The second part of Theorem 3.10.2 of [*"Introduction to representation
theory"* by Etingof, Golberg, Hensel, Liu, Schwender, Vaintrob and Yudovina](http://www-math.mit.edu/%7Eetingof/repb.pdf) states that
if $A$ and $B$ are $k$-algebras ($k$ an algebraically closed field) and $M$ is an irreducible finite dimensional representation of $A\otimes\_k B,$ then $M\cong V\otimes\_k W$ where $V$ and $W$ are finite dimensional irreducible representations of $A$ and $B$ respectively.


My question is about the first part of the remark following this theorem. This remark states that the previous proposition fails for infinite dimensional representations, "e.g. it fails when A is the Weyl algebra
in characteristic zero." I don't see how to construct an irreducible infinite dimensional representation $M$ of $A\otimes B,$
where $A$ is the Weyl algebra, such that $M\ncong V\otimes\_k W$.


(I asked the same question on Math.SE more than one year ago without receiving answers, also after starting bounties)



Infinite dimensional irreducible representations of a tensor product
[Nate's suggestion on math.SE](https://math.stackexchange.com/questions/3349346/infinite-dimensional-irreducible-representations-of-a-tensor-product) works. We'll show that if $A = k[x, \partial\_x]$ and $B = k[y, \partial\_y]$ are both taken to be the Weyl algebra, then the module over $A\_2 = A \otimes B \cong k[x, \partial\_x, y, \partial\_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$.


Explicitly this module $M$ consists of elements of the form $f(x, y) e^{xy}$ where $f$ is a polynomial, with the obvious action by multiplication and differentiation. Abstractly it is the quotient of $k[x, \partial\_x, y, \partial\_y]$ by the left ideal $(x - \partial\_y, y - \partial\_x)$. We can show very straightforwardly that every nonzero element of $M$ generates it, by computing that


$$(\partial\_x - y) x^i y^j e^{xy} = ix^{i-1} y e^{xy}$$


and similarly that


$$(\partial\_y - x) x^i y^j e^{xy} = j x^i y^{j-1} e^{xy}.$$


In other words, this module is the pullback of the usual module $k[x, y]$ under the (edit: inverse of the) automorphism $A\_2 \to A\_2$ sending $x \mapsto x, y \mapsto y, \partial\_x \mapsto \partial\_x - y, \partial\_y \mapsto \partial\_y - x$. Now $k[x, y]$ is irreducible which means so is this module (explicitly, every nonzero element is a generator because we can repeatedly differentiate resp. apply the above maps to get to $1$ resp. $e^{xy}$), and $k[x, y]$ is a tensor product $k[x] \otimes k[y]$.


But we can show that $M$ is not such a tensor product (this property is not invariant under twisting by automorphisms). If it were such a tensor product $V \otimes W$, then a pure vector $v \otimes w$ would have the property that there is some differential operator $a(x, \partial\_x) \in k[x, \partial\_x]$ such that $(a \otimes 1)(v \otimes w) = av \otimes w = 0$; in words, it would satisfy a polynomial differential equation involving $x$ only.


No nonzero element of $M$ has this property. The key point is that every nonzero element of $k[x, \partial\_x]$ makes elements bigger in the lex order: we have


$$x \left( x^i y^j e^{xy} \right) = x^{i+1} y^j e^{xy}$$
$$\partial\_x \left( x^i y^j e^{xy} \right) = ix^{i-1} y^j e^{xy} + x^i y^{j+1} e^{xy}$$


and so formally, writing an arbitrary element $a \in k[x, \partial\_x]$ as a sum $\sum a\_{k, \ell} \partial\_x^k x^{\ell}$, we see that if $\partial\_x^k x^{\ell}$ is the largest monomial in $a$ in the lex order where $x > \partial\_x$, then $a(x^i y^j e^{xy})$ has largest monomial $x^{i + \ell} y^{j + k}$, and in particular it does not vanish, so the same is true if $x^i y^j e^{xy}$ is replaced by any other element with the same largest monomial.


