Диссертация (1137342), страница 18

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Let us accordingly write V (C) = V+ (C) ⊕ V− (C) and denote byΠ± (C) the projectors on the corresponding subspaces. Their explicit form is˛f (z 0 ) dz 01,Π± (C) f (z) =2πi C± ,|z0 |=|z|±0 z 0 − zwhere the subscript of C± indicates the orientation of C. Projectors Π± (C) are simpleinstances of Plemelj operators to be extensively used below.Let us next associate to every trinion T [k] with k = 2, . . . , n − 3 the spaces ofvector-valued functionsM [k][k][k]H[k] =H,+ ⊕ H,− ,H,± = CN ⊗ V± C[k] .=in,outWith respect to the first decomposition, it is convenient to write the elements f [k] ∈H[k] as!![k][k]ffin,−in,+f [k] =⊕.[k][k]fout,+fout,−[k]Here f,± denote N -column vectors which represent the restrictions of analytic and[k]principal part of f [k] to boundary circle C . Now define an operator P [k] : H[k] → H[k]by−1˛[k][k]Ψ+ (z) Ψ+ (z 0 ) f [k] (z 0 ) dz 01[k] [k].(4.19)P f (z) =[k]2πi Cin[k] ∪ Coutz − z0794.

Fredholm determinant and Nekrasov sum representations of isomonodromic tau functions[k]The singular factors 1/ (z − z 0 ) for z, z 0 ∈ Cin,out are interpreted with the followingprescription: the contour of integration is deformed to appropriate annulus (e.g. Ak−1[k][k]for Cin and Ak for Cout ) as to avoid the pole at z 0 = z. Matrix function Ψ[k] (z) is asolution of the 3-point RHP described in the previous subsection. Its normalizationis irrelevant as the corresponding factor cancels out in (4.19).Lemma 4.4. We have P [k]can be explicitly written as![k]fin,−P [k] :[k][k]![k]fin,−7→[k]![k]⊕fout,+fout,−fout,+[k]= P [k] and ker P [k] = Hin,+ ⊕ Hout,− . Moreover, P [k]fin,+⊕[k]2a[k] b[k]c[k] d[k]![k]fin,−[k]!,fout,+where the operators a[k] , b[k] , c[k] , d[k] are defined by1a[k] g (z) =2πi1b g (z) =2πi[k]˛˛[k]Cout˛Ψ+ (z) Ψ+ (z 0 ) g (z 0 ) dz 0,z − z0[k][k]Cin˛[k](4.20b)[k](4.20c)−1[k]z−[k](4.20a)z ∈ Cin ,Ψ+ (z) Ψ+ (z 0 ) g (z 0 ) dz 0,z − z0hi−1[k][k]0Ψ+ (z) Ψ+ (z ) − 1 g (z 0 ) dz 0Cout[k]z ∈ Cin ,,−1[k][k]1d g (z) =2πi[k]z − z0[k]Cin1c g (z) =2πi[k]hi−1[k][k]Ψ+ (z) Ψ+ (z 0 ) − 1 g (z 0 ) dz 0[k]z0z ∈ Cout ,[k]z ∈ Cout .

(4.20d),[k]Proof. Let us first prove that Hin,+ , Hout,− ⊂ ker P [k] . This statement follows from[k][k][k]the fact that Ψ+ holomorphically extends inside Cin and outside Cout , so that theintegration contours can be shrunk to 0 and ∞. To prove the projection property,decompose for example[k]P [k] fout,+1=2πi˛(z) =outh[k][k]−1Ψ+ (z) Ψ+ (z 0 )i[k]− 1 fout,+ (z 0 ) dz 0z − z0[k]Cout1+2πi˛[k][k]Cout ,|z 0 |>|z|fout,+ (z 0 ) dz 0.z − z0[k]The first integral admits holomorphic continuation in z outside Cout thanks to nonsingular integral kernel, and leads to (4.20d), whereas the second term is obviously[k][k]equal to fout,+ .

The action of P [k] on fin,− is computed in a similar fashion.The leftmost and rightmost trinions T [1] and T [n−2] play somewhat distinguishedrole. Let us assign to them boundary spaces[1][1][n−2]H[1] := Hout,+ ⊕ Hout,− ,[n−2]H[n−2] := Hin,+ ⊕ Hin,− ,804.2. Tau functions as Fredholm determinantsand the operators P [k] : H[k] → H[k] with k = 1, n − 2 defined by[1] [1]P fP[n−2] [n−2]f1(z) =2πi1(z) =2πi˛[1][1]Cout−1[1]Ψ+ (z) Ψ+ (z 0 ) f [1] (z 0 ) dz 0,z − z0˛[n−2][n−2]Ψ+(z) Ψ+[n−2]Cin−1(z 0 ) f [n−2] (z 0 ) dz 0.z − z0Analogously to the above, one can show that[1][1][1][1]P [1] : fout,+ ⊕ fout,− 7→ fout,+ ⊕ d[1] fout,+ ,[n−2][n−2][n−2][n−2]P [n−2] : fin,− ⊕ fin,+ 7→ fin,− ⊕ a[n−2] fin,− ,where the operators d[1] , a[n−2] are given by the same formulae (4.20a), (4.20d).

Note[1][n−2]in particular that P [1] and P [n−2] are projections along their kernels Hout,− and Hin,+ .Let us next introduce the total spaceH :=n−2MH[k] .k=1It admits a splitting that will play an important role below. Namely,H = H+ ⊕ H− ,[1][2][2][n−3][n−3][n−2]H± := Hout,± ⊕ Hin,∓ ⊕ Hout,± ⊕ . . . ⊕ Hin,∓ ⊕ Hout,± ⊕ Hin,∓ .(4.21)Combine the 3-point projections P [k] into an operator P⊕ : H → H given by thedirect sumP⊕ = P [1] ⊕ . . . ⊕ P [n−2] .Clearly, we haveLemma 4.5. P⊕2 = P⊕ and ker P⊕ = H− .Another important operator PΣ : H → H is defined using the solution Ψ̂ (z) (defined by (4.17)) of the n-point RHP in a way similar to construction of the projection(4.19):1PΣ f (z) =2πi˛CΣ−1Ψ̂+ (z) Ψ̂+ (z 0 ) f (z 0 ) dz 0,z − z0CΣ :=n−3[[k][k+1]Cout ∪ Cin.(4.22)k=1We use the same prescription for the contours: whenever it is necessary to interpretthe singular factor 1/ (z − z 0 ), the contour of integration goes clockwise around thepole.Let HA be the space of boundary values on CΣ of functions holomorphic on A =Sn−3k=1 Ak .Lemma 4.6.

PΣ2 = PΣ and HA ⊆ ker PΣ .814. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functions[k][k+1]Proof. Given f ∈ HA , the integration contours Cout and Cinin (4.22) can bemerged thanks to the absence of singularities inside Ak , which proves the secondstatement. To show the projection property, it suffices to notice thatPΣ2 f [k]1(z) =(2πi)2‹CΣ−1Ψ̂+ (z) Ψ̂+ (z 00 ) f [k] (z 00 ) dz 0 dz 00.(z − z 0 ) (z 0 − z 00 )Because of the ordering of contours prescribed above, the only obstacle to merging[k][k]Cout and Cin in the integral with respect to z 0 is the pole at z 0 = z.

The result followsby residue computation.Lemma 4.7. PΣ P⊕ = P⊕ and P⊕ PΣ =PΣ .Proof. Similar to the proof of Lemma 4.6. Use that Ψ̂−1 Ψ[k] has no jumps on Γ[k]to compute by residues the intermediate integrals in PΣ P⊕ and P⊕ PΣ .The above suggests to introduce the notationHT := im P⊕ = im PΣ .(4.23)The space HT ⊂ H can be thought of as the subspace ofSfunctions on the union of[k][k][k]with monodromyboundary circles Cin , Cout that can be continued inside n−2k=1 Tand singular behavior of the n-point fundamental matrix solution Φ (z). The onlyexception is the regular singularity at ∞ where the growth is slower.The structure of elements of HT is described by Lemma 4.4.

Varying the positionsof singular points, one obtains a trajectory of HT in the infinite-dimensional Grassmannian Gr (H) defined with respect to the splitting H = H+ ⊕ H− . Note that eachof the subspaces H± may be identified with N (n − 3) copies of the space L2 (S 1 ) offunctions on a circle; the factor n − 3 corresponds to the number of annuli and N isthe rank of the appropriate RHP.We can also writeH = HT ⊕ H − .(4.24)The operator P⊕ introduced above gives the projection on HT along H− . Similarly,the operator PΣ is a projection on HT along ker PΣ ⊇ HA . We would like to express[k][k]it in terms of 3-point projectors. To this end let us regard fin,− , fout,+ as coordinateson HT . Suppose that f ∈ H can be decomposed as f = g + h with g ∈ HT andh ∈ HA .

The latter condition means that[k][k+1]k = 1, . . . , n − 3,hout,± = hin,± ,which can be equivalently written as a system of equations for components of g:[k][k−1][k−1][k][k+1][k+1][k][k−1]gin,− − c[k−1] gin,− − d[k−1] gout,+ = fin,− − fout,− ,[k][k+1]gout,+ − a[k+1] gin,− − b[k+1] gout,+ = fout,+ − fin,+ ,[1][n−2](4.25)where gin,− = 0, gout,+ = 0. The first and second equations are valid in sufficiently824.2. Tau functions as Fredholm determinants[k][k]narrow annuli containing Cin and Cout , respectively. Define!![k][k+1][k]f−fgout,+out,+in,+g̃k =,f˜k =,[k+1][k+1][k]gin,−fin,− − fout,−0 a[k+1]Uk =,k = 1, . .

. , n − 3,d[k]0 [k+1]00b0Vk =,Wk =,k = 1, . . . , n − 4,000 c[k+1]U1 V1 0.0f˜1g̃1 W1 U2 V2.0f˜2 g̃2 ~=,0WU..K=~g=,f.23. ..  .. ....Vn−4 g̃n−3f˜n−300. Wn−4 Un−3.(4.26)The system (4.25) can then be rewritten in a block-tridiagonal form(1 − K) ~g = f~.(4.27)The decomposition H = HT ⊕ HA thus uniquely exists provided that 1 − K is invertible.Let us prove a converse result and interpret K in a more invariant way. Considerthe operators P⊕,+ : H+ → HT and PΣ,+ : H+ → HT defined as restrictions of P⊕and PΣ to H+ . The first of them is invertible, with the inverse given by the projectionon H+ along H− . Hence one can consider the composition L ∈ End (H+ ) defined byL := P⊕,+ −1 PΣ,+ .(4.28)We are now going to make an important assumption which is expected to hold generically (more precisely, outside the Malgrange divisor). It will soon become clear thatit is satisfied at least in a sufficiently small finite polydisk D ⊂ Cn−3 in the variablesa1 , .

. . , an−3 , centered at the origin.Assumption 4.8. PΣ,+ is invertible.Proposition 4.9. For g ∈ H+ , let g̃k and ~g be defined by (4.26). In these coordinates,L−1 = 1 − K.Proof. Rewrite the equation L−1 f 0 = f as P⊕,+ f 0 = PΣ,+ f . Setting f = P⊕,+ f 0 +h,the latter equation becomes equivalent to PΣ h = 0. The solution thus reduces toconstructing h ∈ HA such that (h + P⊕,+ f 0 )− = 0, where the projection is taken withrespect to the splitting H = H+ ⊕ H− . This can be achieved by setting[k][k+1][k+1][k][k+1][k]hout,+ = hin,+ = − (P⊕,+ f 0 )in,+ ,hout,− = hin,− = − (P⊕,+ f 0 )out,− .It then follows that f = f 0 + h+ = (1 − K) f 0 .834. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functionsTau functionDefinition 4.10.

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