## Abstract

In a recent publication, it was shown that a large class of integrals over the unitary group ** U**(

*n*) satisfy nonlinear, non-autonomous difference equations over

*n*, involving a finite number of steps; special cases are generating functions appearing in questions of the longest increasing subsequences in random permutations and words. The main result of the paper states that these difference equations have the

*discrete Painlevé property*; roughly speaking, this means that after a finite number of steps the solution to these difference equations may develop a pole (Laurent solution), depending on the maximal number of free parameters, and immediately after be finite again (‘

*singularity confinement*’). The technique used in the proof is based on an intimate relationship between the difference equations (discrete time) and the Toeplitz lattice (continuous time differential equations); the point is that the Painlevé property for the discrete relations is inherited from the Painlevé property of the (continuous) Toeplitz lattice.

## 1. Introduction

In a recent publication (Adler & van Moerbeke 2003), we have shown that a large class of integrals over the unitary group satisfy difference equations over *n*, involving a finite number of steps; these integrals are motivated by generating functions appearing in questions of the longest increasing subsequences in random permutations and words (see Rains 1998; Tracy & Widom 1999, 2001; Adler & van Moerbeke 2001, 2003, 2004; Baik & Rains 2001; Borodin 2003). The main result of this paper, announced in Adler & van Moerbeke (2003), states that those difference equations, which are also recursion relations, have the *discrete Painlevé property;* roughly speaking, this means that the solution to these difference equations may develop a pole (formal Laurent solution) after a finite number of steps and then immediately become finite again. Moreover, these formal Laurent solutions depend on the *maximal* number of free parameters, which equals ((order of difference equation)−1)×(dimension of phase space), with the poles disappearing after a finite number of steps (‘*singularity confinement*’).

The property of singularity confinement was introduced by Grammaticos *et al.* (1991; see also Suris 1989), and further studied in Grammaticos *et al.* (1999), as a method to find discrete Painlevé systems. They were motivated by the famous Painlevé property for continuous systems (Ince 1944) that movable (initial condition-dependent) singularities be single-valued. They were further motivated to get a classification of discrete Painlevé equations in the style of the situation for the continuous case which they and others have embarked on and had some success. For instance, Clarkson & Webster (2000) used singularity confinement to get the so-called d-PIII equation, whose particular solutions in the limit go to Painlevé III. It should be pointed out that singularity confinement can fail to produce integrability, as was shown by Hietarinta & Viallet (1998), and further tests for integrability have been proposed, such as using ‘algebraic entropy’ by Bellon & Viallet (1999) or using Nevanlinna theory (Ablowitz *et al.* (2000) or Ramani *et al.* (2003)). Thus, in discrete systems, the situation is more complicated than in the continuous situation, which should come as no surprise.

Nonetheless, singularity confinement is still a stiff requirement for a discrete system to pass and quite often (but not all the time) indicates integrability. In discovering that a large class, related to combinatorics, of integrals over the unitary group satisfy discrete recursion relations, it is natural to ask what the nature of these recursion relations might be. Moreover, since from the derivation of these relations they were clearly related to an integrable system called the *Toeplitz lattice*, it was natural to wonder if these discrete relations were integrable or at least have some ‘integrable-like property’, especially since two of the relations coming from combinatorics actually possessed invariants, one case being that of McMillan and the other being a generalization of the McMillan case (McMillan 1971). This paper answers the latter question in the affirmative. Indeed, this huge class of recursion relations coming from unitary integrals and combinatorics possesses the ‘integrable-like Painlevé property’ called singularity confinement as this paper will demonstrate. It would be worthwhile to compute the algebraic entropy of these examples, as it has been hoped that also requiring that the algebraic entropy is zero would suffice for integrability.

The technique used in the proof is new and is based on an intimate relation between the difference equations (discrete time) and the Toeplitz lattice (continuous time differential equations), introduced by Adler & van Moerbeke (2001); the point is that the ‘Painlevé property’ for the discrete relations are inherited from the Painlevé property of the (continuous) Toeplitz lattice. Before making a more precise statement and describing the technique, we recall the basic facts about the Toeplitz lattice and the recursion relations (Adler & van Moerbeke 2001, 2003).

For and , consider the matrix integrals(1.1)where d*M* is Haar measure on , and . Special choices of and lead to generating functions in combinatorics (see Adler & van Moerbeke 2003). Set and . In Adler & van Moerbeke (2003), it was shown that the ratioswith , satisfy the Toeplitz lattice, an integrable Hamiltonian system,(1.2)where *i*=1, 2, 3, …. Moreover, *τ*_{n} is a polynomial expression in the variables *x*_{k} and *y*_{k} and *τ*_{1}The Hamiltonians appearing in (1.2) are given bywhere the matrices *L*_{1} and *L*_{2} are defined by(1.3)and(1.4)where *x*_{0}=*y*_{0}=1. The system admits a reduction, interesting in its own right, obtained by putting *x*_{k}=*y*_{k} for all *k*. We refer to it as the *self-dual Toeplitz lattice*.

In Adler & van Moerbeke (2001), it was shown that the matrix integrals (1.1) satisfy a -algebra of Virasoro constraints which, combined with the Toeplitz lattice equations, lead to difference equations for *x*_{k} and *y*_{k} given by Adler & van Moerbeke (2003), a subset of the cases leading to recursion relations, which we now describe. Given arbitrary polynomialsthe variableswithand , satisfy 2*N*+1-step difference equations , where *x*=(*x*_{1},*x*_{2},…) and *y*=(*y*_{1},*y*_{2},…), and where the polynomials and are defined in terms of the matrices and defined above (denote the derivative of the polynomial *P*_{i} by )(1.5)Looking closely, one observes that these difference equations and are indeed linear in *x*_{k+N} and *y*_{k+N} and can thus be solved in terms of . See appendix A for a proof of this fact.

This paper deals with the difference equations (1.5) for their own sake, without further reference to the special solution and , given by the unitary matrix integrals above. Moreover, we will consider the bi-infinite Toeplitz lattice, which is defined as in (1.2), but with . The recursion relations are then also considered for , with the semi-infinite case obtained by specialization. The bi-infinite Toeplitz lattice will be introduced in §2, where we also discuss the self-dual Toeplitz lattice and the recursion relations.

It came as a surprise that the generic solutions of these (very general) equations (1.5) have the *singularity confinement* property, a fact observed by Borodin (2003) in the very special case of unitary matrix integrals related to the longest increasing sequences of random permutations. In this case, the recursion relation is only a three-step relation. We shall see that although the relations inherit the confinement property from their integrable ancestor, the Toeplitz lattice, they need not, as there are many places where they can easily lose this property. The main result of the paper is to show that this large zoo of examples (1.5), indeed, possess the singularity property, namely the following.

**(singularity confinement: general case).** *For any* *,* *the difference equations* *admit a formal Laurent solution* and *in a parameter λ, having a (simple) pole at k*=*n and λ*=0, *and no other singularities. These solutions depend on* 4*N non-zero free parameters**Setting* , *and* , *the explicit series with coefficients rational in γ read as follows:*

**(singularity confinement: self-dual case).** *For any* , *the difference equations* *admit two*1 *formal Laurent solutions* *in a parameter λ, having a (simple) pole at k*=*n only and λ*=0. *These solutions depend on* 2*N non-zero free parameters**Explicitly, these series with coefficients rational in α are given by*

The proof of theorems 1.1 and 1.2 is by no means direct, but proceeds via the Painlevé analysis for the Toeplitz lattice. As a starting point, the zero locus , of all polynomials and , forms an invariant manifold for the vector field of the Toeplitz lattice with Hamiltonian , by viewing the coefficients of and as constants, except for *u*_{±1}, which moves linearly in time. Explicitly, this vector field is given by(1.6)In the self-dual case, this vector field reduces to(1.7)The first idea is then to restrict the principal balances (formal Laurent solutions depending on the maximal number (dimensional phase space 1) of free parameters, besides time) of (1.6) to these invariant manifolds. We fix *n* and look for a formal Laurent solution to the Toeplitz lattice that has a (simple) pole for *x*_{n} and *y*_{n} only, and we find such a unique family, as given by the following proposition.

*For arbitrary but fixed n*, *the first Toeplitz lattice vector field* *(1.6)* *admits the following formal Laurent solutions*:*whereas for all remaining k such that* ,(1.8)(1.9)*where* *and all a*_{i}, *b*_{i}, *with* *and with* , *are arbitrary free parameters, and with* . *In the self-dual case, it admits the following two formal Laurent solutions, parametrized by ϵ*=±1(1.10)*where a*_{+}, *a*_{−} *and all a*_{i}, *with* *are arbitrary free parameters and* .

Together with time *t*, these parameters are in bijection with the phase space variables; we can put for the general Toeplitz lattice for example for and and . Thus, this formal Laurent solution is the natural candidate to work with (see §3).

It is however, *a priori*, not clear that these formal Laurent solutions can be restricted to the invariant manifold . Indeed, upon introducing a proper time dependence for *u* already mentioned, one has that and satisfy a system of differential equations, as given in the following proposition.

*Upon setting* , *the recursion relations satisfy the following differential equations*:(1.11)*which specialize in the self-dual case* *(1.7)* *to*(1.12)

In addition to propositions 1.3 and 1.4, many other arguments are needed to fine-tune the free parameters, when going from the Laurent solutions of the Toeplitz lattice to the existence of formal Laurent solutions to the difference equations, depending on the announced number of free parameters (see §6). The proof of these facts will be spread over two sections, as the arguments get rather involved; see §5 for the self-dual case and §6 for the case of the general Toeplitz lattice.

This ultimately leads to the proof of the main theorems 1.1 and 1.2.

Denote by *P* the uniform probability on the group of permutations *π*_{k} and by *L*(π_{k}) the length of the largest (strictly) increasing subsequence of *π*_{k}. According to an identity, due to Gessel (1990),The quantities defined for *n*>0 bysatisfy the following three-step relation, found by Borodin (2003):possessing the McMillan invariant (McMillan 1971)with

According to Rains (1998) and Tracy & Widom (1999),andrespectively, where and are *odd permutations* of the order of 2*k* and 2*k*+1 acting on and , respectively. Thensatisfies a five-step recursion relation possessing the invariantwith

## 2. An invariant manifold for the first Toeplitz flow

In this section, we introduce the bi-infinite Toeplitz lattice, in analogy with the semi-infinite Toeplitz lattice, introduced by Adler & van Moerbeke (2001). We also recall the basic formulae related to the invariant manifold that we will introduce below (see Adler & van Moerbeke 2003).

The (bi-infinite) *Toeplitz lattice* consists of two infinite strings of vector fields on the (real or complex) linear space of bi-infinite sequences . The particular vector field that we will be interested in (the ‘first’ Toeplitz vector field) is given by(2.1)The semi-infinite Toeplitz lattice is obtained from it by setting (*x*_{k},*y*_{k})=(0,0) for *k*<0 and (*x*_{0},*y*_{0})=(1,1). The invariant polynomials of the matrices *L*_{1} and *L*_{2}, defined by(2.2)provide two infinite strings of constants of motion and of (2.1), defined by(2.3)The first Toeplitz vector field (2.1) is the Hamiltonian vector field that corresponds towith respect to the Poisson structure defined byand the functions and are all in involution with respect to {.,.}, as follows from a direct computation. As a corollary, all Hamiltonian vector fields and commute. If we denote , whenever this makes sense, then for *i*=1, 2, …,and similarly for , which leads to the following expression for the vector field :(2.4)The vector field has the same form, but with *L*_{1} replaced with *L*_{2}. This is a particular case of a phenomenon that we will refer to as duality. Namely, there is a natural automorphism *σ* of our phase space, given by . It preserves the first Toeplitz vector field (2.1); permutes the Hamiltonians ; permutes the Lax operators as follows: ; and reverses the sign of the Poisson structure. The first Toeplitz vector field (2.1) can be restricted to the fixed point locus of *σ*, which leads to the *self-dual* (bi-infinite) Toeplitz lattice(2.5)All constructions in this paper will be done for this self-dual lattice first, and then for the general Toeplitz lattice. This is not only for pedagogical reasons: even if the ideas that lead to the proofs are similar in both cases, the self-dual lattice, for our purposes, cannot be treated as a particular case of the general Toeplitz lattice, as we will see.

For *i*=1, the equations (2.4) for and for specialize to(2.6)

Fixing 2*N* constants , with *u*_{N}≠0 and *u*_{−N}≠0, we consider the polynomials(2.7)whose derivatives we simply denote by and . They lead to two strings of polynomials2 and in :(2.8)Note that the only elements that appear in these polynomials are the diagonal and next-to-diagonal entries of and for *l*=1,…,*N*. For fixed *u*, we consider the zero locus of all polynomials and ,(2.9)In terms of the variables *x*_{i} and *y*_{i}, the leading terms of and are given bySee appendix A for a precise statement, a few more terms and a proof. We often write Δ_{k} as a shorthand for the vector and *z*_{k} for .

In order to get the corresponding formulae for the self-dual case, we put , so that *σ* permutes *P*_{1} and *P*_{2}, as well as and , hence *P*_{1}=*P*_{2} in the self-dual case, and . Writing and , the polynomials and reduce in that case to(2.10)while its leading terms are now given by(2.11)The zero locus now takes the simple form(2.12)

Following Adler & van Moerbeke (2003), we show that upon introducing a proper time dependence, the polynomials and satisfy a simple set of differential equations, showing that the zero locus (2.9) of these polynomials is a (time-dependent) invariant manifold of the first Toeplitz flow (2.1).

*Let* *be a solution to the first Toeplitz vector field* *(2.1)**,* *to wit**and let* *and* , *where*(2.13)*Then* *and* *satisfy the following differential equations*:(2.14)*In particular*, *is a (time-dependent) invariant manifold of the first Toeplitz flow. In the self-dual case, these differential equations specialize to*(2.15)*Then* *is a (time-dependent) invariant manifold of the first vector field of the self-dual Toeplitz lattice, where* .

We first show that(2.16)where ^{u} is the Hamiltonian vector fieldIt suffices to prove that , the other identity being obtained by duality (indeed, since ). In view of definition (2.8) of , this means that we need to prove that(2.17)According to (2.4), the first equation amounts to(2.18)where we recall that . The proof of (2.18) follows immediately by writing as , and the expression (2.2) for the entries of *L*_{1}. For the second equation in (2.17), the proof is similar.

Note that (2.16) implies that the time-dependent polynomials and are given bywhere can, in view of (2.13), be written asSince the vector field d/d*t* commutes with all the Hamiltonian vector fields and , it follows from these equations and (2.6) thatThis yields the first relation in (2.14). The second equation is obtained by duality.

At points of , all and vanish so the right-hand sides of (2.14) vanish. The unique solution to (2.14) that corresponds to such initial data is the zero solution, . As a consequence, is a time-dependent invariant manifold for the first Toeplitz flow. ▪

## 3. Painlevé analysis of the first Toeplitz flow

In this section, we will show that the first Toeplitz flow admits many families of formal Laurent solutions, a property reminiscent of (finite-dimensional) algebraic completely integrable systems (see Adler *et al.* 2004). They will be used in the §§4–6. We will first consider the self-dual case, which is easier, and then we will consider the full Toeplitz lattice.

### (a) The self-dual Toeplitz lattice

Recall that the first vector field of the self-dual Toeplitz lattice is given by(3.1)

*For any* , *the first vector field* *(3.1)* *of the self-dual Toeplitz lattice admits a formal Laurent solution x*(*t*)*, with only x*_{n}(*t*) *having a pole, given by**where a*_{+},*a*_{−} *and all a*_{i}, *with* , *are arbitrary free parameters; also, ϵ*^{2}=1 *and* . *When* *then* *,* *while* .

We look for formal Laurent solutions *x*(*t*) to (3.1) that have a simple pole for one of the variables (only). To do this, we substitute , with , and into (3.1) for different values of *k*. Taking , we find that , in both cases because needs to cancel the pole coming from *x*_{n}(*t*). Given this, (3.1) with *k*=*n* is given bySince , we deduce from it, on the one hand, that and have opposite signs, so that and that . It follows that and , where *ϵ*^{2}=1. For , the coefficient in *t*^{−1} of (3.1) does not impose any condition on the constant coefficient of *x*_{k}(*t*), which is therefore a free parameter, which we denote as *ϵa*_{k}.

Having determined the first term of the series, we suppose thatwhere all coefficients , with *i*≤*r*, have been determined. We show that (3.1) then yields linear relations on the coefficients . To see that, pick the coefficient in *t*^{r} in (3.1) when *k*≠*n*, while taking the coefficient in *t*^{r−1} when *k*=*n*. This yields the following relations, where ‘known’ means coefficients , with *i*≤*r*:(3.2)This yields a linear system in the unknowns , where , which has upper triangular form when is put at the end. It uniquely determines the coefficients , except when *k*=*n*±1 and *r*=0: the corresponding equations both reduce then to 0=0, so that and are also free parameters; we denote them by . Then the third equation in (3.2) implies that ; also, the first equation is explicitly given by , for . Since for *r*>0 we can solve uniquely for all , we get a formal Laurent solution depending on the free parameters, as indicated. The extra term that is given in the proposition is easily verified. ▪

Note that under the natural correspondence between the phase variables *x*_{k} (with *k*≠*n*) and the free parameters *a*_{k} (*a*± in the case *k*=*n*±1), we have that the number of free parameters on which the coefficients of the series depend is one less than the number of phase variables, a property reminiscent of principal balances for (finite-dimensional) algebraic completely integrable systems (see Adler *et al.* 2004, ch. 6). There are of course also formal Laurent solutions that depend on less free parameters (lower balances), but these will not be used here.

For future reference, we give the first few terms of the formal Laurent series of , which is easily computed from the series given in proposition 3.1,(3.3)The displayed terms are the only ones that will be needed below.

### (b) The full Toeplitz lattice

We will now show that the full Toeplitz lattice also allows such formal Laurent solutions. To make the analogy with the self-dual case transparent, we will vectorize the variables and the equations; namely, we introduce and , for ; the variables *a*_{k} and *b*_{k} will be the free parameters in the formal Laurent series. With these notations, the first Toeplitz vector field (2.1) becomes(3.4)

*For any* , *the vector field* *(3.4)* *of the (general) Toeplitz lattice admits a formal Laurent solution* , *such that only x*_{n}(*t*) *and y*_{n}(*t*) *have a (simple) pole. It is given by**where* *and all* , *with* , *are arbitrary free parameters and where* . *Precisely, the free parameters a*_{n±1} *satisfy the condition* . *The parameters on which the next order term in the series x(t) and y(t) depend is given in* *table 1*.

*In* *§6*, *we will need some extra information on these formal Laurent series, namely that the coefficient in t*^{2} *of z*_{k}, *for* *depends in the following way on* :(3.5)*where* *is independent of a*_{k+2} *and of b*_{k+2}. *In particular*, *depends linearly on a*_{k+2} *and is independent of b*_{k+2}, *while* *depends linearly on b*_{k+2} *and is independent of a*_{k+2}. *This easily follows from the given terms by considering the coefficient of t in* *(3.4)*.

For fixed , we look for formal Laurent solutions to (3.4) where *x*_{n}(*t*) or *y*_{n}(*t*) have a simple pole, and where none of the other variables *x*_{k}(*t*) or *y*_{k}(*t*) have a pole (in *t*). Thus, we substitute and into (3.4) for different values of *k*. For *k*=*n*±1, we find that , because needs to cancel the pole coming from *x*_{n} or from *y*_{n}; we put , so that . The parameters *a*_{n±1} are free, except that *a*_{n+1}*a*_{n−1}≠0. Next, (3.4) with *k*=*n* yieldswhich shows, on the one hand, that and are both different from zero (since at least one of them is supposed to be different from zero), so that also *a*_{n+1}*a*_{n−1}≠0. On the other hand, it shows that and are expressible in terms of *a*_{n+1} and *a*_{n−1} asFor , the coefficient in *t*^{−1} of (3.4) does not impose any condition on the constant coefficient of *z*_{k}(*t*), yielding free parameters for the constant coefficients of *x*_{k} and *y*_{k}, with . We denote these free parameters by . Upon specialization, some of the formulae below may contain *c*_{n+1} or *c*_{n−1}; it is understood that these stand forWe can now proceed as in the second part of the proof of proposition 3.1, namely we suppose thatwhere all coefficients , with *i*≤*r*, have been determined. On the coefficients , , we find linear relations by substituting the above series into (3.4). For *k* such that , it is clear that, as in the self-dual case, is linearly computed in terms of the known coefficients, from the coefficient of *t*^{r}, when substituting the series in (3.4). Therefore, let us concentrate on what happens for . Taking *k*=*n*±1 in (3.4), the coefficient of *t*^{r} yieldsa linear equation in *x*_{n±1} and *y*_{n±1}, which can be written in the compact formwhere is the matrix that governs the linear problemSince , this linear system admits a unique solution, except when *r*=0 (recall that *r*≥0). Before analysing the case *r*=0 further, let us first consider what happens to (3.4) in the remaining case *k*=*n*. As in the self-dual case, we pick the coefficient of *t*^{r−1} in (3.4) to find a linear system that can be written in the compact formwhere the matrix is given bySince , we have again that is determined uniquely, unless *r*=0. Thus, we are done with *r*≥1.

As we have seen, a free parameter may appear in , and , but one has to check whether the corresponding linear equations are consistent. Therefore, we substitute(3.6)in (3.4), which yields for and *t*=0 the homogeneous linear systemwhich is equivalent to(3.7)Thus, upon setting , where *a*_{+} and *a*_{−} are free parameters, we have that . Similarly, for *k*=*n* the substitution of the series (3.6) in (3.4) yields at the level *t*^{−1}These equations are proportional, in view of (3.7). Thus, we havewhere *a* is a free parameter.

The first two terms in the series lead at once to the second and third columns of table 1. In order to obtain the last column, it suffices to list on which parameters the linear term (resp. the constant term) in the right-hand side of depends, when *k*≠*n* (resp. when *k*=*n*). The two leading terms of *x*(*t*) and *y*(*t*) that we computed suffice for doing this. ▪

It is easily verified that the involution *σ* that permutes *x*_{k} and *y*_{k} extends naturally to an involution on the free parameters, given by(3.8)

Note that, altogether, we have besides the free parameters *a*_{k},*b*_{k}, for , which naturally correspond to the variables *x*_{k} and *y*_{k}, five extra free parameters and *a*, which correspond to the remaining six variables *x*_{n±1,}*y*_{n±1} and *x*_{n}, *y*_{n}, which again yields that the number of free parameters, plus time, is equal to the number of phase variables. This count will be important, and rigorous, when we restrict these formal Laurent solutions to certain finite-dimensional submanifolds.

## 4. Tangency to

We have seen that the polynomials and , which define an invariant manifold for the first Toeplitz flow, satisfy a non-autonomous system of linear differential equations, where the time dependence is defined by the latter flow. In a (finite-dimensional) manifold setting, if such differential equations have coefficients that depend smoothly on time, solutions (integral curves) that start out on the invariant manifold will stay on it, by the uniqueness of solutions to differential equations with smooth coefficients and given initial conditions. In the case that we deal with the situation is quite a bit different, because the coefficients develop poles in *t*, for *t*=0, and of course the solutions are only formal Laurent series. As it turns out, the conditions that assure that the formal Laurent solutions ‘stay on the invariant manifold’ are similar to those in the smooth case for the self-dual Toeplitz lattice, but are different in an essential way for the general Toeplitz lattice.

### (a) Tangency in the self-dual case

We start out with the case of the self-dual Toeplitz lattice.

*Let x*(*t*) *denote the formal Laurent solution that is given by* *proposition 3.1*, *and denote* , *where we recall that* . *Then, as formal series in t*,(4.1)*Moreover*, *as a formal series in t, for all* , *as soon as x*(*t*) *is such that*

According to (2.11), involves only the variables *x*_{l} with (2*N*+1-step relation). Since only *x*_{n}(*t*) has a pole, as soon as does not contain *x*_{n}, i.e. if . However, note that (2.15) impliesso that , as the leading term of is non-zero (recall that is a free parameter). This argument can be repeated to yield for all *k*<*n*, and similarly it is shown that for all *k*>*n*. Since satisfies the differential equation (2.15), for *k*=*n*, we have in view of (3.3) thatwhich leads upon integration to (4.1).

Suppose now that *x*(*t*) is such that for all . In view of the first part of the proof, we have that for all . We show that this implies that as a formal series in *t*, for all . We do this by induction on : assuming that for , we show that for . Note that in the case *r*=1, the assumption holds. For , the right-hand side of (2.15) is , by (3.3) and by the assumption, so that , hence , by integration. For , we have from (3.3) that , so that (2.15) yields for that , i.e. . For *k*=*n*, we have that has a double pole, but since we have just shown that the differential equation (2.15) for *k*=*n* leads to and we conclude that , as has to be shown. ▪

### (b) Tangency in the general case

For the full Toeplitz lattice, the tangency condition is rather similar, yet is different in some detail that will turn out to be crucial in §5. We recall that the differential equations that are satisfied by the polynomials and are given by(4.2)

*Let* *denote the formal Laurent solution that is given by* *proposition 3.2*, *and let* , *where u*(*t*) *is given by* *(2.13)*. *Then, as a formal series in t*, *and* *for* . *Also*(4.3)*where* *and* *are both linear combinations of* *and* *(for the explicit formula, see* *(4.9)**); moreover, the latter coefficients are related in the following way*:(4.4)

As in the self-dual case, the polynomials and define 2*N*+1-step relations, so they depend only on the variables *x*_{l} and *y*_{l} with . Only *x*_{n}(*t*) and *y*_{n}(*t*) have a pole, so that and for . Writing (4.2) for as(4.5)and taking as consecutive values in (4.5), we find that and for all , since does not vanish for *t*=0 when . Similarly, and when . So we have that and when and we are left with the case *k*=*n*.

In order to deal with the case *k*=*n*, we write (4.2) as an equation for and in two different ways(4.6)Either of them implies that and that , so we writeand similarly for . In fact, as and have a simple zero, while and have a simple pole, the coefficient of *t*^{−2} in (4.6) leads to the following linear equations:(4.7)where we have written , so thatIt is now sufficient to substitute and in (4.7) to find the coefficient of *t*^{−2} in (4.3). Actually, the latter corresponds to taking the lower sign; equating the two expressions for in (4.7) that correspond to the two signs leads to (4.4); note that this is also the expression that is obtained from the two expressions of in (4.7).

It remains to compute and , which can be done from the coefficient of *t*^{−2} in and in , computed from their differential equations(4.8)Since has a double pole, while and have no pole, the contribution of the first term to the coefficient in *t*^{2} will be linear in and . Since *x*_{n}(*t*) and *y*_{n}(*t*) have a simple pole, while and have a double pole, the contribution of the second term will yield a linear combination of, on the one hand, and which, as we have seen, are themselves linear combinations of and in ; and on the other hand, and which are the unknowns. Explicitly, this linear system is given by(4.9)Since and are linear combinations of and , it follows that each of and is a linear combination of and , as we asserted. ▪

*Suppose that* *is a formal Laurent solution of the first vector field of the Toeplitz lattice, such that* *and* *for all k with* , *and such that, as formal Laurent solutions in t*, and . *Then, as formal Laurent series*, *for all* .

According to (4.4), the hypothesis implies that . In view of proposition 4.2, we have that and for every . We will now proceed by induction on , but in a different way than in the self-dual case: assuming that and for , as well as and , we show that and for , as well as and . Note that the *r*=1 induction assumption needs to be shown at the end of the proof, as only part of it is in the actual hypothesis of the theorem.

For *k* such that , the differential equations (4.2) yield that and , so that and , by integration. So we are left with . Let us writewhich we substitute in(4.10)Remembering that , we pick the coefficient of *t*^{r} in (4.10), which leads to the following linear system:(4.11)Sinceit follows, since , that , and hence that and . It follows that, if we substitute the series in(4.12)then the coefficient of *t*^{r−1} is simply given bySincewe have that , so that and , as shown.

We finally check that our assumptions imply that for *r*=1 the induction hypothesis is valid. According to proposition 4.2, we have that and . Let us write and . Then we need to show that . From (4.11), which is also valid for *r*=0, we conclude that . It was assumed that , i.e. that , so that we can conclude that . In order to obtain a second relation between and , we consider the residue in the first3 equation in (4.12), which reduces to , since . Thus, , as shown. ▪

## 5. Restricting the formal Laurent solutions: the self-dual case

We have seen conditions on that guarantee that solutions *x*(*t*) to the self-dual Toeplitz lattice that start out in the invariant manifold stay in it, formally speaking. In this section, we show how these conditions can be translated into conditions on the formal Laurent solution *x*(*t*) to the first vector field of the self-dual Toeplitz lattice.

### (a) Structure of the polynomials

The polynomials , which define the invariant manifolds , depend on the variable *x*_{n} in a special way, which we will analyse by using the fact that remains pole free (for ) when the formal Laurent series *x*(*t*) are substituted in them, as we have seen in proposition 4.1. Let us denote by the algebra of polynomials in all variables *x*_{k}, where , and by the subalgebra of those polynomials that are independent of . Also, let us denote by the subalgebra of that consists of those elements that can be written as polynomials in and , with , where(5.1)Thus, elements of may depend only on through and . For future use, we give the first few terms of the formal Laurent series of the generators of , as obtained by substituting the series from proposition 3.1 in (5.1)(5.2)It follows that , for any . Note that the polynomials , which both have the property , belong to , since(5.3)The following proposition generalizes this statement.

*For* , *let* *,* *where x*(*t*) *is the formal Laurent solution to the first vector field of the self-dual Toeplitz lattice, constructed in* *proposition 3.1*. *If* then , *i.e. G is a polynomial in*

We suppose that is such that , where . We write *G* as a polynomial in *x*_{n} with coefficients in where . If *l*=0, then we are done. Let us suppose therefore that *l* is minimal, but *l*>0. We will show that this leads to a contradiction. Since each coefficient belongs to , we have that . Thus, the pole that *x*_{n}(*t*) has needs to be compensated by a zero in , i.e. . We show that this implies that . By Euclidean division in , we can write as(5.4)where *K*_{1}, *K*_{2} and *K*_{3} belong to , and where *K*_{3} is of degree 1 at most in *x*_{n+1} and *x*_{n−1}: we can write *K*_{3} aswhere are elements of that are independent of *x*_{n+1} and *x*_{n−1}. Since and , it follows from (5.4) that , and so that the leading terms and of and satisfy . Since the leading terms of all *x*_{k}(*t*), with , and the leading terms of and are all independent, even modulo *ϵ*, it follows that , as and are independent of . Using (5.3), it follows thatwhere , showing that , as promised. Then,with . This contradicts the minimality of *l*. ▪

*For* , *is of the form*(5.5)*i.e. the constant term in* *is a polynomial in the variables*4 *and* *only*.

According to (2.11), depends on only. For , we know from proposition 4.1. that , so that proposition 5.1 yields that depends on *x*_{n} through and only, i.e. is a polynomial in and the *x*_{l} with and *l*≠*n*. Each of these variables is *O*(1), so the constant term in is a polynomial in their leading terms, which are the parameters *a*_{k−n} *a*_{k+N}, *a*_{+} and *a*_{−} (see (5.2)). ▪

It is clear that when , then is independent of *a*_{+} and *a*_{−}, as it cannot contain or . The following lemma deals with the case of , which is slightly harder because develops a pole.

*is of the form**where* *is a polynomial in all its arguments, with a*_{n+N+1} *and a*_{n−N−1} *present (linearly)*.

Consider the following alternative ways of writing :(5.6)*H*_{n} is a polynomial in , because (2.16) implies that , and because is always divisible by (see (2.4)). Also, we have put to obtain the second equality. The first equation in (5.6) implies that , since and , while . The second equation in (5.6) then allows us to conclude that , and hence also that , since *u* is an arbitrary vector of constants. Thus, is, by proposition 5.1, an element of , depending (linearly) on the parameters *u*_{i}.

Summarizing, the constant term in will be given by the constant term in , hence it will depend only on the first two terms of *x*_{n}(*t*) and on the first two terms of *G*_{n}(*t*), where . The latter first two terms can depend only on the first two terms of the variables and *w*_{2} that appear in *G*_{n}; the first two terms of their series can be read off from (5.2), yielding that the constant term in can only depend on . Note that the only dependence on *a*_{n−N−1} can come from the presence of *x*_{n−N}, but (2.11) tells us that *x*_{n−N} appears linearly in , and with a non-zero coefficient. Therefore, the parameter *a*_{n−N−1} is indeed present in the constant term in ; similarly, *a*_{n+N+1} is also present. The leading term of was already determined in proposition 4.1. ▪

### (b) Parameter restriction

We now show that we can tune the free parameters in the formal Laurent solution *x*(*t*) of the self-dual Toeplitz lattice in such a way that for all , as a formal series in *t*. As it turns out, it will be possible to keep 2*N*−1 parameters arbitrary, and the other ones are determined rationally in terms of these. Together with time, it means that the constructed solution depends on 2*N* free parameters, which is the maximum one can hope for in an 2*N*+1-step relation.

*Keeping the* 2*N−*1 *parameters* *arbitrary, the other parameters in the formal Laurent series x(t), given by* *proposition 3.1*, *can be chosen as rational functions of these parameters, so that* , *as a formal series in t, for all* .

In this proof, we will assume that *N*>1. See remark 5.5 below for the adaption to the case *N*=1. According to proposition 4.1, it suffices to determine the parameters in the series *x*(*t*) so that , the constant term in , is zero, for all . Thus, we need to write in terms of the parameters in the series *x*(*t*). We do this for the different values of *k* in a very specific order, as indicated in table 2. The second column indicates which we consider; it is easy to see that we consider all of them (exactly once); it is understood that steps (6)–(8) are absent when *N*=2. We know from (2.11) that for any , depends only on the variables , which yields the third column. It is important to point out that the two written variables, which are the extremal terms, are actually present in , and that these two variables appear linearly (see proposition A.1 in appendix A).

The delicate step is in obtaining the last column; the information displayed in it contains the parameters5 that may appear in , where the underlined term actually does appear, and it appears linearly. Before validating this column in each of the steps, let us first point out how the proposition follows from it. Precisely, we can, in each step, solve for one of the underlined parameters in terms of the non-underlined parameters, as the underlined parameter appears linearly in the equation . Using the previous steps, this yields inductively a rational formula for each of the parameters, in terms of , which remain free. In fact, the variables *a*_{n−2N−i}, with *i*>0 are determined in steps (1)–(3); while *a*_{n} does not exist; the variables *a*_{n+i+1} with *i*>0 are determined in steps (6)–(12); the only other variables are *a*_{−} and *a*_{+}, which are determined in steps (4) and (5).

We now show that in each step the parameters that are indicated in the fourth column of the table indeed appear (linearly) in . This is done by carefully using the leading terms of , as given by proposition A.1. As a general remark, note that (2.11) implies that contains the variables *x*_{k−N} and *x*_{k+N} linearly, but that the behaviour of its coefficients and , evaluated at *t*, depends on *k*, as given in (3.3).

For step (1), we have that have no pole in *t*, so that only their leading coefficients, the parameters , can appear. Since *x*_{n−2N−1} appears (linearly) in , with a coefficient , that is non-vanishing for *t*=0, namely , the parameter *a*_{n−N−1} appears (linearly) in . The same argument works in steps (2) and (3). Step (4) is more interesting because it involves *x*_{n} (linearly). However, *x*_{n} appears only in the leading term of , which we can write, using , as(5.7)Now , and the other factors in (5.7) are finite, non-vanishing, which yields the proposed dependence on the parameters in step (4). For step (5), *x*_{n} may be present in other terms than the leading term in , but in view of proposition 5.1, is a polynomial in and in *w*_{1} and *w*_{2} only. Since their series do not have a pole for *t*=0, we get an eventual dependence on *a*_{+} and *a*_{−}, besides the parameters . Let us show that *a*_{+} actually appears. The leading term in is, according to (2.11),Since it is the only term in that contains *x*_{n+1}, we can write , whereand *P*_{2} is independent of *x*_{n+1}, so *P*_{2} depends only on . Now , sincewhile the other factors that appear in *P*_{1}(*t*) are *O*(1). Since , this implies that , so that *P*_{2} satisfies the hypothesis of proposition 5.1; since *P*_{2} is independent of *x*_{n+1}, we may conclude, as in step (4), that *P*_{2} is independent of *a*_{+}. On the other hand, *P*_{1}(0) depends (linearly) on *a*_{+}, asThe conclusion is that depends (linearly) on *a*_{+}.

We are at step (6). Skip this step and steps (7) and (8) when *N*=2. Proposition 5.1 implies that can depend only on the proposed parameters, and that the dependence comes from the constant terms of the series in (5.2). The dependence of on *a*_{n+2} comes only from the leading term which, at *t*, is *O*(1), since and non-vanishing. It follows that depends on *a*_{n+2} (linearly). The same happens in steps (7) and (8), as the leading term will always contain the product , which is finite and non-zero for *t*=0.

A new phenomenon arises in step (9). Note that we have moved to , keeping for step (10). The leading term of iswhich does not contribute to , since , while all other factors in this term are finite in *t*. Therefore, is independent of *a*_{n+N+1}. To show that depends on *a*_{n+N}, we need to investigate the next term in , the one that contains *x*_{n+N}, because it is the only one that might lead to a dependence on *a*_{n+N}. According to proposition A.1, this term consists of the following three pieces:(5.8)The two terms on the first line of (5.8) do not contribute to , again because both terms contain , and all other terms are finite for *t*=0. The third term however does contribute, when *j*=0, as ; moreover, this term is the only one that involves *a*_{n+N}, so that the latter parameter appears (linearly) in . For step (10), the presence of *a*_{n+N+1} was established in lemma 5.3. Starting from step (11), the leading coefficients do not contain or anymore, so that everything goes smoothly. ▪

When *N*=1, the polynomial that defines the recursion relation reduces toSteps (4)–(9) then get replaced by two steps in which we consider , which allows us to determine *a*_{±}. Indeed, substituting the series *x*(*t*) in yields for the leading term (*t*=0)The other parameters are determined as in the general case.

## 6. Restricting the formal Laurent solutions: the general case

In this section, we will do a similar analysis as the one that has been done for the case of the self-dual Toeplitz lattice in §5.

### (a) Structure of the polynomials and

We first investigate on which parameters the leading term(s) in the polynomials and depend on the free parameters. We denote by the algebra of all the polynomials in the variables *x*_{i} and *y*_{i}, where , while stands for the subalgebra of that consists of all polynomials that do not depend on *x*_{n} and on *y*_{n}. Consider the following four polynomials6:(6.1)For future use, observe that these polynomials are linked by the following identity:(6.2)In fact, both expressions in (6.2) are equal to . We denote by the subalgebra of that consists of all polynomials that can be written in terms of these four polynomials, besides all *x*_{i} and *y*_{i}, with *i*≠*n*. The polynomials *w* have the following series in *t*, when the first few7 terms of the series *x*_{i}(*t*) and *y*_{i}(*t*) that are constructed in proposition 3.2 are substituted in them.(6.3)whereThe formal Laurent series for the other polynomials in (6.1) is found from it by using the automorphism *σ* (see (3.8)), which yields in particular(6.4)It follows that if then , where , with *x*(*t*) and *y*(*t*) as above. We will show that the converse is also true, so that the algebra plays, in the general case, a similar role as in the self-dual case. For this, we need the following lemma.

*Let G be a polynomial in* *that is independent of w*_{2} *and none of whose terms contains x*_{n+1}*y*_{n+1} *or x*_{n−1}*y*_{n−1}. *If* *then G*=0, *as a formal series in t*.

It follows from (6.3) thatwhere*T* is an invertible matrix, since . Let *G* be a polynomial in that is independent of and suppose that . We write , where is a polynomial in the variables *x*_{k} and *y*_{k} with *k*≠*n* only. Note that *g*_{ijk}(0) is independent of *a*,*a*_{+} and *a*_{−}. Therefore, the fact that *T* is invertible and that *a*,*a*_{+} and *a*_{−} are independent free variables implies that *g*_{ijk}(*t*)=*O*(*t*) for any *i*,*j*,*k*. If we assume now, in addition, that *g*_{ijk} does not contain either product *x*_{n+1}*y*_{n+1} or *x*_{n−1}*y*_{n−1}, then it is clear that *g*_{ijk}=0 since the leading terms *a*_{k} of *x*_{k} and *b*_{k} of *y*_{k} are independent (*k*≠*n*), except that . ▪

*For* , *let* , *where* (*x*(*t*)*, y*(*t*)) *is the formal Laurent solution to the first vector field of the Toeplitz lattice, constructed in* *proposition 3.2*. *If G(t)*=*O*(1), *then* , *i.e. G depends only on x*_{n} *and y*_{n} *through the polynomials* *and*

Given , we may write *G* as a polynomial in *x*_{n} and *y*_{n}, with coefficients in ; in fact, writing , we may assume that *G* is independent of *x*_{n} and we writewhere . We suppose that this is done in such a way that *l* is minimal. If *l*=0, then and we are done; assume therefore that *l*>1. We will show that , which is in contradiction with the minimality of *l*, like in the self-dual case. We first show that we may assume that *w*_{2} is absent in *G*_{1}*y*_{n}. If we substitute in the identity (6.2), then we findwhich allows us to replace any term in *G*_{1}*y*_{n} that contains *w*_{2}, or a power of it, by a term of lower degree in *w*_{2}, at the cost of changing *G*_{l−1}, so that we can eventually remove *w*_{2} entirely from the leading coefficient *G*_{l}. Assuming that *G*_{l} does not depend on *w*_{2}, we perform an Euclidean division in ,(6.5)where *K*_{1}, *K*_{2} and *K*_{3} belong to , with *K*_{3} independent of *w*_{2} and not containing *x*_{n−1}*y*_{n−1} or *x*_{n+1}*y*_{n+1}.

Assume now that *G*(*t*)=*O*(1). Since all *G*_{i}(*t*) are *O*(1), as , we must have that *G*_{l}(*t*)=*O*(1), as *y*_{n}(*t*) has a pole. Then (6.5) implies that *K*_{3}(*t*)=*O*(*t*), since . This means that *K*_{3} satisfies the conditions of lemma 6.1, hence that *K*_{3}=0. The identitiesthen imply that , as has to be shown. ▪

As a first application of this proposition, we show how the shown terms in (6.3) can easily be computed. Since *w*_{i}(*t*)=*O*(1), we also have for *i*=1, 2. By proposition 6.2, , in factwhere , with . Since , it follows thatwhich yield after integration the linear terms in (6.3). The same formulae can be used to show that and , which are the *t*^{2} terms in *w*_{1}(*t*) and *w*_{2}(*t*), depend only on the parameters and *a*; the precise formula will not be needed, except that they depend on *c*_{n+3} as follows:(6.6)where the dots are independent of *a*_{n+3} (and of *b*_{n+3}).

The following lemma is the analogue of lemma 5.1 and is proven in exactly the same way.

*If k*≠*n, then the series* *and* *are of the form**where we recall that* *and that* , *and* *are polynomials in their arguments*.

For *k*=*n*, the corresponding result is more complicated and the method of proof is different from the one in the self-dual case (lemma 5.3).

The constant terms and are of the formwhere *A* is an invertible 2×2 matrix and is a polynomial 2-vector that depends on the listed free parameters only. See proposition 4.2 for the leading terms of and .

We will assume in our proof that *N*>2 (see remark 6.5 below). The proof is based on the explicit expression for that is given in proposition A.1 (see appendix A), which we write in the form , whereOur first claim is that . Since and have a double pole, while *x*_{n}(*t*) has a simple pole, . The terms in the above expression that do not involve *x*_{n} or *y*_{n} are also *O*(1), because and for *k*≠*n*. There are a few terms that contain *x*_{n} or *y*_{n} (linearly), but they are all of the form , or , which are both *O*(1). It follows that , and hence that . Thinking of *u* as constants we have, in view of proposition 6.2, that .

Since has a double pole, only the first three terms of and of can contribute to the constant term in ; in view of table 1, this contribution can yield only a dependence on the parameters and *a*.

We now turn to the other terms in *H*_{n} and we use their explicit form to show that they depend only on the listed parameters. Let us first consider the following terms that do not involve *x*_{n} or *y*_{n}:(6.7)Since has a simple zero for *i*=1 and is *O*(1) for *i*>1, we have that and have a simple zero, so we only need to look for the parameters that appear in the first two terms of the coefficients. The former adds nothing new to the latter parameter list. For the coefficients of the first one for example, we read from table 1 that the constant and linear terms of depend only on and , which falls inside the proposed limits. Note in particular that neither nor appears in this term. We arrive similarly at the same conclusion for the other three terms in (6.7). Note that the lowest free parameter that appears is ; it comes from the last term in (6.7).

We now get to the terms that contain *x*_{n} or *y*_{n}. As we have already noted, these terms always come with or . As , we must investigate the first three terms in the remaining factors. For the termwe need to look at , which yields terms with a low index, the lowest coming from the coefficient in *t*^{2} in , to wit *b*_{n−N−1} and *a*_{n−N}. The other three terms that involve *x*_{n} or *y*_{n} can be written asAgain, since 1−*x*_{n}*y*_{n} has a double pole, the first three terms in will contribute to the constant term in . It is clear that *B*_{2} will contain *a*_{n+N+1}, coming from and *b*_{n+N+1}, coming from *y*_{n+N−1}. To know the precise value, it suffices to substitute the relevant coefficients of the formal Laurent series *x*(*t*),*y*(*t*) in the following part of *B*_{2},which gives by using proposition 3.2 and in particular −(*x*_{n}(1−*x*_{n+1}*y*_{n+1}))^{(0)}=*a*_{+}*a*_{n+1} and −(*y*_{n}(1−*x*_{n+1}*y*_{n+1}))^{(0)}=−*a*_{+}*b*_{n−1},(6.8)where the dots are independent of *a*_{n+N+1} and *b*_{n+N+1.} There remains one term in *H*_{n}, namely the lead in term . It does not involve *x*_{n} but does involve 1−*x*_{n+1}*y*_{n+1}, which will also lead to a dependence on *a*_{n+N+1}. Writing *C*(*t*)=*C*_{1}*t*+*C*_{2}*t*^{2}+*O*(*t*^{3}), we have thatwhere the dots are again independent of *a*_{n+N+1} and *b*_{n+N+1}. Summing up, we have that the leading terms in are given byBy duality, the leading terms in are given byWe may conclude that(6.9)whereSince*A* is invertible. ▪

*The above proof breaks down at several places when N*=2. *The polynomial H*_{n} *then reduces to*(6.10)*Using* *(6.6)* *and* *proposition 3.2*, *we find that H*_{n} *depends in the following way on a*_{n+3} *and b*_{n+3}*:**It leads as in the case N*>2 *to* *(6.9)*, *with precisely the same matrix A*.

### (b) Parameter restriction

The parameter restriction works more or less like in the self-dual case, the main difference coming from the fact that in the self-dual case we had to put all , while in the general case the tangency condition is equivalent to the following:

and for all

*k*with ,, and

.

In a sense, the condition replaces the condition , which is redundant because it is a consequence of the other conditions (see proposition 4.3).

*Keeping the 4N*−1 *parameters*8 *arbitrary, the other parameters in the formal Laurent series* , *given by* *proposition 3.2*, *can be chosen as rational functions of these parameters, so that* *and* , *identically in t, for all* .

We give the proof in the case *N*>1 only, leaving the case *N*=1 to the reader (see remark 5.5 for the self-dual *N*=1 case). As in the self-dual case, we summarize the order in which we treat the different equations in a table (table 3). The second column shows which we consider. For , it is clear that each Δ_{k} appears (precisely once). The fact that appears in step , while Δ_{n−1} already appears in step (8), comes from the fact that we consider in step (9a) the coefficient in *t* of (rather than the coefficient in *t*^{0}); similarly, is absent because the nullity of is a consequence of the nullity of the other (proposition 4.3). We know from proposition A.1 that for any ,(6.11)so thatThis leads, with no effort, to the third column of the table. For future use, let us recall that depends (linearly) on *x*_{k−N} and on *x*_{k+N}, while depends (linearly) on *y*_{k−N} and *y*_{k+N}.

Let us now turn, line by line, to the last column, which demands a careful inspection of the polynomials and . In particular, we show that these polynomials depend on the underlined parameter(s) (linearly), in such a way that one can solve for them. In steps (1)–(3), we have that *z*_{n} is absent, so that () depends only on , i.e. on . Now depends on (linearly), but not on , while the opposite is true for , so that we can solve the equation linearly for , and similarly can be solved linearly for in terms of . For *k*=1, this gives (resp. ) in terms of the 4*N*−1 parameters , so that by taking we get recursively in terms of these parameters, for all .

We now get to step (4), which is different because involves *x*_{n} and *y*_{n}. As for , according to proposition A.1, *x*_{n} appears only in the leading term of , which we can write aswhere , as . Therefore, using (6.11),which can be solved linearly for *a*_{−} in terms of the previous parameters ( for ). Using the automorphism *σ* (see (3.8)),so that can be solved linearly for .

For step (5), *x*_{n} and *y*_{n} may be present in several terms in , but in view of proposition 6.2, and are polynomials in and in *w*_{1} and *w*_{2} and their *σ* analogues only. Thus, and depend on their leading terms only, to wit and . It follows that the only *new* parameters that appear at step (5) are *a*_{+} and *a*. Let us show that they appear in such a way that we can solve them (linearly) in terms of the other parameters. We do this as in the self-dual case by isolating the leading term in as given in proposition A.1, namely we write as(6.12)The relation (6.12) was obtained by writing the leading termand throwing the term into . Since and since the first two terms in (6.12) belong to , the last term in (6.12) is also *O*(1) in *t*; since in addition this term does not contain *z*_{n+1}, by proposition 6.2 and (6.1) *x*_{n} and *y*_{n} can only appear in it multiplied by , and so by proposition 3.2 we may conclude that the contribution from this term in will not involve *a*_{+} or *a*. Also, the second term in (6.12), does not contribute to since while all other factors are *O*(1). Thus, the dependence on *a*_{+} and *a* in comes entirely from the first term in (6.12), which in view of proposition 3.2 and (6.3) is given byBy duality,where *σ*(*Ω*) was given in (6.4). Since *Ω* and *σ*(*Ω*) are linearly independent, as linear functions of *a*_{+} and *a*, we can indeed solve and linearly for *a*_{+} and *a* in terms of the other parameters.

Steps9 (6)–(8) are easy, the point being that by proposition A.1, for Let us concentrate on the next steps, which are more exciting. In step (9a), we need to compute the linear term in , where we recall from propositions 4.2 and 6.2 that , hence that this linear term depends only on the constant and linear terms of the elements of . Since , with leading termwe have from proposition 3.2 thatwhere the dots involve only the previous parameters. Therefore, we may solve (linearly) for *a*_{n+N}. Step (9b) is similar to step (9) in the self-dual case; note that we postpone again *Δ*_{n} to the next step. First of all, and so . The leading term in , namely the term , cannot contribute to because it is *O*(*t*), which explains the absence of in . By proposition A.1, can come only from , which appears only once, namely inyielding at *t*=0 a non-zero linear term in , as .

Step (10) is the hardest one, but we dealt with it in lemma 6.4. Note that after this step, we have since the nullity of the previous already implies that (proposition 4.2). Starting from step (11), everything goes smoothly, as for and the leading term of , resp. will produce precisely the new parameter , resp. (linearly). ▪

## 7. Singularity confinement

We have constructed in §§5 and 6 formal Laurent series for the Toeplitz lattice (self-dual and general case) solving the recursion relations in the general case). We will now transform these into solutions of the recursion relations (resp. ), depending on a certain number of free parameters and blowing up for only one (resp. two) variable. We will mainly concentrate on the self-dual case, as the general case is dealt with in precisely the same way.

The main tool to do this transformation is a formal version of the implicit function theorem, which we explain in the case of one variable, the scalar case. Suppose that we have a formal series in *t*,(7.1)one may think, for example, of as a formal solution of a vector field (differential equation ) on the real line, with initial condition . In our case, the functions *f*_{i} will be rational. We want to solve the equation formally, namely we wish to construct the formal series in *t*with the property that , as a formal *t*-series identity. More precisely, we claim that there exist for any unique (rational) functions , such thatwhere is given by (7.1). This is a trivial consequence of a formal version of Taylor's theorem. For example, for *s*=1 we neglect all terms in *t*^{2} and the condition on *g*_{1} becomesso that . For *s*=2, we neglect the terms in *t*^{3}, givingwhich has as a unique solution. Continuing in this way, it is clear that equals , up to a differential polynomial in the , with *j*<*i*. Note that when all are rational functions, the same will be true for all .

Let us apply this to the formal Laurent series that we have constructed for the self-dual Toeplitz lattice, and that yield formal solutions to the recursion relations , where . Recall from proposition 5.4 that these formal Laurent solutions *x*_{k}(*t*) depend on 2*N*−1 parameters , which are the leading coefficients of , namely(7.2)where the higher-order terms are rational functions of the parameters . Besides the parameters *a*_{k}, these functions also depend (polynomially) on the parameters that define the recursion relations, namely , for . The formal implicit function theorem then leads to the following proposition.

*There exist for* *rational functions**such that* , *formally inverts* *(7.2)**,* *i.e.**for* *,* *with* .

We can use these series to replace the free parameters in the series , by , where we think of the latter as (partial) initial conditions to the recursion relation. To do this, one simply substitutes for in each of the series , and rewrites this as a series in *t*; by construction, this simply gives for . For , this yieldswhere we recall that . The functions are rational in *α* and *u*. We will now use the formal implicit10 function theorem again, but in a form which is different from the one explained above: putting , i.e. we putwhich we solve for *t* as a formal series in *λ*,(7.3)where it is important to note that the constant term in this series is absent. Indeed, let us first substitute (7.3) in the series for *a*_{k} that was obtained in proposition 7.1, to get . Then, the latter and *t*(*λ*) are substituted in all *x*_{k}(*t*), to yield series in *λ* whose coefficients are rational functions of (and of ), which take the following form:It may seem that we have reached the final result, but we should not forget that these series are constructed from solutions to the recursion relations , where . However, letting , and using (7.3) to get rid of *t*, we have thatNote that, when it is all worked out, the *x*_{k} are formal power series in *λ* (except *x*_{n} which has a simple pole in *λ*), and their coefficients are rational functions of the initial conditions and of the parameters . Writingleads to our final result.

*The recursion relations* *admit for any* *two*11 *formal Laurent solution* , *depending on* 2*N free parameters* *and λ with x*_{n} *having a (simple) pole for* *,* *and no other singularities. Explicitly, these series with coefficients rational in α are given by*

The corresponding theorem for the recursion relations , which was formulated in §1 (theorem 1.1), follows in the same way, using the formal Laurent solutions *z*(*t*) that solve the recursion relations.

## Acknowledgments

M.A. and P.v.M. acknowledge the support of a National Science Foundation grant no. DMS-04-06287. P.v.M. and P. V. acknowledge the support of a European Science Foundation grant (MISGAM) and a Marie Curie Grant (ENIGMA). P.v.M. acknowledges the support of a NATO, a FNRS and a Francqui Foundation grant.

## Footnotes

One contribution of 15 to a Theme Issue ‘30 years of finite-gap integration’.

↵They are parametrized by

*ϵ*=±1.↵The structure of the matrices

*L*_{1}and*L*_{2}implies that and are indeed polynomials. They are also polynomials (of degree 1) in the variables*u*_{i}, but we often do not mention this, because we think of these variables as parameters.↵Taking the second equation would lead to the same result.

↵Recall that and that

*a*_{n}does not exist; so and*a*_{n}may be thought of as being absent in the list. Thus,*a*_{±}is the natural substitute for .↵Besides the constants

*u*_{1}, …,*u*_{N}that define*P*.↵Recall that

*σ*denotes the involution that permutes all*x*_{i}↔*y*_{i}.↵

*A priori*, one needs to compute an extra term in the series*z*_{k}(*t*) (see proposition 3.2) in order to find the shown terms in (6.3). After proposition 6.2, we will however show how such a cumbersome calculation can be avoided.↵Recall that

*c*_{k}=(*a*_{k},*b*_{k}) and that*a*_{n±1}*b*_{n±1}=1.↵Skip these steps if

*N*=2.↵Call this the formal inverse function theorem, if you wish.

↵Parametrized by

*ϵ*=±1.↵We give in each case the terms that will be used, no more, no less. When

*s*=2, only the first two lines survive; the term on fourth line coincides with the first term on the second line and should only be counted once.↵As in the case of lemma A.1, when

*N*=2 then the term −*u*_{2}*x*_{k}*x*_{k+1}*y*_{k−1}(1−*x*_{k}*y*_{k}), which appears twice, should only be taken into account once.- © 2007 The Royal Society