Диссертация (Isomonodromic deformations and quantum field theory), страница 40

In order to expandthis method for the B-case one has to learn more about the theory of “exotic fermions”on Riemann surfaces, probably along the lines of [FSZ, DVV], and we postpone thisfor a separate publication.Another set of open problems is obviously related with generalization to otherseries and twisted fields related with external automorphisms. Here only the E-casesseem to be straightforward, since standard bosonization can be immediately appliedin the simply-laced case, and there should be not many problems with the fermionconstruction. However, it is not easy to predict what happens in the situation whenKac-Moody algebras at level k = 1 have fractional central charges, and the directapplication of the methods developed in this chapter is probably impossible.

It isstill not very clear, what is the role of these exact conformal blocks in the context ofmulti-dimensional supersymmetric gauge theories, since generally there is no Nekrasovcombinatorial representation in most of the cases. We hope to return to these issuesin the future.Finally, there is an interesting question of possible generalization of our approachto the twisted representations with k 6= 1, which has been already considered in[FSS]. Some overlap with our formulas with sect. 8 of this chapter suggests that suchgeneralization could exist. We hope to return to this problem elsewhere.AppendixIdentities for lattice Θ-functionsHere we present few rigorously proved identities, used to verify representation-theoreticconsiderations at the level of computations of characters.First identity for AN −1 and DN Θ-functions0One can describe the lattices AN −1 , DN and DNin a similar way: NXAN −1 ={k1 , .

. . , kN ki = 0} i=1NXDN ={k1 , . . . , kN ki ∈ 2Z} i=1NX0DN={k1 , . . . , kN ki ∈ 2Z + 1}i=1190(6.205)6.9. Identities for lattice Θ-functionsThe last lattice is actually just DN lattice, but shifted by vector (1, 0, . . . , 0). So allthese definitions can be rewritten asNX(6.206)ki ∈ S}LS = {k1 , .

. . , kN i=1where S ⊆ Z: in our cases it should be chosen to be {0}, 2Z, and 2Z + 1, respectively.Notice also that for S = Z we get the simplest BN lattice.By definitionX1~ 2ΘLS (~v ; q) =q 2 (~v+k)(6.207)k1 +...kN ∈SFor our purposes we need this function computed for the vector13l1 − 1l1 − 31 − l1, r1 +, . . . , r1 +)⊕2l12l12l1l2 − 1l2 − 31 − l2⊕(r2 +, r2 +, . .

. , r2 +) ⊕ ...⊕2l22l22l2lK − 31 − lKlK − 1, rK +, . . . , rK +)⊕(rK +2lK2lK2lK~v = (r1 +(6.208)where l1 + . . . + lK = N . Let us parameterize vector ~k as follows:~k = (n1 , . . . , n1 ) ⊕ . . . ⊕ (nK , . . . , nK ) + ω (l1 ) ⊕ . . . ⊕ ω (lK ) +a1aKa1aKaKa1)+m~ 1 ⊕ ... ⊕ m~K+( , . .

. , ) ⊕ . . . ⊕ ( , . . . ,l1l1lKlK(6.209)where m~ i ∈ Ali −1 , andωa(l) = (l−a aal−a,...,,− ,...,− )llll(6.210)so that the first number is repeated a times, whereas the second one l − a times.Hence, vectors ~k ∈ LS are parameterized by vectors {m~ i ∈ Ali −1 } and integer numbers{ni ∈ Z; ai ∈ Z/li Z}, restricted byKX(ni li + ai ) ∈ S(6.211)i=1The algorithm of decomposition (6.209) works as follows: first we sum up all components of ~k inside each cycle – each number divided by li gives ni , whereas remainder(l )gives ai . Subtracting (ni , . . . , ni ) + ωaii , we are left with the vectors {m~ i } with vanishing sums of components.Now it is easy to see thatΘ(~v + ωa(l11 ) ⊕ ωa(l22 ) ⊕ .

. . ⊕ ωa(lKK ) ; q) = Θ(~v ; q)13Notation ~v ⊕ ~u means (v1 , . . . , vk ) ⊕ (u1 , . . . um ) = (v1 , . . . , vk , u1 , . . . , um ).191(6.212)6. Twist-field representations of W-algebras, exact conformal blocks and character identitieswhich follows from the fact that Θ(~v ; q) = Θ(σ(~v ); q), where σ is a permutation. Forexample, take σa to be a-th power of the cyclic permutation, then: 1−ll−1l + 1 − 2a l + 3 − 2al−1 1−ll − 1 − 2aσa,...,=,,...,,,...,=2l2l2l2l2l2l2ll−11−l,...,+ ωa(l)=2l2l(6.213)(l1 )(l1 )(lK )and therefore any vector ~v + ωa1 ⊕ ωa2 ⊕ . . .

⊕ ωaK can be obtained by severalpermutation of components of ~v , so the corresponding Θ-functions are equal. ThusXaaaa1(~v +m~ 1 ⊕...⊕m~ K +(n1 + l 1 ,...,n1 + l 1 )⊕...⊕(nK + l K ,...,nK + l K ))211KKΘLS (~v ; q) =q2KP(ni li +ai )∈Si=1m~ i ∈QAli −1(6.214)turns into the sum over several orthogonal sublatticesΘLS (~v ; q) =Xq1 (l1 )(ρ̂⊕...⊕ρ̂(lK ) +m~ 1 ⊕...⊕m~ K )22X·m~ i ∈Ali −1KPq12KPi=1ali (ni + l i +ri )2i=(ni li +ai )∈Si=1=KYΘAli −1 (ρ̂(li )KPX; q) ·q i=11(n0i +ri li )22lin01 +...+n0K ∈Si=1(6.215)where1−ll−1 l−3(6.216),,...,)2l2l2lOne can identify the last factor in the r.h.s. with the contribution of zero modes,related to the r-charges [GMtw].ρ̂(l) = (Product formula for AN −1 Θ-functionsApply (6.215) to the simplest case of ΘBN (ρ̂(N ) ; q) with S = ZX n2ΘBN (ρ̂(N ) ; q) = ΘAN −1 (ρ̂(N ) ; q) ·q 2N(6.217)n∈ZUsing definition (6.216) and Jacobi triple product formula we getΘBN (ρ̂(N ) ; q) = qN−1 XYN 2 −124Nqk2+ N −1−2ak22N=a=0 k∈Z=qN 2 −124N∞Y(1 + q1N(k− 12 ))2Xn∈Zn2q 2N =∞Y(6.218)(1 − q n )Nn=1k=1as well as∞Y11(1 + q N (k− 2 ) )2∞Yn=1k=1192n(1 − q N )(6.219)6.10.

Exotic bosonizationsSubstituting into (6.217) one obtains∞QΘAN −1 (ρ̂(N ) ; q) = qN 2 −124N(1 − q k )Nk=1∞Q=k(1 − q N )η(q)N(6.220)1η(q N )k=1or the product formula [Mac] for ΘAN −1 (ρ̂(N ) ; q), where the r.h.s. is expressed in termsof the Dedekind functions. Substituting this into (6.215) we get it in its final formXΘLS (~v ; q) =q12NP(vi +ki )2i=1=k1 +...+kN ∈SKYη(q)li1i=1X·η(q li )KPqi=11(ni +li ri )22li(6.221)n1 +...+nK ∈SAn identity for DN and BN Θ-functionsHere we show how ΘDN (~v∗ ; q) can be simplified if ~v∗ contains at least one component1. One has then2X1~ 2ΘDN (~v∗ ; q) = ΘDN (( 12 , v2 , .

. . , vn ); q) =q 2 (~v∗ +k) =(6.222)k1 +...+kn ∈2Z1= ΘDN ((− 2 , v2 , . . . , vn ); q) = ΘDn (~v∗ − (1, 0, . . . , 0); q)Since for the lattices DN t {DN − (1, 0, . . . , 0)} = BN , it follows from (6.222) thatΘDN (~v∗ ; q) = 21 ΘBN (~v∗ ; q)(6.223)Exotic bosonizationsHere we present some details of the bosonization procedures, used in the main text.NS × RConsider, first, construction [Ber, BBT] relating pair (of N S and R!) fermions to atwisted boson 14X Jr√ Xa2n+1=i2= φ(ξ)φ̃(t) = ir(6.224)rt(2n + 1)ξ 2n+11n∈Zr∈Z+ 2with differently normalized oscillator modes [aM , aN ] = M δM +N,0 (M, N ∈ 2Z + 1).Compute the correlator−hφ(ξ)φ(ζ)i = 2Xha2n+1 a2m+1 iξ−2n−1 −2m−1ζ= −2∞X(ζ/ξ)2n+1n=0ζ= 2 log 1 −ξ2n + 1=ζ2ξ+ζ− log 1 − 2 = − log= −[φ+ (ξ), φ− (ζ)]ξξ−ζ14It is more convenient to use in this section coordinate ξ =around 0 maps ξ to −ξ.193√(6.225)t, so analytic continuation in t6. Twist-field representations of W-algebras, exact conformal blocks and character identitiesassuming |ξ| > |ζ|.

Now introduce11η̂(ξ) = √ : eiφ(ξ) := √ eiφ− (ξ) eiφ+ (ξ)22(6.226)1η̂(ξ)η̂(ζ) = e−(φ− (ξ)+φ− (ζ)) ei(φ+ (ξ)+φ+ (ζ)) e−[φ+ (ξ),φ− (ζ)] =21ξ−ζ= : ei(φ(ξ)+φ(ζ)) :2ξ+ζ(6.227)so that for |ξ| > |ζ|while for |ξ| < |ζ|1 i(φ(ξ)+φ(ζ)) ζ − ξ:e:2ξ+ζIt means that OPE of the η̂-fields has fermionic nature:η̂(ζ)η̂(ξ) =(6.228)1ξ+ζ1 ξ + ζ (φ(ξ)−φ(ζ))ζ:∼:e+ reg.

∼+ reg.2ξ −ζ2ξ −ζξ−ζP ηkand in the anticommutator of components η̂(ξ) =ξkη̂(ξ)η̂(−ζ) =˛{ηk , (−1)l ηl } =˛ζ l−1 dζζ(6.229)k∈Zζξ k−1 dξ = δk+l,0ξ−ζone gets unusual sign factor.It is interesting to point our that the Ramond zero mode η02 =representation˛√dξ iφ− (ξ) iφ+ (ξ)2η0 =ee=ξ2= 1 − 2a−1 a1 + a2−1 a21 − (a−3 + a3−1 )(a3 + a31 ) + .

. .9(6.230)12has bosonic(6.231)For example, the action of this operator on low-level vectors gives√√√2η0 · |0i = |0i,2η0 · a−1 |0i = −a−1 |0i,2η0 · a2−1 |0i = a2−1 |0i(6.232)√√12412η0 · a−3 |0i = a−3 |0i − a3−1 |0i,2η0 · a3−1 |0i = − a−3 |0i − a3−1 |0i33331 −2Here in the second line one gets the matrix 13with the eigenvalues ±1.−4 −1We also haveη0 ηk = −ηk η0 , k 6= 0(6.233)√so one can identify 2η0 = (−1)F −F0 , where F is fermionic parity.

Generally, algebra,generated by {ηk }, has two representations with the vacua |0i± , such that η0 |0i± =±|0i± . One can also take direct sum of such representations: bosonization formula inthis representation looks asσ1η̂(ξ) = √ eiφ− (ξ) eiφ+ (ξ)2194(6.234)6.10. Exotic bosonizationsExistence of this bosonization at the level of characters gives us obvious identity∞Y∞Y1=(1 + q k )2k+11−qk=0k=1(6.235)Notice that above consideration actually concerns R and N S fermions because onecan construct two combinationsX η2p1√ (η̂(z) − η̂(−z)) == iψ̂N S (z)zp21p∈Z+ 2(6.236)X η2n1√ (η̂(z) + η̂(−z)) == ψ̂R (z)zn2n∈ZthenJ(z) =X Jp√ 1 ∗ √ψ̂ ( z)ψ̂( z) = iψN S (z)ψR (z) =zz p+1p∈Z+ 21Xη2q η2nJp = i(6.237)n+q=p√√1here t ψ̂N S ( t) and t− 2 ψ̂R ( t) are usual Ramond and Neveu-Schwarz fermions.Here we consider fermion corresponding to the branch point of type [l]− .

Thismeans that we should have− 12η(z)η(σ(w)) ∼1,z−w(6.238)and such monodromy that η(e4πil z) = ±η(z). Let us use the construction form (6.10.1)11z− 2η(z) = √ η̂(z 2l )2lTherefore11(6.239)1z − 2 w− 2w 2l1η(z)η(σ(w)) ∼11 ∼2lz−wz 2l − w 2lSo final construction states that one should have(6.240)111z− 2η(z) = σ1 √ eiφ− (z 2l ) eiφ+ (z 2l )2 l(6.241)R×RLet us take two Ramond fermions ψ (1) , ψ (2) and introduce X ψn1ψ(z) = √ ψ (1) (z) + iψ (2) (z) =n+ 122n∈Z z X ψn∗1ψ ∗ (z) = √ ψ (1) (z) − iψ (2) (z) =n+ 122n∈Z z195(6.242)6. Twist-field representations of W-algebras, exact conformal blocks and character identitiesSince there are two zero modes ψ0∗ and ψ0 , one expects to have four vacua |0i, ψ0 |0i,ψ0∗ |0i, ψ0∗ ψ0 |0i.We can mimic expansion (6.242) using fractional powersXψ(z) =p∈Z+ 21ψN S,pz,p+ 1 +σXψ ∗ (z) =∗ψNS,p12p∈Z+ 21z p+ 2 −σ(6.243)∗.

It means that after standardwith σ = 12 , i.e. ψn = ψN S,n− 1 and ψn∗ = ψNS,n+ 122bosonizationψ(z) = e−iφ− (z) e−iφ+ (z) e−Q z −J0 , ψ ∗ (z) = eiφ− (z) eiφ+ (z) eQ z J0J0 |0i = σ|0i = 21 |0i(6.244)one gets ψ0∗ |0i = 0, and only one half of the vacuum states survive. To identify this√ (1)representation with something well-known, consider the eigenvectors 2ψ0 |0i± =√ (1)±|0i± of 2ψ0 = ψ0 + ψ0∗ :1|0i+ = √ (|0i + ψ0 |0i),2Acting by√ (2)2ψ0 = i(ψ0∗ − ψ0 ) one gets√ (2)2ψ0 |0i+ = |0i− ,i|0i− = √ (|0i − ψ0 |0i)2√(2)2ψ0 |0i− = |0i+(6.245)(6.246)The character of such module is given by2∞Y11n∈Z(1 + q k )2 = q − 8 Q∞k=11 2q 2 (n+ 2 )P(6.247)(1 − q k )k=1where in the l.h.s.