aa r X i v : . [ m a t h - ph ] J a n ON THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION STEFAN HOROCHOLYN
Abstract.
We derive a formula for the signature of the symmetrized Stokesmatrix S + S T for the tt ∗ -Toda equation. As a corollary, we verify a conjectureof Cecotti and Vafa regarding when S + S T is positive definite, reminiscentof a formula of Beukers and Heckmann for the generalized hypergeometricequation. The condition S + S T > tt ∗ equation; we show that the Stokes matrices S satisfying thiscondition are parameterized by the points of an open convex polytope. Introduction
The classical Stokes phenomenon for meromorphic ODE has begun to play animportant role in geometry, notably in singularity theory, Frobenius manifolds andmirror symmetry. For a (real) Stokes matrix S , the symmetrized matrix S + S T arises in the context of Frobenius manifolds and the tt ∗ equation (e.g. [6, 7, 13, 14]).The tt ∗ equation is a system of nonlinear PDEs which appeared in the work ofCecotti and Vafa [2, 3, 4] on the classification of supersymmetric field theories inphysics. It is a special case of the harmonic map equation in differential geometryfor maps from a surface to a noncompact symmetric space [11]. Dubrovin [5]showed that it admits an isomonodromic deformation interpretation, as well asa zero-curvature formulation. This leads to a Riemann-Hilbert correspondencebetween (local) solutions and monodromy data of a meromorphic ODE. Clarifyingthis correspondence is a subject of current research activity relating several fieldsof mathematics, including Hodge theory and algebraic geometry.There are very few examples where solutions can be found. A special case of the tt ∗ equation, introduced by Cecotti and Vafa, and studied mathematically by Guest-Its-Lin [9, 10] and Mochizuki [15], are the tt ∗ -Toda equation. This is, essentially,the well-known Toda field equation (2-dimensional periodic Toda lattice), althougheven in this case the existence of the solutions predicted by Cecotti and Vafa wasproved only recently (in the aforementioned references).This article was motivated by the conjectures of Cecotti and Vafa regarding thesymmetrized Stokes matrix S + S T , in the case of the tt ∗ -Toda equation. We shallgive a necessary and sufficient condition for S + S T to be positive definite, and asimple formula for the signature of S + S T , in general, which is reminiscent of aformula of Beukers and Heckmann for the generalized hypergeometric equation [1].Let us now state the tt ∗ -Toda equation and explain the relevant Stokes matrix.The equations are:(1.1) 2( w i ) zz = − e w i +1 − w i ) + e w i − w i − ) , w i : C ∗ → R subject to two further conditions:(1) the “anti-symmetry” condition: w i + w n − i = 0; and(2) the radial condition: w i = w i ( | z | ). We assume that w i = w i + n +1 for all i ∈ Z . In what follows, we write n + 1 = 2 m or n + 1 = 2 m + 1 (i.e. m := ⌊ n +12 ⌋ ).This system is the compatibility condition for the linear system: (cid:26) Ψ z = (w z + λ W)Ψ , Ψ z = ( − w z + λ W T )Ψ , where: w = diag ( w , . . . , w n ) , W = e w − w . . . e w n − w n − e w − w n . If we write x = | z | , then the radial version of (1.1) is the compatibility conditionfor a linear system, which may then be transformed to (see Equation 1.4 of [10]):(1.2) Ψ ζ = (cid:16) − ζ W − ζ x w x + x W T (cid:17) Ψ , Ψ x = (cid:16) xζ W + xζ W T (cid:17) Ψ , where ζ = λz .The equation for Ψ ζ in (1.2) is a meromorphic linear ODE in the complex variable ζ , with poles of order two at both ζ = 0 and ζ = ∞ . The Stokes matrices at thesetwo poles are equivalent, so we shall only consider the Stokes matrix at ζ = ∞ , anddenote it by S . By the general theory of isomonodromic deformations (e.g. [8]),Stokes matrices S correspond to local solutions near 0 (i.e. defined on intervals ofthe form (0 , ε )) of the tt ∗ -Toda equation. Further details and explanation may befound in [9, 10], where S is computed in terms of the asymptotic behaviour of thefunctions w i .It was conjectured by Cecotti and Vafa that the condition S + S T > tt ∗ -Toda equation is globally defined on C ∗ (i.e. such that ε = ∞ ). This was confirmed in [9, 10, 11, 15], and in Theorem5.6 of [10], a stronger result (also suggested by Cecotti and Vafa) was shown: anecessary and sufficient condition for the local solution of the tt ∗ -Toda equation tobe globally defined on C ∗ is that the eigenvalues of the monodromy S ( S − ) T areof unit length.It is, therefore, of interest to describe the set of such Stokes matrices explicitly,and this is our first main result. For such Stokes matrices, we prove the followingexplicit characterization of the signature of S + S T , showing that they form an openconvex polytope described by simple equations, and we expect this result to be ofuse in future investigations of the tt ∗ -Toda equation: Theorem: S + S T has the same signature as the diagonal matrix: diag (( − n +1 p ( π ) , . . . , ( − n +1 p ( π n )) . Here, π k are the n + 1 roots of x n +1 − ( − n +1 , and the real polynomial p ( x ) is thecharacteristic polynomial of a certain matrix R satisfying ( − n R n +1 = SS − T . Corollary: S + S T > − n +1 p ( π k ) > for all k , and the set of suchStokes matrices is in 1-1 correspondence with an open, convex polytope of R m . Our second main result is a formula for the sign of p ( π k ) when the eigenvaluesof S ( S − ) T are unimodular. We refer the reader to Corollary 2.12 for the precise N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 3 statement of this result. This characterizes the signature of S + S T in terms ofthe configurations of the eigenvalues of R with respect to the roots π k on the unitcircle.These results are given in Section 2 for a conveniently defined, “idealized Stokesmatrix” S. In Section 3, we explain the precise relation between this “idealizedStokes matrix”, and the “actual” Stokes matrices of [9, 10].Notational remark: In this paper, N shall denote the natural numbers, Z the in-tegers, Z ≥ the non-negative integers, R the real numbers, C the complex numbers,and C ∗ = C \ { } the complex plane punctured at the origin. For a matrix A, itstranspose is denoted A T , and A − T will denote the inverse of A T .2. Main Results
Let R ∈ SL n R be given by:(2.1) R := − p n − · · · · · ... ... I n − ... − p · · · · ·− ǫ · · · , where ǫ := ( − n , and where the characteristic polynomial p ( x ) of R, p ( x ) = x n + n − X k =1 p k x k + ǫ , is signed-palindromic :(2.2) p ( x ) = ( − x ) n p (cid:0) x (cid:1) ⇔ p n − k = ǫp k ∀ ≤ k ≤ n − . In addition, let S be the upper-triangular Toeplitz matrix:(2.3) S := ǫp · · · ǫp n − ǫp n − ǫp n − ... . . . . . . . . . ...... . . . . . . 1 ǫp · · · · · · . Proposition 2.1. − ǫ R n = SS − T . Proof:
We show that − ǫ R n has an upper-lower-triangular decomposition: − ǫ R n = UL , where U and L have only 1s on the diagonal, and in doing so, will show that U = Sand L − = S T . We first apply − ǫ R n and UL to the flag: F : ⊂ F ⊂ F ⊂ · · · ⊂ F n − ∼ = R n , F k := < e n , e n − , . . . , e n − k +1 , e n − k > R , where e k is the k th canonical unit vector. By inspection, L fixes F , so it suffices tocompare respective applications of − ǫ R n and U to F . STEFAN HOROCHOLYN
The proof is facilitated by two observations: the first is that R e k = e k − forall 2 ≤ k ≤ n , which implies that e k = R n − k e n for all 1 ≤ k ≤ n . The secondobservation is that, by Cayley-Hamilton:R n + n − X k =1 p k R k + ǫ I n = 0 , which, taken together with the first observation, yields:R n e n = − p n − e − · · · − p e n − − ǫ e n . The proof proceeds as follows: applying − ǫ R n and U to e n ∈ F yields: ǫp n − e + · · · + ǫp e n − + e n = − ǫ R n e n = U e n = u ,n e + · · · + u n − ,n e n − + e n , which determines the last column of U. Next, to determine the second-last column,we observe that, on the one hand:R n F = R n F + R n < e n − > R , and on the other hand, < R e > R = R n F by the above observations. Hence, itfollows that: − ǫ R n e n − = ǫ R (cid:0) p n − e + p n − e + · · · + p e n − + ǫ e n (cid:1) ≡ ǫ R (cid:0) p n − e + · · · + p e n − + ǫ e n (cid:1) mod R n F = ǫp n − e + · · · + ǫp e n − + e n − . By comparing this with the application of U to e n − ∈ F :U e n − = u ,n − e + · · · + u n − ,n − e n − + e n − , we see that we have determined the second-last column of U. Continuing thisprocess, at the k th step ( k ≥ F k − :R n F k − = R n F k − + R n < e k > R , and by the above observations:R n F k − = R k − < e k − , e k − , . . . , e > R . Hence, taking e n − k +1 ∈ F k − , it follows that: − ǫ R n e n − k +1 = ǫ R k − (cid:0) p n − e + · · · + p n − k +1 e k − + p n − k e k + · · · + p e n − + ǫ e n (cid:1) ≡ ǫ R k − (cid:0) p n − k e k + · · · + p e n − + ǫ e n (cid:1) mod R n F k − = ǫp n − k e + · · · + ǫp e n − k + e n − k +1 . Comparing this expression with that obtained from application of U:U e n − k +1 = u ,n − k +1 e + · · · + u n − k,n − k +1 e n − k + e n − k +1 , we see that we have determined the ( n − k + 1) th column of S, for all 1 ≤ k ≤ n .Evidently, we have deduced that u ij = ǫp i − j , from which we conclude that U = S.Now, we repeat the above, but for the matrix − ǫ R − n = L − U − and the flag:˜ F : ⊂ ˜ F ⊂ ˜ F ⊂ · · · ⊂ ˜ F n − ∼ = R n , ˜ F k := < e , . . . , e k +1 > R , which is fixed by U − . It follows that e k +1 = R − k e for all 0 ≤ k ≤ n −
1, fromwhich it follows, by Cayley-Hamilton and (2.2), that:R − n e = − p n − e n − · · · − p e − ǫ e . We then deduce that L − = S T , as was to be shown. (cid:4) N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 5 For the next proposition, let ω := e πin , and define the matrix Π ǫ ∈ GL n R by:(2.4) Π ǫ := (cid:20) I n − ǫ (cid:21) . Then the characteristic polynomial of Π ǫ is x n − ǫ , and hence, its eigenvalues are: π k := ( ω k , ≤ k ≤ n − , n = 2 m ,ω k + , ≤ k ≤ n − , n = 2 m + 1 . Proposition 2.2.
The eigenvalues of ǫ (S + S T ) are p ( π ) , . . . , p ( π n − ) . Proof:
Since p ( x ) satisfies (2.2), it is then evident that ǫ (S + S T ) = p (Π ǫ ), andhence, S + S T commutes with Π ǫ , so they may be diagonalized simultaneously. Toeach π k , then, we let v k denote the corresponding eigenvector of Π ǫ :(2.5) v k := 1 n (1 , π k , π k , . . . , π n − k ) T , ≤ k ≤ n − , It then follows that ǫ (S + S T ) v k = p ( π k ) v k for each k . (cid:4) Corollary 2.3.
S + S T has the same signature as the diagonal matrix diag ( ǫp ( π ) , . . . , ǫp ( π n − )) . In particular: • S + S T is positive definite iff ǫp ( π k ) > for all k . • The number of zero eigenvalues is the number of common eigenvalues of R and Π ǫ . (cid:4) As before, let us write n = 2 m for even n , and n = 2 m + 1 for odd n . We recallthat a complex number of unit norm is said to be unimodular . Proposition 2.4. (cf. [4] , [10] ) If S + S T is positive definite, then the eigenvaluesof R are unimodular. Moreover, the set of all R such that S + S T is positive definiteis in 1-1 correspondence with the bounded convex region of R m defined by: P := m \ k =0 { ( p , . . . , p m ) ∈ R m | ǫp ( π k ) > } . Proof: SS − T preserves the symmetric bilinear form S + S T [4]: (cid:0) SS − T (cid:1) T (S + S T ) (cid:0) SS − T (cid:1) = S + S T . Hence, SS − T is orthogonal, and all eigenvalues of SS − T are unimodular. As a result,by Proposition 2.1, the eigenvalues of R must be unimodular, as well.Next, we establish that the set of all R defined as in (2.1), such that S + S T ispositive definite, is in 1-1 correspondence with P : • By Corollary 2.3, ǫp ( π k ) > k , and thus, the entries p , . . . , p m ofeach such R determine a unique point ( p , . . . , p m ) ∈ P . • Conversely, given a point P := ( p , . . . , p m ) ∈ P , and defining R , S accord-ing to (2.1), (2.2) and (2.3) via the components of P , it follows by thedefinition of P and by Corollary 2.3 that S + S T is positive definite, andhence, by the first assertion, all eigenvalues of R are unimodular.Since the entries p , . . . , p m of R are the elementary symmetric polynomials ofthe eigenvalues of R, all of which lie, by assumption, in the compact set S , P isconsequently contained in the continuous image of a compact set, and hence, isbounded. STEFAN HOROCHOLYN
Lastly, as each inequality ǫp ( π k ) > R m , and that theintersection of any collection of convex sets is convex, we see that P is convex. (cid:4) Taking S + S T non-degenerate, henceforth, we shall now consider the dependenceof the signature σ of S + S T on the eigenvalues of R, when R has only unimodulareigenvalues. By Proposition 2.2 , we note that σ is constant with respect to anyvariation (within S ) of an eigenvalue of R such that the eigenvalue does not passthrough a root of x n − ǫ . Hence, σ is a function of only the number of eigenvaluesof R between each root of x n − ǫ . When n = 2 m + 1, the conjugate symmetry ofthe eigenvalues implies that σ is also a function of the number of eigenvalues in thearc { e iθ | θ ∈ [0 , π m +1 ) } . We now introduce some notation to assist in discussingthis: Definition 2.5.
Assume R has only unimodular eigenvalues e ± iθ j , ≤ j ≤ m .(For n = 2 m + 1 , we do not include the guaranteed eigenvalue z = 1 in this list.)When n = 2 m , the configuration ρ of R is defined to be ρ = ( ρ , . . . , ρ m ) ∈ Z m ≥ such that ρ k := n j θ j ∈ (cid:0) ( k − πm , kπm (cid:1)o . For n = 2 m + 1 , the configuration ρ ofR is defined to be ρ = ( ρ , . . . , ρ m ) ∈ Z m +1 ≥ such that: ρ k := n j θ j ∈ (cid:2) , π m +1 (cid:1)o , k = 0 , n j θ j ∈ (cid:0) (2 k − π m +1 , (2 k +1) π m +1 (cid:1)o , ≤ k ≤ m . Necessarily, the sum of the components of ρ is always m , for all n , and we saythat two matrices (for the same n ) have the same configuration whenever theirconfiguration sequences agree.To find σ in terms of ρ , we shall apply Descartes’ Rule of Signs to a polynomialwith only real roots uniquely derived from p ( x ), and then relate this to a formulaproved via an adaptation of the argument of [1, Theorem 4.5]. The result will thenbe, in principle, a formula to determine σ from only the entries of S.2.1. Recall that Descartes’ Rule of Signs is the following classical result, whoseproof we omit: Proposition 2.6. (cf. [12, Theorem 6.2d] ) Let p ( x ) ∈ R [ x ] have degree n ∈ N , withnon-zero leading coefficient a n , and let ν denote the number of sign changes in thesequence of non-zero coefficients of p ( x ) , starting with a n and listed in decreasingorder of the power of x . If r denotes the number of real, positive roots of p ( x ) ,where each root is counted according to its algebraic multiplicity, then ν − r is evenand non-negative. To refine this for polynomials with only real roots, we first prove:
Lemma 2.7.
Let p ( x ) = P nk =0 a k x k ∈ R [ x ] be non-zero, and let ν and µ be thenumber of sign changes in the decreasing sequence of non-zero coefficients of p ( x ) and p ( − x ) , respectively. Then ν + µ ≤ n , and equality holds iff a k = 0 for all k . Proof:
Let σ ( a k , a k − ) denote the number of sign changes in the 2-term sequence( a k , a k − ), allowing this to be 0 when at least one of a k or a k − are 0. Then by This was observed in [4, p.27] in the context of the general tt ∗ -equation. N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 7 definition of ν , ν = P nk =1 σ ( a k , a k − ). Now, by inspection: χ k := σ ( a k , a k − ) + σ (( − k a k , ( − k − a k − ) = ( , a k = 0 and a k − = 0 , , a k = 0 or a k − = 0 , so summing over all k , it follows that P nk =1 χ k = µ + ν . But the left-hand side isan n -term summation of 1s and 0s, so µ + ν ≤ n , and µ + ν = n iff all n terms ofthe sum are 1, iff a k = 0 for all 0 ≤ k ≤ n . (cid:4) Corollary 2.8. If p ( x ) ∈ R [ x ] has only real roots, then ν = r . Proof:
Let ν and µ be defined as in the lemma, and let r and s be the number ofpositive roots of p ( x ) and p ( − x ), respectively. (Clearly, s is the number of negativeroots of p ( x ).) Then by Descartes’ Rule applied to both p ( x ) and p ( − x ), there are α, β ∈ Z ≥ such that ν − r = 2 α and µ − s = 2 β .First, assume that p ( x ) has only non-zero real roots, so that n = r + s . Then n = ν + µ − α + β ), so by the lemma and non-negativity of α and β , α = β = 0.Now suppose that p ( x ) has only real roots, and t of them zero. Then p ( x ) x − t has only non-zero real roots, and the number of its positive roots is also r , so bythe previous assertion, if ν ′ is the number of sign changes of p ( x ) x − t , then ν ′ = r .But evidently, ν = ν ′ , which proves the assertion. (cid:4) Now, consider the following: for any p ( x ) ∈ R [ x ] satisfying (2.2), it may beshown by induction that there is a unique monic ˜ p ( x ) ∈ R [ x ] such that:(2.6) ( p ( x ) = x m ˜ p (cid:0) x + x (cid:1) , n = 2 m ,p ( x ) = ( x − x m ˜ p (cid:0) x + x (cid:1) , n = 2 m + 1 , (For n = 2 m +1, we note that p (1) = 0 by (2.2), so after factoring p ( x ) = ( x − q ( x ),we see that q ( x ) satisfies (2.2), and thus, the even case factorization applies.)As all eigenvalues of R are unimodular, ˜ p ( x ) has one root 2 cos θ j for each con-jugate pair of roots e ± iθ j of p ( x ). This motivates the following: Proposition 2.9.
Given ˜ p ( x ) as in (2.6), let ˜ p [0] ( x ) := ˜ p ( x + 2) for all n , and: ˜ p [ k ] ( x ) := ( ˜ p (cid:0) x + 2 cos kπm (cid:1) , ≤ k ≤ m , n = 2 m , ˜ p (cid:0) x + 2 cos (2 k − π m +1 (cid:1) , ≤ k ≤ m + 1 , n = 2 m + 1 . Denote by ν k the number of sign changes in the sequence of non-zero coefficients of ˜ p [ k ] ( x ) , as in Descartes’ Rule. Then the configuration ρ of R satisfies: ν k − ν k − = ( ρ k , ≤ k ≤ m , n = 2 m ,ρ k − , ≤ k ≤ m + 1 , n = 2 m + 1 . Conversely, ν k is given by: ν k = (P kj =1 ρ j , ≤ k ≤ m , n = 2 m , P k − j =0 ρ j , ≤ k ≤ m + 1 , n = 2 m + 1 . Proof:
Evidently, when n = 2 m , ρ k is the number of roots of ˜ p ( x ) in the interval (cid:0) kπm , ( k − πm (cid:1) , and when n = 2 m + 1, ρ k is the number of roots of ˜ p ( x ) in: ( (2 cos π m +1 , , k = 0 , (cid:16) (2 k +1) π m +1 , (2 k − π m +1 (cid:17) , ≤ k ≤ m . STEFAN HOROCHOLYN
On the other hand, by construction of the ˜ p [ k ] ( x ), the number of positive roots r k of ˜ p [ k ] ( x ) is the same as the number of roots of ˜ p ( x ) strictly greater than 2 cos kπm when n = 2 m , and when n = 2 m + 1, it is the same as the number of roots of ˜ p ( x )strictly greater than 2, for k = 0, or 2 cos (2 k − π m +1 for all other k . Moreover, all m roots of ˜ p ( x ) are real, and thus, ν k = r k by Corollary 2.8. Applying Descartes’ Ruleto ˜ p [ k ] ( x ) and ˜ p [ k − ( x ), it then follows that: ν k − ν k − = ( ρ k , ≤ k ≤ m , n = 2 m ,ρ k − , ≤ k ≤ m + 1 , n = 2 m + 1 . Conversely, given ρ , it follows from the above that, for all k : ν k = (P kj =1 ρ j , ≤ k ≤ m , n = 2 m , P k − j =0 ρ j , ≤ k ≤ m + 1 , n = 2 m + 1 . (cid:4) For notational convenience, we shall always denote the sequence of the numberof sign changes of ˜ p [ k ] ( x ) by: ν := ( (0 , ν , . . . , ν m − , m ) , n = 2 m , (0 , ν , . . . , ν m , m ) , n = 2 m + 1 . Remark 2.10.
For the matrix R with characteristic polynomial p ( x ) = x m + 1,S + S T = 2I m , and the configuration is ρ = (1 , , . . . , ν = (0 , , , . . . , m − , m ). Hence, by connectedness of P in Propo-sition 2.4, for any R with only unimodular eigenvalues when n = 2 m , S + S T > iff its configuration is (1 , , . . . , iff its sequence of sign-change num-bers is (0 , , . . . , m − , m ). Similarly, for the matrix R with characteristic poly-nomial p ( x ) = x m +1 −
1, we have S + S T = 2I m +1 , ρ = (0 , , , . . . , ν = (0 , , , , . . . , m − , m ), and this is the only configuration for which S + S T > T > iff the eigenvalues e ± iθ j of R interlace with the roots π k of ǫ (including theguaranteed root z = e i when n = 2 m + 1): ( < θ < πm < θ < πm < · · · < ( m − πm < θ m < π , n = 2 m , − π m +1 < < π m +1 < θ < π m +1 < · · · < (2 m − π m +1 < θ m < π , n = 2 m + 1 . σ using the sequence ν of sign-change numbers of the ˜ p [ k ] ( x ). Before proving the formula, we first note:(1) Since the characteristic polynomial p ( x ) of R satisfies (2.2), it follows thatdet( x I n − R − T ) = p ( x ), as well.(2) Letting D := I n − Π ǫ R T , we remark that D has rank 1, and for all x ∈ C n ,D x = (cid:0) e ∗ n (S + S T ) x (cid:1) e n . Note, as well, that (Π ǫ R T ) = I n .(3) [1] If a rank one n × n matrix M acts on C n as M x = w ( x ) u for some linearform w and for some u ∈ C n , then det(I n + M) = 1 + w ( u ).(4) Letting v k be defined as in (2.5), we note that P n − k =0 π k v k = Π ǫ e = ǫ e n .We now prove what is essentially a special case of Theorem 4.5 of [1]: Proposition 2.11.
For R with only unimodular eigenvalues such that S + S T isnon-degenerate, denote the eigenvalues of R as z k := e πiθ k , where θ k ∈ [0 , , for ≤ k ≤ n − . When n = 2 m + 1 , we take z m = 1 , so θ m = 0 . If n j := { k | θ k < arg π j } for each ≤ j ≤ n − , then sgn ( p ( π j )) = ( − n j − j . N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 9 Proof:
We adapt the method of proof of Theorem 4.5 of [1] as follows: Expand-ing p ( x ) and x n − ǫ into their complex linear factors, we use the above remarks toobtain: n − Y k =0 ( z k − x )( π k − x ) − = det(R − x I n ) det(Π ǫ − x I n ) − = det(R − T − x I n ) det(Π ǫ − x I n ) − = det (cid:0) (R − T Π − ǫ − x Π − ǫ )(I n − x Π − ǫ ) − (cid:1) = det (cid:0) ( − D + I n − x Π − ǫ )(I n − x Π − ǫ ) − (cid:1) = det (cid:0) I n − D(I n − x Π − ǫ ) − (cid:1) = 1 − e ∗ n (S + S T )(I n − x Π − ǫ ) − e n = 1 − e ∗ n (S + S T )(Π ǫ − x I n ) − Π ǫ e n . Using the expression for e n in terms of the v k then yields: n − Y k =0 ( z k − x )( π k − x ) − = 1 − n − X k,j =0 π j π k π j − x v ∗ k (S + S T ) v j = 1 − n − X j =0 π j π j − x v ∗ j (S + S T ) v j . Taking the residue at x = π j , and inserting ǫ = i Q n − k =0 ( π k ) z − k , we find that: ǫ v ∗ j (S + S T ) v j = − ǫ ( z j − π j ) π − j Y k = j ( z k − π j )( π k − π j ) − = − i n − Y k =0 π k z − k ! ( z k − π j ) π − j Y k = j ( z j − π j )( π k − π j ) − = − i z j π j − π j z j Y k = j z k π j − π j z k π k π j − π j π k − = 2 sin π ( θ j − arg π j ) Y k = j sin π ( θ k − arg π j )sin(arg π k − arg π j ) . The sign of the denominator is ( − j , by inspection, and the sign of the numeratoris ( − n j , by definition of n j . Thus, sgn ( p ( π j )) = ( − n j − j , by Proposition 2.2. (cid:4) Corollary 2.12.
Let ν be the sequence of sign-change numbers of ˜ p [ k ] ( x ) , and let S + S T have n + positive and and n − negative eigenvalues. Then for n = 2 m : ( n j = ν j , ≤ j ≤ m ,n m − j = 2 m − ν j , ≤ j ≤ m − , and for n = 2 m + 1 : ( n j − ν j +1 , ≤ j ≤ m ,n m − j − m − ν j +1 , ≤ j ≤ m − . Note that z k z n − k = − k , since θ k ∈ [0 ,
1) and the z k come in conjugate pairs, exceptfor z m = 1 when n = 2 m + 1. A similar statement holds for the π k . Consequently, for all j : sgn ( p ( π j )) = ( ( − ν j − j , n = 2 m , ( − ν j +1 − ( j +1) , n = 2 m + 1 , and thus, for all n : n + − m − (P m − j =1 ( − ν j − j , n = 2 m , − P mj =1 ( − ν j − j , n = 2 m + 1 . Proof:
The relations between n j and ν j follow from the definition of n j andfrom Proposition 2.9. Note that, when n = 2 m + 1, n j ≥ j due to theguaranteed root z m = 1 of p ( x ). By Proposition 2.11, the formula for sgn ( p ( π j ))then follows immediately. Since n + + n − = n whenever S + S T is non-degenerate,it follows by the above formula for sgn ( p ( π j )) that: ǫ (2 n + − n ) = ǫ ( n + − n − ) = ( ( − ν + ( − ν m − m + 2 P m − j =1 ( − ν j − j , n = 2 m , ( − ν m +1 − ( m +1) + 2 P mj =1 ( − ν j − j , n = 2 m + 1 . When n = 2 m , ν j = j for j = 0 and j = m , and hence: n + = m + 1 + m − X j =1 ( − ν j − j . When n = 2 m + 1, ν m +1 = m for j = m , and hence: n + = m + 1 − m X j =1 ( − ν j − j . (cid:4) Remark 2.13.
We now provide some sample calculations. When n = 4 (cf. [10]),the relation between σ = ( n + , n − ) and ν = (0 , ν ,
2) is: ν σ (2 ,
2) (4 ,
0) (2 , n = 6, the signatures may be tabulated as follows: ν (cid:31) ν ,
2) (2 ,
4) (4 ,
2) (2 , × (4 ,
2) (6 ,
0) (4 , × × (4 ,
2) (2 , × × × (4 , Application to the tt ∗ -Toda equation Now we shall apply the results of the previous section to the symmetrized Stokesmatrices of the tt ∗ -Toda equation, which were calculated in [9, 10, 15]. Let usconsider the pole at infinity (of order 2) of Equation (1.2), for arbitrary n + 1. InSection 4 of [9], the case n + 1 = 4 was treated in detail. The same method appliesfor any n ≥
3, so we shall just summarize the results briefly.Recall that we wish to determine the Stokes data at ζ = ∞ for the ODE:(3.1) Ψ ζ = (cid:16) − ζ W − ζ x w x + x W T (cid:17) Ψ , N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 11 where w and W are defined before (1.2). If η := ζ − , then we may re-write this as:Ψ η = (cid:18) − η x W T + O (cid:0) η (cid:1)(cid:19) Ψ . Letting ω := e πin +1 , d n +1 := diag (1 , ω, . . . , ω n ), and Ω := (cid:0) ω ij (cid:1) ni,j =0 , we may usethe matrix P ∞ := diag ( e w , . . . , e w n ) Ω − to diagonalize W T as:W T = P ∞ d n +1 P − ∞ . Then by Proposition 1.1 of [8], and reverting back to ζ , we see that there exists aunique formal solution Ψ ∞ f of (3.1) of the form:Ψ ∞ f = P ∞ I n +1 + X j ≥ ψ ∞ j ζ − j e Λ log η + x ζ d n +1 . It may then be verified, by direct substitution into (3.1), that Λ = 0. By Theorem1.4 of [8], there is then a unique holomorphic solution Ψ to (3.1), with asymptoticexpansion Ψ ∞ f , on any Stokes sector based at ζ = ∞ .There are 2( n + 1) Stokes rays, given by all ζ ∈ C ∗ satisfying:(3.2) cos (cid:0) arg ζ + arg( ω j − ω k ) (cid:1) = 0 , where ω := e πin +1 . As fundamental Stokes sectors, when n + 1 = 2 m , we take:Ω ∞ = { ζ ∈ C ∗ | − π < arg ζ < π + πn +1 } , Ω ∞ = { ζ ∈ C ∗ | π < arg ζ < π + πn +1 } , and when n + 1 = 2 m + 1, we take:Ω ∞ = { ζ ∈ C ∗ | − π − π n +1) < arg ζ < π + π n +1) } , Ω ∞ = { ζ ∈ C ∗ | π − π n +1) < arg ζ < π + π n +1) } . The Stokes matrix S ∞ is defined by Ψ ∞ = Ψ ∞ S ∞ , where Ψ ∞ is the canonicalsolution with prescribed asymptotics on Ω ∞ , and where the analytic continuationof Ψ ∞ to Ω ∞ is taken in the positive direction.Letting Π := (cid:0) I n (cid:1) , and using the symmetries of (1.2), as in Section 4 of [9], wefind that: S ∞ = ( (cid:0) Q ∞ Q ∞ n +1 Π (cid:1) m Π − m , n + 1 = 2 m , (cid:0) Q ∞ Q ∞ n +1 Π (cid:1) m Q ∞ Π − m , n + 1 = 2 m + 1 , and for all n , the inverse of the monodromy of Ψ ∞ is:S ∞ S ∞ = (cid:0) Q ∞ Q ∞ n +1 Π (cid:1) n +1 . Here, the matrices Q ∞ k are the “Stokes factors” of S ∞ and S ∞ , defined with respectto the Stokes sectors Ω ∞ k +1 = e kπi Ω ∞ for all k ∈ n +1 Z (i.e. Ψ ∞ k + n +1 = Ψ ∞ k Q ∞ k ).As in Section 5 of [10], we may convert to real matrices ˜S ∞ k and ˜Q ∞ k by conju-gating by (cid:0) diag (1 , ω, . . . , ω n ) (cid:1) r . Taking ε := ( − n +1 and:(3.3) R ε := (cid:20) I n − ε (cid:21) , R := ˜Q ∞ ˜Q ∞ n +1 R ε , For n + 1 = 2 m , r = , and for n + 1 = 2 m + 1, r = m + 1. (3.4) J := ( R mε = (cid:0) I m − I m (cid:1) , n + 1 = 2 m , R mε (cid:0) ˜Q ∞ (cid:1) − = Π m (cid:0) ˜Q ∞ (cid:1) − , n + 1 = 2 m + 1 , we then obtain:(3.5) S := ˜S ∞ = R m J − , SS − T = ˜S ∞ ˜S ∞ = − ε R n +1 . We are now almost in the situation of Section 2 of this article. Due to the choiceof formal solutions made in [9], the Stokes matrices ˜S ∞ k are elements of the group: { A ∈ SL n +1 R | A T J A =
J } , but the “idealized” Stokes matrices of Section 2 are not, in general. Therefore, to beable to apply the obtained results to S + S T , a further transformation is required.To find the correct transformation, we need explicit expressions for the matrices˜Q ∞ and ˜Q ∞ n +1 . Their non-zero entries can be deduced from: Lemma 3.1.
The diagonal entries of ˜Q ∞ and ˜Q ∞ n +1 are , and the other entriessatisfy the rule (for ≤ i = j ≤ n + 1 ): ( n + 1 = 2 m ) ( arg( ω i − − ω j − ) = nπn +1 mod 2 π ⇒ ( ˜Q ∞ ) i,j = 0 , arg( ω i − − ω j − ) = ( n − πn +1 mod 2 π ⇒ ( ˜Q ∞ n +1 ) i,j = 0 . ( n + 1 = 2 m + 1) ( arg( ω i − − ω j − ) = (2 n +1) π n +1) mod 2 π ⇒ ( ˜Q ∞ ) i,j = 0 , arg( ω i − − ω j − ) = (2 n − π n +1) mod 2 π ⇒ ( ˜Q ∞ n +1 ) i,j = 0 . Proof:
We follow the proof of Lemma 4.4 of [9]. For the complex Stokes factorsQ ∞ k , k ∈ n +1 Z , we have:Q ∞ k = lim ζ →∞ (Ψ ∞ k ) − Ψ ∞ k + n +1 = lim ζ →∞ e − ζx d n +1 (cid:0) I n +1 + O ( ζ ) (cid:1) e ζx d n +1 , and hence, (Q ∞ k ) ii = 1 for all i . On the other hand, for ( i, j ) such that 1 ≤ i = j ≤ n + 1, the entry (Q ∞ k ) ij = 0 so long as there is a path ζ t → ∞ in Ω k ∩ Ω k + n +1 suchthat Re ζ t ( ω j − − ω i − ) ≤
0. Since Ω k ∩ Ω k + n +1 is a sector of angle π , it followsthat (Q ∞ k ) ij is necessarily zero only if ( ω j − − ω i − ) Ω k ∩ Ω k + n +1 overlaps with theclosed half-plane { Re ζ ≤ } . But this occurs iff :arg( ω i − − ω j − ) = ( (2 n +1 − ( n +1) k ) πn +1 mod 2 π , n + 1 = 2 m , (4 n +3 − n +1) k ) π n +1) mod 2 π , n + 1 = 2 m + 1 . It then follows by the definition of the ˜Q ( ∞ ) k that their entries satisfy the sameconditions, and by substituting k = 1 and k = 1 n +1 , the assertion follows. (cid:4) Consequently, when n + 1 = 2 m , the entries must satisfy are: ( arg( ω i − − ω j − ) / π m = 2 m − m ⇒ ( ˜ Q ∞ ) i,j = 0 , arg( ω i − − ω j − ) / π m = 2 m − m ⇒ ( ˜ Q ∞ n +1 ) i,j = 0 , and when n + 1 = 2 m + 1, we instead have: ( arg( ω i − − ω j − ) / π m +1) = 4 m + 1 mod 8 m + 4 ⇒ ( ˜ Q ∞ ) i,j = 0 , arg( ω i − − ω j − ) / π m +1) = 4 m − m + 4 ⇒ ( ˜ Q ∞ n +1 ) i,j = 0 . Hence, the potentially non-zero entries are:
N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 13 • n + 1 = 2 m : ( ˜ Q ∞ ) m − k, k , ( ˜ Q ∞ ) m +1+ k, m − k , ( ˜ Q ∞ n +1 ) m − − k, k and( ˜ Q ∞ n +1 ) m + k, m − k , for k ≥ • n + 1 = 2 m + 1: ( ˜ Q ∞ ) m +1 − k, k , ( ˜ Q ∞ ) m +2+ k, m +1 − k , ( ˜ Q ∞ n +1 ) m − k, k ,and ( ˜ Q ∞ n +1 ) m +1+ k, m +1 − k for k ≥ n + 1 = 2 m that for all k ≥ ( ( ˜ Q ∞ ) m − k, k + ( ˜ Q ∞ ) m +1+ k, m − k = 0 , ( ˜ Q ∞ n +1 ) m − − k, k + ( ˜ Q ∞ n +1 ) m +1+ k, m − − k = 0 . For convenience of notation, let us define p m := ( ˜ Q ∞ n +1 ) m, m and:(3.6) − p m − k − := ( ˜ Q ∞ n +1 ) m − − k, k , − p m − k − := ( ˜ Q ∞ ) m − k, k . Then ˜Q ∞ and ˜Q ∞ n +1 are the block-matrices:˜Q ∞ = ˜L ∞ (cid:0) ˜L ∞ (cid:1) − T , ˜Q ∞ n +1 = ˜L ∞ n +1 p m E m,m (cid:0) ˜L ∞ n +1 (cid:1) − T , where: ˜L ∞ = I m − ℓ − X k =0 p m − k − E m − k, k , ℓ := ⌊ m ⌋ , ˜L ∞ n +1 = I m − ℓ − X k =0 p m − k − E m − − k, k , ℓ := ⌊ m − ⌋ . When n + 1 = 2 m + 1, we instead have (see Appendix): ( ˜ Q ∞ ) m +1 − k, k + ( ˜ Q ∞ n +1 ) m +1+ k, m +1 − k = 0 , ( ˜ Q ∞ ) m +2+ k, m +1 − k + ( ˜ Q ∞ n +1 ) m − k, k = 0 , and looking ahead to Proposition 3.4, it will be convenient to define:(3.7) p m − k := ( − m ( ˜ Q ∞ ) m +1 − k, k , p m − k − := ( − m − ( ˜ Q ∞ n +1 ) m − k, k . Then: ˜Q ∞ = ˜L ∞ (cid:0) ˜L ∞ n +1 (cid:1) − T , ˜Q ∞ n +1 = ˜L ∞ n +1 (cid:0) ˜L ∞ (cid:1) − T , where: ˜L ∞ = I m +1 + ℓ ′ − X k =0 p m − k E m +1 − k, k , ℓ ′ := ⌊ m +12 ⌋ , ˜L ∞ n +1 = I m + ℓ ′ − X k =0 p m − k − E m − k, k , ℓ ′ := ⌊ m ⌋ . To facilitate the next few propositions, we introduce the permutation matrix∆ := P m − k =0 E m − − k, k and the block-matrix:(3.8) F := " L [ m ] U [ m ] , n + 1 = 2 m , " L [ m +1] U [ m ] , n + 1 = 2 m + 1 , where L [ m ] and U [ m ] are defined as follows (here, ℓ = ⌊ m ⌋ and ℓ = ⌊ m − ⌋ ):L [ m ] := I m + ℓ − X k =0 εp m − k − k X j =0 E m − k + j,j +1 + ℓ − X k =0 εp m − k − k X j =0 E m − k − j,j +1 , U [ m ] := " ∆L [ m − ∆ . For example, when m = 4:L [4] = εp εp εp εp , U [4] = εp εp . When m = 1, we define L [1] := 1 and U [1] := 1, and for the degenerate cases m ≤ [ m ] be the empty (0 ×
0) matrix. This allows us to state the followinglemma:
Lemma 3.2.
The following block-matrix identities hold for all m ≥ : (3.9) U [ m ] L [ m ] T = εp · · · εp m − εp m − . . . . . . εp n − . . . . . . ... εp . (3.10) F ˜Q ∞ = " L [ m − U [ m +1] , n + 1 = 2 m , " L [ m ] U [ m +1] , n + 1 = 2 m + 1 , (3.11) F ˜Q ∞ ˜Q ∞ n +1 = " L [ m − U [ m +2] , n + 1 = 2 m , " L [ m − U [ m +2] , n + 1 = 2 m + 1 . Proof:
These follow directly from (3.8), (3.6), and (3.7). (cid:4)
To make the connection with Section 2, we now define R as in Equation (2.1)using the entries p , . . . , p m defined via (3.6) or (3.7), and by defining p n +1 − k := εp k .Then: N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 15 Proposition 3.3.
R = F R F − . Hence: p ( x ) := det( x I n +1 − R ) = x n +1 + n X k =1 p k x k + ε , p n +1 − k = εp k . Proof:
Using (3.11), we obtain RFR − ε = F ˜Q ∞ ˜Q ∞ n +1 . (cid:4) Proposition 3.4.
Let
S := F S F T . Then: S = εp · · · εp n − εp n . . . . . . εp n − . . . . . . ... εp . Proof:
Letting J − p := F J − F T , we find, by definition (3.5) of S and Lemma3.3, that S = R m J − p . Hence, by inspection when n + 1 = 2 m , and by (3.10) when n + 1 = 2 m + 1:J − p = " − ε L [ m ] U [ m ] T U [ m ] L [ m ] T , n + 1 = 2 m , " − ε L [ m ] U [ m ] T U [ m +1] L [ m +1] T , n + 1 = 2 m + 1 . Since R e k = e k − for all 1 ≤ k ≤ n (cf. proof of Proposition 2.1), we see thatR m J − p e = e , and:R m J − p e k = p k − e + · · · + p e k − + e k , (cid:26) ≤ k ≤ m , n + 1 = 2 m ≤ k ≤ m + 1 , n + 1 = 2 m + 1 . This determines the first m ( m + 1) columns of S; we now determine the remaining m columns by applying S to the flag: F : ⊂ F ⊂ F ⊂ · · · ⊂ F n ∼ = R n +1 , F k := < e n +1 , . . . , e n +1 − k > R . From (3.9), we observe that − ε J − p e n +1 = e m , and that: − ε J − p e n +1 − k − e m − k ∈ < e m , . . . , e m − k +1 > R , ∀ ≤ k ≤ m − . As a result, for all 0 ≤ k ≤ m − − ε J − p F k = < e m , . . . , e m − k > R = (cid:26) R m F k , n + 1 = 2 m , R m +1 F k , n + 1 = 2 m + 1 , and hence: R m J − p F k = − ε R n +1 F k = U F k , where U was found in the proof of Proposition 2.1. Therefore, the last m columnsof S are the last m columns of U, and this concludes the proof. (cid:4) Theorem 3.5. S + S T has the same signature as diag ( εp ( π ) , . . . , εp ( π n )) . Proof:
By Proposition 3.4, S + S T and S + S T are congruent via F T . Since realsymmetric matrices are diagonalizable, congruence implies that they have equalrank and signature, and thus, the theorem follows by Propositions 3.3 and 2.2. (cid:4) To conclude, we state what this means in terms of solutions to the tt ∗ -Todaequation (1.1). It was shown in [9, 10, 11, 15] that solutions w i : C ∗ → R arein one-to-one correspondence with real numbers γ i satisfying γ i − γ i − ≥ − i , where 2 w i ( z ) ∼ γ i log | z | as | z | →
0. When n + 1 = 2 m , the correspondingeigenvalues of R are exp (cid:0) ± iπn +1 ( γ j + 2 j + 1) (cid:1) , 0 ≤ j ≤ m −
1, with0 ≤ πn +1 ( γ + 1) ≤ πn +1 ( γ + 3) ≤ · · · ≤ πn +1 ( γ m − + 2 m − ≤ π . The condition S + S T > n + 1) th roots of unity:0 < γ + 1 < < γ + 3 < < · · · < m − < γ m − + 2 m − < m , and this means that | γ j | < j = 0 , . . . , m − Appendix: The Stokes factor matrices for the case n + 1 = 2 m + 1We present the details of the n + 1 = 2 m + 1 case, some of which require separatecalculation from the n + 1 = 2 m case. We first recall, from Section 3 and (3.2),that the potentially non-zero entries of Q ∞ and Q ∞ n +1 are:( Q ∞ ) m +1 − k, k , ( Q ∞ ) m +2+ k, m +1 − k , ( Q ∞ n +1 ) m − k, k , and ( Q ∞ n +1 ) m +1+ k, m +1 − k , for k ≥ ∞ and Q ∞ n +1 have the followingblock structure:Q ∞ = (cid:20) L U (cid:21) , Q ∞ n +1 = " L n +1 U n +1 , where ( ℓ := ⌊ m +12 ⌋ , ℓ := ⌊ m ⌋ ):L := I m +1 + ℓ − X k =0 α m − k E m +1 − k, k , L n +1 := I m + ℓ − X k =0 α m − k − E m − k, k , and where the upper-triangular matrices U and U n +1 will be expressed in termsof L n +1 and L , respectively, using the following symmetries. Let ω := e (cid:0) πin +1 (cid:1) ,d n +1 := diag (1 , ω, . . . , ω n ), Π be as before, and ∆ := P n − k =0 E n − k, k . Recall thatfor all k ∈ n +1 Z :(1) Z n +1 -symmetry: Q ∞ k + n +1 = ΠQ ∞ k Π − .(2) “Anti-symmetry”: Q ∞ k +1 = d − n +1 (cid:0) Q ∞ k (cid:1) − T d n +1 .(3) Reality (at ζ = ∞ ): Q ∞ k = (∆Π) − (cid:0) Q ∞ n +1 n +1 − k (cid:1) − ∆Π.Then by Z n +1 -symmetry, and anti-symmetry for k = 1:Π m Q ∞ n +1 Π − m = Q ∞ = d − n +1 (cid:0) Q ∞ (cid:1) − T d n +1 , and hence: Q ∞ n +1 = (cid:0) d n +1 Π m (cid:1) − (cid:0) Q ∞ (cid:1) − T (cid:0) d n +1 Π m (cid:1) . Moreover, by the reality condition for k = 1, and Z n +1 -symmetry for k = nn +1 :Q ∞ = (∆Π) − (cid:0) Q ∞ nn +1 (cid:1) − (∆Π) = (∆Π) − Π − (cid:0) Q ∞ n +1 (cid:1) − Π(∆Π) = ∆ (cid:0) Q ∞ n +1 (cid:1) − ∆ . N THE STOKES MATRICES OF THE tt ∗ -TODA EQUATION 17 The block structure of Q ∞ n +1 then implies that:Q ∞ = U n +1 − T L n +1 − T , and hence: U = L n +1 − T , U n +1 = L − T . In addition, using the above identities, we deduce that: (cid:0) Π m ∆ (cid:1) Q ∞ (cid:0) Π m ∆ (cid:1) − = d n +1 (cid:0) Q ∞ (cid:1) ∗ d − n +1 ⇔ (cid:26) L T1 = diag (1 , ω, . . . , ω m ) L ∗ diag (1 , ω, . . . , ω m ) − , U T1 = diag (1 , ω, . . . , ω m − ) U ∗ diag (1 , ω, . . . , ω m − ) − . Using the entry-wise expression of L and U , this implies that for all k ≥ α m − k = α m − k ω m − k , α m − k − = α m − k − ω m − k − . Consequently, we see that: α m − k = | α m − k | ( ω − ) m − k , α m − k − = | α m − k − | ( ω − ) m − k − . We now wish to determine the exponent r for which the following matrices arereal-valued: d rn +1 Q ∞ d − rn +1 , d rn +1 Q ∞ n +1 d − rn +1 , such that d rn +1 Π = Πd rn +1 . Evidently, the non-zero entries of these matrices are: | α m − k | ω ( m − k )( r − ) , | α m − k − | ω ( m − k − r − ) , and these are real iff for all k :( m − k )( r − ) ≡ n +12 , ( m − k − r − ) ≡ n +12 . Of the two solutions r = and r = m + 1, only r = m + 1 satisfies d rn +1 Π = Πd rn +1 .As such, we define:˜Q ∞ := d m +1 n +1 Q ∞ d − m − n +1 , ˜Q ∞ n +1 := d m +1 n +1 Q ∞ n +1 d − m − n +1 , which implies that:˜L := diag (1 , ω, . . . , ω m ) m +1 L diag (1 , ω, . . . , ω m ) − m − = I m +1 + ℓ − X k =0 ( − m | α m − k | E m +1 − k, k , ˜L n +1 := diag (1 , ω, . . . , ω m − ) m +1 L n +1 diag (1 , ω, . . . , ω m − ) − m − = I m + ℓ − X k =0 ( − m − | α m − k − | E m − k, k . ♠ References [1] F. Beukers and G. Heckman,
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