$\newcommand{\ad}{\operatorname{ad}}$As my recent post (here) did not receive any answers yet, I thought I would ask a similar question in which I'm also interested.

Let $A=$ $^{k \langle x,y\rangle }\Big/_{(yx-xy-1)}$ be the Weyl Algebra over a field $k$ of characteristic $0$. For an element $q(x,y) \in A$, we have the operator $\ad_q(h):=[q,h]=qh-hq$. Also, we say that $\ad_q$ is locally nilpotent if $\forall h \in A$ $\exists n_h \in \mathbb{N}$ such that $\ad_q^{n_h}(h)=0$. My question is simply:

For which $q$ is $\ad_q$ locally nilpotent?

If $q \implies \ad_q$ locally nilpotent, I say that $q$ is good. Here's what I have:

  • It's simple to show that $q(x,y)=r(x)+s(y)$ is good if $\deg(r)=0,1$ or $\deg(s)=0,1$.
  • If $q$ is a polynomial in $\theta=xy$, I can show that $q$ is good iff $q$ is constant.

As $\ad_q$ is a derivation (it satisfies the Leibniz rule), it might be useful to notice that $q$ is good iff $\exists n_x,n_y$ such $\ad_q^{n_x}(x)=0$ and $\ad_q^{n_y}(y)=0$.

I conjecture that the polynomials of the form $r(x)+s(y)$ with $\deg(r)=0,1$ or $\deg(s)=0,1$ are in fact the only good polynomials, but currently I'm not seeing a direct way to prove this. Any ideas/hints? Thank you!


1 Answer 1


Dixmier gives a complete characterization of ad-locally nilpotent elements of $A=A_1$ in his paper "Sur les algèbres de Weyl", Bulletin de la S. M. F., tome 96 (1968), Theorem 9.1: An element $q \in A_1$ is ad-locally nilpotent iff it belongs to the orbit of the constant coefficient differential operators under the automorphism group of $A_1$.

  • $\begingroup$ Just to be explicit, this automorphism group is generated by "triangular" automorphisms, which have the form $x \mapsto x,\quad y \mapsto y + p(x)$ or $x \mapsto x + p(y),\quad y \mapsto y$, where $p(-)$ is an arbitrary polynomial. This is a big group... $\endgroup$ Commented Apr 8, 2015 at 15:08
  • 1
    $\begingroup$ Thank you for your answer. Also, I would like to add that this strongly resembles Rentschler's theorem in the polynomial algebra $A=k[x,y]$, that says that every locally nilpotent derivation of $A$ is conjugate of a derivation like $p(x) \frac{d}{dy}$ by a tame $k$-automorphism. $\endgroup$
    – u1571372
    Commented Apr 17, 2015 at 0:09

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