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Let $A$ be a (possibly nonunital) Banach *-algebra, and $H$ be a Hilbert space. If $\pi: A \to B(H)$ is a *-homomorphism, i.e. a representation, then why must $\pi$ be equivalent to a direct sum of cyclic representations? A typical Zorn's Lemma argument justifies that $\pi(A)\xi_i$ gives an orthogonal decomposition of $H$ for some family of $\xi_i$ but these are not necessarily cyclic subspaces wrt $\pi(A)$ because $\xi_i$ might not be in such a space. (This argument can be found in Takesaki volume 1 p. 41 Prop 9.17)

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Do you want to add the assumption of nondegeneracy? – Jonas Meyer Sep 13 '12 at 3:22
I do, I forgot that. No Zorn's lemma argument will go as far as even what I've done without nondegeneracy. – Jeff Sep 13 '12 at 3:31
The notion of cyclic subspace wrt $\pi(A)$ generated by $\xi \in H$ is not lost without the identity operator. It is just the closure of $(\pi(A)\xi + \mathbb{C}\xi)$. I don't see how to use this to solve my question either though. – Jeff Sep 13 '12 at 18:25
Actually I take that last comment back. Without identity, neither the space in my last comment nor just $\pi(A)\xi$ are cyclic it seems? At least not with cyclic vector $\xi$ – Jeff Sep 13 '12 at 21:28
up vote 1 down vote accepted

I just got it! A typical Zorn argument shows that for some nonzero family of $\xi_i$, $H$ can be decomposed into an orthogonal direct sum of $\overline{\pi(A)\xi_i}$. Therefore, for a given i, $\xi_i$ breaks down into $\sum_{j} x_{ij}$ where $x_{ij} \in \overline{\pi(A)\xi_j}$. Thus, we have

$\xi_i-x_{ii} \in \sum_{j \neq i} \overline{\pi(A)\xi_j}$

Now, for any $a \in A$ we have

$\pi(a)(\xi_i-x_{ii}) \in \sum_{j \neq i} \overline{\pi(A)\xi_j} \cap \overline{\pi(A)\xi_i}=0$

By nondegeneracy, this shows that $\xi_i=x_{ii} \in \overline{\pi(A)\xi_i}$. The rest from here goes as it does when there's a unit in $A$.

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