Диссертация (Isomonodromic deformations and quantum field theory), страница 7

Therefore one can say thatMg4 (θ 1 , θ 2 ; σ; θ 3 , θ 4 )/H = Mg3 (θ 1 , θ 2 , −σ) × Mg3 (σ, θ 3 , θ 4 ) .(2.23)The torus action is free, so locally it looks as a product (actually it is true evenglobally because the fibration (M1 , M2 , M3 ) 7→ (M1 , M2 , M3 )/G is trivial: we cangive an algebraic parametrization for one representative from each conjugacy class).Therefore we have the equality for the open subsets (denoted by ≈):Mg4 (θ 1 , θ 2 ; σ; θ 3 , θ 4 ) ≈ Mg3 (θ 1 , θ 2 , −σ) × H × Mg3 (σ, θ 3 , θ 4 ) .(2.24)The above considerations suggest the following choice of coordinates on Mg4 :• Gluing parameters σ: rank g items.• Invariant functions on Mg3 × Mg3 (for example, tr M1 M2−1 , tr M3−1 M4 ).

Theyare invariant with respect to the action of “relative twists”: we have 2 dim Mg3such functions.• Relative twist parameters, which change under the twist (for example, tr M2 M3−1 ,tr M2−1 M3 ), rank g items. These coordinates will be denoted by β ∈ h.This procedure is schematically depicted in Fig.4.6 for the sl3 case, where dim M3sl3 =2, dim Msl4 3 = 8. The coordinates on each copy of Msl3 3 are denoted by µ, ν.

Theindices {1, 2, 3, 4} of the matrices are replaced by {0, t, 1, ∞}Figure 2.1: Coordinates on Msl4 3 : eight = two σ’s + two β’s + µ0t + ν0t + µ1∞ + ν1∞Pants decomposition for MgnSuppose that the coordinates on Mgn−1 are chosen via the pants decomposition. Splitthe matrices into two groups and defineSn−3 = M1 . . . Mn−2 ,−1Mgn = {(M1 , . . . , Mn−2 , Sn−3), (Sn−3 , Mn−1 , Mn )}/G == {(M1 , . . . , Mn−2 , e−2πiσ ), (e2πiσ , Mn−1 , Mn )}/H ≈ Mgn−1 × H × Mg3 .(2.25)Iteratively repeating this procedure, one is led to the following choice of coordinateson Mgn :212. Isomonodromic τ -functions and WN conformal blocks• (n − 3) rank g gluing parameters σ i ,• (n − 3) rank g relative twist parameters β i ,•n−2Pdim Mg3 (σ i−1 , θ i+1 , −σ i ) 3-point moduli of flat connections (here we identifyi=1σ 0 = θ 1 and σ n−2 = −θ n ).Anticipating the result, let us mention that these coordinates are convenient fromthe CFT point of view: σ i will parametrize intermediate charges in the conformalblock and β i will be the Fourier transformation parameters.

This description wasshown to be valid in the sl2 case [GIL12], [ILTe] and was recently demonstrated tohold for slN case with dim Mg3 = 0 [GavIL]. From a more conceptual point of view,this decomposition illustrates that all extra parameters in the τ -function expansioncome from the 3-point functions.Iterative solution of the Schlesinger systemIn order to study the generic Schlesinger system, let us follow the approach proposedin the original paper of M. Jimbo [Jimbo] and in [SMJ, part 2].Let us take the 4-point Schlesinger system and fix the singularities to be 0, t, 1, ∞.The system becomest∂t A0 = [At , A0 ] ,t[At , A1 ] ,t∂t A1 =t−111[At , A1 ] .∂t At = − [At , A0 ] −tt−1(2.26)Fixing the integral of motion A∞ = −A0 − At − A1 , one obtainst∂t A0 = [A0 , A1 + A∞ ] ,t∂t A1 = t(1 − t)−1 [A0 + A∞ , A1 ] .(2.27)The isomonodromic τ -function is defined by∂t log τ =11tr At A0 +tr At A1 .tt−1(2.28)Let us study the solution of the system (2.27) for the case when A1 (t) is finite inthe limit t → 0: A1 (t) = A1 (0) + O(t>0 ).

Under this assumption we havet∂t A0 (t) = [A0 , A∞ + A1 (0) + O(t>0 )] .If the last term were absent, then the solution would be A0 = t−A∞ −A1 (0) Ã0 tA∞ +A1 (0) .Therefore it is natural to introduceB = −A1 (0) − A∞ = lim(A0 (t) + At (t)) ,t→0−BÃ0 (t) = t22A0 (t)tB ,(2.29)2.3. Iterative solution of the Schlesinger systemwhere Ã0 (t) has a well-defined limit as t → 0.

We see that in view of its definitionB describes the total monodromy around 0 and t in the limit t → 0. Since thedeformation is isomonodromic, this monodromy is constant and is given by M0 Mt =M0t ∼ e2πiB . This allows to make the identificationB = σ.(2.30)Our system then becomest∂t Ã0 (t) = [Ã0 (t), t−σ (A1 (t) − A1 (0))tσ ] ,t∂t A1 = t(1 − t)−1 [tσ Ã0 (t)t−σ + A∞ , A1 (t)] .(2.31)Here we see an operator tadσ , which produces some fractional powers of t. It isconvenient to impose the condition that (σ, σ) 1, or at least that for all roots αone has |(σ, α)| < 12 . This allows to organize the terms of the expansion in powers oft according to their order of magnitude in the asymptotic behavior.

If necessary, onecan perform an analytic continuation of the solution from the region with small σ.We know that in the Lie algebra the operator tadσ acts bytσ Eα t−σ = t(σ,α) Eα ,tσ Hα t−σ = Hα ,(2.32)where α ∈ g∗ is a root and Eα , Hα are the elements of the Cartan-Weyl basis. Let usdefine a grading on the space of monomialsdeg[tk+(σ,w) ] = (k, w) ,where w ∈ Qg is an element of the root lattice Qg =rankgLZαi of g.

It is useful toi=1define a filtrationQ0g ⊂ Q1g ⊂ Q2g ⊂ . . . ⊂ Qg(2.33)on this root lattice, which is recursively constructed as follows: Q0g = {0}, Q1g is theset of all roots of g and 0, andQi+1= {x + y|x ∈ Qig , y ∈ Q1g } = Q1g + . . . + Q1g .gAlso define the double filtration Vn,m on the space of all fractional-power series:tk+(σ,w) ∈ Vn,m ⇔ (k ≥ n) ∧ (w ∈ Qmg ),Vn+1,m ⊂ Vn,m ,Vn,m ⊂ Vn,m+1 .(2.34)Each term of the filtration is generated by these monomials.

This definition turns outto be useful because of the propertiest· : Vn,m → Vn+1,m ,tadσ : Vn,m → Vn,m+1 ,Vn1 ,m1 · Vn2 ,m2 → Vn1 +n2 ,m1 +m2 .23(2.35)2. Isomonodromic τ -functions and WN conformal blocksOne can also see that the degrees present in Vn+1,m+k are larger then in Vn,m if σis sufficiently small (∀α ∈ Q1g : |(σ, α)| < k1 ). We also define a slightly ambiguousnotation Vn,w by(2.36)tk+(σ,w) ∈ Vn,w ⇔ (k ≥ n) .Now we have all the ingredients that are necessary for the construction of aniterative solution of the system (2.31). Our initial data will be given by the triple ofmatrices σ, Ã0 (0) and A1 (0). Symbolically, the system (2.31) can be written asÃ0 (t) = F0 (Ã0 (t), A1 (t)) ,A1 (t) = F1 (Ã0 (t), A1 (t)) ,(2.37)where “affine” bilinear (in the sense f (x, y) = axy + bx + cy + d) functions F0 , F1have the following properties:F0 : Vn0 ,m0 × V0,0 → 0 ,F0 : Vn0 ,m0 × Vn1 ,m1 → Vn0 +n1 ,m0 +m1 +1 ⊂ Vn0 +n1 ,∞ ,F1 : Vn0 ,m0 × Vn1 ,m1 → Vn0 +n1 +1,m0 +m1 +1 + Vn1 +1,m1 ⊂ Vn1 +1,∞ .(2.38)Let us substitute into (2.37) the expressionsÃ0 (t) = Ã0 (0) +A1 (t) = A1 (0) +kt∞Xk=1∞Xk=1kk kÃ0 (t), t A1 (t) ∈tk Ãk0 (t) ,tk Ak1 (t) ,(2.39)Vk,∞ .From (2.38) we immediately see that (2.31) takes the form≤kÃk0 (t) = f0k (Ã<k0 (t), A1 (t)) ,<kAk1 (t) = f1k (Ã<k0 (t), A1 (t)) .(2.40)Because of the ≤ sign our strategy of solving will be to compute first Ak1 (t), and thensubsequently determine Ãk0 (t).

One can also write down explicit formulas for bilinearsf1k and f0k , which are immediate (though cumbersome) consequences of the system(2.31).Now let us determine which powers (k, w) will be actually present in the solution.This will be done again iteratively, using only the properties (2.38):• Taking Ã0 (0) ∈ V0,0 and A1 (0) ∈ V0,0 , and computing F1 , we get an element ofV1,1 , thereforeA1 ∈ V0,0 + V1,1 + . . .• Take Ã0 (0) ∈ V0,0 and A1 ∈ V0,0 + V1,1 + . .

., then Ã0 ∈ V0,0 + V1,2 + . . .• For Ã0 ∈ V0,0 + V1,2 + . . . and A1 ∈ V0,0 + V1,1 + . . . one finds A1 ∈ V0,0 + V1,1 +V2,3 + . . .242.3. Iterative solution of the Schlesinger system• Setting Ã0 ∈ V0,0 + V1,2 + . . . and A1 ∈ V0,0 + V1,1 + V2,3 + . . . yields Ã0 ∈V0,0 + V1,2 + V2,4 . . .• ...Continuing this procedure one finds the following structureÃ0 (t) ∈∞XVk,2k ,k=0A1 (t) ∈ V0,0 +∞X(2.41)Vk,2k−1 .k=1It is easy to check that these spaces are stable under the action of (F0 , F1 ) describedby the rules (2.38). This is somewhat similar to the statement that the cone is stableunder the addition operation.Indeed, let us try to find an element of Ã0 (t) lying in Vk,2k+1 .

For this one wouldneed n0 + n1 ≤ k, m0 + m1 ≥ 2k, so m0 + m1 ≥ 2(n0 + n1 ). Since m1 ≤ 2n1 − 1 forn1 6= 0 (when F0 vanishes) and m0 ≤ 2n0 , such an element cannot exist. Similarly,for A1 , let us take an element lying in Vk,2k . One then needs to satisfy the constraintsn1 ≤ k − 1, m1 ≥ 2k (impossible) or n0 + n1 + 1 ≤ k and m0 + m1 + 1 ≥ 2k, whichimplies m0 + m1 ≥ 2n0 + 2n1 + 1. But m1 ≤ 2n1 and m0 ≤ 2n0 , therefore one cannotget such an element neither.Now let us compute the τ -function and try to understand in which elements ofthe filtration does it lie.

Since we havet∂t log τ (t) = − tr [t−σ (A1 + A∞ )tσ Ã0 + Ã20 ] + t(1 − t)−1 tr [(A1 + A∞ + tσ Ã0 t−σ )A1 ] ,(2.42)naively it could be a term in V0,1 . However, computing the constant part one findst∂t log τ (t) = tr (B Ã0 − Ã20 ) + . . . = tr (At A0 ) + . . . =111= tr (At + A0 )2 − tr A20 − tr A2t + .

. . =222111= (σ, σ) − (θ 0 , θ 0 ) − (θ t , θ t ) + . . . ,222(2.43)where Aν ∼ θ ν . For convenience, let us introduce the notation111χ = (σ, σ) − (θ 0 , θ 0 ) − (θ t , θ t )222(2.44)The terms present in tr (t−σ A1 (t)tσ Ã0 (t)) that are closest to the boundary originate∞Pfrom the constant part of A1 (t).