prev

next

out of 15

Published on

23-Dec-2016View

215Download

2

Embed Size (px)

Transcript

<ul><li><p>Invent mathDOI 10.1007/s00222-013-0486-8</p><p>Self-adjoint commuting ordinary differentialoperators</p><p>Andrey E. Mironov</p><p>Received: 31 May 2012 / Accepted: 1 October 2013 Springer-Verlag Berlin Heidelberg 2013</p><p>Abstract In this paper we study self-adjoint commuting ordinary differen-tial operators of rank two. We find sufficient conditions when an operator offourth order commuting with an operator of order 4g + 2 is self-adjoint. Weintroduce an equation on potentials V (x),W(x) of the self-adjoint operatorL = (2x + V )2 + W and some additional data. With the help of this equa-tion we find the first example of commuting differential operators of ranktwo corresponding to a spectral curve of higher genus. These operators havepolynomial coefficients and define commutative subalgebras of the first Weylalgebra.</p><p>Mathematics Subject Classification (2010) 37K10 37K20</p><p>1 Introduction</p><p>The problem of finding commuting differential operators is a classical prob-lem of differential equations (for the first results see [13]). In the case of</p><p>This paper was completed in the Hausdorff Research Institute for Mathematics (Bonn).The author is grateful to the Institute for hospitality. This work was also partiallysupported by grant 12-01-33058 from the Russian Foundation for Basic Research; grantMD-5134.2012.1 from the President of Russia; and a grant from Dmitri ZiminsDynasty foundation.A.E. Mironov (B)Sobolev Institute of Mathematics, Novosibirsk, Russiae-mail: mironov@math.nsc.ru</p><p>A.E. MironovLaboratory of Geometric Methods in Mathematical Physics, Moscow State University,Moscow, Russia</p></li><li><p>A.E. Mironov</p><p>operators of rank greater than one, this problem has not been solved untilnow. In this paper we study self-adjoint commuting ordinary differential op-erators. One of the main results of this paper is the following. We find anexample of commuting differential operators of rank two corresponding tospectral curves of higher genus.</p><p>If two differential operators</p><p>Ln = nx +n2</p><p>i=0ui(x)</p><p>ix, Lm = mx +</p><p>m2</p><p>i=0vi(x)</p><p>ix</p><p>commute, then there is a nonzero polynomial R(z,w) such thatR(Ln,Lm) = 0 (see [3]). The curve defined by R(z,w) = 0 is called thespectral curve. This curve parametrizes common eigenvalues of the operators.If</p><p>Ln = z, Lm = w,then (z,w) . For almost all (z,w) the dimension of the space ofcommon eigenfunctions is the same. The dimension is a divisor of m and n.The greatest common divisor of the orders of all the operators in the maximalcommutative ring containing Ln and Lm is called the rank.</p><p>Commutative rings of ordinary differential operators were classified byKrichever [4, 5] (algebraic aspects of this theory see in [6, 7]). The ring isdetermined by the spectral curve and some additional spectral data. If therank is one, then the spectral data define commuting operators by explicitformulas (see [4]). In the case of operators of rank greater than one thereare the following results. Krichever and Novikov [8, 9] using the method ofdeformation of Tyurin parameters found operators of rank two correspond-ing to elliptic spectral curves. These operators were studied in the papers[1018]. Mokhov [19], found operators of rank three also corresponding toelliptic spectral curves. Besides this there are examples of operators of rankgrater than one corresponding to spectral curves of genus 2, 3 and 4 (see[2023]).</p><p>The main results of this paper are the following. We consider a pairL4,L4g+2 of commuting differential operators of rank two whose spectralcurve is a hyperelliptic curve of genus g</p><p>w2 = Fg(z) = z2g+1 + c2gz2g + + c0. (1)Operators L4 and L4g+2 satisfy the equation (L4g+2)2 = Fg(L4). The curve has a holomorphic involution</p><p> : , (z,w) = (z,w).Common eigenfunctions of L4 and L4g+2 satisfy the second order differentialequation [5]</p></li><li><p>Self-adjoint commuting ordinary differential operators</p><p> (x,P ) = 1(x,P ) (x,P ) + 0(x,P )(x,P ),P = (z,w) . (2)</p><p>The coefficients 0(x,P ),1(x,P ) are rational functions on with 2g sim-ple poles depending on x, 0 has also an additional simple pole at infinity.These functions satisfy Krichevers equations (see below). To find operatorsL4,L4g+2 it is enough to find 0, 1.</p><p>It is not difficult to prove that if 1 is invariant under the involution , thenthe operator L4 is self-adjoint. S.P. Novikov has proposed the conjecture thatthe inverse is also true. In this paper we prove this conjecture.</p><p>Theorem 1 The operator L4 is self-adjoint if and only if1(x,P ) = 1</p><p>(x, (P )</p><p>). (3)</p><p>At g = 1 Theorem 1 was proved by Grinevich and Novikov [10].Let us assume that the operator L4 is self-adjoint</p><p>L4 =(2x + V (x)</p><p>)2 + W(x),then the functions 0, 1 have simple poles at some points</p><p>(i(x),</p><p>Fg</p><p>(i(x)</p><p>)) , 1 i g.In the next theorem we find the form of 0(x,P ),1(x,P ) and introduce anequation on V,W,i and Fg(z).</p><p>Theorem 2 If operator L4 is self-adjoint, then</p><p>0 = 12Q</p><p>Q+ w</p><p>Q V, 1 = Q</p><p>Q,</p><p>where</p><p>Q = (z 1(x)) (z g(x)</p><p>).</p><p>Function Q satisfies the equation4Fg(z) = 4(z W)Q2 4V</p><p>(Q</p><p>)2 + (Q)2 2QQ(3)+ 2Q(2V Q + 4VQ + Q(4)), (4)</p><p>where Q,Q,Q(k) mean xQ,2xQ, kxQ.</p><p>From Theorem 2 we get the following corollary.</p><p>Corollary 1 The function Q satisfies the linear equationL5Q =</p><p>(5x + 2</p><p>V, 3x</p><p> + 2z W V , x)Q = 0, (5)</p></li><li><p>A.E. Mironov</p><p>where A,B = AB + BA. Potentials V,W have the form</p><p>V =(</p><p>(Q)2 2QQ(3) 4Fg(z)4(Q)2</p><p>)z=j</p><p>,</p><p>W = 2(1 + + g) c2g.(6)</p><p>The functions 1(x), . . . , g(x) satisfy the equations(Q)2 2QQ(3) 4Fg(z)</p><p>4(Q)2</p><p>z=j</p><p>= (Q)2 2QQ(3) 4Fg(z)</p><p>4(Q)2</p><p>z=k</p><p>. (7)</p><p>At g = 1 the formula (6) gives</p><p>V = (W)2/4 W W /2 4F1((W + c2)/2)</p><p>(W )2.</p><p>Let me recall some elements of the spectral theory of finite-gap one dimen-sional Schrdinger operators (see [24]). Let L2 be a finite-gap operator</p><p>L2 =(2x + u(x)</p><p>) = z,</p><p> be the hyperelliptic spectral curve (1). The function = (x,P ), P =(z,w) has g zeros (j (x),w(j (x))) . The polynomial Q satisfiesthe equations</p><p>4Fg(z) = 4(z u)Q2 (Q</p><p>)2 + QQ,L3Q =</p><p>(3x + 2z u, x</p><p>)Q = 0. (8)</p><p>The potential u has the form</p><p>u = 2(1 + + g) c2g.From (8) follows Dubrovins equations</p><p> j = 2i</p><p>Fg(j )</p><p>k =j (k j ).</p><p>Thus, Theorem 2 and Corollary 1 are generalizations of corresponding state-ments for Schrdinger operators, in particular (7) are analogs of Dubrovinsequations, the operator L5 is an analog of L3, potentials W and u are definedby the same formulas.</p><p>To find self-adjoint operators L4,L4g+2 it is enough to solve (4). In thispaper we find partial solutions of the equation for arbitrary g. These solutionscorrespond to operators with polynomial coefficients.</p></li><li><p>Self-adjoint commuting ordinary differential operators</p><p>Theorem 3 The operator</p><p>L4 =</p><p>(2x + 3x3 + 2x2 + 1x + 0</p><p>)2 + g(g + 1)3x, 3 = 0commutes with a differential operator L4g+2 of order 4g + 2. The operatorsL</p><p>4, L</p><p>4g+2 are operators of rank two. The spectral curve of operators L4,</p><p>L4g+2 is a hyperelliptic curve with explicitly computed coefficients (see (23)).</p><p>Remark For arbitrary g we prove in Appendix that for general parameters0, . . . , 3 the right hand side of (23) has no multiple roots of order greaterthan two, so the spectral curve has no cuspidal singularities. Moreover, it canbe checked that for g = 1, . . . ,15 the spectral curve defined by (23) is non-singular. It seems plausible that the spectral curve is non-singular for every g.However I have no proof of this fact at the moment.</p><p>If g = 1, 1 = 2 = 0, 3 = 1, then the operators L4,L4g+2 coincide withthe famous Dixmier operators [25] whose spectral curve is an elliptic curve.Operators L4,L</p><p>4g+2 define commutative subalgebras in the first Weyl alge-</p><p>bra A1.Theorem 3 means that the equation</p><p>Y 2 = X2g+1 + c2gX2g + + c0has nonconstant solutions X,Y A1 for some ci . It is easy to see that thegroup Aut(A1) preserves the space of all such solutions. It would be veryinteresting to describe the orbits of Aut(A1) in the space of solutions and todescribe the action Aut(A1) on spectral data. This gives a chance to compareEnd(A1) and Aut(A1) (the Dixmier conjecture is: End(A1) = Aut(A1)).</p><p>In Sect. 2 we recall the method of deformations of Tyurin parameters. InSects. 35 we prove Theorems 13. In Appendix we analyse (23).</p><p>2 Operators of rank l > 1</p><p>Common eigenfunctions of commuting differential operators are BakerAkhiezer functions. Let me recall the definition of the BakerAkhiezer func-tion at l > 1 [5]. We take the spectral data</p><p>{,q, k, , v,(x)</p><p>},</p><p>where is a Riemann surface of genus g, q is a fixed point on , k is a localparameter near q ,</p><p>(x) = (0(x), . . . ,l2(x))</p></li><li><p>A.E. Mironov</p><p>is a set of smooth functions, = 1 + + lg is a divisor on , v is a set ofvectors</p><p>v1, . . . , vlg, vi = (vi,1, . . . , vi,l1).The pair (, v) is called the Tyurin parameters. The Tyurin parameters definea stable holomorphic vector bundle on of rank l and degree lg with holo-morphic sections 1, . . . , l . The points 1, . . . , lg are the points of the lineardependence of the sections</p><p>l(i) =l1</p><p>i=1vj,ij (i).</p><p>The vector-function = (1, . . . ,l) is defined by the following properties.1. In the neighbourhood of q the vector-function has the form</p><p>(x,P ) =( </p><p>s=0s(x)k</p><p>s</p><p>)0(x, k),</p><p>where 0 = (1,0, . . . ,0), i(x) = (1i (x), . . . , li (x)), the matrix 0 satis-fies the equation</p><p>d0dx</p><p>= A0, A =</p><p>0 1 0 . . . 0 00 0 1 . . . 0 0. . . . . . . . . . . . . . . . . .</p><p>0 0 0 . . . 0 1k1 + 0 1 2 . . . l2 0</p><p>.</p><p>2. The components of are meromorphic functions on \{q} with the sim-ple poles 1, . . . , lg , and</p><p>Resij = vi,jResil, 1 i lg, 1 j l 1.For the rational function f (P ) on with the unique pole of order n at q thereis a linear differential operator L(f ) of order ln such that</p><p>L(f )(x,P ) = f (P )(x,P ).For two such functions f (P ), g(P ) operators L(f ), L(g) commute.</p><p>The main difficulty to construct operators of rank l > 1 is the fact thatthe BakerAkhiezer function is not found explicitly. But the operators can befound by the following method of deformation of Tyurin parameters.</p><p>The common eigenfunctions of commuting differential operators of rank lsatisfy the linear differential equation of order l</p><p>(l)(x,P ) = 0(x,P )(x,P ) + + l1(x,P )(l1)(x,P ).The coefficients i are rational functions on [5] with simple polesP1(x), . . . ,Plg(x) , and with the following expansions in the neighbour-hood of q</p></li><li><p>Self-adjoint commuting ordinary differential operators</p><p>0(x,P ) = k1 + g0(x) + O(k),j (x,P ) = gj (x) + O(k), 0 < j < l 1,</p><p>l1(x,P ) = O(k).Let k i(x) be a local parameter near Pi(x). Then</p><p>j = ci,j (x)k i(x) + di,j (x) + O</p><p>(k i(x)</p><p>).</p><p>Functions cij (x), dij (x) satisfy the following equations [5].</p><p>Theorem 4ci,l1(x) = i (x),di,0(x) = vi,0(x)vi,l2(x) + vi,0(x)di,l1(x) vi,0(x),di,j (x) = vi,j (x)vi,l2(x) vi,j1(x) + vi,j (x)di,l1(x) vi,j (x),</p><p>j 1,where</p><p>vi,j (x) = ci,j (x)ci,l1(x)</p><p>, 0 j l 1, 1 i lg.</p><p>To find coefficients of the operators it is enough to find i .</p><p>3 Proof of Theorem 1</p><p>In the case of operators of rank two the common eigenfunctions of L4 andL4g+2 satisfy (2). In the neighbourhood of q we have the expansions</p><p>0 = 1k</p><p>+ a0(x) + a1(x)k + O(k2</p><p>),</p><p>1 = b1(x)k + b2(x)k2 + O(k3</p><p>).</p><p>(9)</p><p>Functions 0, 1 have 2g simple poles P1(x), . . . ,P2g(x), and by Theorem 4</p><p>0(x,P ) = vi,0(x)i (x)</p><p>k i(x) + di,0(x) + O(k i(x)</p><p>), (10)</p><p>1(x,P ) = i (x)</p><p>k i(x) + di,1(x) + O(k i(x)</p><p>), (11)</p><p>di,0(x) = v2i,0(x) + vi,0(x)di,1(x) vi,0(x). (12)Let be the hyperelliptic spectral curve (1), q = , k = 1</p><p>z. Let us find</p><p>coefficients of the operator of order four corresponding to the meromorphicfunction z with the pole of order two in q</p><p>L4 = z.</p></li><li><p>A.E. Mironov</p><p>Lemma 1 The operator L4 = 4x + f2(x)2x + f1(x)x + f0(x) has the fol-lowing coefficients:</p><p>f0 = a20 2a1 2b1 a0 , f1 = 2(b1 + a0</p><p>), f2 = 2a0.</p><p>Operator L4 is self-adjoint if and only if b1 = 0, herewith L4 = (2x +V (x))2 + W(x), where V (x) = a0(x), W = 2a1(x).Proof From (2) it follows that the fourth derivative of is</p><p>(4) = (20 + 1 0 + 0(21 + 2 1</p><p>) + 0)</p><p>+ (31 + 2 0 + 1(20 + 3 1</p><p>) + 1) .</p><p>With the help of (2) and the last equality we rewrite L4 = z in the formP1 + P2 = z,</p><p>where</p><p>P1 = f0 + f20 + 20 + 1 0 + 0(21 + 2 1</p><p>) + 0 ,P2 = f1 + f21 + 31 + 2 0 + 1</p><p>(20 + 3 1</p><p>) + 1 .This gives</p><p>P1 = z = 1k2</p><p>, P2 = 0. (13)From (9) we have</p><p>P1 1k2</p><p>= f2 + 2a0k</p><p>+ (f0 + a0(f2 + a0) + 2(a1 + b1</p><p>) + a0) + O(k) = 0,</p><p>P2 =(f1 + 2</p><p>(b1 + a0</p><p>)) + O(k) = 0.From here we find the coefficients of L4.</p><p>Operator L4 is self-adjoint if f1 = f 2, i.e. at b1 = 0. Lemma 1 is proved. If 1 satisfies the condition (3), then</p><p>1 =</p><p>s>1b2sk</p><p>2s,</p><p>hence, by Lemma 1 L4 is self-adjoint.Let us prove the inverse part of Theorem 1. We assume that L4 is self-</p><p>adjointL4 = L4 = 4x + f2(x)2x + f 2(x)x + f0(x).</p><p>If 1,2 Ker(L4 z), then1L42 2L41</p><p>= x(1</p><p>2 2 1 </p><p>( 1 2 2 1</p><p>) + f2(1</p><p>2 2 1</p><p>)) = 0.</p></li><li><p>Self-adjoint commuting ordinary differential operators</p><p>Hence, on the space Ker(L4 z) the following skew-symmetric bilinear form(1,2) = 1 2 2 1 </p><p>( 1 2 2 1</p><p>) + f2(1</p><p>2 2 1</p><p>)</p><p>is defined. Let 1(x,P ),2(x,P ) satisfy (2). Using i =</p><p>(0 + 21 + 1</p><p>) i +</p><p>(01 + 0</p><p>)i</p><p>and (2) we get(1,2) =</p><p>(1</p><p>2 2 1</p><p>)(f2 + 20 + 21 + 1</p><p>).</p><p>Since 1,2 satisfy the second order differential equation (2), we have(1,2) = e</p><p>1(x,z,w)dxg1(z,w)</p><p>(f2(x) + 20(x, z,w)</p><p>+ 21 (x, z,w) + 1(x, z,w))</p><p>= g2(z,w),where g1(z,w), g2(z,w) are some functions on . Let us represent 1 in theform</p><p>1(x, z,w) = G1(x, z) + wG2(x, z),where G1,G2 are rational functions on . Let</p><p>G1(x, z) =</p><p>G1(x, z)dx, G2(x, z) =</p><p>G2(x, z)dx,</p><p>then</p><p>eG1(x,z)ewG2(x,z) = g2(z,w)g1(z,w)(f2 + 20 + 22 + 1)</p><p>.</p><p>From the last identity it follows that for arbitrary x = x1, x = x2 the functioneG1(x1,z)G1(x2,z)ew(G2(x1,z)G2(x2,z))</p><p>is a rational function on . This is possible only ifG2(x1, z) G2(x2, z) = 0,</p><p>or, equivalently, G2 = 0. Hence, 1 = G1(x, z). This means that 1 is invari-ant under the involution . Thus, Theorem 1 is proved.</p><p>4 Proof of Theorem 2</p><p>Assume that 1 is invariant under , then by (9)(11) we have</p><p>0 = H1(x)z 1(x) + +</p><p>Hg(x)</p><p>z g(x) +w(z)</p><p>(z 1(x)) (z g(x))+ (x),</p><p>1(x,P ) = 1(x)</p><p>z 1(x) g(x)</p><p>z g(x),</p></li><li><p>A.E. Mironov</p><p>where Hi(x), (x) are some functions. In the neighbourhood of q the function0 has the expansion</p><p>0 = 1k</p><p>+ +(1 + + g + c2g2</p><p>)k + O(k2).</p><p>Hence, by Lemma 1V = , W = 2(1 + + g) c2g. (14)</p><p>Thus</p><p>0 = Q1Q</p><p>+ wQ</p><p> V (x), 1(x,P ) = Q</p><p>Q,</p><p>where Q1 = Q1(x, z) is some function. Let us substitute 0, 1 into (13).From P2 = 0 we get Q1 = Q+s2 , where s is a constant. From P1 = z weget</p><p>s2 4sw + 4w2 4(z W)Q2 + 4V (Q)2 (Q)2 + 2QQ(3) 2Q(2V Q + 4VQ + Q(4)) = 0.</p><p>The last identity is possible only if s = 0 because Q is a polynomial in z.Theorem 2 is proved.</p><p>To get (5) one should differentiate (4) in x and divide the result by Q. Toget the formula (6) one should substitute z = j into (4).</p><p>5 Proof of Theorem 3</p><p>Let us consider the equations (4) where V,W are potentials of the operator L44Fg(z) = 4</p><p>(z g(g + 1)3x</p><p>)Q2 4(3x3 + 2x2 + 1x + 0</p><p>)(Q</p><p>)2</p><p>+ (Q)2 2QQ(3) + 2Q(2(33x2 + 22x + 1)Q</p><p>+ 4(3x3 + 2x2 + 1x + 0)Q + Q(4)). (15)</p><p>We prove that the nonlinear equation (15) has a polynomial solutionQ(x, z) of degree g in z and degree g in x for some polynomial Fg(z). Afterthat we prove that</p><p>0 = 12Q</p><p>Q+</p><p>Fg(z)</p><p>Q (3x3 + 2x2 + 1x + 0</p><p>), (16)</p><p>1 = Q</p><p>Q(17)</p><p>satisfy (12) for the curve w2 = Fg(z). The fun...</p></li></ul>