Representations of the Weyl algebra

(This post is a solution to problem 1.26 from Etingof, Goldberg and Hensel’s text on representation theory).

In this post we will study the representation theory of the Weyl algebra $A=k\langle x,y\rangle/(xy-yx-1)$ , where $k$ is algebraically closed.

If the characteristic of the field $k$ is zero, then there are no non-trivial finite dimensional representations of $A$ , since in such a representation $x, y$ must act as matrices and the identity $[x,y]=1$ cannot hold since the commutator of two matrices has null trace, while the identity does not.

It is easy to prove by induction that $y^jx=xy^j+jy^{j-1}$ . Suppose then that $I$ is a bilateral ideal of $A$ containing a non-zero element $p(x,y)$ , in which we will suppose that both variables appear with non-zero coefficients. The previous identity shows that $xp-px=p'(y)$ (in fact, $p'$ is the derivative of $p$ with respect to $y$ evaluated at $x=1$ ), so we may always find a polynomial $q$ depending on only one variable in any ideal. This shows that the only non-zero ideal is in fact the whole algebra, since if $q(x)\in I$ then so does $yq-qy = q(1)$ which is a unit in the algebra. This in turn provides another proof of the fact that there are no non-trivial finite dimensional representations, since such a representation is given by a map of algebras $A\to \mathrm{End}_k(V)$ , and since the latter is finite-dimensional we necessarily have non-trivial kernel. Since the kernel is an ideal in $A$ , it follows that $A$ , and in particular its unit, must act as $0$ , which proves that $V=0$ .

Suppose now that our base field has positive characteristic $p$ . The commutation relation easily shows that $x^p, y^p$ are central, since $y^px = xy^p +py^{p-1}=xy^p$ . We claim that the center is exactly the ideal $(x^p, y^p)$ . If not, suppose $p$ is a polynomial not contained in this ideal. Without loss of generality we will assume that there is a term of the form $x^iy^j$ with $p$ does not divide $j$ . Then $xp-px = p'(y)$ where $p'$ is a non-zero polynomial and so $p$ is not central.

Let us now find all the irreducible finite dimensional representations of $A$ .
Taking traces in the identity $xy-yx=1$ , we obtain that any finite dimensional representation $V$ must be of dimension $np$ . Suppose $v\in V$ is an eigenvector of $y$ ; say $yv = \lambda v$ . We claim that the $x$ -cyclic space generated by $v$ is an $A$ -submodule, since it is obviously $x$ -stable and

$y\sum \mu_i x^iv = \sum( \mu_i\lambda x^iv + i\mu_i x^{i-1}v).$

By Schur’s lemma we know that, since $x^p, y^p$ are central and $V$ is irreducible, their action on $V$ is scalar. Therefore $x^pv=\mu v$ and so $V=\langle v,xv,\dots, x^{p-1}v\rangle$ . Moreover, these vectors are linearly independent, since the representation is of dimension $np$ and $v\neq 0$ .

In conclusion, any finite dimensional representation $V$ must be of dimension $p$ and there is a basis $\{v,xv,\dots, x^{p-1}v\}$ where the action of $x$ and $y$ is given by matrices

$x=\begin{bmatrix} 0&0&0&\dots&\mu\\1&0&0&\dots&0\\ 0&1&0&\dots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&0&1&0\end{bmatrix}, y=\begin{bmatrix} \lambda&1&0&\dots&0\\0&\lambda&2&\dots&0\\ 0&0&\lambda&\ddots&0\\ \vdots&\vdots&\ddots&\ddots&p-1\\ 0&0&0&0&\lambda\end{bmatrix}$

and conversely different choices of $\lambda,\mu$ gives rise to non-isomorphic irreducible finite dimensional representations.

1 Response to Representations of the Weyl algebra

Alberto Riccardi says:

May 11, 2021 at 8:02 am

In the proof that the Weyl algebra is simple, I simply cant undestand the way, by commuting a polynomial in x and y, with (say) x, you can always get a polynomial in one variable.