We associate with the regionD the Hilbert space H = L2(∂D, | dz|), which, by virtue of the form ofD =.
a∈ADa∪.
b∈BDb, is the direct sums of several copies of L2(S1). We also need the Cauchy boundary operators
C±: H → H±, C±[ f](z) =
"
∂D
f(ζ ) dζ
(ζ− z±)2iπ, (B.13) which projects onto the space H+ of analytic functions on D. Remember that D is dis-connected, so an analytic function onD is really collection of analytic functions on each
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of theDc’s. Quite opposite instead, sinceD−is connected, an analytic function in H−is a “single” analytic function. The following simple remark will be useful in the sequel.
Remark B.2. If ∞ ∈ D− then none of the disks Dc is unbounded: then H+ consists of analytic functions in each of the (finite) disks and H− consists of analytic functions on the complement that tend to zero at infinity. Thus, the constant functions belong to H+. Conversely, if ∞ ∈ D+then one of the disksDc is a disk at infinity; then H+ con-sists of analytic functions that vanish at z = ∞ and H− of all analytic functions in complement C \ D. In particular, the constant functions belong to H−, contrary to the
previous case. !
Notation: In the sequel, we shall use H, H± to denote also the space of (row) vector functions or matrix-valued functions with entries in H, H±; the meaning (scalar or vector or matrix) should be clear within the context.
Consider the finite-dimensional vector subspaces of H−
V := C−[H+S−1], W := C−[H+S], (B.14)
where H+= {row vectors of analytic functions in D+} and S(z) := Γ (z)D(z) as before. It is clear that V and W are finite-dimensional vector spaces because in each diskDC we have S(z) = Γ (z)zLcc−Kc. Thus, both S(z) and S(z)−1 have at most finite-order poles and either the points b ∈ B or a ∈ A (respectively).
Remark B.3. If Γ is only a formal germ of analytic function, then the Cauchy projectors shall be understood as the projector onto the Laurent tail, and in that case C+:= Id + C−. We shall nevertheless continue using it as an integral operator for clarity.
For example, in all the cases discussed in Section 3, the contour Σ extends to
∞, but nonetheless Γ (z) admits a formal expansion at z = ∞ independent of the sector.
Therefore, Γ does define a formal germ of analytic function at ∞. !
Proposition B.3. Given V, W as in (B.14), consider the map
G : V −→ W, v6−→ C−[vS].
(B.15)
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Then the matrix R(z) = R(A B L K
)(z) exists (and thus is unique) if and only if G is invertible.
In this case, the inverse is
G−1: W −→ V, w6−→ C−
NwS−1R−1O R.
(B.16)
!
Proof. This is a classical result that we prove for the interest of the reader. First of all we need to verify that G indeed maps to W; to this end let v = C−[h+S−1] ∈ V; then
C−[vS] = C+[vS] − vS = C+[vS] − C−[h+S−1]S =
∈H+
K LI J
C+[C−[h+S−1]S] + h+−C+[h+S−1]S.
Taking the projection C−of both sides (and recalling that C−◦ C−= −C−) leaves the left-side invariant and thus
C−[vS] = C−[C+[h+S−1]S] ∈ C−[H+S] = W.
As for the claims proper, we prove the two implications below:
[⇐] Suppose that G : V → W is invertible and let L(z) the matrix G−1[C−[S]] (mean-ing, the result of adjoining all the rows of G−1[C−[etjS]]). Then we claim that R(z) = 1 − L(z) is the solution and R+(z) = R(z)S is a (formal) germ of analytic function in D+. To see it we need to verify that R+∈ H+(in the sense of matrices with entries in H+). But this is clear
C−[(1 − L)S] = C−[S] − C−[LS] = C−[S] − C−[S] = 0
since by assumption G(L) = C−[LS] = C−[S] ∈ V (row-wisely). We remark (leaving the ver-ification to the reader) that the above argument does not rely in the least on ∞ being on the “exterior” ofD. The only difference between the two cases is that the constant func-tions belong to H+ if ∞ is in D− and to H− if ∞ ∈ D+. We, however, still need to show that both R, R+are analytically invertible.
Note that, within Dc, det R+(z) = det R(z) det S(z) = det R(z) det Γ (z)zTr(Lc−Kc) and since Γ is (formally) analytic inD+and thanks to the zero index condition (B.1) then the number of zeroes (with multiplicities) of det R+inD+is the same as the number of zeroes of det R inD−; for this reason it is sufficient to show that det R(z) ̸= 0 for z ∈ D−. Suppose det R(z0)= 0 at some point z0∈ D−. Then there is a nonzero constant row-vector f such
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that fR(z) vanishes at z0. Thus, φ−(z) :=z−zf0R(z) is analytic at z = z0and thus belongs to 0). This concludes the proof of the sufficiency. Note that the uniqueness follows now from Proposition B.1.
[⇒] Now suppose that the matrix R = R(B A K L
) exists: we claim that the inverse is given by (B.16). Indeed, using C+− C−= Id and setting R+:= RS ∈ H+ we have (The statement R+= RS ∈ H+is by definition of R, second bullet in Definition B.1.)
G[C−[wR−1+ ]R] = C−N
Thus, G is inverse and the inverse is written by (B.16). "
As we shall promptly see, the matrix representation of G : V → W is precisely the characteristic matrix in Definition 2.2: to this end we fix appropriate bases in V, W
V = P
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The reasons for the shifts δa∞, δb∞in (B.19), (B.20) is the fact that if ∞ ∈ D+(i.e., it belongs to A ∪ B) then the constant functions belong to H−(see Remark B.2).
Proposition B.4. The matrix G representing G = C−[•S] : V → W in the bases (B.19), (B.20) is the characteristic matrix in Definition 2.2
G(b,ν,k);(a,µ,ℓ)= resz=ares
ζ=b
eTµΓ−1(z)Γ (ζ )eνdzdζ
(za)ℓ−δa∞(z − ζ )(ζb)kb,ν+1−k−δb∞. (B.21)
! Proof. The proof is a straightforward exercise using equation (B.22) of Lemma B.2 (immediately hereafter) and using that D(z) = z−Kb bwhen z ∈ Db. "
Lemma B.2. [1] Let w ∈ W = C−[H+S]; then its components in the basis (B.20) are given by
reszb=0w(z) · eνzk−δb b∞dzb
zb
, 1 ≤ k ≤ kb,ν. (B.22)
[2] Let v ∈ V = C−[H+S−1]; then its components in the basis (B.19) are given by
zresa=0v(z) · Γ (z)eµzℓa−δa∞dza
za
, 1 ≤ ℓ ≤ ℓa,µ. (B.23)
!
Proof. It is sufficient to verify resz=bwb′,ν′,k′(z) · eνzk−δb b∞dzb
zb = δbb′δkk′δν′ν res
z=ava′,µ′,ℓ′(z) · Γ (z)eµzℓa−δa∞dza
za = δaa′δℓℓ′δµ′µ. The verification of the first is absolutely straightforward and left to the reader. To verify the second, we observe that the quantity to be computed is the one below
resz=ares
ζ=a′
eTµ′Γ−1(ζ )
ζa′ℓ′−δa′∞(ζ− z)Γ (z)eµzℓa−δa∞dza
za
.
Clearly, this is zero whenever a ̸= a′because the result of the residue in ζ = a′is analytic at z = a (and vanishes at a if a = ∞). If a = a′
resz=ares
ζ=a
eTµ′Γ−1(ζ )
ζaℓ′−δa∞(ζ − z)Γ (z)eµzℓa−δa∞dza
za
= resz=ares
ζ=aeTµ′
(Γ−1(ζ )Γ (z) − 1)
ζaℓ′−δa∞(ζ− z) eµzℓa−δa∞dza
za + resz=ares
ζ=a
eTν′eµzℓa−δa∞
ζaℓ′−δa∞(ζ − z) dza
za
.
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The first term gives zero because the result of the residue in ζ is an analytic function in z (and vanishing at a = ∞ if a = ∞) and ℓ ≥ 1, and the second term clearly yields
δµ′µδℓℓ′. "
B.3.1 Proof of Theorem 2.1
The proof is simply the collection of Proposition B.3 (stating that the R(z) exists if and only if G is invertible), together with the fact that the characteristic matrix G(A B
L K ) is nothing else but the matrix representation of G in a special basis (Prop. B.4).
B.3.2 Variation of det G
We now assume that Γ (z) is allowed to vary smoothly (while remaining in formal space H+ of the regionD, of course) with respect to some additional manifold of parameters, which we shall denote generically by t. Then, clearly, the matrix G representing G in the bases (B.19), (B.20) also inherits a dependence on these t. In addition to that, the matrix G depends on the positions of the points A ∪ B: we shall denote this dependence generically with c.
We want to express the logarithmic derivative of det G(t, c) with respect any deformation (of Γ or of the positions c) in terms of an integral. Before proceeding we consider the general.
Lemma B.3. Consider the linear map G = C−[•S] of H−sending V to W, and fix bases (Or any other bases depending smoothly on the parameters.)(B.19), (B.20) {vj}, {ws} (we use generic indices j, s to label elements of the basis) in V, W (one can also use any other bases depending smoothly on the parameters). Let ∂ denote the derivative with respect to any of the deformation parameters the bases depend on; then the following maps define endomorphisms of V, W;
V∂[vj] = (G−1∂G)vj+ ∂vj−
-k
GkjG−1∂wk, (B.24)
W∂[ws] = ∂GG−1ws− ∂ws+ G
-j
G−1js∂vj. (B.25)
Moreover, if [Gij] is the matrix representative of G : V → W, then
∂ln det G = TrVV∂= TrWW∂. (B.26)
!
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Proof. Differentiating the relation Gvj=,
kGkjwk with respect to ∂, we obtain (we denote by a dot˙the derivative ∂)
Gv˙ j+ G ˙vj= For the other case, we multiply by G−1js both sides of (B.27) and sum over j to obtain
GG˙ −1ws+ G
Tracing this last over W on both sides gives the second statement. "
For simplicity we shall consider now separately the two types of deformations, starting with those of Γ ; we shall denote by ∂ any derivative (infinitesimal deformation or vector field on the parameter space t) of Γ .
Proposition B.5. Let ∂ denote any infinitesimal deformation of Γ (z, t) and let S = Γ (z, t)D(z) where D is as in (B.4). Consider G = C−[•S] as an operator mapping H− to
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Remark B.4. The expression TrG−1∂G would be the variation of the determinant of G if this operator had a determinant; in general, however, G−1∂G is a T ¨oeplitz operator on an infinite dimensional space and in particular not a trace-class perturbation of the
identity, and thus there is no notion of determinant. !
Proof of Proposition B.5. [1] Let h−∈ H−. Using Id = C+− C−we have
C−[C−[h−∂S]R+−1]R + C−[h−∂SS−1]
= −C−[h−∂S]R+−1R + C+[C−[h−∂S]R−1+ ]R − h−∂SS−1+ C+[h−∂SS−1]
= −C+[h−∂S]S−1+ C+[C−[h−∂S]R−1+ ]R + C+[h−∂SS−1].
Since the left side is in H−, we take the projection C− (which only affects it by a minus sign) and thus
C−[C−[h−∂S]R+−1]R + C−[h−∂SS−1] = C−[C+[h−∂S]S−1] − C−[C+[C−[h−∂S]R+−1]R]. (B.33)
The first term on the right side of (B.33) clearly belongs to V = C−[H+S−1] already; as for the second term, we know that R = 1 + L and that (the rows of) L ∈ C−[H+S−1] and then we have
C−[H+R] = C−[H+C−[H+S−1]] = C−N
−H+S−1+ H+C+[H+S−1]O
= C−[H+S−1].
Therefore, the trace is a finite-dimensional trace on V = C−[H+S−1].
[2] The operator in (B.31) is (using ((B.16)) C−N
C−[h−∂S]R−1+ O
R + C−N
h−∂SS−1O
= −C−N
h−∂SR+−1O
R +✭✭✭✭✭✭✭✭✭ C−N
C+[h−∂S]R+−1O
R + C−N
h−∂SS−1O
= −C−
Nh−∂SS−1R−1O R + C−
Nh−∂SS−1O .
Writing out the Cauchy projectors explicitly we find
−C−[h−∂SS−1R−1]R + C−[h−∂SS−1] =
"
∂D
dξ 2iπ
"
∂D
dζ 2iπ
h−(ζ )∂S(ζ )S−1(ζ )*
1 − R−1(ζ )R(ξ)+ (ζ− ξ−)(ξ− z−) .
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It is easily seen that the integration is independent of the order because the integrand is analytic at ξ = ζ . Moreover,
limζ→ξ
1 − R−1(ζ )R(ξ) ζ − ξ = limζ
→ξR−1(ζ )R′(ξ )= R−1(ζ )R′(ζ ).
It then follows that the trace is given by (B.32). "
Corollary B.1. Under the same assumptions and notation of Proposition B.5, we have
∂ln det G =
"
∂D
dz 2iπTr*
R−1R′∂SS−1+
=
"
∂D
dz 2iπTr*
R−1R′∂Γ Γ−1+
. !
Proof. We have seen in Proposition B.5 that
"
Tr(R−1R′∂SS−1) dz
2iπ =TrH−(G−1∂G + C−[•∂SS−1]) = TrH−(G−1∂G + M∂).
We have also seen in the same proposition that the trace is saturated on V = C−[H+S−1], namely
TrH−(G−1∂G + M∂)= TrV(G−1∂G + M∂).
The basis (B.20) in W is independent of the deformations of Γ ; on the other hand the basis of V (B.19) is obtained by choosing some vector h+(z) ∈ H+ independent of the deformation and projecting it like v = C−[h+S−1] (with different choices of h+(z) as dis-played in (B.19), but always independent of deformations of Γ ). Then
∂v= C−[h+∂S−1] = −C−[h+S−1∂SS−1]. (B.34)
Since ∂SS−1= ∂Γ Γ−1∈ H+then we can continue
(B.38) = −C−[(C+[h+S−1] − C−[h+S−1])∂SS−1] = C−[C−[h+S−1]∂SS−1] = M∂[v].
Thus,
"
Tr(R−1R′∂SS−1)dz 2iπ
P roposition B.5[2]
= TrH−(G−1∂G + M∂)P roposition B.5[1]
= TrV*
G−1∂G + M∂+
= TrV(V∂)Prop. B.3= TrCm*
G−1∂G+
= ∂ ln det G. "
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Remark B.5. The major considerations we have made apply equally well (without almost any change) if we replace the locally defined diagonal rational function D(z) by any locally defined (i.e., not necessarily diagonal) rational function Q(z). In particular,
• The finite dimensionality of V = C−[H+Q−1Γ−1], W = C−[H+ΓQ] is preserved.
• Assuming that the total index of det Q(z) around D is zero, also Proposi-tion B.1 (with the obvious modificaProposi-tion of (B.2)), ProposiProposi-tion B.3, Lemma B.3, Proposition B.5, Corollary B.1 remain valid. In particular in Lemma B.3 the choice of bases (B.19), (B.20) would need modification (depending on the cases), but the result is really of general nature and it is noncommittal as to the specific choice of bases.
The reason why this is interesting is because the process of “soliton insertion” in the literature is accomplished by means of choosing a Q(z) of the form 1 + N(z) with N(z) a triangular rational matrix. Only the part that follows after this remark (i.e., the depen-dence on the position of the poles) would require some more significant modification.!
We now turn our attention to the dependence of G(t, c) on the positions of the points in A, B. Suppose ∂ denotes the deformation of the position of a point b ∈ B; we note that we shall consider disjoint union of A, B. This means that if z = a = b is at the same time a pole and a zero of D(z) (clearly, in different entries along the diagonal!), we consider the deformation of the position of the pole independently from that of the position of the zero.
Theorem B.1. Under any deformation ∂ of Γ , and of the position of poles/zeroes of D, we have
∂ln det G =
"
∂DTr(R−1R′∂SS−1)dz 2iπ +
"
∂DTr(Γ−1Γ′∂D D−1)dz
2iπ, (B.35) where S(z; t, c) = Γ (z; t)D(z; c) and G = G(A B
L K
) is the characteristic matrix of
Proposi-tion B.4 (and thus of DefiniProposi-tion 2.2). !
This has the immediate corollary.
Corollary B.2. The anomaly of the Malgrange differential under Schlesinger transfor-mations is given by
Ω
=
∂; [M], CA B
L K D>
− Ω (∂; [M]) = ∂ ln det G(A B L K
). (B.36)
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The proof of the corollary is simply the comparison of Theorem B.2 and
Proposition B.2. !
Proof of Theorem B.2. The case of deformations of Γ with fixed poles (i.e., with ∂ D ≡ 0) was already addressed in Corollary B.1 and thus we only need to address a deformation of the positions of the poles/zeroes of D(z) fixing Γ (∂Γ ≡ 0). We separate the proof in two cases, according to which type of pole we are moving (i.e., pole b ∈ B of D or pole a ∈ A of D−1). We now need to compute the trace TrWW∂. Looking at (B.38), we observe that the last term, C−[∂h+ΓD], is a strictly triangular matrix and thus with zero trace; indeed
∂bh(b,µ,k)+ = kh(b,µ,k−1)+ . Therefore, the only term that contributes to the trace is the first in (B.38). Using (B.22) for the components along the chosen basis (B.20), the trace is com-puted below most a simple pole, hence
(B.44) =
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=
"
∂DTr(S−1R−1(R′S + RS′)∂bD D−1)dz 2iπ
=
"
∂DTr(R−1R′Γ ∂bD D−1Γ−1)dz 2iπ +
"
∂DTr(Γ−1Γ′∂bD D−1) dz 2iπ +
"
∂DTr(D−1D′∂bD D−1)dz 2iπ.
Note that the last term is zero because D−1D′∂bD D−1=zKbb2 for z ∈ Da and 0 in the other disks, and the residue vanishes. Thus, we have
Tr(∂bGG−1)=
"
∂DTr(R−1R′Γ ∂bD D−1Γ−1)dz 2iπ +
"
∂DTr(Γ−1Γ′∂bD D−1)dz
2iπ. (B.41)
Motion of poles a ∈ A of D−1; this time the basis of W is independent of the defor-mation, and thus we shall use Proposition B.3 in the form of (B.24); we then need to find the action of V∂. The basis (B.19) is obtained by projection of elements of H+ of the form h(a,µ,ℓ)+ = eTµzaLa−ℓEµµ so C−[h(a,µ,ℓ)+ S−1] = eTµC−[z−ℓa Γ−1(z)] and 0 < ℓ ≤ (La)µµ= ℓa,µ. Clearly, the number of such vectors is TrLa. For such a v = C−[h+S−1] (we drop the super-scripts) we have
V∂= G−1∂G[v] + ∂v = G−1∂G[v] − C−[h+S−1∂SS−1] + C−[∂h+S−1] (B.42)
= G−1∂G[v] + M∂[v] − C−[C+[h+S−1]Γ ∂ D D−1Γ−1] + C−[∂h+S−1]. (B.43)
Looking at (B.43), we observe that the last term, C−[∂h+D−1Γ−1] is a strictly triangular matrix and thus with zero trace; indeed ∂ah(a,µ,ℓ)+ = ℓh(+a,µ,ℓ−1).
Therefore, the only the first three terms in (B.43) contribute to the trace; by Proposition B.5 of these, the first two are precisely (B.32) and thus we need only to eval-uate the trace of the third term. Note that the component of any vector v ∈ V along the element va,µ,ℓis computed by (B.23) and thus the trace is computed
Tr(∂GG−1)Lemma B.3= TrV(B.46)
= TrV(G−1∂aG + M∂a)− -N µ=1
ℓa,µ
-ℓ=1
zresa=0C−[C+[h(a,µ,ℓ)+ S−1]Γ ∂aD D−1Γ−1]Γ (z)eµzaℓdza
za
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Proposition B.5
Likewise before, in the sum (B.44) only the term ℓ = 1 is nonzero and since then the poles involved in the residue are all simple that term is computed as follows:
− res
This completes the proof. "
The Theorem B.2 and Corollary B.2 are the main ingredient of the proof of Theorem 2.2; however, in Theorem 2.2 the diagonal matrix is globally defined and not piecewise. We need to see which additional term arises in Theorem B.2 if D(z) is replaced by a globally defined diagonal rational function with the same zeroes/poles.
We shall now denote by Dloc(z) the diagonal matrix that was denoted by D thus far.
Theorem B.2. Let R(z) = R(A B L K
)(z) be Schlesinger transformation of Definition B.1. Let
˜S(z) = Γ (z)D(z) and D(z) the diagonal rational function D(z) =!
c∈B∪AzcLc−Kc. If ∂ denotes any deformation of Γ or of the zeroes/poles of D, then
∂ln det G =
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Proof. To be clear, D(z) and Dlocare related as follows (χSdenotes the indicator function
so that, within a given disk Dc0 the matrix T(z) is analytic and given by T(z) =
!
c;c̸=c0zLcc−Kc, z ∈ Dc0. Let ∂ be a derivative of the position of any zero/pole of D(z) (so
∂Γ ≡ 0), then (When differentiating Dloc(z) w.r.t. the position of a pole, we understand it for fixedD; for example if b is a point of B then Dloc−1∂bDloc=z−bKbχDb.)
where S(z) = Γ (z)Dloc(z), namely, it has exactly the same meaning as in Theorem B.2.
Use now the fact that R+(z) := R(z)S(z) is analytic; then plug R(z) = R+(z)S−1(z) = R+(z)Dloc−1(z)Γ−1(z) into (B.47) and simplify (using the cyclicity of the trace) to obtain
"
Therefore, the two relative τ differentials differs by the explicit term
" It is easy to see that the right side of (B.48) is expressed as the sum of the residues of
Tr(D′D−1∂D D−1). "
We thus can finally prove Theorem 2.2.
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Proof of Theorem 2.2
[1] This point is obvious from the very definition of R and simple computations.
[2]We compute the integrand in ω(∂; [ ˜M]). To this end we use the following identities
˜
Γ−1Γ˜′= D−1Γ−1R−1R′ΓD + D−1Γ−1Γ′D + D−1D′,
∂(D−1MD)D−1M−1D = D−1∂MM−1D − ∂ DD−1+ D−1MD−1∂DM−1D, Γ−∂MM−1Γ−−1= ∂Γ+Γ+−1− ∂Γ−Γ−−1.
(B.49)
Therefore, plugging in and simplifying, using the cyclicity of the trace several times and (B.49), we find (below ∆ denotes the jump operator ∆( f) = f+− f−):
Tr[ ˜Γ−−1Γ˜−′∂(D−1MD)D−1M−1D]
= Tr[Γ−−1Γ−′∂MM−1+ R−1R′∆*
∂(ΓD)D−1Γ−1+
+ ∆(Γ−1Γ′∂D D−1)+
− M−1M′∂D D−1+ D−1D′(∂MM−1− ∂ DD−1+ MD−1∂DM−1)].
Now, if we have#
Σ∆F2iπdz and F has some poles outside of Σ then this reduces, by the Cauchy theorem, to the sum of the residues of F . This concludes the proof of the second part.
[3] From Theorem B.3 we have
-c
resz=cTr(R−1R′∂(ΓD)D−1Γ−1)+ resz=cTr(Γ−1Γ′∂D D−1)
= ∂ ln det G −
-c
resz=cTr(D−1D′D−1∂D) dz
The Vandermonde-like expression is simply the explicit evaluation of the last term. Note that none of the terms has residue at ∞, which explains the exclusion in the final for-mula. This concludes the proof.
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