Диссертация (1137342), страница 25

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We haveZ̃ ∗bif (ν|Y 0 , Y ) =Y(−ν)q0∗ + 1iiQ×Qi,ji,j2Yν −1 (ν)q∗ + 1iiQ0∗ν − qi − p∗j Qi,j0∗∗qi − q j − νi,jY(−ν)p∗ + 1i2Yiν −1 (ν)p0∗ + 1 ×i22i∗ν + p0∗i + qj∗p0∗i − pj + ν,whereno{qi∗ } = (L − 1/2) , (q1 − 1) , . . . , (qd−1 − 1) , (q^−1),dnog ,{p∗i } = (p1 + 1) , . . . , (pd + 1) , 1/2 ^0{qi0∗ } = (L − 1/2) , (q10 − 1) , . . . , qd0 0 −1 − 1 , (q^d0 − 1) , 0g0g{p0∗i } = (p1 + 1) , . . . , (pd0 + 1) , 1/2 ,and d, d0 denote the number of boxes on the main diagonals of Y, Y 0 . The abovenotation means that one has either to simultaneously include or not to include thecoordinates tilded in the same way.

These numbers are included in the case when bothof them are positive (it implies that qd 6= 21 or qd0 0 6= 12 ). Fig. 4.15 below illustrates thedifference between these two cases.We may now consider one by one four possible options, namely: i) qd 6= 12 , qd0 0 6= 12 ;ii) qd = qd0 0 = 12 ; iii) qd 6= 12 , qd0 0 = 12 ; iv) qd = 12 , qd0 0 6= 21 .

For instance, for qd 6= 12 ,1144.5. Relation to Nekrasov functionsFigure 4.15: Possible mutual configurations of main diagonals of Y , Y ∗ ; qd = 12 (left) and qd 6= 12 (right).qd0 0 6=12after massive cancellations one obtainsd0dd0 d YYYY1Z̃ ∗bif111−10=(−ν)Lν (ν)L(−ν)1×ν + pi +−ν + pi +22−ν + qi0 − 12ν + qi − 21Z̃ bifi=1i=1i=1i=1QQQQ111100(ν − L) (ν + L)i ν − qi + 2i ν − 2 + qii ν + pi + L + 2i ν − L − 2 − piQ 0Q 0 1Q×Q=111ν2i L + 2 − qi − νi qi − L − 2 − νi pi + 2 + νi − 2 − pi + νQQ(ν − L − 21 − pi ) i (ν + p0i + L + 12 )Z∗Q 0= bif ,= (1 − ν)L (1 + ν)L Qi11Z bifi (L + 2 − qi − ν)i (qi − L − 2 − ν)where the first line of the first equality corresponds to the ratio of diagonal parts andthe second to non-diagonal ones. The proof in the other three cases is analogous.

Corollary 4.36.Z bif = Z̃ bif for arbitrary Y, Y 0 ∈ Y iff Z bif = ±Z ∗bif for diagrams1with {qi } = 2 , . . . , L − 21 (that is, for the diagrams containing a large square on theleft).Lemma 4.37. The equality Z bif = Z̃ bif holds for given diagrams Y, Y 0 ∈ Y with alarge square iff it holds for the diagrams with a large square and one deleted box.Proof. Suppose that we have added one box to the ith row of Y 0 . The only boxesj =2Figure 4.16: A pair of Young diagrams (red andgreen) with a large square (black) and added box(blue square).whose contribution to Z bif depends on the added box lie on its left in the diagram Y 0and above it in the diagram Y , see Fig. 4.16.

The contribution from the boxes on theleft (green circles) was initially given byZ leftbif = Q(ν)p0 +L+ 1i20j≥ĵ (pi − pj + ν) · (ν)ĵ−i+1115,4. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functionswhere ĵ = min [{j|pj + j ≤ p0i + i + 1} ∪ {L}] (notice that we can move ĵ in the range0where pQj +j = pi +i+1). The contribution from the boxes on the top (red circles) wastopZ bif = j<ĵ (−ν + pj − p0i − 1). After addition of one box (blue square) it transformsQ0into Z ∗topbif =j<ĵ (−ν + pj − pi ), whereas the previous part becomesZ ∗leftbif = Q(ν)p0 +L+ 3i0j≥ĵ (pi2− pj + 1 + ν) · (ν)ĵ−i+1.The ratio of the transformed and initial functions is then given byQQ 0Q00p0i + L + 21 + νp0i + L + 12Z ∗bifj<ĵ (pi − pj + ν)j (pi − pj + ν)j≥ĵ (pi − pj + ν)QQQ==.000Z bifj (pi − pj + 1 + ν)j≥ĵ (pi − pj + 1 + ν)j<ĵ (−ν + pj − pi − 1)On the other hand, the ratio Z̃ ∗bif /Z̃ bif is easier to compute since the addition ofone box to the ith row of Y 0 simply shiftsp0i 7→ p0i + 1.

From (4.111) 1 one coordinate,and the large square condition {qi } = 2 , . . . , L − 21 it follows thatQ 0p0i + 12 + L + νZ̃ ∗bif1Z ∗bifj (pi − pj + ν)0Q= pi + + ν × 0 1=,02Z bifpi + 2 + νZ̃ bifj (pi − pj + 1 + ν)which finishes the proof.Using two inductive procedures described above, any pair of diagrams Y, Y 0 ∈ Ycan be reduced to equal squares, in which case the statement of Theorem 4.34 can bechecked directly.Step 3Let us move to the third part of our plan and proveTheorem 4.38. Z bif ν Q0 , Y 0 ; Q, Y = C ν Q0 − Q Z bif ν + Q0 − QY 0 , Y .Proof.

It is useful to start by computing Z bif for the “vacuum state”n1 31o(α)Q, ,...,Q −, qα = ∅for Q(α) > 0,pα = pα :=2 22n1 31o(α)Q, , . . . , −Q −pα = ∅, qα = qα :=for Q(α) < 0.2 22One obtainsQ0QQ0 Q 0YYYYZ̃ bif ν pQ , ∅; pQ , ∅ = (−1)Q(Q+1)/2ν −1 (ν)i(−ν)i(ν + i − j)−1 =i=1i=1i=1 j=1Q0Q(Q+1)/2= (−1)QQY Γ (ν + i) YΓ (i − ν) Y Γ (ν − j + 1)=Γ (ν + 1) i=1 Γ (−ν) j=1 Γ (ν − j + Q0 + 1)i=1G (1 + ν + Q0 ) G (1 − ν + Q)(−1)Q(Q+1)/2G (ν + 1) G (ν + Q0 + 1 − Q)=0G (1 + ν)G (1 − ν) Γ (ν + 1)Q Γ (−ν)Q G (ν + 1 − Q) G (ν + Q0 + 1)G (1 + ν + Q0 − Q)Q(Q+1)/2 G (1 − ν + Q)= (−1).0G (1 + ν − Q) G (1 − ν) Γ (1 + ν)Q Γ (−ν)Q=1164.5.

Relation to Nekrasov functionsUsing the recurrence relationG (1 − ν) π QG (1 − ν + Q)= (−1)Q(Q−1)/2,G (1 + ν − Q)G (1 + ν) sin πνand the reflection formula Γ (−ν) Γ (1 + ν) = − sinππν , the last expression can be rewritten as 0G (1 + ν + Q0 − Q)C (ν|Q0 − Q) := Z̃ bif ν pQ , ∅; pQ , ∅ =.0G (1 + ν) Γ (1 + ν)Q −QNext let us rewrite the expression for Z̃ bif (ν|Y 0 , Q0 ; Y, Q) for charged Young diagrams in terms of uncharged ones. To do this, we will try to understand how thisexpression changes under the following transformation, shifting in particular all particle/hole coordinates associated to Y 0 :p0i 7→ p0i + 1,qi0 7→ qi0 − 1,ν 7→ ν − 1.It should also be specified that if we had q 0 = 21 , then this value should be droppedfrom the new set of hole coordinates; if not, we should add a new particle at p0 = 12 .Looking at Fig.

4.12, one may understand that this transformation is exactly the shiftQ0 7→ Q0 + 1 preserving the form of the Young diagram.Now compute what happens with Z̃ bif (ν|Y 0 , Q0 ; Y, Q). One should distinguish twocases:1. If there is no hole at q 0 =12in (Y 0 , Q0 ), then it follows from (4.111) thatZ̃ bif (ν − 1|Q0 + 1, Y 0 ; Q, Y )=Z̃ bif (ν|Q0 , Y 0 ; Q, Y )Q1YY −ν + pi + 1ν−+qν000i22= Qi× ν |p | ν −|q | = ν Q −Q .11νν + qi − 2 ii ν − 2 − pii2.

Similarly, if there is a hole at q 0 =12to be removed, thenZ̃ bif (ν − 1|Q0 + 1, Y 0 ; Q, Y )=Z̃ bif (ν|Q0 , Y 0 ; Q, Y )Q1Y ν − pi − 1 Yν0i ν − 2 + qi−1|p0 | −|q0 |+12= ν Q −Q .=ν Q×νν11νν − 2 + qij ν − 2 − pjiiThe computation of the shift of Q is absolutely analogous thanks to the symmetryproperties of Z̃ bif .IntroducingZ bif (ν|Q0 , Y 0 ; Q, Y )Z̃ ?bif (ν|Q0 , Y 0 ; Q, Y ) =,C (ν|Q0 − Q)it is now straightforward to check thatZ̃ ?bif (ν − 1|Q0 + 1, Y 0 ; Q, Y )Z̃ ?bif (ν + 1|Q0 , Y 0 ; Q + 1, Y )== 1,Z̃ ?bif (ν|Q0 , Y 0 ; Q, Y )Z̃ ?bif (ν|Q0 , Y 0 ; Q, Y )1174. Fredholm determinant and Nekrasov sum representations of isomonodromic tau functionsand therefore Z̃ ?bif (ν|Q0 , Y 0 ; Q, Y ) = Z̃ ?bif (ν + Q0 − Q|0, Y 0 ; 0, Y ).

Finally, combiningthis recurrence relation with C (ν|0) = 1, one obtains the identityZ̃ bif (ν|Q0 , Y 0 ; Q, Y )= Z̃ bif (ν + Q0 − Q|Y 0 , Y ) ,0C (ν|Q − Q)which is equivalent to the statement of the theorem.Step 4At this point, we have already shown that~00~ 0 ~0,mZY~Y ,m(T ) = ±ei(δ1 η+ −δ1 η− )m +i(δ1 η+ −δ1 η− )m ẐY~Y ,Q,~Q (T ) .000It remainsto check the signs in the reference limit described above.

Note that sgn Ẑ = 1, since sgn (C (ν|Q0 , Q)) = 1 and sgn (Z bif (ν|Y 0 , Y )) = 1 as ν → ∞.Everywhere in this subsection the calculations are done modulo 2.First let us compute the sign of the non-diagonal part of Z. To do this, one hasto fix the ordering asxI : p0+ + σk−1 , p0− − σk−1 , −q+ − θk + σk , −q− − θk − σk ,00y I : − q++ σk−1 , −q−− σk−1 , p+ − θk + σk , p− − θk − σk .The variables in each of these groups are ordered as p1 , p2 , . . . where p1 > p2 > . . .This giveslsgn Z|non−diag = |p0− | · |q0+ | + |q+ | · |q0+ | + |q0− | + |p+ | + |q− | · |q0+ | + |q0− | + |p+ | + |p− | ++|q+ | (|q+ | − 1) |q− | (|q− | − 1) |p+ | (|p+ | − 1) |p− | (|p− | − 1)++++2222+|q0+ | · (|q0− | + |p+ | + |p− |) + |q0− | · (|p+ | + |p− |) + |p+ | · |p− |.Using the charge balance conditions|p+ | − |q+ | = |q− | − |p− | = m,|p0+ | − |q0+ | = |q0− | − |p0− | = m0 ,the above expression can be simplified tolsgn Z|non−diag = m + m0 + m|p+ | + m0 |p0+ | + |p+ | + |p− |.Next compute the sign of the diagonal part, X|p0 | − |q0+ | + |q+ | − |p− |0lsgn Z|diag =(p0− + q++ q+ + p− ) + −.2Combining these two expressions, after some simplification we getlsgn (Z) = X X XX111100q+ ++q+ ++p− ++p− ++m.2222|p+ |·|q+ |+|p0+ |·|q0+ |+1184.5.

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