i∫0−∞dxf(ix)=0 Adding up, we then have∑n=0∞f(n)=12f(0)+∫0∞dxf(x)+i∫0∞dxf(ix)−f(−ix)e2πx−1 which is the Abel-Plana formula.It is not apriori clear to us why the application of the Abel-Plana formula turns the low temperature expansion into a high temperature expansion, but it works this way in all explicit examples that we have encountered.Appendix CThe plethystic dualization methodHere we describe the plethystic dualization method that was used in [8] to compute the index I(β)=e−βEI˜SBI˜reg where E is the Casimir energy. We separate the generating function into a singular and a regular part at β=0. The singular part goes into the Stefan-Boltzmann factor I˜SB as shown in the main text. The regular part is treated as follows,lnI˜reg(β)=∑n∈Z∫ϵ∞dsse2πinsβfreg(s) We shall assume that freg(s)=−freg(−s) and so freg(s)/s does not have a simple pole at s=0. We putμ(n,s)=dsse2πinsβfreg(s) where we notice that μ(−n,s)=μ(n,−s). We then consider the following rewritings∑n∈Z∫ϵ∞μ(n,ns)=∑n=1∞∫ϵ∞μ(n,s)+∑n=1∞∫ϵ∞μ(−n,s)+∫ϵ∞μ(0,s)=∑n=1∞(∫−∞−ϵμ(n,s)+∫ϵ∞μ(n,s))+12(∫−∞−ϵμ(0,s)+∫ϵ∞μ(0,s))=∑n=1∞∫Rμ(n,s)+12∫Rμ(0,s)−∑n=1∞∫−ϵϵμ(n,s)−12∫ϵϵμ(0,s) We next look at the third and fourth terms,−∑n=1∞∫−ϵϵμ(n,s)−12∫−ϵϵμ(0,s)=−12∑n∈Z∫−ϵϵμ(n,s)=−12∑n∈Z∫−ϵϵdsse2πinsβfreg(s)=−β2[1sfreg(s)]|s=0 We thus have three terms to compute,lnI˜reg=A+B+C whereA=∑n=1∞∫Rdsse2πinsβfreg(s)B=12∫Rdssfreg(s)C=−β2[1sfreg(s)]|s=0 By noting that freg(s)=−freg(−s) implies freg(0)=0, we can writeC=−β2lims→0[freg(s)−freg(0)s]=−β2lims→0∂sfreg(s) We now see thatC=βE where E is the Casimir energyE=−12∂βfreg(β)|β=0 Then eC cancels against e−βE.Let us try to illustrate this method by dualizing the Dedekind eta function. We start by rewriting its corresponding generating function asf(β)=e−β1−e−β=coshβ22sinhβ2−12 Since the first term is antisymmetric, we can apply the plethystic dualization method on this term. The plethystic sum is given by(C.1)∑n=1∞1nf(nβ)=∑n=1∞(12ncoshnβ2sinhnβ2−12n) We note that although the whole expression is convergent, being equal to −∑n=1∞log(1−e−nβ), the sum ∑n=1∞−12n is divergent and has to be regularized if we shall be able to separate the two terms. But let us ignore this, and just apply the plethystic dualization method on the first term. From the singular piece, we get the Stefan-Boltzmann term∑n∈Z∫ϵ∞dsse2πins/β1s=π26β and from the regular piece we get the perturbative and nonperturbative contributions12∫−∞∞dsscoshs2sinhs2=∑n=1∞12n∫−∞∞dsscoshs2sinhs2e−4π2kn/β=∑n=1∞1ne−4π2kn/β respectively. The perturbative contribution cancels against the second term in the plethystic sum (C.1), leaving us with∑n=1∞1nf(nβ)=π26β+∑k,n=1∞1ne−4π2knβ The correct answer should have in addition a log-term 12logβ, which we are missing.To make this computation rigorous, we may regularize the divergences by replacing f(β) with f(β)e−ϵβ2 and then at the end take ϵ→0. 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