The Problem: Let $A$ be a Weyl Algebra, generated by two elements $x, y$ with the relation
$$xy – yx – 1 = 0$$
Suppose that char$k = p$.
(b) What is the center of $A$?
Hint: Show that $x^p$ and $y^p$ are central elements.
(c) Find all irreducible finite dimensional representations of $A$.
Hint. Let $V$ be a irreducible finite dimensional representation of $A$, and $v$ be an eigenvector of $y$ in $V$. Show that $\{v, xv, x^2v,…x^{p-1}v\}$ is a basis of $V$.
For part (b) the proposition of this section part (i) tells us that A basis for the Weyl algebra $A$ is $\{x^iy^j, i,j \geq 0\}$. Therefore, to use the hint, we just need to show that $x^px^iy^j = x^iy^jx^p$. In the proof of the proposition they use the $yx – xy = 1$ to interchange $x$ and $y$ at the cost of error terms that have smaller number of letters $x,y$ then the original word. This seemed to be useful for this part until I realized that it didn't allow me to use that char$k = p$. When I see char$k = p$, the first thing I think of is $(x+y)^p = x^p + y^p$ but that's about all that pops to my mind.
For part (c), taking the hint we know that $v$ satisfies $\rho(y)v = k_0v$ for some $k_0 \in k$. I'm confused because $x \in A$ while $v \in V$ so now sure what $xv$ would mean. Are we viewing $V$ as a vector space over $A$ or over $k$?
Best Answer
Part a: It would be helpful to have proven as a warm-up exercise that the Weyl algebra is simple in characteristic $0$. I will take the liberty of slightly rewriting the Weyl algebra as $k[x, \partial]/(\partial x - x \partial - 1)$ because I like thinking about its action on $k[x]$ by differential operators. If you prove simplicity using a method like the one in this math.SE answer a key step is the following.
In the Weyl algebra we have by definition that $[\partial, x] = 1$, so $[\partial, -]$, as a derivation, behaves exactly like the partial derivative with respect to $x$ on the subalgebra generated by $x$ (that's why I like calling it $\partial$); proving this amounts to using the lemma to prove by induction that
$$[\partial, x^n] = nx^{n-1}.$$
Proof. The first half is just a restatement of the definition of centrality; for the second half we use Lemma 1 to show that if $[z, a] = 0$ and $[z, b] = 0$ then $[z, ab] = 0$, so $\text{ker}([z, -])$ is a subalgebra. $\Box$
So to find central elements it's necessary and sufficient to find elements $f$ such that $[f, x] = [f, \partial] = 0$.
Now in characteristic $p$ a funny thing can happen which is that $n$ can be equal to zero if $p \mid n$, and in particular we have that
$$[\partial, x^p] = 0.$$
Switching the roles of $\partial$ and $x^p$ and writing this identity as $[x^p, \partial] = 0$, this tells us that the derivation $[x^p, -]$ vanishes on $\partial$, and since it clearly vanishes on $x$ then by Lemma 2 $x^p$ is central. Exactly the same argument (we could even use an automorphism of the Weyl algebra to formalize this) gives $[\partial^p, x] = 0$ and so $\partial^p$ is central.
So the center contains $k[x^p, \partial^p]$. We now show that this is exactly the center. Write an arbitrary element of the center as
$$f = \sum f_{ij} x^i \partial^j.$$
This is central iff $[f, x] = [f, \partial] = 0$ as we mentioned above. We were thinking of this condition in terms of the derivation $[f, -]$ but we can switch roles again and think of it as $[x, f] = [\partial, f] = 0$: this condition is equivalent to the derivations $[x, -]$ and $[\partial, -]$ both vanishing on $f$, which is a useful observation because it suggests that to do the computations we apply Leibniz some more. This gives
$$[x, f] = \sum f_{ij} x^i [x, \partial^j] = \sum f_{ij} x^i (-j \partial^{j-1})$$
and using that we know $\{ x^i \partial^j \}$ is a basis for the Weyl algebra we conclude that this is equal to zero iff each coefficient $-j f_{ij}$ is equal to zero iff for any $j$ either $p \mid j$ or $f_{ij} = 0$. Similarly
$$[\partial, f] = \sum f_{ij} [\partial, x^i] \partial^j = \sum f_{ij} (ix^{i-1}) \partial^j$$
we conclude that this is equal to zero iff each coefficient $i f_{ij}$ is equal to zero iff for any $i$ either $p \mid i$ or $f_{ij} = 0$. Hence if $f_{ij} \neq 0$ then $p | i, j$ which says exactly that $f \in k[x^p, \partial^p]$ as desired.
The idea to use $(x + y)^p = x^p + y^p$ is a sensible thing to try but unfortunately this identity only holds if $x$ and $y$ commute and so in the noncommutative world it's necessary to do something else.
Part b: $xv$ is shorthand for the action of $x$ on $v$ coming from the module structure. Very formally, a finite-dimensional representation $V$ of $A$ is a finite-dimensional vector space equipped with an algebra homomorphism $\rho : A \to \text{End}(V)$, and then $xv$ is shorthand for $\rho(x)(v)$.
The hint to use eigenvectors implies that we're working over an algebraically closed field $k$ so we'll assume this. Taking the hint, we'll let $v \in V$ be an eigenvector of $x$ (I have switched the variables, it doesn't matter though) and try to say something about the vectors $v, \partial v, \partial^2 v, \dots$. By definition we have
$$xv = \lambda v$$
which gives
$$x(\partial v) = (\partial x - 1)v = \partial (\lambda v) - v = \lambda (\partial v) - v.$$
What this says is that the image of $\partial v$ in the quotient $V / \text{span}(v)$ is an eigenvector of $x$ with the same eigenvalue $\lambda$ again. This is equivalent to a slightly more annoying statement about $x$ acting on the invariant subspace $\text{span}(v, \partial v)$ by the Jordan block $\left[ \begin{array}{cc} \lambda & -1 \\ 0 & \lambda \end{array} \right]$ but I prefer my reformulation which I think is a bit cleaner and more conceptual.
Note that it follows that $v$ and $\partial v$ are linearly independent. Otherwise one would be a multiple of the other, say $\partial v = cv$, and then we would have $x(\partial v) = xcv = \lambda cv = \lambda (\partial v)$ which would give $v = 0$.
Generally we have $[x, \partial^n] = -n \partial^{n-1}$ as explained previously which gives
$$x(\partial^n v) = (\partial^n xv - n \partial^{n-1})v = \lambda (\partial^n v) - n \partial^{n-1} v$$
which gives that the image of $\partial^n v$ in the quotient $V / \text{span}(v, \dots \partial^{n-1} v)$ is an eigenvector with the same eigenvalue $\lambda$ again, and we again have linear independence as long as $n \le p-1$ so that $n \partial^{n-1} \neq 0$. Once we hit $n = p$ the above becomes
$$x(\partial^p v) = \lambda (\partial^p v)$$
so we can no longer rule out the possibility that $\partial^p v$ is a scalar multiple of $v$, and in fact by Schur's lemma $\partial^p$ must act by a scalar $\lambda'^p$ (written this way because $\partial$ must have an eigenvalue $\lambda'$ and so $\partial^p$ must act by $\lambda'^p$). At this point we're done: $\text{span}(v, \dots \partial^{p-1} v)$ is invariant under the action of both $x$ and $\partial$, so by irreducibility it must be all of $V$, and since the vectors are linearly independent they must be a basis.
The scalar $\lambda'$ (equivalently, $\lambda'^p$, since the Frobenius map $x \mapsto x^p$ is invertible over an algebraically closed field of characteristic $p$) can be chosen freely; we can check explicitly that the result is an irreducible module $V_{\lambda, \lambda'}$ of the Weyl algebra by checking that the single relation $[\partial, x] = 1$ is satisfied, which we can check on the basis we just very conveniently computed.
So we've shown that the irreducible representations of the Weyl algebra are $p$-dimensional and parameterized by their central character (the homomorphism $k[x^p, \partial^p] \to k$ from the center to $k$ describing the scalars by which every element acts, by Schur's lemma): every possible central character occurs, and there's a unique irrep $V_{\lambda, \lambda'}$ with a given central character. (This reflects that in characteristic $p$ the Weyl algebra is an Azumaya algebra over its center.)
Note that we could have anticipated that the dimensions of the irreps needed to at least be divisible by $p$. There's a famous argument that shows that the Weyl algebra has no nonzero finite-dimensional representations in characteristic $0$, because if $XD - DX = 1$ where $X, D$ are two $n \times n$ matrices then taking traces of both sides gives $\text{tr}(XD - DX) = 0 = \text{tr}(1) = n$. So in characteristic $0$ this gives $n = 0$ and in characteristic $p$ this gives $p \mid n$. The above argument involving eigenvectors gives a different proof that there are no finite-dimensional representations in characteristic $0$ because it exhibits an arbitrarily long sequence of linearly independent vectors (which then give an infinite-dimensional representation).