most recent 30 from http://mathoverflow.net 2013-05-24T09:53:51Z http://mathoverflow.net/feeds/question/110429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110429/existence-of-irreducible-subspace anton 2012-10-23T13:36:01Z 2013-02-09T13:05:19Z

<p>Let $H$ denote a Hilbert space and let $\cal A$ be a subalgebra of the algebra ${\cal B}(H)$ of all bounded operators on $H$ such that $\cal A$ consists of compact operators only and such that each vector $v\in H$ lies in the closure of ${\cal A}v$. </p> <p>Is it true that there must exist an irreducible subspace for $\cal A$? </p>

http://mathoverflow.net/questions/110429/existence-of-irreducible-subspace/110460#110460 Nik Weaver 2012-10-23T19:04:23Z 2012-10-23T19:04:23Z

<p>Does "subalgebra" mean "C*-subalgebra"? If so then yes, see Arveson, <em>An Invitation to C*-Algebras</em> Theorem 1.4.4.</p>

http://mathoverflow.net/questions/110429/existence-of-irreducible-subspace/110479#110479 Nik Weaver 2012-10-23T21:23:29Z 2012-10-23T21:23:29Z

<p>Well, if ${\cal A}$ can be non self-adjoint then I think there are easy counterexamples. Let $(e_n)$ be the standard basis of $l^2({\bf N})$ and let ${\cal A}$ be the closed span of the operators $e_n \otimes e_m$ for $n \leq m$ (where $(e_n \otimes e_m) v = \langle v, e_n\rangle e_m$). The condition that any $v$ lie in the closure of ${\cal A}v$ holds because $\sum_1^N e_i \otimes e_i \to I$ strongly.</p> <p>It's easy to see that the closed invariant subspaces for ${\cal A}$ are precisely the subspaces <code>${\rm span}\{e_n: n\geq N\}$</code> for $N \in {\bf N}$, of which there is no nonzero minimum.</p>