Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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

share|cite|improve this question

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 $\xi =0$.

share|cite|improve this answer
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 :)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.