Non-surjective representation of infinitely-generated simple module (see: Jacobson Density/Burnside's theorems)

Question

The book I'm working through covers (on its way to the Artin-Wedderburn theorem) the Jacobson Density theorem:

Let $L$ be an irreducible left $R$-module, with $D := \text{End}_{R-}(L)^{\text{op}}$ (so $L$ is a $(R,D)$-bimodule).
Let $v_1,\dots,v_n \in L$ be elements that are linearly independent over $D$.
Then for any set $w_1,\dots,w_n \in L$, there exists $r \in R$ such that $rv_i = w_i$ for each $i$.

It has the following corollary:

Under the above assumptions, note that $L$ is a right $D$-vectorspace (since $D$ is a division ring by Schur's Lemma).
Then if $L$ is finite-dimensional over $D$, the associated representation $\lambda: R \to \text{End}_{-D}(L)$ is surjective.

(This just follows by picking a (finite) basis of $L$ over $D$, let $f \in \text{End}_{-D}(L)$, and seeing where $f$ sends the basis elements; then JDT says $f$ acts on the basis as multiplication by some $r$.)

We also have Burnside's Theorem as a corollary: when $R$ is an algebra over algebraically-closed $\mathbb{k}$, by Schur's Lemma we have $\text{End}_{R-}(L)^{\text{op}} \cong \mathbb{k}$ for finite-dimensional $L$; and so we get the statement $\lambda: R \to \text{End}_{\mathbb{k}}(L)$ is surjective.

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The question, then: what's a counterexample to this in the non-finite-dimensional case? That is, what, if any, is an example of a ring $R$ and irreducible $R$-module $L$, with $D := \text{End}_{R-}(L)^{\text{op}}$, where the associated representation $R \to \text{End}_{-D}(L)$ is not surjective?

Per JDT, necessarily $L$ must be infinite-dimensional over $D$. Bonus points if $R$ is an algebra over an algebraically-closed field, so as to relate to Burnside's theorem specifically, but I'd be happy with any example.

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For the sake of showing I made an effort: frankly, I think my main obstacle is not knowing good examples of infinitely-generated irreducible modules. Also, I saw this question, which is almost what I was looking for, but the example the sole answer gives is a semisimple module, not simple.

I started with $\mathbb{C}[x]$-modules, but eventually realized that, since isomorphism classes of irreducible modules are in bijection with maximal ideals, and the only maximal ideals of $\mathbb{C}[x]$ are of the form $(x - a)$ for $a \in \mathbb{C}$, we get not just finite-dimensional but in fact one-dimensional irreducibles (corresponding to 1x1 Jordan blocks); and, in fact, the nullstallensatz says (if I'm thinking of this correctly?) that the only maximal ideals of $\mathbb{C}[x_1,\dots,x_n]$ are of the form $(x_1 - a_1,\dots,x_n - a_n)$ with, again, a one-dimensional quotient.

The natural next place to go would be $\mathbb{C}[x_1,x_2,x_3,\dots]$ in countable variables, but I'm not really sure what maximal ideals "look like" here. Other answers on this site suggest I can get non-nullstallensatz-like maximal ideals in uncountable-many variables via something like $R := \mathbb{C}[\{X_i\}_{i \in \mathbb{C}(t)}] \to \mathbb{C}(t)$, $X_i \mapsto i$, which is surjective onto a field and so has maximal kernel; but then it seems to me that, since $R$ just acts by multiplying and then quotient'ing, $D := \text{End}_{R}(\mathbb{C}(t))$ should (I think?) be the same as $\text{End}_{\mathbb{C}(t)}(\mathbb{C}(t)) \cong \mathbb{C}(t)$, and then $\lambda: R \to \text{End}_{-D}(\mathbb{C}(t)) \cong \mathbb{C}(t)$ is, in fact, surjective; if I did that right, then a similar issue should occur with any other field extensions.

I tried finding some other infinite-dimensional irreducible modules -- Wikipedia suggests some Lie algebra stuff? But unfortunately I don't know anything about representations of Lie algebras. There's also my professor's perennial "weird" module example, the Weyl algebra's action by x and d/dx on $\mathbb{C}[x]$, but I did my best to work through the calculations there and I think I found that the endomorphisms of $\mathbb{C}[x]$, as a subring of $\text{End}_{\mathbb{C}}(\mathbb{C}[x])$ -- which is the column-finite countably-indexed matrices $\mathbb{C}^{\mathbb{N} \times \mathbb{N}}$ -- that commute with the matrices $X$ and $Y$ corresponding to x and d/dx... are only multiples of the identity, ie $\text{End}_{A_1}(\mathbb{C}[x]) \cong \mathbb{C}$; and another answer on this site suggests the only nontrivial ideal of $\mathbb{C}^{\mathbb{N} \times \mathbb{N}}$ is the set of finite-rank maps, whereas both $X$ and $Y$ are infinite-rank, meaning the representation map $A_1 \to \text{End}_{\mathbb{C}}(\mathbb{C}[x]) \cong \mathbb{C}^{\mathbb{N} \times \mathbb{N}}$ is surjective. So... I'm stuck. (Sidenote: I'd also welcome comments that any of my analyses in the above paragraphs are incorrect in some way.)

Answer 1

You almost got it. The Weyl algebra $A = \mathbb{C}[x, \partial]$ acting on $M = \mathbb{C}[x]$ is in fact an example. Your calculation that $\text{End}_A(M) \cong \mathbb{C}$ is correct (this also follows from Dixmier's lemma), but your claim that the action map $A \to \text{End}_{\mathbb{C}}(M)$ is surjective is incorrect; the easiest way to see this is that $A$ is countable-dimensional but $\text{End}_{\mathbb{C}}(M)$ is uncountable-dimensional. Your observation about ideals of this ring is irrelevant; the image of this map is a subring, not an ideal.

Maybe more interestingly, we can explicitly exhibit an element of $\text{End}_{\mathbb{C}}(M)$ which is not in the image of the action map, but which is in the "closure" of the image in some sense. Namely, the action map only allows us to apply polynomials in the differential operator $\partial$, but in fact we can apply arbitrary power series in $\partial$: for example, we can apply the power series $e^{\partial t}$ for any $t \in \mathbb{C}$, which acts via translation:

$$e^{\partial t} f(x) = f(x + t).$$

This is a formal version of Taylor's theorem. It's a nice exercise to show that none of these operators are in the Weyl algebra (except when $t = 0$, of course). You can also check that they are linearly independent, and there are uncountably many of them, so all but countably many of them can't be in the Weyl algebra.

Many other examples can be produced by taking $A$ to be the universal enveloping algebra $U(\mathfrak{g})$ of a finite-dimensional semisimple Lie algebra $\mathfrak{g}$ and taking $M$ to be an infinite-dimensional simple Verma module. The same countability argument can be used; if $\mathfrak{g}$ is finite-dimensional then $U(\mathfrak{g})$ is countable-dimensional, so the action map can't be surjective if $M$ is an infinite-dimensional simple module, and such modules already exist for, say, $\mathfrak{g} = \mathfrak{sl}_2$. The Weyl algebra example is similar; it's a quotient of the universal enveloping algebra of the $3$-dimensional Heisenberg algebra.

An example that might be easier to understand is to take $M$ to be any countable-dimensional vector space and to take $A \subset \text{End}(M)$ to be scalar multiples of the identity + the finite rank operators. Here it's easy to show that $M$ is irreducible over $A$ (since $A$ acts transitively on $M \setminus \{ 0 \}$), and the same countability argument as above shows that the inclusion $A \hookrightarrow \text{End}(M)$ isn't surjective, or you can explicitly exhibit a linear operator that isn't a scalar multiple of the identity + a finite rank operator. For example any such operator has many eigenvectors, so $x, \partial$ in the Weyl algebra acting on $\mathbb{C}[x]$ are both counterexamples.