Non degenerate representations for C*-algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:36:13Z http://mathoverflow.net/feeds/question/56394 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56394/non-degenerate-representations-for-c-algebras Non degenerate representations for C*-algebras Alessandro Gentile 2011-02-23T13:54:45Z 2011-02-23T18:45:11Z <p>Hi!</p> <p>While studying C*-algebras I found 2 different definitions for non degenerate representations (<em>-homomorphisms $\pi:\mathcal{A} \rightarrow B(\mathcal{h})$ where $\mathcal{A}$ is a C</em>-algebra and $B(\mathcal{h})$ is the space of bounded linear operators on some Hilbert space $\mathcal{h}$):</p> <p>1) For every non-zero $\xi \in \mathcal{h}$ there exists $a \in \mathcal{A}$ such that $\pi(a)\xi \neq 0$;</p> <p>2) The set ${\pi(a)\xi \quad a \in \mathcal{A}, \xi \in \mathcal{h}}$ is dense in $\mathcal{h}$.</p> <p>Are they equivalent?</p> <p>Thanks, Alessandro</p> http://mathoverflow.net/questions/56394/non-degenerate-representations-for-c-algebras/56396#56396 Answer by Jan Jitse Venselaar for Non degenerate representations for C*-algebras Jan Jitse Venselaar 2011-02-23T14:14:13Z 2011-02-23T14:27:06Z <p>Yes they are. This is Proposition I.9.2 in Theory of Operator Algebras I by Takesaki.</p> <p>Short proof:</p> <p>2) => 1): suppose $\pi(a)\xi = 0$ for all $a$. Then $(\pi(a)\eta|\xi) = 0$ for all $\eta\in h$ and $a\in\mathcal{A}$ hence $\xi=0$.</p> <p>1) => 2): Take $\xi \in h$ orthogonal to all $\pi(a) \eta$. Then from $(\xi| \pi(a^* a) \xi)= 0$ for all $a$ it follows that $\xi =0$.</p> http://mathoverflow.net/questions/56394/non-degenerate-representations-for-c-algebras/56409#56409 Answer by Stefan Waldmann for Non degenerate representations for C*-algebras Stefan Waldmann 2011-02-23T15:50:16Z 2011-02-23T18:45:11Z <p>In fact, for unital $C^*$-algebras non-degeneracy just means $\pi(1) = 1$. In the non-unital case there is even a sharper statement than your item (2): One can find for every $\phi$ and every $\epsilon > 0$ another vector $\psi$ and a <em>positive</em> algebra element $a \in \mathcal{A}^+$ with $$\phi = \pi(a)\psi \quad \textrm{and} \quad \|\phi - \psi\| &lt; \epsilon.$$ This is nice as it shows that we do not just get a dense subspace and we get in some sense as close as possible to $\pi(1) = 1$. I found this in Blackadars encyclopedia book in Theorem II.5.3.7 and in II.6.1.5. Might be worth a look :)</p>