Диссертация (1137342), страница 23
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The proof of three other identities issimilar.The Cauchy determinant in (4.70) remains invariant upon simultaneous translation[k][k]of all xı and y by the same amount. Let us use this to replace the notation (4.69)in the case N = 2 by(p + σk−1 ,ı ≡ (p, ) ∈ Ik−1 ,x[k](4.85a)ı :=−p − θk + σk ,ı ≡ (−p, ) ∈ Jk ,(−q + σk−1 , ≡ (−q, ) ∈ Jk−1 ,y[k] :=(4.85b)q − θk + σk , ≡ (q, ) ∈ Ik .Define a notation for the chargesmk := |( · , +) ∈ Ik |−|( · , +) ∈ Jk | = |( · , −) ∈ Jk |−|( · , −) ∈ Ik | ,k = 1, . . .
, n−3,and combine them into a vector m := (m1 , . . . , mn−3 ) ∈ Zn−3 . We will also writeσ := (σ1 , . . . , σn−3 ) ∈ Cn−3 and further define[k]η := (η1 , . . . , ηn−3 ) ,eiηk :=d∞,−[k].(4.86)d∞,+The parameters η provide the remaining n − 3 local coordinates on the space MΘ ofmonodromy data. The main result of this section may now be formulated as follows.1044.4. Rank two caseTheorem 4.30. The isomonodromic tau function of the Garnier system admits thefollowing multivariate combinatorial expansion:τGarnier (a) = const ·−θ2a1 0n−3Y−θ2ak kk=1X×eim·ηm∈Zn−3X−2θk θlak×1−al1≤k<l≤n−22n−3Y Y~ ,mY ak (σk +mk ) +|Y~k | n−2k−1[k]T,Z Y~k−1k ,mkak+12 k=1k=1Y~1 ,...,Y~n−3 ∈YY(4.87)where Y~k stands for the pair of chargeddiagrams associated to (Ik , Jk ), the total Youngnumber of boxes in Y~k is denoted by Y~k , and~Y,mk−1Z Y~k−1,mkkQ=T [k] =Y(θk + σk−1 + 0 σk )p+ 1 Y2p − 21 ! (2σk−1 )p+ 1 (−p,)∈J(p,)∈Ik−12Q×Y2k−10(1 − θk − σk−1 + 0 σk )p− 12×p − 12 ! (1 − 2σk−1 )p− 10 =±0Q(θk + σk−1 − σk )p+ 1 Y0 =± (1 − θk + σk−1 + σk )p− 212×1! (1 + 2σk )p− 1p − 12 ! (−2σk )p+ 1p−2(p,)∈I22kY[k][k][k][k]xı − xy − yı0 =±(−p,)∈JkY×Q0 =±ı,∈Ik−1 tJk ;ı<ı,∈Jk−1 tIk ;ı<YY[k]x[k]ı − y.ı∈Ik−1 tJk ∈Jk−1 tIk(4.88)Proof.
Consider the product in thePfirst line of (4.70). The balance conditions |Ik | =k−1I,Jk−1|Jk | imply that the factors such as e j=0 θk in (4.84) cancel out from Z Ik−1T [k] .k ,JkThe factors of the form ± also compensate each other in the product of elemen[k] ±1tary determinants in (4.65). The factors d∞, in (4.84c) and (4.84d) produce theexponential eim·η in (4.87).The total power in which the coordinate ak appears in (4.70) is equal toXXXX2mk σk − 2mk−1 σk−1 −p−p+p+p=(p,)∈Ik−1(−p,)∈Jk−1(−p,)∈Jk(p,)∈Ik = 2mk σk + m2k + Y~k − 2mk−1 σk−1 + m2k−1 + Y~k−1 .The last equality is demonstrated graphically in Fig. 4.13. The prefactor in the firstline of (4.87) comes from two sources: i) the shifts of (initially traceless) Garniermonodromy exponents Θk by −θk 1 making one of their eigenvalues equal to 0 and ii)the prefactor Υ (a) from Theorem 4.11.In the Appendix, we show that the formula (4.88) can be rewritten in terms ofNekrasov functions.
In the Painlevé VI case (n = 4), this transforms Theorem 4.30into Theorem 4.2 of the Introduction.1054. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functionsQ>0Q<0Figure 4.13: A charged Maya diagram m and the associatedpartition Y (m) for positive and negative charges Q (m).
Giventhe positions p (m) = (p1 , . . . , pr ) and q (m) = (−q1 , . . . , −qs ) ofparticles and holes, the red and green areas represent the sums2PPPPpk andqk . We clearly havepk + qk = Q(m)+ |Y (m) |2in both cases.Hypergeometric kernelRecall that the matrices Θ0 , . . . , Θn−1 are by convention diagonal with eigenvaluesdistinct modulo non-zero integers.
However, all of the results of Section 4.2 remainvalid if the diagonal parts corresponding to the degenerate eigenvalues are replacedby appropriate Jordan blocks.In this subsection we will consider in more detail a specific example of this typeby revisiting the 4-point tau function.
We will thus follow the notational conventionsof Subsection 4.2.5. Fix n = 4, N = 2 and assume furthermore that the monodromyrepresentation ρ[L] : π1 (P1 \ {0, t, ∞}) → GL (2, C) associated to the internal trinionT [L] is reducible, whereas its counterpart ρ[R] : π1 (P1 \ {0, 1, ∞}) → GL (2, C) for theexternal trinion T [R] remains generic. For instance, one may set000 0Θ0 = S =,Θt =,0 −2σ1 0so that the monodromy matrices M0 , Mt can be assumed to have the lower triangularform1010M0 =,Mt =,M0 Mt = e2πiS .−2πiκe−2πiσ e−4πiσ2πiκe2πiσ 1(4.89)[L]The solution Ψ (z) of the appropriate internal 3-point RHP may be constructedfrom the fundamental solution of a Fuchsian system002σ ,%t∂z Φ[L] = Φ[L] (4.90)−z (z − t)zwith a suitably chosen value of the parameter %.
Taking into account the diagonalmonodromy around ∞, such a solution Φ[L] (z) on C\R≥0 can be written as1Φ[L] (z) = %t (−z)−2σ−1l2σ1 + 2σ0tz(−z)−2σ1 = C̃0 % (−z)−2σl−1−2σ2σ0zt(−z)−2σ,(4.91)1064.4. Rank two casewhere la (z) := 2 F11 + a, 1 ; z , and the modified connection matrix C̃ is lower02+atriangular:10.C̃0 = π%t−2σ−1sin 2πσThe monodromy matrix around 0 is clearly equal to M0 = C̃0 e2πiΘ0 C̃0−1 .
This allowsto relate the monodromy parameter κ to the coefficient ρ of the Fuchsian system(4.90) asκ = %t−2σ .(4.92)The 3-point RHP solution Ψ[L] (z) inside the annulus A is thus explicitly given by1010.%tΦ[L] (z) = (4.93)Ψ[L] (z) =−l2σ zt 10 (−z)2σA(2σ + 1) zThis formula leads to substantial simplifications in the Fredholm determinant representation (4.48) of the tau function τJMU (t). It follows from from the structure of(4.93) and (4.49b) that% zt l2σ zt − zt0 l2σ zt00d−+ (z, z ) =(4.94)1 + 2σz − z0is the only non-zero element of the 2 × 2 matrix integral kernel d (z, z 0 ) (note thatthe lower indices here are color and should not be confused with half-integer Fouriermodes).
This in turn implies that the only entry of a (z, z 0 ) contributing to thedeterminant is[R][R][R][R]Ψ+− (z) Ψ++ (z 0 ) − Ψ++ (z) Ψ+− (z 0 )1a+− (z, z ) =.det Ψ[R] (z 0 )z − z00(4.95)Therefore, (4.48) reduces toτJMU (t) = det (1 − a+− d−+ ) .(4.96)The action of the operators a+− , d−+ involves integration along a circle C ⊂ A.The kernel a+− (z, z 0 ) extends to a function holomorphic in both arguments insideC. Therefore in the computation of contributions of different exterior powers to thedeterminant one may try to shrink all integration contours to the branch cut B :=[0, t] ⊂ R. The latter comes from two branch points 0, t of d−+ (z, z 0 ) defined by(4.94).Lemma 4.31.
Let |<σ| < 12 . For m ∈ Z≥0 , denote Xm = Tr (a+− d−+ )m . We haveXm = Tr KFm ,where KF denotes an integral operator on L2 (B) with the kernelσKF (z, z 0 ) = −κ (zz 0 ) a+− (z, z 0 ) .107(4.97)4. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functionsProof. Let us denote by Bup and Bdown the upper and lower edge of the branch cutB. After shrinking of the integration contours in the multiple integral Ik to B, theoperators a+− , d−+ should be interpreted as acting on W = L2 (Bup ) ⊕ L2 (Bdown )instead of L2 (C). Here L2 (Bup,down ) arise as appropriate completions of spaces ofboundary values of functions holomorphic inside DC \B, where DC denotes the diskbounded by C.
The space W can be decomposed as W = W+ ⊕ W− , where theelements of W+ are continuous across the branch cut, whereas the elements of W−have opposite signs on its two sides:W± = {f ∈ W : f (z + i0) = ±f (z − i0) , z ∈ B} .We will denote by pr± the projections on W± along W∓ .Since a+− (z, z 0 ) is holomorphic in z, z 0 inside C, it follows that im a+− ⊆ W+ ⊆ker a+− . Therefore Xk remains unchanged if a+− is replaced by pr+ ◦ a+− ◦ pr− . Thisis in turn equivalent to replacing d−+ by pr− ◦ d−+ ◦ pr+ . Given f = g ⊕ g ∈ W+ withg ∈ L2 (B), the action of d−+ on f is given byˆ t1[d−+ (z, z 0 − i0) − d−+ (z, z 0 + i0)] g (z 0 ) dz 0 =(d−+ f ) (z) =2πi 0ˆ tttl2σ z0 +i0− l2σ z0 −i0%t=g (z 0 ) dz 0 .2πi (1 + 2σ) 0z 0 (z − z 0 )An importantconsequence of the lower triangular monodromy is that the jump ofl2σ zt0 on B yields an elementary function, cf (4.91):l2σt0z + i0− l2σt0z − i0 0 2σ+1z= −2πi (2σ + 1).tSubstituting this jump back into the previous formula and using (4.92), one obtainsˆ t 02σz g (z 0 ) dz 0,z ∈ DC \B.(d−+ f ) (z) = κz0 − z0Next we have to compute the projection pr− of this expression onto W− .
Writepr− ◦ d−+ f = h ⊕ (−h), with h ∈ L2 (B). Thenh (z) =1[(d−+ f ) (z + i0) − (d−+ f ) (z − i0)] = πiκz 2σ g (z) ,2z ∈ B.Finally, write a+− ◦ pr− ◦ d−+ f as g̃ ⊕ g̃ ∈ W+ . It follows from the previous expressionfor h (z) thatˆ t2σg̃ (z) = −κa+− (z, z 0 ) z 0 g (z 0 ) dz 0 ,z ∈ B.0The minus sign in front of the integral is related to orientation of the contour C inthe definition of a.
We have thereby computed the action of a+− ◦ pr− ◦ d−+ on W+ .Raising this operator to an arbitrary power k ∈ Z≥0 and symmetrizing the factorsz 0 2σ under the trace immediately yields the statement of the lemma.1084.4. Rank two caseTheorem 4.32. Given complex parameters θ1 , θ∞ , σ satisfying previous genericityassumptions, letσ + θ1 + θ∞ , σ + θ1 − θ∞θ1σϕ (x) := x (1 − x) 2 F1;x ,(4.98a)2σ1 + σ + θ1 + θ∞ , 1 + σ + θ1 − θ∞θ11+σ;x .(4.98b)ψ (x) := x(1 − x) 2 F12 + 2σDefine the continuous 2 F1 kernel byK̃F (x, y) :=ψ (x) ϕ (y) − ϕ (x) ψ (y),x−yand consider Fredholm determinant D (t) := det 1 − λK̃F (0,t) ,λ ∈ C.(4.99)(4.100)Then D (t) is a tau function of the Painlevé VI equation with parametersθ~ = (θ0 = σ, θt = 0, θ1 , θ∞ ).
The conjugacy class of monodromy representation for theassociated 4-point Fuchsian system is generated by the matrices (4.89) ande−2πiθ1M1 =i sin 2πσM∞cos 2πθ∞ − e−2πiσ cos 2πθ1s−1 e−2πiσ [cos 2πθ∞ − cos 2π (θ1 − σ)],se2πiσ [cos 2π (θ1 + σ) − cos 2πθ∞ ]e2πiσ cos 2πθ1 − cos 2πθ∞(4.101a)e−2πiθ∞cos 2πθ1 − e−2πiσ cos 2πθ∞s−1 [cos 2π (θ1 − σ) − cos 2πθ∞ ]== M1−1 e−2πiS .s [cos 2πθ∞ − cos 2π (θ1 + σ)]e2πiσ cos 2πθ∞ − cos 2πθ1i sin 2πσ(4.101b)where2(θ1 + σ)2 − θ∞,λ=κ2σ (2σ + 1)Γ (1 − 2σ) Γ (θ1 + σ + θ∞ ) Γ (θ1 + σ − θ∞ )s=−.Γ (1 + 2σ) Γ (θ1 − σ + θ∞ ) Γ (θ1 − σ − θ∞ )(4.102)(4.103)Proof. To prove that D (t) is a Painlevé VI tau function with λ and κ related by(4.102), it suffices to combine the determinant representation (4.96) with Lemma 4.31,and substitute into the formula (4.95) for a+− (z, z 0 ) explicit hypergeometric expressions (4.76).−1The formula (4.101b) follows from M∞ = C∞ e2πiΘ∞ C∞, where C∞ is obtainedfrom the connection matrix (4.81) by replacements (θk , σk−1 , σk ) → (θ1 , σ, −θ∞ ).
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