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Hi!

While studying C*-algebras I found 2 different definitions for non degenerate representations (-homomorphisms $\pi:\mathcal{A} \rightarrow B(\mathcal{h})$ where $\mathcal{A}$ is a C-algebra and $B(\mathcal{h})$ is the space of bounded linear operators on some Hilbert space $\mathcal{h}$):

1) For every non-zero $\xi \in \mathcal{h}$ there exists $a \in \mathcal{A}$ such that $\pi(a)\xi \neq 0$;

2) The set $\{\pi(a)\xi \quad a \in \mathcal{A}, \xi \in \mathcal{h}\}$ is dense in $\mathcal{h}$.

Are they equivalent?

Thanks, Alessandro

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2 Answers 2

up vote 5 down vote accepted

Yes they are. This is Proposition I.9.2 in Theory of Operator Algebras I by Takesaki.

Short proof:

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 $\xi=0$.

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 $\xi =0$.

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Thank you for helping me! –  Alessandro Gentile Feb 23 '11 at 15:36

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 positive algebra element $a \in \mathcal{A}^+$ with \begin{equation} \phi = \pi(a)\psi \quad \textrm{and} \quad \|\phi - \psi\| < \epsilon. \end{equation} 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 :)

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