Simple modules of the Weyl algebra

Question

Let $\mathbb{F}$ be a field and W be the Weyl algebra, as the algebra over $\mathbb{F}$ generated by $a,b$ with relation $ab-ba=1$.

The description of simple modules over the Weyl algebra over algebraically closed fields of characteristic $p$ is known, they have dimension $p$ over $\mathbb F$ and they are in one to one correspondence with pairs of matrices of the form

$$M_a:=\begin{bmatrix} 0 & 0 & 0 & \cdots & \mu \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & 1 & 0 \end{bmatrix}, \quad M_b:=\begin{bmatrix} \lambda & p-1 & 0 & \cdots & 0 \\ 0 & \lambda & p-2 & \ddots & \vdots \\ 0 & 0 & \lambda & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & 1 \\ 0 & 0 & \cdots & 0 & \lambda \end{bmatrix}$$

for $\mu,\lambda \in \mathbb{F}$.

Let $\mathbb{F}=\mathbb{Z}_p$ be the field of $p$ elements and assume that $M$ is a module such that the action of $a$ and $b$ is invertible. Then, condition $ab-ba=1$ implies that $\dim_{\mathbb{F}}(M)$ is a multiple of $p$. So, assume that $\dim_{\mathbb{F}}(M)=p$. Is it true that if $M$ is simple then $M$ is given by a pair of matrices as above?

For algebraically closed $\mathbb F$, the answer is yes, according to ferngaston.wordpress.com/2016/06/30/… . For $\mathbb F = \mathbb Z_p$, I suspect that is no, since I don't think the eigenvalues have to lie in $\mathbb Z_p$. — darij grinberg
– darij grinberg, Commented Dec 14, 2024 at 20:55

LSpice · Accepted Answer · 2025-01-14 02:23:44Z

The answer to your particular question is Yes.

First of all, whenever you have representation over $\mathbb F_p$ you have a representation over its algebraic closure. First claim that when you suppose $\dim M = p$ then it is simple over algebraic closure $\mathbb F$. The reason for that is that dimension $p$ is minimal possible so over algebraic closure it's impossible to have non-trivial submodule.

Next claim is that the eigenvalues of $M_b $ lie in $ \mathbb F_p $ in that case. The characteristic polynomial equals to $(x -\lambda)^p$. So $\lambda^p \in \mathbb F_p$ thus $\lambda \in \mathbb F_p$.

Fix the basis over algebraic closure $e_1, \dots, e_p$ for which $M_a$, $M_b$ has desired form. I claim that $e_i$ are in $\mathbb F_p$ (up to constant). First of all $e_1$ is unique (up to constant) eigenvector for $M_b$ with eigenvalue $\lambda$ from $\mathbb F_p$. So it could be chosen as $\mathbb F_p$ vector. Then $e_2, \dots, e_p$ are also $\mathbb F_p$ vectors, as they are of the form $M_a^k e_1$. Then $\mu$ also in $\mathbb F_p$. So the matrices in basis $e_1, \dots, e_p$ has desired form.

Although the converse, that all simple modules have dimension $p$, is not true because if one does not restrict dimension, $M_b$ could have more than one eigenvalue, all of them lying outside $\mathbb F_p$.

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Simple modules of the Weyl algebra

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