Диссертация (Isomonodromic deformations and quantum field theory), страница 13

We are going to prove in next section, thatdefinitions (3.87) are indeed self-consistent and also consistent with (3.90), whichfollows from the generalized Hirota bilinear relations, satisfied by the monodromyvertex operators.Actually, we have four different matrix kernels (3.87) with the indices αβ, αβ̇, α̇β,α̇β̇, corresponding to all possible combinations of different regions.

When we changefrom one region to another one, then we have to change the basis of solutions, andthis transition can be given by some matrix Cα̇α .The expansion of these kernels, e.g. for 0 < z, w < 1Kαβ (z, w) =X11δαββ+hθ|Vν (1)ψ̃ α −p ψ−q|σiz p− 2 +σα wq− 2 −σβ =z − w p,q>0Xδαβαβ p− 12 +σα q− 12 −σβ=+Kpqzwz − w p,q>0(3.91)or at z, w > 1Kα̇β̇ (z, w) =Xδα̇β̇11+hθ|ψ̃pα̇ ψqβ̇ Vν (1)|σiz −p− 2 +θα̇ w−q− 2 −θβ̇ =z − w p,q>0Xδα̇β̇α̇β̇ −p− 12 +θα̇ −q− 12 −θβ̇K̃pqz=+wz − w p,q>0(3.92)give the corresponding matrix elements for the fermionic modes. The corresponding) are in fact defined up to the factors sα s−1and K̃pαβmatrix elements (Kpαββ whichα ,qβα ,qβcomes from the ambiguity in normalization of the vertex operator. For any three monodromy matrices M0 M1 M∞ = 1 one can fix all their invariant functions (e.g.

traces)and diagonalize M∞ , but then one possible transformation survives: a simultaneousconjugationMi 7→ S −1 Mi S(3.93)by diagonal S = diag(s1 , . . . , sN ). This gives vertex operators, actually different bysJ 0 factor with corresponding multiplicative renormalization of their matrix elements.For special vertex operators with ν = νN ej these matrix elements can be expressedin terms of the products (3.50), for exampleβKpαβ= hθ|Vν (1)|pα , qβ ; σi = hθ|Vν (1)ψ̃ α −pα ψ−q|σi =α ,qββ1f1,α (σ, θ + ν, pα )f2,β (σ, θ + ν, qβ )=pα + qβ − σα + σβK̃pαβ= hpα , qβ ; θ|Vν (1)|σi = hθ|ψ̃pαα ψqββ Vν (1)|σi =α ,qβ1=−f1,α (θ + ν, σ, pα )f2,β (θ + ν, σ, qβ )pα + qβ − θα + θβWe shall return to discussion of special case below in sect. 3.4.3.54(3.94)3.4. Vertex operators and Riemann-Hilbert problemThe general formula for 2n-point fermionic correlator is given by the Wick formula0hθ|dα̇N YYψ̃α̇θ (zα̇,i )ψα̇θ (wα̇,i )Vν (1)N YdαYψ̃ασ (zα,i )ψασ (wα,i )|σi =(3.95)α=1 i=1α̇=1 i=1Kαβ (zα,i , wβ,j ) Kαβ̇ (zα,i , wβ̇,j )= hθ|Vν (1)|σi · detKα̇β (zα̇,i , wβ,j ) Kα̇β̇ (zα̇,i , wβ̇,j )On the punctured unit circle |z| = |w| = 1, z 6= 1, w 6= 1 one hasXKα̇β (z, w) =Cα̇α Kαβ (z, w)αKαβ̇ (z, w) =XKαβ (z, w) C −1β(3.96)β̇βIt follows then from (3.95), that there are two operator identitiesXψ̃α̇θ (z)Vν (1) = Vν (1)Cα̇α ψ̃ασ (z)αψα̇θ (z)Vν (1)= Vν (1)Xψασ (z)(C −1 )αα̇(3.97)αActually, these identities are enough to define the operator Vν (1).

The simplestquantity to compute is˛dz r− 1 +σασ−1Vν (1)ψα,r Vν (1) =Vν (1)ψασ (z)Vν (1)−1z 2(3.98)2πi|z|=1Using (3.97) one can rewrite this equivalently as˛Xdz r− 1 +σα θσ−1βVν (1)ψα,r Vν (1) =Cαz 2ψβ (z) =2πiβ|z|=1˛=Xβ,sCαβ|z|=1Xdz r−s−1+σα −θβ θzψβ,s =Cαβ2πiβ,s=Xβ,sCαβˆ2πdφ 2πi(r−s+σα −θβ )φ θeψβ,s =2π(3.99)0−i(e2πi(σα −θβ ) − 1) θψr − s + σα − θβ β,sIn principle, this formula includes all possible information about Vν (t). Now it is easyto proveTheorem 3.2. Vν (t) is a primary field of the conformal WN ⊕ H algebra with thehighest weights uk (ν).Proof: First we notice that due to (3.97) and to the definitions (3.85), (3.65) onehasUkθ (z)Vν (t) = Vν (t)Ukσ (z)55(3.100)3.

Free fermions, W-algebras and isomonodromic deformationsin the region |z| = t, z 6= t. This means that Uk (z) are actually single-valued operators(with trivial monodromies). Actually, we have already proved in Theorem 3.1 thatstates |σi are highest weight vectors, souk (σ)hθ| . . . Uk (z)|σi =+ less singular hθ| . . . |σi(3.101)zkand, since (3.88) is symmetric under the permutation of the singular points, one canalso conclude, that for a different pointuk (ν)hθ| .

. . Uk (z)Vν (t) . . . |σi =+ less singular hθ| . . . Vν (t) . . . |σi (3.102)(z − t)kso (Uk,n>0 Vν )(t) = 0, and it means, that Vν (t) is just a primary field.Generalized Hirota relationsNow consider any operator O with linear adjoint action on fermionsXXOO−1 ψα,r O =Rrα,sβψβ,s , O−1 ψ̃α,−r O =ψ̃β,−s (RO )−1sβ,rαs,β(3.103)s,βwhich is generally a relabeling of a GL(∞) transformation for a single fermion.

Itleads to a standard statement of commutativity of two operators in H ⊗ HXXψα,−r ⊗ ψ̃α,r O ⊗ Oψα,−r ⊗ ψ̃α,r =O⊗O(3.104)r,αr,αwhich is an operator form of the bilinear Hirota relation [MJD/KvdL, AZ].Let us now point out, that we have already introduced by (3.97) a particularsubclass of general transformations (3.103)XXCα̇α ψ̃α (z)(C −1 )αα̇ ψα (z) , V −1 ψ̃α̇ (z)V =V −1 ψα̇ (z)V =(3.105)ααwhere C and C −1 can be now interpreted as monodromy matrices: one can consider(3.105) as a linear relation between two analytic continuations of the fermionic fieldsδαβat |z| = 1 towards z → ∞ and z → 0, preserving the OPE ψ̃(z)α (z)ψβ (z 0 ) = z−z0 +. . ..An immediate consequence of (3.105) isTheorem 3.3. The Fourier modes of the bilinear operatorsXX IkI(z) =ψα (z) ⊗ ψ̃α (z) =z k+1αk∈Z†I (z) =XαX I†kψ̃α (z) ⊗ ψα (z) =k+1zk∈Z(3.106)commute with Vν (t) ⊗ Vν (t) in the senseIkθ · Vν (t) ⊗ Vν (t) = Vν (t) ⊗ Vν (t) · IkσθI † k · Vν (t) ⊗ Vν (t) = Vν (t) ⊗ Vν (t) · Ikσ56(3.107)3.4.

Vertex operators and Riemann-Hilbert problemProof: First we notice thatI θ (z) · Vν (t) ⊗ Vν (t) = Vν (t) ⊗ Vν (t) · I σ (z)(3.108)holds at |z| = t, z 6= t, due to (3.97)Xψαθ (z) ⊗ ψ̃αθ (z) · Vν (t) ⊗ Vν (t) = Vν (t) ⊗ Vν (t)αX(C −1 )β̇α Cαγ̇ ψβ̇σ (z) ⊗ ψ̃γ̇σ (z) =α,β̇,γ̇= Vν (t) ⊗ Vν (t)Xψβ̇σ (z) ⊗ ψ̃β̇σ (z)β̇(3.109)To continue this equality to z = t one has just to check that I (z) · Vν (t) ⊗ Vν (t) isregular.

Due to the symmetry of (3.88) this is the same as to check that I σ (z)·|σi⊗|σiis regular. Since,θσI (z) · |σi ⊗ |σi =σX X ψα,n|σi1α=Xαn<0σψα,−1 |σi2⊗z n+ 2 +σα⊗σX ψ̃α,m|σi1m<0σψ̃α,−1 |σi2z m+ 2 −σα=+ O(z)this expression is regular, this completes the proof.Let us notice that we have also got the equalitiesσIk≥0· |σi ⊗ |σi = 0,(3.110)σI † k≥0 · |σi ⊗ |σi = 0(3.111)while, for example†I−1|θi ⊗ |θi =Xψ̃α,−1/2 ⊗ ψα,−1/2 |θi ⊗ |θi =αI−1 |θi ⊗ |θi =XX|1α , θi ⊗ | − 1α , θiαψα,−1/2 ⊗ ψ̃α,−1/2 |θi ⊗ |θi =αX(3.112)| − 1α , θi ⊗ |1α , θiαbut†hθ| ⊗ hθ| · I−1= hθ| ⊗ hθ| · I−1 = 0(3.113)We shall see below, that existence of extra bilinear operator relations lead actually tothe infinite number of Hirota-like equations for the τ -function.Let us also notice that operator tL0 belongs to the quasigroup, but it does notcommute with Ik :(3.114)tL0 I(z)t−L0 = tI(tz)which means that tL0 Ik t−L0 = t−k .

So, in principle,vertex operator can contain someQL0factors ti , but in such a combination with ti = 1.Now we are ready to prove, that the correlation functions (3.86) (and in fact anycorrelation function hθ ∞ |Oψ̃α (z)ψβ (w)|θ 0 i = hθ ∞ |Vθn−2 (tn−2 ) . .

. Vθ1 (t1 )ψ̃α (z)ψβ (w)|θ 0 iwith two fermions) can be decomposed into two correlation functions with a single573. Free fermions, W-algebras and isomonodromic deformationsfermion insertion. In addition to (3.112), (3.113) one has to compute commutator ofthis operator with ψ̃ ⊗ ψ using the contour integral representation˛˛hidx XI−1 , ψ̃α (z) ⊗ ψβ (w) =  + ψγ (x) ⊗ ψ̃γ (x) · ψ̃α (z) ⊗ ψβ (w) =x γzw˛˛Xdxdx Xδγαδγβ=⊗ ψ̃γ (x)ψβ (w) +ψγ (x)ψ̃α (z) ⊗=x γ x−zx γx−wzw11= · 1 ⊗ ψ̃α (z)ψβ (w) + · ψβ (w)ψ̃α (z) ⊗ 1zw(3.115)Inserting this operator identity inside the correlation functions, and using (3.112),(3.113) we get0 = hθ ∞ | ⊗ hθ ∞ | · I−1 · O ⊗ O · ψ̃α (z) ⊗ ψβ (w) · |θ 0 i ⊗ |θ 0 i =X= hθ ∞ | ⊗ hθ ∞ | · O ⊗ O · ψ̃α (z) ⊗ ψβ (w)| − 1γ , θ 0 i ⊗ |1γ , θ 0 i+γ+11−z w(3.116)hθ ∞ |O|θ 0 i · hθ ∞ |Oψ̃α (z)ψβ (w)|θ 0 iThe first term in the r.h.s.

is equal to the bilinear combination of the correlationfunctions with a single fermion insertion, so one gets finallyhθ ∞ |Oψ̃α (z)ψβ (w)|θ 0 ihθ ∞ |O|θ 0 i =zw Xhθ ∞ |Oψ̃α (z)| − 1γ , θ 0 ihθ ∞ |Oψβ (w)|1γ , θ 0 i=z−w γ(3.117)which for O = Vν (1) gives the relation between (3.90) and (3.87). Substituting hereδαβthe OPE ψ̃α (z)ψβ (w) = z−w+ reg. and taking residue at z → w one also proves thatmatrices in (3.90) are indeed inverse to each other.Riemann-Hilbert problem: hypergeometric exampleA hypergeometric solution to the Riemann-Hilbert problem with three singular pointsat z = 0, 1, ∞ can be given by the following formulasz β F(α, β, ν|z)−z 1+β C(α, β, ν)F(α, 1 + β, ν|z)φ(z) =,−z 1−β C(α, −β, ν)F(α, 1 − β, ν|z)z −β F(α, −β, ν|z)z −β F(−α, −β, −ν|z)z 1+β C(α, β, ν)F(α, 1 + β, −ν|z)−1φ (z) = 1−βz C(α, −β, ν)F(α, 1 − β, −ν|z)z β F(α, β, −ν|z) α+β−ν where we have introduced F(α, β, ν|z) = 2 F1 −α+β−ν,z for a standard hyper2β(−α−β+ν)(α−β+ν)geometric function and the constant C(α, β, ν) =.2β(2β+1)These formulas give solution to the linear system (3.88) with the residues in thefollowing conjugacy classes:A0 ∼ θ 0 = σ = diag(β, −β),A∞ ∼ θ ∞ = θ = diag(α, −α)A1 ∼ θ 1 = ν = diag(2ν, 0)58(3.118)3.4.

Vertex operators and Riemann-Hilbert problemAccording to (3.86), (3.87)hθ|Vν ψ̃α (z)ψβ (w)|σi =[φ(z)φ(w)−1 ]αβz−w(3.119)It means, for example, that in order to study the matrix elements with ψ1 , ψ̃1 oneneeds to consider the functionb 11 (z, w) = z −β wβ [φ(z)φ(w)−1 ]11 = F(α, −β, −ν|z)F(α, β, ν|w)−KQ(α + 0 β + ν)−,0 =±14β 2 (4β 2 − 1)(3.120)zwF(α, 1 − β, −ν|z)F(α, 1 + β, ν|w)b 11 (z, z) = 1 becomes a non-trivial bilinear relationAlready the simplest fact, that Kfor the hypergeometric function. However, our claim is much stronger: this functionis almost as nice as (3.40) since its expansion (3.91) is given byb 11 (z, w)1K=−z−wz−w∞X2β(α − β + ν)a+1 (−α + β + ν)a+1 (−α + β − ν)b+1 (α + β − ν)b+1 b a−zw(a+b+1)(−α+β−ν)(α+β−ν)a!b!(2β)b+1 (−2β)a+1a,b=0K(z, w)11 =(3.121)and it is indeed a generation function of the matrix elements we are interested in.One can substitute here a = q − 12 , b = p − 12=h(α, −α)|V(2ν,0) (1)ψ1,−p ψ̃1,−q |(β, −βi =2β(α + β − ν)q+ 1 (−α + β − ν)q+ 1 (α − β + ν)p+ 1 (−α − β + ν)p+ 1222(p + q)(−α + β − ν)(α + β − ν)(p −1)!(q2−2(3.122)1)!(2β)q+ 1 (−2β)p+ 1222and compare this formula with (3.48)pY (θβ0 − θ1 )p+ 1 Yθβ − θ1f1,1 (θ, θ 0 , p)f2,1 (θ, θ 0 , q)12q=×1p+q(p − 2 )! βθβ0 − θ1 β6=1 (θβ − θ1 )p+ 12pY (θ1 − θβ0 )q+ 1 Yθ1 − θβ112q××=1p+q(q − 2 )! βθ1 − θ0 β6=1 (θ1 − θβ )q+ 1β=(θ1 − θ2 )(−θ1 +(3.123)2θ10 )p+ 1 (−θ12(p + q)(−θ1 + θ10 )(θ1 − θ20 )(p+ θ20 )p+ 1 (θ1 − θ10 )q+ 1 (θ1 − θ20 )q+ 1222− 12 )!(q − 21 )!(θ2 − θ1 )p+ 1 (θ2 − θ1 )q+ 122It is easy to see, that after the appropriate identificationθ1 = β, θ2 = −β, θ10 = α + ν, θ20 = −α + ν(3.124)the r.h.s.’s in two last formulas coincide exactly.In addition to the hypergeometric case another explicit example can be providedby the exact conformal blocks, considered in [GMtw].

We are planning to consider itin detail elsewhere.593. Free fermions, W-algebras and isomonodromic deformationsIsomonodromic tau-functions and Fredholm determinantsIsomonodromic tau-functionFirst we need to prove the simpleLemma 3.4. Monodromies of ψβ (w) and ψ̃α (z) in the matrix elementshY 0 , n0 , θ|Vν (1)ψ̃ασ (z)ψβσ (w)|Y , n, σi(3.125)do not depend on n, Y , n0 , Y 0 .Proof: All these matrix elements can be obtained from (3.95) by certain contour integration, producing fermionic modes from the fermionic fields. However, in(3.95) due to the Wick theorem factorization, all contributions have the factorizedform Kαγ (z, •) × .