Classifying irreducible finite-dimensional representations of the $q$-Weyl algebra I have that $A = \langle x, y : yx = qxy \rangle$ is the $q$-Weyl algebra, with basis $x^iy^j$, $i, j \in \mathbb{Z}$. The base field is $\mathbb{C}$.
I need to classify the irreducible finite-dimensional representations. Assuming $V$ is such a module, I have shown that $\{v, xv, \dots, x^{m-1}v\}$ is a basis, where $v$ is a $\lambda$-eigenvector of $y$. I have also shown (using Schur's lemma)  that $x^m, y^m$ act by scalars, $\mu$ and $\lambda^m$ respectively. I now need to show that the isomorphism class of $V$ is determined by the scalars that $x^m, y^m$ act by.
All I know is that another representation $W$ is isomorphic to $V$ if and only if there is an intertwining $\phi: V \to W$ and that $V \cong W$. Using the basis given above, I can write the matrix of $y$ as $\text{diag}[\lambda, q\lambda, \dots, q^{m-1}\lambda]$. I can also write the matrix of $x$ as having nonzero entries on the subdiagonal and the top right $m,1$ entry.
Now I am trying to figure out how else to use the basis to classify $W$, but am not sure where to go from here. It seems like there are many possibilities, since all that needs to be checked are that $x$ and $y$ act in the same way. Somehow I feel like I need to use that $V \cong W$, but cannot see how to proceed from there. I thought that I could write out an isomorphism $\phi$ using the basis of $V$ but all I can tell is that $\phi(x^kv)$ has to correspond to some other basis element of $W$. I am mostly lost from here, any guide would be great!
 A: Here are my notes on the problem. For each $q \in \mathbb{C}^\times$, define $A_q = \mathbb{C} \langle x^{\pm 1}, y^{\pm 1} \mid yx = qxy \rangle$, and let $V$ be a finite-dimensional representation of $A_q$.


*

*The existence of a finite-dimensional representation $V$ implies that $q$ is a root of unity. So from here on assume that $q$ is a primitive $m$th root of unity for some $m \geq 1$.

*The operators $x^m$ and $y^m$ commute (and in fact generate the centre of $A_q$). Hence if $V$ is indecomposable there are a unique pair $\lambda, \mu \in \mathbb{C}^\times$ such that $V$ is a $\lambda$-generalised eigenspace for $x^m$ and a $\mu$-generalised eigenspace of $y^m$.

*The operator $x$ induces an isomorphism of eigenspaces $V(\nu) \xrightarrow{\sim} V(q \nu)$, where $V(\nu)$ denotes the $\nu$-eigenspace of $y$ (and the same holds for generalised eigenspaces).  This implies that the dimension of any finite-dimensional representation is a multiple of $m$.

*Fix a particular $m$th root $\nu = \sqrt[m]{\mu}$ of $\mu$. The indecomposable representation $V$ has $y$-eigenspaces $V(\nu), \ldots, V(q^{m - 1} \nu)$, and multiplication by $x$ gives isomorphisms between these eigenspaces. In particular, multiplication by $x^m$ gives an invertible endomorphism of $V(\nu)$ whose only eigenvalue is $\lambda$, and so we pick some $v \in V(\nu)$ such that $x^m v = \lambda v$. Then $v, xv, \ldots, x^{m-1}v$ generate a subrepresentation, which must be irreducible since it has dimension $m$ and every representation of $A_q$ has dimension a multiple of $m$.


Now, say we have two irreducible representations $W$ and $V$ which are of the same "central type" $(\lambda, \mu)$. By the discussion above, we can choose a particular root $\nu$ such that $\nu^m = \mu$, and start defining an intertwining operator $\phi \colon W \to V$ by choosing any nonzero vector $w$ in the one-dimensional $y$-eigenspace $W(\nu)$ and sending it to any nonzero vector $v$ in $V(\nu)$. If we want to extend $\phi$ to an intertwining operator our hand is forced: for exampe $xw$ must be sent to $xv$ because
$$ \phi(xw) = x \phi(w) = xv.$$
Therefore define $\phi$ to be the map taking the basis $(w, xw, \ldots, x^{m-1}w)$ to the basis $(v, xv, \ldots, x^{m-1}v)$. This map is well-defined since those are indeed bases, we just need to check that it is an isomorphism of representations. Therefore we need to verify
$$ \phi(x \cdot x^i w) = x \cdot \phi(x^i w) \text{ for all } i = 0, \ldots, m-1, \text{ and}$$
$$ \phi(y \cdot x^i w) = y \cdot \phi(x^i w) \text{ for all } i = 0, \ldots, m-1.$$
Writing down the actions of $x$ and $y$, the first equation becomes $\phi(x^{i + 1} v) = x^{i + 1} w$, which is immediate for $i \neq m - 1$ from the definition of $\phi$, and for $i = m - 1$ is just checking that $x^m v = x^m w$ which is true since $x^m$ acts by the scalar $\lambda$ on both. The second equation should just be verifying that each basis is an eigenbasis of $y$, and we are sending vectors to a vector with the same eigenvalue.
