surjectivity of irreducible representation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:14:37Z http://mathoverflow.net/feeds/question/83705 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83705/surjectivity-of-irreducible-representation surjectivity of irreducible representation zroslav 2011-12-17T14:14:30Z 2011-12-17T15:13:44Z <p>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.</p> <p>The standard textbooks on representation theory don't contain answer on this and googling doesn't help.</p> <p>The interesting case for me is an irreducible representation of universal enveloping of semisimple Lie algebra.</p> <p>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.</p> <p>Upd: I've forgot: the ground field is algclosed of char 0.</p> http://mathoverflow.net/questions/83705/surjectivity-of-irreducible-representation/83706#83706 Answer by Vladimir Dotsenko for surjectivity of irreducible representation Vladimir Dotsenko 2011-12-17T14:24:50Z 2011-12-17T14:24:50Z <p>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.</p> http://mathoverflow.net/questions/83705/surjectivity-of-irreducible-representation/83707#83707 Answer by George McNinch for surjectivity of irreducible representation George McNinch 2011-12-17T14:25:15Z 2011-12-17T14:25:15Z <p>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. </p> <p>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.</p>