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I don't know how to show the following: Let $A$ be an associative algebra (not necessary finite-dimensional) and $p\colon A\to End(V)$ be it irreducible finite-dimensional representation. Then $p$ in general is not surjective.

The standard textbooks on representation theory don't contain answer on this and googling doesn't help.

The interesting case for me is an irreducible representation of universal enveloping of semisimple Lie algebra.

Why am I asking: I am reading articles (of G.I. Olshaskiy) on centralizers of Lie subalgebras and trying to understand if $U(gl_{n+m})^{gl_m}$ is a sufficient object to consider or not.

Upd: I've forgot: the ground field is algclosed of char 0.

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  • $\begingroup$ Many things written about Olshankii's construction, there is nice survey by A. Molev arxiv.org/abs/math/0211288 and it is discussed there section 2.13 page 19. There is also book by Molev with the same name. Molev's texts are usually well-written. $\endgroup$
    – Alexander Chervov
    Commented Dec 18, 2011 at 17:00
  • $\begingroup$ @Alexander: Oh, no: Molev didn't write about what I asked. Also Olshanskiy's paper about it is much better (for me) than Molev's survey. For example I still don't know how to check the first equality in this survey in the way that Molev said (if you know how to prove this identity in universal enveloping algebra in this way - let me know). Instead of that I read Olshanskiy's paper where he proved this (in another way). $\endgroup$
    – zroslav
    Commented Dec 20, 2011 at 21:48
  • $\begingroup$ Molev's paper has it's own minuses. I'll read now Molev's book and I hope that this book is much more better (I don't think however that Molev's paper is very bad: he stated lots of interesting that made Yangians much more clear for me - this survey was the second source about Yangians I read, the first was Molev-Nazarov-Olshanskiy paper) $\endgroup$
    – zroslav
    Commented Dec 20, 2011 at 22:01

2 Answers 2

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What's the ground field? Of course if it's $\mathbb{R}$ and $A=\mathbb{R}[t]/(t^2+1)$, then the regular module is irreducible, but the corresponding $p$ is not surjective. Over an algebraically closed field it's true even for infinite-dimensional $A$ though.

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  • $\begingroup$ Why this is true for algebraically closed field? $\endgroup$
    – zroslav
    Commented Dec 17, 2011 at 15:15
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    $\begingroup$ Because it's called Jacobson's density theorem. Check Theorem 2.5 (a) in arxiv.org/abs/0901.0827 for a proof. $\endgroup$
    – darij grinberg
    Commented Dec 17, 2011 at 15:48
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    $\begingroup$ Actually for algebraically closed fields it is Burnside's theorem of which Jacobson's theorem is a generalization. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 17, 2011 at 17:00
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Let $D \ne k$ be a central division algebra of finite dimension over a (commutative) field $k$ say with $\dim_k D = n^2$. Now view $V = D$ as a left module over itself; then $V$ is an irreducible $D$-module.

Since $\dim \operatorname{End}_k(D) = n^4 > n^2$, the image of the mapping $D \to \operatorname{End}_k(V) = \operatorname{End}_k(D)$ is a proper subalgebra.

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