4
$\begingroup$

I have been told, that there is a theorem about irreducible representation of $C^*$-algebras, but I have troubles finding it. It is also possible, that this theorem is consequence of some theorem I've already come across, but I did not realize that. From what I have understand, the theorem should state something like this:

Let $\pi: A \to B(\mathscr{H})$ be a representation. If this representation is irreducible, than for every two points in $\mathscr{H}$ there is a path from one to the other.

If needed I can provide the definitions of irreducible representations or some equivalent theorems.

$\endgroup$
2
  • 3
    $\begingroup$ What do you mean by path? I presume you mean that for any pair of vectors $x,y\in H$ there exists $a\in A$ such that $y = \pi(a)x$. A representation of $\pi$ with this property is called algebraically irreducible. Kadison's transitivity theorem asserts that every irreducible representation of a C*-algebra is also algebraically irreducible. $\endgroup$
    – Phoenix87
    May 7, 2015 at 12:26
  • $\begingroup$ Yes, that's what I meant by path, at least I think, I am still finding my way around these terms (sometimes having trouble to rephrase geometrical intuition in algebraic sense and vice versa). Thanks for the theorem. I've looked it up and try to understand that. $\endgroup$
    – quapka
    May 7, 2015 at 12:53

1 Answer 1

1
$\begingroup$

The Kadison Transivity theorem is just your answer. See for example (Murphy page 150).

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .