COMPACTNESS of FLOQUET ISOSPECTRAL SETS for the MATRIX HILL's EQUATION 1. Introduction This Paper Considers Some Questions Of

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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 128, Number 10, Pages 2933–2941 S 0002-9939(00)05634-3 Article electronically published on April 7, 2000

COMPACTNESS OF FLOQUET ISOSPECTRAL SETS FOR THE MATRIX HILL’S EQUATION

ROBERT CARLSON

(Communicated by Hal L. Smith)

Abstract. Let M(Q)denotethesetofselfadjointK × K potentials for the matrix Hill’s equation having the same Floquet multipliers as −D2 + Q. Elementary methods are used to show that M(Q) has compact closure in the space of continuous matrix valued functions.

1. Introduction This paper considers some questions of inverse spectral theory for the matrix Hill’s equation (1) −Y 00 + Q(x)Y = λY, Y ∈ CK ,λ∈ C,Q(x)=Q(x +1).

The K × K matrix function Q : R → MK (C) is assumed to have J ≥ 0 continuous derivatives. The basic Floquet theory may be described as in the scalar case. For each λ ∈ C the equation (1) has a 2K dimensional space of solutions on which translation by 1 acts linearly. The Floquet multipliers ξk(λ) are the eigenvalues of this linear map. Introduce the K ×K matrix solutions C(x, λ),S(x, λ) of (1) with the initial data 0 0 C(0,λ)=IK ,C(0,λ)=0K,S(0,λ)=0K,S(0,λ)=IK . With respect to the basis given by the columns of C(x, λ)andS(x, λ)the2K × 2K matrix C(1,λ) S(1,λ) Ψ = 1 C0(1,λ) S0(1,λ)

represents the translation operator. The Floquet variety FQ consists of the pairs 2 (λ, ξ) ∈ C satisfying the characteristic polynomial det(ξI − Ψ1(λ)) = 0. Hill’s equation (1) will have periodic, respectively antiperiodic, eigenvalues λn where at least one of the multipliers ξk(λn) is 1, respectively −1. In the scalar case (K =1)thesetM(Q) is usually deﬁned to be the set of potentials having the same periodic and antiperiodic spectra as Q. By the Wronskian identity the determinant of the translation map is 1, so the eigenvalues in the scalar case may be described using the discriminant ∆(λ)=C(1,λ)+S0(1,λ), which is just the trace of the translation map. The periodic and antiperiodic eigenvalues are respectively the roots of ∆ − 2and∆+2.

Received by the editors November 10, 1998. 2000 Mathematics Subject Classiﬁcation. Primary 34A55; Secondary 34L40. Key words and phrases. Hill’s equation, inverse spectral theory, KdV.

c 2000 American Mathematical Society 2933

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Since the discriminant is an entire function with order of growth 1/2 [9, p. 20], Hadamard’s Theorem [1, p. 206] implies that ∆(λ) may be recovered up to a constant factor from the periodic or antiperiodic eigenvalues. The constant may also be determined [9, p. 22]. Thus in the scalar case being isospectral for periodic eigenvalues is equivalent to having the same set of Floquet multipliers for all λ ∈ C. This Floquet isospectrality is the condition we will consider in the matrix case. The development of the elaborate and beautiful inverse spectral theory for the scalar Hill’s equation [5, 10, 11, 13] received a strong impetus from the discovery [8] that solutions of the KdV equation with initial data Q evolve in M(Q). As in the scalar case, the matrix equation

000 0 0 Qt = −(1/4)Q +(3/4)QQ +(3/4)Q Q

dL maybeexpressedintheLaxform dt =[L, A], and spatially periodic solutions of the nonlinear evolution equation lead to Floquet isospectral operators L(t). Except for [2], which considers spectral theoretic characterizations of constant matrix potentials, it appears that almost none of the inverse spectral theory for the matrix Hill’s equation has been developed. The main goal of this work is to establish the compactness result for M(Q) contained in Theorem 1.1. To be precise about the diﬀerentiability of the potentials, deﬁne

5 Q = {Q : R → MK (C)| Q(x) ∈ C (R),Q(x)=Q(x +1)}.

Theorem 1.1. The set of self adjoint potentials Q(x) ∈Qwith the same Flo- quet variety has compact closure in the space of continuous K × K matrix valued functions.

The limited results currently available for matrix potentials contrast sharply with the detailed analyses of the scalar case. More refined compactness results were used already in [8]. Generalizing earlier results for finite gap potentials, McKean and Trubowitz [11] proved that the set of C∞ scalar potentials with a fixed periodic spectrum is a product of (generally infinitely many) circles. By an explicit con- struction, this result is extended to potentials in L2[0, 1] in [3]. Alternate methods [6] provided a similar result for potentials with a square integrable derivative. In [7] the L2 potentials are given a global set of coordinates which simultaneously exhibit all isospectral sets as products of circles. A thorough analysis of isospectral sets for potentials in L2[0, 1] with a variety of generalized periodic boundary conditions is in [12].

2. Preliminary results Estimates describing solutions of the equation (1) as |λ|→∞may be established using techniques commonly seen in the scalar case. Except for some minor notational changes, and the rearrangement of some terms which do not commute in the vector case, our estimates and their proofs are very similar to those appear- ing in [4]. The reader may consult this reference for the proofs of Lemma 2.1 and Lemma 2.2. √ We begin by establishing some notational conventions. For λ ∈ C let ω = λ, the root chosen continuously for −π0. The

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imaginary part of ω is denoted by =ω. A element Y ∈ CK will have the usual norm   y1 XK   | | | |2 1/2 . Y =[ yk ] ,Y=  .  , k=1 yK and a K × K matrix Q will have the operator norm kQk =sup|QY |. |Y |=1

The K × K identity matrix is IK , and the zero matrix is 0K . Estimates for solutions of (1) may be developed by using the model equation −Y 00 = λY . This equation has a basis of 2K solutions which are the columns −1 of the K × K diagonal matrix valued functions cos(ωx)IK , ω sin(ωx)IK .By using the variation of parameters formula a solution of (1) satisfying Y (0,λ)=α, Y 0(0,λ)=β,withα, β ∈ CK , may be expressed as a solution of the integral equation Z x (2) Y (x, λ)=cos(ωx)α + ω−1 sin(ωx)β + ω−1 sin(ω[x − t])Q(t)Y (t, λ) dt. 0 Diﬀerentiation with respect to x gives Z x Y 0(x, λ)=−ω sin(ωx)α +cos(ωx)β + cos(ω[x − t])Q(t)Y (t, λ) dt. 0 An iteration scheme based on (2) and integration by parts provides the desired expansion of Y (x, λ), together with error estimates, when Q(x) is suﬃciently dif- ferentiable. For J ≥ 0letCJ denote the Banach space of K × K matrix valued functions Q(x) whose components have J continuous derivatives on [0, 1]. Equip CJ with the norm

(j) kQkJ =maxkQ (x)k, 0 ≤ j ≤ J, 0 ≤ x ≤ 1.

It will be notationally convenient to deﬁne Aj(x)=Q(x)Aj (x)andBj(x)= Q(x)Bj(x). Lemma 2.1. Suppose that Q ∈ CJ ,andY (x, λ) is the solution of (1) satisfying Y (0,λ)=α and Y 0(0,λ)=β,with|α| + |β|≤1, 0 ≤ x ≤ 1,and|ω|≥1.Then K there are C valued functions Aj (x) and Bj(x), such that XJ −j −J−1 |=ω|x |Y (x, λ) − ω [cos(ωx)Aj (x)+sin(ωx)Bj (x)]| = O(ω e ). j=0

For 0 ≤ j ≤ J the coeﬃcients Aj(x),Bj (x) satisfy

A0(x)=α, A1(x)=0, Z Z x t −2 −1 −1 A2(x)=2 [Q(x) − Q(0)]α − 2 Q(t)[β +2 Q(s)αds] dt, 0 0 Z x −1 B0(x)=0,B1(x)=β +2 Q(t)αdt, B2(x)=0, 0

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and for j ≥ 3 the coeﬃcients satisfy the recursion relations b(j−X2)/2c − k −2k−2 A(2k) −A(2k) Aj (x)= ( 1) 2 [ j−2k−2(x) j−2k−2(0)] k=0

Z b − c x (jX3)/2 − −1 B − k −2k−3 B(2k+1) −B(2k+1) 2 j−1(t) dt + ( 1) 2 [ j−2k−3(x) j−2k−3(0)], 0 k=0

b(j−X3)/2c − − k −2k−3 A(2k+1) A(2k+1) Bj(x)= ( 1) 2 [ j−2k−3(x)+ j−2k−3(0)] k=0

Z b − c x (jX2)/2 −1 A − k −2k−2 B(2k) B(2k) +2 j−1(t) dt + ( 1) 2 [ j−2k−2(x)+ j−2k−2(0)]. 0 k=0 The function Y 0(x, λ) has a similar expansion obtained from that of Y (x, λ)by termwise diﬀerention. Lemma 2.2. Under the hypotheses of Lemma 2.1 there are CK valued functions Cj(x) and Dj(x) such that JX−1 0 −j −J |=ω|x |Y (x, λ) − ω [cos(ωx)Cj (x)+sin(ωx)Dj(x)]| = O(ω e ). j=−1

The coefficients Cj(x),Dj (x) satisfy C−1(x)=0, D−1(x)=−α,and 0 0 − − Cj(x)=Aj(x)+Bj+1(x),Dj(x)=Bj(x) Aj+1(x),j=0,...,J 1. From the computed initial coefficients and the recursion relations we can see that half of the coefficients vanish.

Lemma 2.3. If j is odd, the coefficients Aj(x) and Cj(x) are 0.Ifj is even, the coefficients Bj(x) and Dj(x) are 0. The asymptotic expansions for Y (x, λ), and Y 0(x, λ) specialize to give expansions 0 0 Rfor the matrix functions C(1,λ), C (1,λ), S(1,λ), and S (1,λ). Defining Q0 = 1 0 Q(t) dt, and using the periodic continuity condition Q(1) = Q(0), the following formulas are obtained for J ≥ 2: −1 −1 (3) C(1,λ)=cos(ω)IK +2 ω sin(ω)Q0 Z Z 1 t −2−2ω−2 cos(ω) Q(t) Q(s) ds dt + O(ω−3e|=ω|), 0 0

−1 −1 −2 S(1,λ)=ω sin(ω)IK − 2 ω cos(ω)Q0 Z Z h 1 t i +2−2ω−3 sin(ω) 2Q(0) − Q(t) Q(s) ds dt + O(ω−4e|=ω|), 0 0

0 −1 C (1,λ)=−ω sin(ω)IK +2 cos(ω)Q0 Z Z h 1 t i +2−2ω−1 sin(ω) 2Q(0) + Q(t) Q(s) ds dt + O(ω−2e|=ω|), 0 0

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0 −1 −1 S (1,λ)=cos(ω)IK +2 ω sin(ω)Q0 Z Z 1 t −2−2ω−2 cos(ω) Q(t) Q(s) ds dt + O(ω−3e|=ω|). 0 0 3. Trace of the Floquet matrix

The above expansions may be used to develop asymptotic expressions for trΨ1(λ). Since the trace is independent of the selected basis, the coefficients of these expansions depend only on the set of Floquet eigenvalues. An analysis of these coefficients provides the proof of Theorem 1.1. It will be convenient to assume that the matrix Q0 is diagonal, Q0 =diag[q1,...,qK ]. Of course a simple linear change of variables reduces (1) to this case when Q0 is similar to a diagonal matrix. Rather than working directly with Ψ (λ) it is easier to consider the similar matrix 1 C(1,λ) ωS(1,λ) IK 0K IK 0K Ψ(λ)= −1 0 0 = −1 Ψ1(λ) ω C (1,λ) S (1,λ) 0K ω IK 0K ωIK whose entries have a more uniform description in the style of (3). We may write XJ −J−1 |=ω| (3.a) Ψ(λ)= ψj(λ)+O(ω e ), j=0 with −j −j A B ψj(λ)=2 ω [ψj cos(ω)+ψj sin(ω)], A B × where the coefficients ψj ,ψj are constant K K matrices. By Lemma 2.2 the coefficients Cj and Dj for the solutions C(x, λ)andS(x, λ) may be computed directly from the coefficients Aj = Aj (1,λ)andBj = Bj(1,λ), and the derivatives 0 0 0 0 Aj = Aj(1,λ)andBj = Bj(1,λ). Using the subscripts C, S to indicate the source function C(x, λ)orS(x, λ), we find that −j A AC,j AS,j+1 2 ψj = 0 0 , AC,j−1 + BC,j AS,j + BS,j+1 −j B BC,j BS,j+1 2 ψj = 0 − 0 − . BC,j−1 AC,j BS,j AS,j+1 It will be convenient to introduce the following notation: Z Z Z 1 t1 tm−1 Im(Q)= Q(t1) Q(t2)··· Q(tm) dtm ...dt1. 0 0 0 For the following calculations it is assumed that Q(x) has 5 continuous derivatives on R,sothatQ(j)(0) = Q(j)(1) for j =0,...,5. The first few terms in the expansion of Ψ(λ)are A IK 0 B 0 IK ψ0 = ,ψ0 = , 0 IK −IK 0 A 0 −Q0 B Q0 0 ψ1 = ,ψ1 = , Q0 0 0 Q0 A −I2(Q)0 B 02Q(0) − I2(Q) ψ2 = ,ψ2 = , 0 −I2(Q) 2Q(0) + I2(Q)0

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2938 ROBERT CARLSON   − − 0 QR(0)Q0 Q0Q(0)  − 1 2  A  0 Q (t)+I3(Q)  ψ3 = − −  , QR(0)Q0 Q0Q(0) 0 1 2 − + 0 Q (t) I3(Q)   − 0 − 2Q (0)R + Q(0)Q0 Q0Q(0) 0  1 2 −  B  + 0 Q (t) I3(Q)  ψ3 =  0 −  , 02Q (0)R Q(0)Q0 + Q0Q(0) 1 2 − + 0 Q (t) I3(Q)

A ψ4 =   0 0 Q0Q (0) + Q (0)Q0  −   R +I2(RQ)Q(0) QR (0)I2(QR)0  1 t 1 t  − Q(t) Q2(s) − Q2(t) Q(s)   0 0 R 0 0   +I (Q)+ 1 Q(t)Q0(t)   4 0   − 0 − 0  ,  Q (0)Q0 Q0Q (0)   −   0+R Q(0)R I2(Q) IR2(Q)Q(0)R   − 1 2 t − 1 t 2  0 Q (t) 0 Q(s)R 0 Q(t) 0 Q (s) 1 0 +I4(Q)+ 0 Q(t)Q (t)

B ψ4 =   2 0 0 6Q (0) − Q0Q (0) + Q (0)Q0  − (2) − −   0 2Q (0)R I2(Q)Q(0) Q(0)I2(Q)   1   + Q(t)Q0(t)+I (Q)   R R0 R 4 R   − 1 Q(t) t Q2(s) − 1 Q2(t) t Q(s)  0 0 0 0   2 0 − 0  .  2Q (0) + Q (0)Q0 Q0Q (0)   − −   Q(0)I2(Q) I2(RQ)Q(0) 0   − (2) − − 1 0  R2Q (0)R I4(Q) R 0 Q(tR)Q (t) 1 2 t 1 t 2 + 0 Q (t) 0 Q(s)+ 0 Q(t) 0 Q (s) Lemma 3.1. Suppose that Q(x) has 5 continuous derivatives on R, and is periodic with period 1. The following expressions are determined by the Floquet variety FQ: Z h 1 i 2 I1 =tr[Q0], I2 =tr[I2(Q)], I3 =tr Q (t) dt − I3(Q) , 0 Z Z Z Z Z h 1 t 1 t 1 i 2 2 0 I4 =tr − Q(t) Q (s) − Q (t) Q(s)+I4(Q)+ Q(t)Q (t) , 0 0 0 0 0 and Z Z Z h 1 1 1 0 2 3 0 I5 =tr 2 [Q (x)] +4 Q + Q0 Q(t)Q (t) dt 0 0 0 Z Z Z Z Z Z 1 x t 1 x t −2 Q2(x) Q(t) Q(s) − 2 Q(x) Q2(t) Q(s) 0 0 0 0 0 0 Z Z Z 1 x t i 2 −2 Q(x) Q(t) Q (s)+2I5(Q) . 0 0 0

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Proof. The function trΨ(λ) depends only on the Floquet multipliers. Taking the trace of (3.a) gives XJ h i −j −j A −j −j B −J−1 |=ω| trΨ(λ)= 2 ω cos(ω)tr(ψj )+2 ω sin(ω)tr(ψj ) + O(ω e ). j=0 By virtue of the elementary integral formulas Z T T − 1 sin(2T ) − sin(2) cos2(ω) dω = + , 1 2 4 Z T sin2(T ) − sin2(1) sin(ω)cos(ω) dω = , 1 2 A B the coeﬃcients tr(ψj )andtr(ψj ) may be recovered from trΨ(λ) by recursively computing Z T Xj−1 A 2 j j − ψj = lim 2 ω [trΨ(λ) trψl(λ)] cos(ω) dω, T →∞ T − 1 1 l=0 and Z T Xj−1 B 2 j j − ψj = lim 2 ω [trΨ(λ) trψl(λ)] sin(ω) dω. T →∞ T − 1 1 l=0 A B Consequently, the traces of ψj and ψj are Floquet spectral invariants. The B last in the list of invariants requires the computation of trψ5 ,anuglyexercisein elementary algebra too long to record here.

Compactness of M(Q) is established by mixing a standard method [10] with some additional matrix analysis.

Proof of Theorem 1.1. We begin with some elementary matrix identities. A simple computation shows that for K × K matrices A and B,tr[AB]=tr[BA]. Applying this identity to a product of n matrices gives

(3.b) tr[A1 ···An−1An]=tr[AnA1 ···An−1], which shows that the trace of a product is invariant under cyclic permutation of the factors. Examining the diagonal entries of a product, we also ﬁnd that diag[AB]=diag[(AB)T ] = diag[BT AT ]=diag[B∗A∗]. If A and B are self adjoint, then (3.c) diag[AB]=diag[BA]. Analysis of the spectral invariants of Lemma 3.1 begins with Z Z Z Z 1 t 1 t tr[I2(Q)] = tr Q(t) Q(s) ds dt =tr [ Q(s) ds]Q(t) dt, 0 0 0 0 by (3.b). Thus Z Z Z 1 t 1 d 2 2 2 2tr[I2(Q)] = tr [ Q(s) ds] dt =tr[ Q(s) ds] =tr[Q0]. 0 dt 0 0

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The invariant I3 contains I3(Q). Integration by parts gives Z Z Z 1 t t I3(Q)=Q0I2(Q) − [ Q(s)][Q(t) Q(s) ds] dt. 0 0 0 Using (3.b) gives Z Z Z Z Z 1 t t 1 t d 3 3 3tr [ Q(s)][Q(t) Q(s) ds] dt =tr [ Q(s)] dt =tr[Q0], 0 0 0 0 dt 0 and 1 tr[I (Q)] = tr[Q I (Q)] − tr[Q3]. 3 0 2 3 0 R R ∗ t ∗ t Since Q(x)=Q (x), it follows that ( 0 Q) = 0 Q. Equation (3.c) yields Z Z Z Z 1 t 1 t diag [ Q(s) ds]Q(t) dt = diag Q(t) Q(s) ds dt, 0 0 0 0 or Z Z 1 t d 2 < 2 diag [ Q(s) ds] dt =2 (diag[I2(Q)]) = diag[Q0]. 0 dt 0

Since Q0 is diagonal and real, 1 X 1 X <(tr[Q I (Q)]) = q3, and <(tr[I (Q)]) = q3. 0 2 2 k 3 6 k k k In addition X X X X 1 K | q3|≤ |q |q2 ≤ [ + q2]q2 ≤ [ +trI (Q)]trI (Q), k k k 4 j k 4 2 2 k k k j

which provides a bound for <(trI3(Q)). When Q = Q∗, Z Z 1 X 1 2 2 tr Q = |Qij | , 0 i,j 0 R I 1 2 which is nonnegative. Since 3 is fixed and the real part of I3(Q) is bounded, 0 Q is bounded. That is, we have an L2 bound for self adjoint matrix potentials Q(x) with a fixed Floquet variety. R The compactness result will follow from bounds on [Q0]2 based on the invariant I .Notefirstthat 5 Z Z Z 1 1 1 1 kQ(x)k≤C (tr[Q∗(x)Q(x)])1/2 ≤ C ( +tr[Q2(x)]). 1 1 4 R 0 0 0 1 k k ∈ Since 0 Q(x) is bounded by some constant C2,thereisanx [0, 1] such that kQ(x0)k≤C2. A pointwise estimate follows from Z Z x 1 0 0 2 kQ(x)k = kQ(x0)+ Q (t) dtk≤C()+tr [Q (t)] ,>0. x0 0 These observations lead to Z 1 sup kQ(x)k≤C()+tr [Q0]2. 0≤x≤1 0

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2 2 For self adjoint matrices Q(x)wealsohavekQ(x)k ≤ C3tr[Q ]. Using the con- stancy of I it is then straightforward to establish bounds on 5 Z 1 tr [Q0(x)]2. 0 Such estimates imply that the magnitudes of the entries of matrix potentials Q(x) with the same Floquet variety are uniformly bounded, and form an equicontinuous family, which establishes compactness of the closure in the space of continuous k k matrix valued potentials with the norm sup0≤x≤1 Q(x) . References

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Department of Mathematics, University of Colorado at Colorado Springs, Col- orado Springs, Colorado 80933 E-mail address: [email protected]

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