5
$\begingroup$

If $A$ is a commutative algebra over an algebraically closed field $k$, and $\rho:A \rightarrow End(V)$ is an irreducible representation of $A$ (where, a priori, $V$ may be infinite dimensional), can we conclude that $V$ must be one-dimensional? This is easy to show if we assume $V$ is finite dimensional. It is also true if $A$ is a $C^*$-algebra, $V$ is a (complex) Hilbert space, $End(V)$ denotes bounded operators, $\rho$ preserves conjugation (but is not required to be continuous), and irreducible means no closed subrepresentations, using some spectral theory. But is it true in the algebraic sense with no further restrictions on $A$ or $V$?

Edit: Given Dag Oskar Madsen's comment, I will need to place some restrictions on $A$... what if $A$ is finitely-generated?

Double Edit: Faisal's comment takes care of $A$ finitely-generated (and countably generated) over $\mathbb{C}$.

$\endgroup$
  • 8
    $\begingroup$ What about $A=k(X)$, the field of rational functions? $\endgroup$ – Dag Oskar Madsen Oct 23 '13 at 20:27
  • 5
    $\begingroup$ Working with $A=\mathbb{C}[T]$, taking $V$ to be a Banach space, and requiring subrepresentations to be closed gives the invariant subspace problem $\endgroup$ – John Wiltshire-Gordon Oct 23 '13 at 20:46
  • 5
    $\begingroup$ At the other end of the spectrum, there is a version of Schur's lemma (due to Dixmier) that works if $A$ is of countable dimension over $k$ and if $k$ is uncountable (and algebraically closed). $\endgroup$ – Faisal Oct 23 '13 at 20:51
5
$\begingroup$

If $A$ is finitely generated, as in your first edit, then this is much more elementary than the result of Dixmier that Faisal mentioned. A simple $A$-module would have the structure of a field extension of $k$ that is finitely generated as a $k$-algebra (since it's a quotient of $A$), but if $k$ is algebraically closed then there are no such extensions.

$\endgroup$
  • $\begingroup$ @QiaochuYuan $A$ is commutative. $\endgroup$ – Jeremy Rickard Oct 23 '13 at 22:23
  • $\begingroup$ A cyclic $A$-module is $A/I$ for some ideal $I$. If $I$ is not maximal, and properly contained in another proper ideal $J$, then $A/J$ is a non-zero proper quotient of $A/I$, so $A/I$ is not simple. I'm not sure what you mean about $A=k(t)$: the only simple module has annihilator the zero ideal, which is maximal since $A$ is a field. $\endgroup$ – Jeremy Rickard Oct 23 '13 at 22:41
  • $\begingroup$ Oh, hmm. I got badly confused somehow. My mistake. But I think this might be less elementary than Dixmier's lemma; doesn't the last step require the weak Nullstellensatz? $\endgroup$ – Qiaochu Yuan Oct 23 '13 at 22:45
  • $\begingroup$ Yes, you're probably right. I have to admit that I hadn't looked up the proof of Dixmier's lemma, and assumed it was harder than that. Still, my answer shows that (given the weak Nullstellensatz) the uncountability of $k$ is unnecessary. $\endgroup$ – Jeremy Rickard Oct 23 '13 at 23:11
  • $\begingroup$ The uncountability of $k$ in my formulation is only there to guarantee that $\dim A < |k|$. If we're only concerned with f.d. $A$, then this last inequality is automatic given that $k$ is alg closed. $\endgroup$ – Faisal Oct 23 '13 at 23:40
4
$\begingroup$

Here are the details on Faisal's suggestion in the comments.

Lemma (Dixmier): Let $k$ be an algebraically closed field, let $A$ be a $k$-algebra with $\dim A < |k|$, and let $V$ be a simple left $A$-module. Then $D = \text{End}_A(V) \cong k$.

Proof. By Schur's lemma, $D$ is a division algebra over $k$. Then $V$, as a $D$-module, is a sum of copies of $D$, and in particular $\dim D \le \dim V \le \dim A < |k|$. Since $k$ is algebraically closed, if $D$ is strictly larger than $k$ then it must contain a transcendental $t$, hence $k(t) \subseteq D$. But the elements $\frac{1}{t - a}, a \in k$ in $k(t)$ are linearly independent, hence $\dim k(t) \ge |k|$; contradiction. $\Box$

If $A$ is commutative, the image of $A$ in $\text{End}(V)$ is contained in $D$, so $A$ acts by scalars and the conclusion follows. (Incidentally, this gives a short proof of the weak Nullstellensatz over uncountable (algebraically closed) fields.)

$\endgroup$
2
$\begingroup$

As John Wiltshire-Gordon suggested in a comment, this is a generalization of the invariant subspace problem. It follows that for Banach spaces the answer is negative and for Hilbert spaces the answer is open.

$\endgroup$
  • 1
    $\begingroup$ It is usual to CW an answer if it is just repeating somebody else's comment as an answer. $\endgroup$ – Benjamin Steinberg Oct 24 '13 at 0:01
  • $\begingroup$ Right, I've fixed it. Sorry about that. $\endgroup$ – Vít Tuček Oct 24 '13 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.