I know that for a representation $\pi$ of a $^*$-algebra $\mathcal{A}$ on a Hilbert Space $\mathcal{H}$, if $\pi$ is irreducible then it is cyclic.
Is the reverse implication also valid - i.e. is a cyclic representation necessarily irreducible?
My definitions are:
- Irreducible: the only closed subspaces of $\mathcal{H}$ that are $\pi$-invariant are $\{0\}$ and $\mathcal{H}$
- Cyclic: there is a vector $v\in \mathcal{H}$ such that the linear subspace $\pi(\mathcal{A})(v)$ is dense in $\mathcal{H}$.
No, here's a counterexample : take $T$ the $2\times2$ diagonal matrix with diagonal $1$ and $2$ on $C^2$. Writing the canonical basis $(e_1,e_2)$, you see that $v:=e_1+e_2$ is a cyclic vector for (the algebra generated by) $T$. But the eigenvectors give one-dimensional invariant vector spaces. (btw, for $C^*$ algebras, nondegenerate cyclic representations decompose into direct sums of irreducible spaces, see any book on $C^*$ algebras like Murphy's).