Centralizers of Rank One in the First Weyl Algebra
aa r X i v : . [ m a t h . R A ] J a n Centralizers of rank one in the firstWeyl algebra.
Leonid Makar-Limanov
Abstract
Centralizers of rank one in the first Weyl algebra have genus zero
Mathematics Subject Classification (2000):
Primary 16S32.
Key words:
Weyl algebra, Centralizers.Take a ∈ A (the first Weyl algebra). Rank of the centralizer C ( a ) isthe greatest common divisor of the orders of elements in C ( a ) (orders asdifferential operators).This note contains a proof of the following.Theorem. If the centralizer C ( a ) of a ∈ A (defined over the field C ofcomplex numbers) has rank 1 then C ( a ) can be embedded into a polynomialring C [ z ].The classical works of Burchnall and Chaundy where the systematic researchof commuting differential operators was initiated are also devoted primarilyto the case of rank 1 but the coefficients of the operators considered by themare analytic functions. Burchnall and Chaundy treated only monic differen-tial operators which doesn’t restrict generality if the coefficients are analyticfunctions. Situation is completely different if the coefficients are polynomial.Before we proceed with a proof, here is a short refresher on the first Weylalgebra.Definition. The first Weyl algebra A is an algebra over a field K gen-erated by two elements (denoted here by x and ∂ ) which satisfy a relation1 x − x∂ = 1. When characteristic of K is zero it has a natural representationover the ring of polynomials K [ x ] by operators of multiplication by x andthe derivative ∂ relative to x . Hence the elements of the Weyl algebra can bethought of as differential operators with polynomial coefficients. They canbe written as ordinary polynomials: a = P c i,j x i ∂ j with ordinary additionbut a more complicated multiplication. If characteristic of K is zero thenthe centralizer of any element a ∈ A \ K is a commutative subalgebra of A . This theorem which was first proved by Issai Schur in 1904 (see [S]) hassomewhat entertaining history which is described in [ML].Given ρ, σ ∈ Z it is possible to define a weight function on A by w ( x ) = ρ, w ( ∂ ) = σ, w ( x i ∂ j ) = ρi + σj, w ( a ) = max w ( x i ∂ j ) | c i,j = 0 for a = P c i,j x i ∂ j and the leading form ¯ a of a by ¯ a = P c i,j x i ∂ j | w ( x i ∂ j ) = w ( a ).One of the nice properties of A which was used by Dixmier in his seminalresearch of the first Weyl algebra (see [D]) is the following property of theleading forms of elements of A : if ρ + σ > a, b ] = { ¯ a, ¯ b } for a, b ∈ A where [ a, b ] = ab − ba and { ¯ a, ¯ b } = ¯ a ∂ ¯ b x − ¯ a x ¯ b ∂ is the standard Poisson bracketof ¯ a, ¯ b as commutative polynomials (¯ a ∂ etc. are the corresponding partialderivatives) provided { ¯ a, ¯ b } 6 = 0.The main ingredient of the considerations below is this property of theleading forms.To make considerations clearer the reader may use the Newton polygonsof elements of A . The Newton polygon of a ∈ A is the convex hull ofthose points ( i, j ) on the plane for which c i,j = 0. The Newton polygons ofelements of A are less sensible than the Newton polygons of polynomials intwo variables because they depend on the way one chooses to record elementsof A but only those edges which are independent of the choice will be used.First case a = ∂ n + P ni =1 a i ∂ n − i .Consider the leading form α of a which contains ∂ n , is not a monomial,and has non-zero weight. This is possible if a C [ ∂ ]. (If a ∈ C [ ∂ ] then C ( a ) = C [ ∂ ].)Since the leading forms of the elements from C ( a ) are Poisson com-mutative with α they are proportional to the fractional powers of α (as acommutative polynomial). Because the rank of C ( a ) is 1 we should have α = c ( ∂ + c x k ) n . Hence there exists an automorphism which makes the2ewton polygon of a smaller (say, area wise), and after an automorphism ψ : x → x, ∂ → ∂ + p ( x ) we have ψ ( a ) ∈ C [ ∂ ]; therefore C ( a ) = C [ ψ − ( ∂ )].Second case a = x m ∂ n + P ni =1 a i ∂ n − i , m > C ( a ) are proportional to thefractional powers of α (as a commutative polynomial) as long as α is theleading form of a relative to weights ρ, σ of x and ∂ provided ρ + σ > α is not zero. Because of that and since the rank is assumedto be one n divides m .We have now two possibilities for α of a non-zero weight: it is c ( x d ∂ + c x k ) n and either k ≥ d or k < d −
1. ( ρ + σ = 0 if k = d − N ( a ) of a on a plane where the x axis is horizontal and the ∂ axis is vertical then N ( a ) has the vertex ( m, n ).This vertex can belong to two edges: left, which corresponds to the form c ( x d ∂ + c x k ) n , k < d − c ( x d ∂ + c x k ) n , k ≥ d . The left edge also can be parallel to the bisectrix of the firstquadrant, in this case ρ + σ = 0. It is also possible that these edges coincide,or that N ( a ) consists just of the vertex ( m, n ).In any of these case a has a non-trivial zero weight form. It can be themonomial cx m ∂ n , or c ( x d ∂ + c ) n , or, if m = n , a form for which ρ + σ = 0.Lemma 1. If a has the leading form of weight zero then C ( a ) is a subringof a ring of polynomials in one variable.Proof. Assume that the weights for which a has the leading form of weightzero are ρ for x and σ for ∂ where ρ, σ ∈ Z and relatively prime (and ρ + σ ≥ b ∈ C ( a ) has a non-zero leading form ¯ b of weightzero (relative to ρ, σ ) because a zero weight form and a non-zero weightform cannot commute i.e. if ρ + σ > ρ + σ = 0 then ¯ a ∈ C [ x∂ ] and only elements of C [ x∂ ]commute with it (see the Remark below). Hence the restriction map b → ¯ b is an isomorphism. An algebra generated by all ¯ b is a subalgebra of x − σ ∂ ρ if ρ > x σ ∂ − ρ if ρ < ✷ x i ∂ i = ( t + i − t + i − . . . ( t + i − i ) where t = x∂ is easy to check by induction since xt = xx∂ = x ( ∂x −
1) = tx − x = ( t − x .General case. a = a ( x ) ∂ n + P ni =1 a i ∂ n − i .In this case a = α n , α ∈ C [ x ]. We may assume that α C and that α (0) = 0 (applying an automorphism x → x + c, ∂ → ∂ if necessary). Theleft leading edge of the Newton polygon containing a point ( m ′ , n ) (where m ′ = ord( a )) corresponds to either a form of weight zero or c ( x d ∂ + c x k ) n where 0 < k < d − x m ′ ∂ n andthe Lemma 1 shows that C ( a ) is isomorphic to a subring of C [ z ].Since the proof of the Theorem turned out to be too simple and too shortwe can complement it by an attempt to describe the rank one centralizersmore precisely. In the first case it is already done, the centralizer is isomor-phic to C [ z ]. Actually if z = ∂ + p ( x ) for an appropriate p ( x ) ∈ C [ x ] then C ( a ) = C [ z ].It would be interesting to describe a for which C ( a ) C [ z ] and the secondcase will provide us with such examples. Recall that in this case we have twopossibilities for the leading form α of a non-zero weight if ρ + σ >
0: it is c ( x d ∂ + c x k ) n and either k ≥ d or k < d − a on a plane where the x axis is horizontal and ∂ axis is vertical may havethe right edge containing the vertex ( m, n ) which corresponds to the leadingform with k ≥ d and the left edge containing this vertex which correspondsto the leading form with k < d − k ≥ d an automorphism x → x, ∂ → ∂ − c x k − d makes the Newtonpolygon of a smaller and after several similar steps we will obtain the Newtonpolygon with the right edge parallel to the bisectrix of the first quadrant. Theleading form which corresponds to this edge cannot be treated as a commuta-tive polynomial. If these form has positive weight when ρ = 1 , σ = − C ( a ) are proportional to fractional powersof this form but as an element of A (not as a commutative polynomial).If the left edge of this Newton polygon is not parallel to the bisectrix wecan consider the centralizer of a in C [ ∂, x, x − ] and proceed with automor-4hisms x → x, ∂ → ∂ − c x k − d where k − d < −
1. Hence there exists anautomorphism x → x, ∂ → ∂ + q ( x ) of C [ ∂, x, x − ] such that ψ ( a ) = x m − n p ( t )where t = x∂ (here q ( x ) is a Laurent polynomial while p ( t ) is a polynomial).Of course C ( a ) = ψ − ( C ( x m − n p ( t )) and the rank of C ( x m − n p ( t )) is one.Since the rank is 1 we can find an element b ∈ D (the skew field of fractionsof A ) commuting with a = x m − n p ( t ) of the form x d − r ( t ) where r ( t ) ∈ C ( t )and deg( r ) = 1.If r is a polynomial then C ( a ) = C [ x d − r ]; if d = 1 then C ( a ) = C [ t ]; if r is not a polynomial and d > x d − r are polynomials.In the last case r ( t ) ∈ C ( t ) but r ( t ) r ( t + d − . . . r ( t + ( k − d − ∈ C [ t ](observe that tx = x ( t +1)). We can reduce this to r ( t ) r ( t +1) . . . r ( t + k − ∈ C [ t ] by rescaling t and r .By shifting t if necessary we may assume that one of the roots of r ( t ) is 0and represent r as a product r r where all roots and poles of r are in Z andall roots and poles of r are not in Z . It is clear that r ( t ) r ( t + 1) . . . r ( t + k − ∈ C [ t ] and r ( t ) r ( t + 1) . . . r ( t + k − ∈ C [ t ]. Since deg( r ) = 1 anddeg( r i ) ≥ k deg( r i ) ≥ r i is equal to zeroand r i ( t ) r i ( t + 1) . . . r i ( t + k − ∈ C for this r i . But then r i ( t ) = r i ( t + k )which is impossible for a non-constant rational function. Since r (0) = 0 and r = 0 we see that r is a constant and all roots and poles of r are in Z .We can assume now that 0 is the largest root of r and write r = ts ( t )where s ( t ) = Q pi =1 ( t + λ i ) Q pi =1 ( t + µ i ) ∈ C ( t ) \ C , λ i ∈ Z , µ i ∈ Z , and 0 ≤ λ ≤ λ ≤ · · · ≤ λ p , µ ≤ µ ≤ · · · ≤ µ p . If µ < r ( t ) r ( t + 1) . . . r ( t + k −
1) would havea pole at t = − µ . Hence µ > s ( t ) are negative integerswhile all zeros of s ( t ) are non-positive integers.A fraction t + λ i t + µ i can be presented as f i ( t ) f i ( t +1) if λ i < µ i or as f i ( t +1) f i ( t ) if λ i > µ i (recall that λ i = µ i ; indeed, t + dt = ( t +1)( t +2) ... ( t + d ) t ( t +1) ... ( t + d − if d >
0, take the reciprocalfraction if d < s ( t ) can be written as s ( t ) s ( t +1) s ( t +1) s ( t ) , s i ( t ) ∈ C [ t ]. Write s ( t ) = s ( t ) s ( t ) , s ( t ) = s ( t ) s ( t ) where s ( t ) is the greatestcommon divisor of s ( t ) , s ( t ). Then s ( t ) = s ( t ) s ( t ) s ( t +1) s ( t +1) s ( t +1) s ( t +1) s ( t ) s ( t ) = s ( t ) s ( t +1) s ( t +1) s ( t ) .A polynomial s ( t ) cannot have positive roots. Indeed, the largest such rootcouldn’t be canceled by a root of s ( t ) or s ( t + 1) and this root would be aroot of s ( t ) (which doesn’t have positive roots).Now, q ( t ) = r ( t ) r ( t + 1) . . . r ( t + k −
1) = t ( t + 1) . . . ( t + k − s ( t ) s ( t + k ) s ( t + k ) s ( t ) is a polynomial. If s ( t ) C then all roots of s ( t + k ) are less then 1 − k and the smallest root of s ( t + k ) would be a pole of q ( t ) (recall that s ( t )5nd s ( t ) are relatively prime). Hence s ( t ) ∈ C , s ( t ) = s ( t +1) s ( t ) , and q ( t ) = t ( t + 1) . . . ( t + k − s ( t + k ) s ( t ) .We can uniquely write s ( t ) = Q i ∈ I φ k,p i ( t + i ), where φ k,p ( t ) = Q pj =0 ( t + jk ), and all p i are maximal possible. Then s ( t + k ) s ( t ) = Q i ∈ I t + i + p i k + kt + i and t + i = t + i + p i k + k for all i , i ∈ I because of the maximality of p i .Hence I ⊂ { , . . . , k − } and each i is used at most once.As we have seen, all roots of s ( t ) are of multiplicity 1 and since ( xr ) N = x N Q N − i =0 ( t + i ) s ( t + N ) s ( t ) the elements ( xr ) N ∈ C [ x ] for sufficiently large N .Therefore the rank of C ( xr ) is one. Observe that the rank is not stableunder automorphisms: the rank of C ( φ ( xr )) where φ ( x ) = x + t M , φ ( t ) = t is M + 1.We understood the structure of C ( a ) when a = x m − n p ( t ). Are theresubstantially different examples of centralizers of rank one for which C ( a ) = C [ a ]? Consider the case of an order 2 element commuting with an order 3element which was completely researched in the work [BC] of Burchnall andChaundry for analytic coefficients. They showed (and for this case it is astraightforward computation) that monic commuting operators of orders 2and 3 can be reduced to A = ∂ − ψ ( x ) , B = ∂ − ψ ( x ) ∂ − ψ ′ ( x ) where ψ ′′′ = 12 ψψ ′ i.e. ψ ′′ = 6 ψ + c and ( ψ ′ ) = 4 ψ + c ψ + c (a Weierstrassfunction). The only rational (even algebraic) solution in this case is (up toa substitution) ψ = x − when c = c = 0. (If ψ is a rational functionthen the curve parameterized by ψ, ψ ′ has genus zero, so 4 ψ + c ψ + c =4( ψ − λ ) ( ψ − µ ) and ψ is not an algebraic function of x if λ = µ .) Thecorresponding operator A = x − ( t − t + 1) = ( x − t − t ) is homogeneous.In our case we have A = f ( x ) ∂ + f ( x ) ∂ + f ( x ). (The leading formfor w ( x ) = 0 , w ( ∂ ) = 1 must be the square of a polynomial.) Here arecomputations for this case. A = ( f ∂ ) − f f ′ ∂ + f ( x ) ∂ + f ( x ) = [ f ∂ + ( f f − f ′ )] − ( f f − f ′ ) ′ f − ( f f − f ′ ) + f . Denote f ∂ + ( f f − f ′ ) by D .Then A = D − φ ( x ) where φ = ( f f − f ′ ) ′ f + ( f f − f ′ ) − f ∈ C ( x ).Analogously to Burchnall and Chaundry case if there is an operator oforder 3 commuting with A then it can be written as B = D − φD − φ ′ f (this follows from [BC] but will be clear from the condition [ A, B ] = 0 aswell). In order to find an equation for φ we should compute [ A, B ]. Ob-serve that [
D, g ( x )] = g ′ f, [ D , g ] = 2 g ′ f D + ( g ′ f ) ′ f, [ D , g ] = 3 g ′ f D +6( g ′ f ) ′ f D +(( g ′ f ) ′ f ) ′ f . Hence [ A, B ] = − D , φD + φ ′ f ]+2[ D − φD, φ ] = − φ ′ f D + ( φ ′ f ) ′ f ) D + ( φ ′ f ) ′ f D + (( φ ′ f ) ′ f ) ′ f ] + 2[3 φ ′ f D + 3( φ ′ f ) ′ f D +(( φ ′ f ) ′ f ) ′ f ] − φφ ′ f = ( − φ ′ f +6 φ ′ f ) D +( − φ ′ f ) ′ f +6( φ ′ f ) ′ f ) D − (( φ ′ f ) ′ f ) ′ f +2(( φ ′ f ) ′ f ) ′ f − φφ ′ f = (( φ ′ f ) ′ f ) ′ f − φφ ′ f . Therefore (( φ ′ f ) ′ f ) ′ = 12 φφ ′ , ( φ ′ f ) ′ f =6 φ + c , ( φ ′ f ) ′ φ ′ f = 6 φ φ ′ + c φ ′ , ( φ ′ f ) = 4 φ + 2 c φ + c and we have aparameterization of an elliptic curve. Since f, φ ∈ C ( x ) this curve must havegenus 0, i.e. 4 φ + 2 c φ + c = 4( φ − λ ) ( φ − µ ).Take z = φ ′ f φ − λ ) . Then φ − µ = z and φ ′ f = 2 z ( z − δ ) where δ = λ − µ .Hence φ ′ = 2 zz ′ , zz ′ f = 2 z ( z − δ ) and z ′ f = z − δ .Assume that δ = 0. Since we can re-scale f and z as f → δf, z → δz ,let us further assume that δ = 1. Then z ′ f = z − R dzz − = R dxf .Recall that f ∈ C [ x ]. Since 2 R dzz − = ln z − z +1 all zeros of f have multiplicity1 and R dxf = ln( Q i ( x − ν i ) c i ) where { ν i } are the roots of f and c i = f ′ ( ν i ).Therefore z − z +1 = c Q i ( x − ν i ) c i where c = 0 and z = c Q i ( x − ν i ) ci − c Q i ( x − ν i ) ci .Now it is time to recall that φ = ( f f − f ′ ) ′ f + ( f f − f ′ ) − f where f, f , f ∈ C [ x ] and thus f φ = f ( z + µ ) ∈ C [ x ]. Because of that zf = c c Q i ( x − ν i ) ci − c Q i ( x − ν i ) ci Q ( x − ν i ) ∈ C [ x ] which is possible only if the rational function1 − c Q i ( x − ν i ) c i doesn’t have zeros.We can write Q i ( x − ν i ) c i as Q j ( x − ν j ) cj Q k ( x − ν k ) ck where c j , c k ∈ Z + . Then Q k ( x − ν k ) c k − c Q j ( x − ν j ) c j ∈ C which is possible only if c = 1.Since z = Q k ( x − ν k ) ck + Q j ( x − ν j ) cj Q k ( x − ν k ) ck − Q j ( x − ν j ) cj we see that z ∈ C [ x ]. So to produce a 2 , f = z − z ′ . If f, z aregiven then A = ( f ∂ + ψ ) − z − µ ) , B = ( f ∂ + ψ ) − z − µ )( f ∂ + ψ ) − zz ′ f is a commuting pair for any ψ ∈ C [ x ] (indeed, f ψ ∈ C [ x ] and ψ + f ψ ′ ∈ C [ x ],hence ψ ∈ C [ x ]). Constant µ = − since we assumed that λ − µ = 1 and2 λ + µ = 0 because the equation is ( φ ′ f ) = 4 φ + 2 c φ + c .Here is a series of examples: z = 1+ x n , f = xn (2+ x n ) , φ = (1+ x n ) + = x n (2 + x n ) + which correspond to A = [ xn (2 + x n ) ∂ + ψ ] − x n (2 + x n ) + )Even the simplest one, A = [ x (2 + x ) ∂ ] − x (2 + x ) + ] cannot be madehomogeneous.It seems that a complete classification of (2 ,
3) pairs of rank one is adaunting task. Our condition on z is that z assumes values ± z ′ = 0.Let us call such a polynomial admissible. We can look only at reduced monicpolynomials z ( x ) = x n + a x n − + . . . because a substitution x → ax + b preserves admissibility. Also λ n z ( λ − x ) preserves admissibility if deg( z ) = n λ n = 1.Examples above are just one value case. Say, an admissible cubic poly-nomial is x − · − x . If z = ( x − ν ) i ( x + ν ) j + 1 then it is admissiblewhen ν i + j = ( − i − − i − j ( i + j ) i + j i i j j . If a composition h ( g ( x )) is admissiblethen g ′ = 0 and h ′ = 0 should imply that h ( g ( x )) = ±
1. Hence h ( x ) shouldbe an admissible function. As far as g is concerned g ′ = 0 should imply thatthe value of g belongs to the preimage of ± h which is less restrictiveif this preimage is large. Because of that it is hard to imagine a reasonableclassification of all admissible polynomials.On the other hand z ≡ z ′ ) for z = x n + a x n − + · · · + a n leads to n − n − n . Say, for n = 4 all admissible polynomials are x ± x + ax + a , a = 64; x − a x + 2 √ a x + a , a = 64.Remark. The number of admissible polynomials of a given degree isfinite. Indeed consider first n − a , . . . , a n . They are satisfied if z ≡ c (mod z ′ ) where c ∈ C . If one ofthe components of the variety defined by these equations is more than one-dimensional then (by Affine Dimension Theorem) its intersection with thehypersurface given by the last homogeneous equation will be at least one-dimensional while condition z ≡ z ′ ) is satisfied only by z = x n (recall that we are considering only reduced monic polynomials).If δ = 0 then ( φ ′ f ) = 4( φ − λ ) and ( φ − λ ) − / = ± R dxf is a rationalfunction.Lemma 2. If f ∈ C [ x ] and R dxf is a rational function then f is a mono-mial, i.e. f = a ( x − b ) d . Proof. If g ′ = f for g ∈ C ( x ) then g = hf , h ∈ C [ x ] since the poles of g are the zeros of f and if the multiplicity of a zero of f is d then the corre-sponding pole of g has the multiplicity d −
1. An equality g ′ = f can berewritten as h ′ f − hf ′ = f . If deg( h ) > h ′ f ) > deg( f ). Hencethe leading coefficients of polynomials h ′ f and hf ′ are the same. This is pos-sible only when deg( h ) = deg( f ). Therefore there exists a c ∈ C for which I was unable to find a published proof for this observation. This proof is a result ofdiscussions with J. Bernstein and A. Volberg h − cf ) < deg( f ). Since ( h − cf ) ′ f − ( h − cf ) f ′ = f we can concludethat deg( h ) = 1 for h = h − cf . Changing the variable we may assumethat h = c x and then c ( f − xf ′ ) = f which is possible only if f = ax d . ✷ Hence when δ = 0 we may assume that f = x d . If d = 0 then this isthe first case and A is a homogeneous operator up to an automorphism. If d > A is a homogeneous operator up toan automorphism of C [ x − , x, ∂ ].These computations show that a description of the structure of central-izers of rank one in A is sufficiently challenging. Can the ring of regularfunctions of a genus zero curve with one place at infinity be realised as acentralizer of an element of A ? Here is a more approachable relevant ques-tion: is there an element of A ∈ D \ A for which p ( A ) ∈ A for a given apolynomial p ( x ) ∈ C [ x ]?Remarks. It seems that the definition of rank of a centralizer as thegreatest common divisor of the orders of its elements appeared in a work ofWilson [W].There are many papers discussing commuting differential operators, pro-viding examples of such pairs, and applications of such pairs to PDE. Aninterested reader can find a rather exhaustive reference list in the recentlypublished work [BZ] of Burban and Zheglow.Acknowledgments.The author is grateful to the Max-Planck-Institut f¨ur Mathemtik in Bonnwhere he worked on this project in July – August of 2019. He was alsosupported by a FAPESP grant awarded by the State of Sao Paulo, Brazil.References.[BZ] Burban, I. and Zheglov, A. Fourier-Mukai transform on Weirstrass cu-bics and commuting differential operators, Internat, J. Math. 29 (2018) no.10, 1850064, 46 pages. 9BC] Burchnall J. L. and Chaundy T. W. Commutative ordinary differentialoperators. Proc. London Math. Soc 21 (1923), 420 – 440.[D] Dixmier J. Sur les algebres de Weyl, Bull. Soc. Math. France 96 (1968),209 – 242.[ML] Makar-Limanov, L. Centralizers in the quantum plane algebra. Studiesin Lie theory, 411–416, Progr. Math., 243, Birkh¨auser Boston, Boston, MA,2006.[S] Schur, I. ¨Uber vertauschbare lineare Differentialausdr¨ucke, Berlin Math.Gesellschaft, Sitzungsbericht 3 Archiv der Math., Beilage (3), 8 (1904), 2–8.[W] Wilson, George Algebraic curves and soliton equations. Geometry today(Rome, 1984), 303–329, Progr. Math., 60, Birkh¨auser Boston, Boston, MA,1985. Department of Mathematics, Wayne State University, Detroit, MI 48202, USA;Department of Mathematics & Computer Science, the Weizmann Institute of Science,Rehovot 76100, Israel.
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