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2 Elementary Orbits in the Hessenberg Case

i k n Everywhere below we denote by ejk an elementary (δαδβ)α,β=1 and by ej a column vector j n Rn (δα)α=1 of the standard basis in . Denote by J an n × n matrix with 1s on the first sub-diagonal and 0s everywhere else. Let b+, n+, b−, n− be, resp., algebras of upper triangular, strictly upper triangular, lower triangular and strictly lower triangular matrices. Denote by H the set J + b+ of upper Hessenberg matrices. For any matrix A we write its decomposition into a sum of lower triangular and strictly upper triangular matrices as

A = A− + A0 + A+

T and define A≥0 = A0 + A+, A≤0 = A0 + A−, Asym = A+ + A0 + A+. A linear Poisson structure on H is obtained as a pull-back of the Kirillov-Kostant structure b∗ b b∗ on −, the dual of −, if one identifies − and H via the trace form. A Poisson bracket of two functions f1,f2 on H then reads

{f1,f2}(X)= hX, [(∇f1(X))≤0, (∇f2(X))≤0]i , (2.1) where we denote by hX,Y i the trace form Trace(XY ) and gradients are computed w.r.t. this form. Symplectic leaves of the bracket (2.1) are orbits of the coadjoint action of the group B− of lower triangular invertible matrices:

O X0 = {J + (AdnX0)≥0 : n ∈ B−} . (2.2)

Following [8], consider a family of orbits whose members are parameterized by increasing se- quences of natural numbers I = {i1,... ,ik : 1 < i1 <...

O MI = ∪ν∈R XI +ν1 ⊂H , (2.3) where

k−1

XI = e1i1 + eij ij+1 + J . (2.4) j X=1

2 O We call orbits XI +ν1 elementary. ′ The set MI of elements of the form −1 −1 −1 X = (J + D)(1 − Ck) (1 − Ck−1) · · · (1 − C1) , (2.5) where D = diag(d1,... ,dn)

ij −1

Cj = cαeα,α+1 , (2.6) α i − =Xj 1 is dense in MI . The following formulae express entries xlm (l

(dl + cl−1)cl · · · cm−1, ij−1

−1 X = (J + D)(1 + U1)(1 − U2) , (2.9) where n−1 n−1 U1 = (1 − ǫα)cαeα,α+1 ,U2 = ǫαcαeα,α+1 (2.10) α=1 α=1 X X n In what follows, we will also use a sequence of integers (νi)i=1 defined by i−1 νi = i(1 − ǫi) − ǫβ (2.11) β X=1 It is easy to check that

j if i = ij for some 0 < j ≤ k ν = . (2.12) i − i−1 ǫ otherwise  β=1 β It follows from (2.12) that P

(i) sequences (ǫi) and (νi) uniquely determine each other.

(ii) If we define, for every i ∈ {1,... ,n}, a set Ni = {να : α = 1,... ,i} and a number i ki = max{j : ij ≤ i} = i − β=1 ǫβ, then P i i Ni = {ki − i + 1,... ,ki − 1, ki} = {1 − ǫβ,... ,i − ǫβ} (2.13) βX=1 βX=1 1This was suggested to us by Yu. Suris [14]

3 3 Solution of the Inverse Problem

Assume that all di 6= 0 and then define the moment sequence S = (si, i ∈ Z) of X: T i si = si(X)= e1 X e1. (3.1)

Our goal is to express coefficients ci, di in terms of S. In fact, only a segment sk+1−n,... ,sn+k will be needed. Let p(λ) = (p0(λ) := 1,p1(λ),... ,pn−1(λ)) be a solution of the truncated eigenvalue problem i (p(λ)X)i = xαipα−1(λ)+ pi(λ)= λpi−1(λ), i = 1,... ,n − 1 (3.2) α=1 X or, equivalently,

(p(λ)(J + D)(1 + U1))i = λ ((1 − U2)p)i (λ), i = 1,... ,n − 1 . (3.3)

Clearly, such solution exists for every λ and is uniquely defined. Moreover, each pi(λ) is a monic polynomial of degree i. We can re-write (3.3) as a 3-term recursion for polynomials pi(λ) :

pi+1(λ)+ bi+1pi(λ) + (1 − ǫi)aipi−1(λ)= λ (pi(λ) − ǫicipi−1(λ), ) i = 0,... ,n − 1 , (3.4) where

bi = di + (1 − ǫi−1)ci−1, ai = dici . (3.5)

Lemma 3.1 For any upper X, polynomials pi(λ) defined by (3.2) satisfy

pi(X)e1 = ei+1, i = 0,... ,n − 1. (3.6)

Proof. Since i Xei = xαieα + ei+1 α=1 X and, by (3.2), i Xpi−1(X)= xαipα−1(X)e1 + pi(X)e1 , α=1 Xn n one concludes that sequences of vectors (ei)i=1 and (pi−1(X)e1)i=1 are defined by the same recur- rence relations. ✷ L T i Lemma 3.2 For i = 1,... ,n − 1, a subspace i generated by vectors (eα )α=1 coincides with a T να i subspace generated by vectors (e1 X )α=1, where να are defined in (2.11). In other words, for some constants γi 6= 0 T j e1 X if i = ij T − L γiei = i 1 (mod i−1)). (3.7) T − β=1 ǫβ ( e1 X otherwise P Moreover, for i = 2,... ,n,

i−1 1−ǫβ i−1 d cβ T i(1−ǫi)− ǫβ ǫi(i−1) β=2 β γi = e X β=1 ei == (−1) c1d1 . (3.8) 1 ǫi P Q(d1 · · · di)

4 Proof. For any X given by (2.5), (2.9) ,consider an upper

−1 −1 −1 −1 V = D(1 − Ck) (1 − Ck−1) · · · (1 − C1) = D(1 + U1)(1 − U2) .

(If D is invertible, then V is an upper triangular factor in the Gauss factorization of X.) T By (2.6), el Cj = 0 for l ≤ ij−1 and l ≥ ij. Thus, for j = 0,... ,k,

eT V d eT 1 C −1 1 C −1 d eT 1 C −1 L ij−1 = ij−1 ij−1 ( − k) · · · ( − 1) = ij−1 ij−1 ( − j) (mod ij −1) d c c eT L . = ij−1 ij−1 · · · ij −1 ij (mod ij −1)

T L Similar argument shows that el V ∈ ij −1 for l < ij−1. This implies

T j T j L j−1 L e1 V = e1 V (mod ij −1)= β=0 diβ ciβ · · · ciβ+1−1 eij (mod ij −1)

ij −1  1−ǫβ  c d c Qd eT L . = 1 1 β=2 β β ij (mod ij −1) (3.9)  Q  On the other hand, for l ∈ {ij−1,... ,ij − 1}, define m ≥ 0 to be a number such that l + β = ij+β−1 for β = 1,... ,m and l + m + 1 < ij+m. In other words, l + m + 1 is the smallest index greater than l that does not belong to the index set I. Then

T −1 T −1 m+1 −1 T L el V = el (1 − Cj) · · · (1 − Ck)D = (−1) cl · · · cl+mdl+m+1 el+m+1(mod l+m) . (3.10)  Let us denote by J = {l1 <...

Combining (3.9), (3.12) with (2.12), one concludes that

T νi T L e1 X = γiei (mod i−1) , (3.13) where γi is defined by (3.8). This implies the statement of the lemma. ✷

Corollary 3.3

i i T α e1 X pi(X)e1 = 0, 1 − ǫβ ≤ α ≤ i − ǫβ (3.14) βX=1 βX=1

5 T νl Proof. By (3.7), e1 X ei+1 = 0 for l = 1,... ,i. But, by Lemma 2.1, ei+1 = pi(X)e1. Then (3.14) follows from (2.13). ✷ Define Toeplitz matrices

(l) i Ti = (sl+α−β)α,β=1 (3.15) and Toeplitz determinants

(i+m− i ǫ ) (m) β=1 β Z ∆i = det Ti (m ∈ ) . (3.16) P Let us also define polynomials

sl sl+1 · · · sl+i ··· ··· ··· ··· (l)   Pi (λ) = det sl−i+1 sl−i+2 · · · sl+1 . (3.17)      1 λ · · · λi      (l) (l) Pi is a of the (i+1)×(i+1) matrix obtained from Ti+1 by replacing the last row with (1, λ, . . . , λi). The following simple lemma is reminiscent of the construction of classical orthogonal polynomials on the real line and the unit circle (see, e.g. [1]) and will be useful for us in what follows.

Lemma 3.4 Let X be with a moment sequence (3.1). For m ∈ Z, define a Laurent polynomial

m (l) R(λ)= λ Pi (λ) . (3.18) Then

T α e1 R(X)X e1 = 0 (α = l + 1 − m − i,... ,l − m) (3.19) T l−m+1 i (l+1) e1 R(X)X e1 = (−1) det Ti+1 (3.20) T l−m−i (l) e1 R(X)X e1 = det Ti+1 . (3.21)

i m+β i+β Proof. R(λ) can be written as β=0 Rβλ , where Rβ is equal to (−1) times the obtained by deleting the last row and (β + 1)st column in (3.17). Then P i T α e1 R(X)X e1 = Rβsm+α+β . (3.22) βX=0 If α is in the range specified in (3.19), the right-hand side of (3.22) becomes a determinant in which two of the rows coincide. Equalities (3.20), (3.21) are obtained in a similar way. ✷ Lemmas above imply the following

(0) Proposition 3.5 Assume that ∆i 6= 0 for i = 1,... ,n. Then polynomials defined by formulas

i 1 (i− β=1 ǫβ) pi(λ)= Pi (λ) . (3.23) (0) P ∆i satisfy (3.2, 3.3, 3.4).

6 i l Proof. If pi(λ)= l=0 pilλ is a solution of 3.2, 3.3, 3.4), then by Corollary 3.3 and (3.1), we have

P i i T α e1 X pi(X)e1 = 0 (1 − ǫβ ≤ α ≤ i − ǫβ ) . (3.24) βX=1 βX=1 (0) If ∆i 6= 0, Lemma 3.4 implies that (3.23) is a unique monic polynomial of degree i that satisfies (3.24). Therefore it has to coincide with the unique solution of (3.2). This completes the proof. ✷ Now we will be able to express parameters ci, di in terms of Toeplitz determinants (3.16). First, we need the following elementary lemma.

Lemma 3.6 Let X be an arbitrary upper Hessenberg matrix and let D0 = 1 and Di (i = 1,... ,n) denote the left upper i × i principal minor of X. Then

i Di = (−1) pi(0) , (3.25) where pi(λ) are polynomials defined by (3.2).

Proof. The Laplace expansion w.r.t ith column leads to the following recursion for Di:

i α Di = (−1) xαiDα−1 α=1 X or i α−1 Di + (−1) xαiDα−1 = 0 , α=1 X which coincides with the recursion for pi(0) obtained when one sets λ = 0 in (3.2). ✷

Theorem 3.7

∆(1)∆(0) ∆(ǫi+1)∆(1−ǫi) d = i i−1 , c = − i+1 i−1 . (3.26) i (0) (1) i (0) (1) ∆i ∆i−1 ∆i ∆i

(1) ∆i Proof. First note that, by (3.23), the right-hand side of (3.25) is equal to (0) , whereas, due to ∆i (2.5), the left-hand side is d1 · · · di. This immediately implies the first of the formulae (3.26). Secondly, it is clear from (3.8), that parameters ci are uniquely determined by di and γi. On the other hand, (3.8), together with (3.6), gives

T νi+1 γi+1 = e1 X pi(X)e1 . (3.27)

i i i+1 Recall that, if ǫi+1 = 1 then νi+1 = − β=1 ǫβ and i − β=1 ǫβ = i + 1 − β=1 ǫβ. Otherwise, ν i i ǫ i i+1 ǫ i+1 = + 1 − β=1 β = + 1 − β=1 βP. Comparing (3.27)P with equalities (3.20),P (3.21) and then using (3.16), one obtains P P i 1−ǫβ (0) d cβ ∆ iǫi+1 β=2 β i(1−ǫi+1) i+1 γi+1 = (−1) c1d1 = (−1) (3.28) ǫi (0) (d1 · · · di+1) +1 Q ∆i

7 Now, to finish the proof, it is enough to check that when one substitutes expressions (3.26) into the right-hand side of (3.8), the result agrees with (3.28). It is not hard to see that if dβ, cβ are defined by (3.26) then (ǫ ) ∆ β+1 ∆(0) 1−ǫβ β+1 β−1 dβ cβ = − , (3.29) (ǫβ) (0) ∆β ∆β

(ǫ2) (ǫi) ∆2 ∆i while d1c1 = − (1) . Therefore, the numerator in the right-hand side of (3.8) is equal to (0) and ∆1 ∆i−1 (1) ǫi ∆i denominator is equal to (0) . For both possible values of ǫi, this implies that formulae (3.8), ∆i (3.26) and (3.28) agree.   ✷ One of the consequences of Lemma 3.2 is an existence of the unique sequence of Laurent poly- nomials r0(λ), . . . , rn−1(λ) of the form i νi+1 νβ ri(λ)= λ + riβλ . (3.30) β X=1 such that T T e1 ri−1(X)= γiei (i = 1,... ,n) (3.31)

Next proposition shows that ri can be conveniently described by formulas similar to (3.23). Proposition 3.8

i+1 iǫi+1 1− β=1 ǫβ i+1 (−1) λ (i− β=1 ǫβ) ri(λ)= P Pi (λ) (3.32) (0) P ∆i

Proof. First note, that by (2.13), (3.30) can be re-written as i i+1 1− ǫβ i ri(λ)= λ β=1 aiβλ . (3.33) P Xβ=1 Next, by (2.11), νi+1 coincides with the lowest (resp. highest) degree in ri(λ) if ǫi+1 = 1 (resp. T T ǫi+1 = 0). Finally, due to Lemma 3.1, the claim that e1 ri(X) is proportional to ei+1 is equivalent to a property T l e1 ri(X)X e1 = 0 (l = 0,... ,i − 1) (3.34) This property is satisfied by Lemma 3.4. ✷ Corollary 3.9 Define a bilinear form (, ) on C[λ, λ−1] by i j (λ , λ )= si+j (i, j ∈ Z) .

Then functions p0(λ),... ,pn−1(λ); r0(λ), . . . , rn−1(λ) form a bi-orthogonal system w. r. t. (, ), i. e. i−1 1−ǫβ j β=2 dβ cβ (ri(λ),pj(λ)) = δ c1d1 . (3.35) i ǫi Q(d1 · · · di) Proof. Follows immediately from Lemmas 3.1, 3.2 and Propositions 3.4, 3.8. ✷

8 4 Symmetric Case

It is natural to ask how should results of the previous section be modified, if initially one identifies b∗ S − with a space of symmetric matrices rather than with H. In this case, a co-adjoint orbit of B− through X0 ∈ S is described as O X0 = {(AdnX0)sym : n ∈ B−} , (4.1)

T where Asym = A≥0 + (A>0) . A set MI is still defined by (2.3, but the definition of XI should be changed as follows:

k−1

XI = e1i1 + ei11 + (eij ij+1 + eij+1ij ) . (4.2) j X=1 ′ An open dense subset MI ⊂ MI , that we are going to study, consists of elements of the form

T −1 T −1 X = (1 − U2 ) (1 + U1 )D(1 + U1)(1 − U2) , (4.3) where, as before, D = diag(d1,... ,d2) and U1,U2 are defined by (2.10). Matrix entries of X then are found to be

l−1 2 cl · · · cm−1(dl + dα(cα · · · cl−1) ), ij−1 < l ≤ m ≤ ij  α i − xlm = xml =  =Xj 1 (4.4)  d c · · · c , i = l

lj −1 m−l 1 1 2 (−1) cl · · · cm−1( + (cm · · · cα) ), lj−1 ≤ l ≤ m < lj  dm dα (X−1) = (X−1) = α=m +1 lm ml  m−l X 1  (−1) cl · · · clj −1 d , lj−1 = l

(i+1+m−2 i ǫ ) D(m) β=1 β Z i = det Ti (m ∈ ) . (4.6) P

9 Proposition 4.1

i − (i−1)ǫi 1− β=1 ǫβ i 1 (−1) λ (i−2 β=1 ǫβ−ǫi) ri−1(λ)= P Pi−1 (λ) . (4.7) D(0) P i−1

Proof. We can argue exactly as in the proof of Proposition 3.8. Note, however, that since X is symmetric, ri−1(X)e1 = γiei. This means that condition (3.34) that guarantees (3.31) has to be replaced by T να e1 ri−1(X)X e1 = 0 (α = 1,... ,i − 1) or, equivalently,

i−1 i−1 T l e1 ri−1(X)X e1 = 0 (l = 1 − ǫβ,... ,i − 1 − ǫβ) . βX=1 βX=1

Function ri−1 defined by (4.7) satisfies this condition for exactly the same reason that function (3.32) satisfies (3.34). ✷

Corollary 4.2

D(0) γ2 = (−1)i−1 i (4.8) i D(0) i−1

Proof. It follows from (3.7), (3.31) and symmetricity of X that

2 T νi γi = e1 ri−1(X)X e1 . (4.9)

Then (4.8) follows from Lemma 3.4, (2.11) and (4.7). ✷ Remark. Note that, as one would expect, in the case of the classical moment problem (I = {2, 3,... ,n}, X is tri-diagonal), a condition that ensures that the right-hand side of (4.8) is positive n is a positive definiteness of the (si+j−2)i,j=1. Similarly, in the case I = {n} which relevant in the study of the peakons lattice and was comprehensively studied in [2], one comes to n a conclusion that the Hankel matrix (s2−i−j)i,j=1 is positive definite. ′ Let us consider now an m × m sub-matrix Xm of X ∈ MI obtained by deleting (n − m) last rows and columns. It is clear from (4.4), that Xm does not depend on parameters cm,... ,cn−1, dm+1,... ,dn.

α T Lemma 4.3 Let sα(Xm)= e1X e1 . then

m m sα(Xm)= sα(X) (2 − 2 ǫβ ≤ α ≤ 2m + 1 − 2 ǫβ) . (4.10) βX=1 βX=1

Proof. For l > 0, an expression for sl(X) in terms of matrix entries of X reads sl(X) =

x1α1 xα1α2 · · · xαl−11. By (4.4), many of the terms in this sum are zero. Moreover, if α ,...,α − 1 Xl 1 10 l < 2k + 1, were k is the cardinality of I, one can find among the non-zero terms the one q α ,... ,α x2 , l q were max( 1 l−1) reaches its maximum. This term is equal to iβ−1,iβ if = 2 and β=1 q Y x x2 , l q m > i < l j iqiq iβ−1,iβ if = 2 + 1. This implies that if j and 0 ≤ 2 + 1 then the expression β Y=1 for sl(X) involves only entries of Xm and, therefore, sl(X)= sl(Xm). Since the largest j such that m m m > ij is given by m − β=1 ǫβ, (4.10) is satisfied for α = 0,... , 2m + 1 − 2 β=1 ǫβ. Similarly, one can use (4.5) to conclude that if q is the largest index such that lq

Proposition 4.4 P

m 1 (m+1−2 ǫβ) det(λ − X )= P β=1 (λ) . (4.11) m (0) m+1 Dm P

m m−1 i Proof. Let = λ + i=0 amiλ . Then the Hamilton-Cayley theorem implies

P m−1 sα+m(Xm)+ amisα+i(Xm)=0 (α ∈ Z) . (4.12) i X=0

By Lemma 4.3, (4.12) remains valid if we replace sα+i(Xm) with sα+i = sα+i(X) for i = 0,... ,m, m m as long as 2 − 2 β=1 ǫβ ≤ α ≤ m + 1 − 2 β=1 ǫβ . This means that, after the right multi- m (m+1−2 β=1 ǫβ) plication of the matrixP used in the definitionP (3.17) of Pm+1 by the matrix m−1 P (1 + β=1 eβ+1,m+1), one gets a matrix of the form

P m (m+1−2 β=1 ǫβ) Tm 0 P m−1 " 1 λ · · · λ det(λ − Xm) # and (4.11) follows. ✷ It drops out immediately from (4.11) and (4.3) that

(1) Dm d · · · d = . (4.13) 1 m (0) Dm Now the analogue of Theorem 3.7 for matrices (4.3) drops out immediately from (4.13), (4.8) and (3.8).

Theorem 4.5

1 (0) (1) 2 ǫ −ǫ D(1)D(0) −D D D(1) i+1 D(0) 1 i i i−1 i+1 i−1 i+1 i−1 di = , ci = . (4.14) D(0)D(0)  D(1)  D(0) D(1) i i−1 i i+1 ! i−1 !

11 5 Conclusion

The results of this paper can be used to make several constructions used in the study of integrable lattices of Toda type more explicit. For example, in [8] we proved that for any I, J there exists ′ ′ ′ ′ a bi-rational Poisson map from MI to MJ , that intertwines the Toda flows on MI and MJ . A ′ construction we gave was by induction and expressions for matrix entries of elements of MJ in ′ terms of matrix entries of elements of MI were very involved. Now we can give explicit formulas for this map using just (3.1) and Theorem 3.7. Similarly, Theorem 4.5 can be used to simplify ( in the case of elementary orbits ) the construction of the Poisson map from the Kostant-Toda to symmetric Toda flows proposed in [4]. A solution to the inverse problem can also be useful in avoiding blow-ups of the Toda flows by switching from one elementary orbit to another, which, in turn, may become important in the context of LU type algorithms for computing eigenvalues of Hessenberg matrices. A natural question that arises in this connection is to give an intrinsic description of all sequences (si) from ′ which an element of MI can be restored for at least one I. Because of possible implications in the coding theory, it would be also interesting to study solvability of the inverse problem over a finite field (cf. [7] where the connection with the coding theory was observed in the tri-diagonal case). Another direction of possible investigation that we plan to pursue in the future is an extension ′ of our results from MI to MI (to this end, some of the genericity assumptions will have to be relaxed) and then to more general co-adjoint orbits. In the latter case, one has to restore and element of the orbit from a collection of rational functions. In the case of generic co-adjoint orbits this has been done in [6, 12].

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