The Malgrange Form and Fredholm Determinants

We consider the factorization problem of matrix symbols relative to a closed contour, i.e., a Riemann-Hilbert problem, where the symbol depends analytically on parameters. We show how to define a function $\tau$ which is locally analytic on the space of deformations and that is expressed as a Fredholm determinant of an operator of"integrable"type in the sense of Its-Izergin-Korepin-Slavnov. The construction is not unique and the non-uniqueness highlights the fact that the tau function is really the section of a line bundle.


Introduction
We shall consider the following prototypical matrix Riemann-Hilbert problem (RHP) on the unit circle Σ (or any smooth closed simple contour): Γ + (z; t) = Γ − (z; t)M (z; t), ∀ z ∈ Σ, Γ(∞) = 1. (1.1) We shall denote by D ± the interior (+) and the exterior (−) regions separated by Σ. Define H + to be the space of functions that are in L 2 (Σ, |dz|) and extend to analytic functions in the interior. We will use the notation H + = H + ⊗ C r (i.e., vector-valued such functions). The vectors will be thought of as row-vectors. We also introduce the Cauchy projectors C ± : L 2 (Σ, |dz|) → H ± : It is well known [14] that is invertible, in which case the inverse is given by Moreover the operator is Fredholm and dim ker(T S ) − dim coker(T S ) = ind Σ det M = 0.
There is no reasonable way, however, to define a "determinant" of T S as it stands. Such a function of t would desirably have the property that the RHP (1.1) is not solvable if and only if this putative determinant is zero.
While this is notoriously impossible in this naive form, we now propose a proxy for the notion of determinant, in terms of a simple Fredholm determinant.
The Malgrange one-form. As Malgrange explains [15] and this can be written as the following integral Here, and below, δ denotes the exterior total differentiation in the deformation space S: The Malgrange form is a logarithmic form in the sense that it has only simple poles on a co-dimension 1 analytic submanifold of the deformation space S and with positive integer Poincaré residue along it; this manifold is precisely the exceptional "divisor" (Θ) ⊂ S (the Malgrange divisor) where the RHP (1.1) becomes non solvable, i.e., where some partial indices of the Birkhoff factorization become non-zero.
Closely related to (1.2) is the following one-form, which we still name after Malgrange: It is also a logarithmic form with the same pole-divisor; indeed one verifies that which is an analytic form of the deformation parameters t ∈ S. In [2] the one-form (1.3) was posited as an object of interest for general Riemann-Hilbert problems (not necessarily on closed contours) and its exterior derivative computed (with an important correction in [3], which is however irrelevant in the present context). It was computed (but the computation can be traced back to Malgrange himself in this case) that It appears from (1.4) that this two form δω M is not only closed, but also smooth on the whole of S, including the Malgrange-divisor. As such, it defines a line-bundle L over S by the usual construction: one covers S by appropriate open sets U α where δω M = δθ α ; on the overlap U α ∩ U β the form θ α − θ β is also exact and one defines then the transition functions by g αβ (t) = exp θ α − θ β . Then a section of this line bundle is provided by the collection of functions τ α : U α → C such that Since ω M is a logarithmic form and each θ α is analytic in the respective U α , the functions τ α (t) have zero of finite order precisely on the Malgrange divisor (Θ) ⊂ S (under appropriate transversality assumptions, the order of the zero is the dimension of Ker T S ). Our goal is to provide an explicit construction of the τ α 's in terms of Fredholm determinants of simple operators of the Its-Izergin-Korepin-Slavnov "integrable" type [13]. Their definition is recalled in due time.

Construction of the Fredholm determinants
The construction carried out below is not unique, and also only local in the deformation space S; this is however not only not a problem, but rather an interesting feature, as we will illustrate in the case of SL 2 (C). The non-uniqueness is precisely a consequence of the fact that we are trying to compute a section of the aforementioned line bundle.
Preparatory step. The assumption that M (z; t) ∈ GL n can be replaced without loss of generality with M (z; t) ∈ SL n ; this is so because of the assumption on the index of det M . Indeed we can solve the scalar problem y + (z) = y − (z) det M (z), y(∞) = 1 and then define a new RHP where Γ ± (z) := Γ ± diag y −1 ± , 1, . . . and hence the new matrix jump for Γ is M (z) = diag(y − (z), 1, . . . )M (z) diag y −1 + (z), 1, . . . with det M ≡ 1. For this reason, from here on we assume M ∈ SL n (C).
We define an elementary matrix (for our purposes) to be a matrix of the form 1 + cE jk , with j = k, where E jk denotes the (j, k)-unit matrix. Proof . We recall that given any matrix M ∈ SL n , there is a permutation Π of the columns such that the principal minors (the determinants of the top left square submatrices) do not vanish, and hence we can write it as where L, U are lower/upper triangular matrices with unit on the diagonal and D = diag(x 1 , . . . , x n ) is a diagonal matrix (see for example [8, Vol. 1, Chapter II]). Denote q = det M j,k j,k≤ the principal minors; these are the nested minors of the original matrix M alluded to in the statement. The matrices L, U are rational in the entries of M and with denominators that are monomials in the q 's. Now, both L, U can clearly be written as products of elementary matrices whose coefficients are polynomials in the entries of L, U (respectively) and so it remains to show that we can write D as product of elementary matrices.
To this end we observe the 'LULU' identity (there is a similar 'ULUL' identity) x 0 0 1 As D ∈ SL n , we can represent it in terms of product of embedded SL 2 matrices using the root decomposition of SL n : and then embed the LULU identity for each factor. Finally, also permutation matrices can be written as product of elementary matrices embedding appropriately the simple identity; This concludes the proof.
Let now M (z) be an SL n matrix valued function, analytic in a tubular neighbourhood N (Σ) of Σ, and let q (z), = 1, . . . , n − 1 be the nested minors alluded to in the Lemma so that they are not identically zero. Since the entries are analytic in N (Σ) we can slightly deform the contour Σ to a contour Σ that avoids all zeroes of every principal minor q (z). The resulting RHP is "equivalent" to the original in the sense that the solvability of one implies the solvability of the other. Note also that this deformation can be done in a piecewise constant way locally with respect to t ∈ S. Thus we have Corresponding to this factorization we can define an equivalent RHP with jumps on R contours Σ 1 , . . . , Σ R , with Σ 1 = Σ and Σ j+1 in the interior of Σ j and all of them in the joint domain of analyticity of the scalar functions a ν (z) (see Fig. 1) which may have poles only at the zeroes of the principal minors q (z) of M (z). This is accomplished by "extending" the matrix Γ − (z) to the annular regions By doing so we obtain the following relations The piecewise analytic matrix function Θ(z) whose restriction to D ν coincides with the matrices Θ ν (2.1), satisfies a final RHP This type of RHP is of the general type of "integrable kernels" and its solvability can be determined by computing the Fredholm determinant of an integral operator of L 2 ( Σ ν , |dz|) R ν=1 L 2 (Σ ν , |dz|) with kernel (we use the same symbol for the operator and its kernel) where χ ν (z) is the projector (indicator function) on the component L 2 (Σ ν , |dz|). Indeed, as explained in [4,10,13], the Fredholm determinant det(Id −K) is zero if and only if the RHP (2.2) is non-solvable and moreover the resolvent operator R = K(Id −K) −1 of K has kernel

The SL 2 case
We would like to express the Malgrange one-form directly in terms of the τ function (Fredholm determinant). Rather than obscuring the simple idea with the general case, we consider in detail the SL 2 case. Let M (z; t) be analytic in (z, t) ∈ N (Σ) × S and with values in SL 2 (C). Using the general scheme above, we have the following factorizations .

Fredholm determinants for dif ferent factorizations
By using the relationship (2.1) between Γ − (extended to an analytic function on Ext(Σ)∪N (Σ)), we can re-express it in terms of the Malgrange one-form of the original problem (1.1); we use the fact that we can deform the contours back to Σ = Σ 1 by Cauchy's theorem. Plugging (2.1) appropriately in (3.2) and using Leibnitz rule, after a short computation we obtain if the factorization has only three term, then we set F 4 ≡ 1).
The cases (3,4,5). The last cases (3.1) (3,4,5) lead essentially to a RHP with a triangular jump; it suffices to re-define Γ by Γ 0 1 −1 0 for z ∈ Int(Σ). If the index of b is zero, ind Σ b = 0,then the solution can be written explicitly in closed form and it is interesting to compute the Fredholm determinant associated to our factorization of the matrix. We will show Proposition 3.1. In the cases (3.1) (3,4,5) and under the additional assumption that ind Σ b = 0, the τ function given by det(Id L 2 (∪Σ j ) −K) and K as in (2.3), equals the constant in the strong Szegö formula, given by [1,5] where now the Toeplitz operator is for the scalar symbol b(z) and H + is the Hardy space of scalar functions analytic in D + and β j are the coefficients of ln b(z) in the Laurent expansion centered at the origin The same applies in the case that the jump is triangular (b ≡ 0 and/or c ≡ 0) under the assumption ind Σ a = 0 and replacing b with a in the above formulas.
Proof . From a direct computation of the term θ (ρ) in (3.3), we find and the solution of the RHP is explicit, Thus we can write explicitly the Malgrange form: In this integral, z is integrated on a slightly "larger" contour Σ − . The Fredholm determinant satisfies δ ln τ (3,4,5) = ω M + θ (3,4,5) = 2 We now show that Indeed, the exchange of order of integration of one of the addenda (and relabeling the variables) yields the other term plus the residue on the diagonal, Therefore, in conclusion, we have (we fix the overall constant of τ by requiring it to be 1 for b ≡ 1) If we write a Laurent expansion of ln b (3.4) the formula (3.5) gives the explicit expression which is also the formula for the second Szegö limit theorem for the limit of the Toeplitz determinants of the symbol b(z), and it is known to to be the Fredholm determinant of an operator [1,5] τ = det where now the Toeplitz operator is for the scalar symbol b(z) and H + is the Hardy space of scalar functions analytic in D + .
Remark 3.2. The index assumption is only necessary for the case (3.1) (5) or (3.1) (1) when b ≡ 0 ≡ c, because in these situations the RHP separates into two scalar problems. However, the assumption ind Σ b = 0 for cases (3,4) or ind Σ a = 0 for the triangular case is not necessary as we now show (in the latter form). Consider the RHP By defining we are lead to the RHP in standard form which is case (1) with c ≡ 0 (or essentially cases (3,4) up to a multiplication by piecewise constant matrices). Even if a(z) = z n and thus ind Σ a = n, this problem is still generically solvable in terms of appropriate "orthogonal polynomials" {p (z)} ∈N defined by the "orthogonality" property Σ p (z)p k (z)µ(z)dz = h δ k .
Then the solution of the Y -problem is written as and the solvability depends only on the condition [6] det Σ z +j−2 µ(z)dz n j, =1 = 0.

(3.6)
Since ∂ ln τ is a closed differential, the exterior derivative of θ (1) must be opposite to the one of the first term, which is given by (1.4). Let us verify this directly; to this end we compute the exterior derivative of θ (1) . A straightforward computation yields The exterior derivative of ω M is given by (1.4), in which we can insert the explicit expression of M (z; t); after a somewhat lengthy but straightforward computation we find that thus confirming that the differential δ ln τ (1) is indeed closed (of course it must be, since the tau function is a Fredholm determinant!). The case (2) is analogous with the replacements b ↔ c and a ↔ d.

Determinants for dif ferent choices of Σ
The factorizations (3.1) (1,2) require that we deform the contour Σ so that a(z) (or d(z)) does not have any zero on Σ. This leads to completely equivalent RHPs of the form (1.1) but not entirely equivalent RHP when expressed in the form (2.2). Note that (3.1) (3,4,5) do not suffer from this ambiguity, because b(z) cannot have any zeroes in N (Σ) since we assumed analyticity of the jump matrix M (z; t).
Consider the factorization (3.1) (1) (with similar considerations applying to the other factorization); in general a(z; t) has zeroes in its domain of analyticity, and their positions depend on t. Therefore it may be necessary, when considering the dependence on t, to move the contour so that certain zeroes are to the left or to the right of it because a zero may sweep across N (Σ) as we vary t ∈ S. So, let Σ, Σ be two contours in the common domain of analyticity of M (z; t) and such that a(z; t) has no zeroes on either one and Σ ⊂ Int( Σ) (see Fig. 2).
Denote with τ (1) and τ (1) the corresponding Fredholm determinants of the operators defined as described; following the same steps as above. Our goal is to show that Theorem 3.3. The ratio of the two Fredholm determinants is given by Note that the evaluation of c at the zeroes of a cannot vanish because det M ≡ 1.
Proof . From the formula (3.6) we get Here Γ − means the analytic extension of the solution Γ to the region Ext(Σ) ∪ N (Σ). Since the integrand of the first term is holomorphic in the region bounded by Σ and Σ, it yields a zero contribution by the Cauchy's theorem and we are left only with the second term, which is computable by the residue theorem; We are assuming that M (z; t) is analytic and also that det M ≡ 1. Now note that a zero v(t) of a(z; t) in general depends on t; suppose that a(z; t) = (z − v(t)) k (C k (t) + O(z − v(t))); then (note that c(v(t); t) cannot be zero because det M ≡ 1 and z = v is already a zero of a).
as explained in [2,3]. The obstacle is not the issue of factorization but the fact that the resulting RHP of the IIKS type leads to an operator K (2.3) which is not of trace-class (and not even Hilbert-Schmidt, which would be sufficient in order to construct a Hilbert-Carleman determinant). Nonetheless, the two form δω M defines a line bundle as explained and therefore it is possible to compute the "transition functions" of the line-bundle; this is precisely what is accomplished (in different setting and in different terminology) in recent works [11,12] and the transition functions can be expressed in terms of explicit expressions, analogously to what we have shown here in this general but simplified setting.