Let $G$ be a torsion free group and $\alpha$ be a nonzero element in its complex group algebra. Assume that $\mathfrak A$ is the Banach subalgebra of $\ell^1(G)$ generated by $\alpha$. Is it possible to extend a nonzero representation of $\mathfrak A$ (on a Hilbert space) to all of $\ell^1(G)$? What is the situation if we consider $G$ to be amenable?
1 Answer
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No. Take $\alpha=g$, a group element, and consider a nontrivial one dimensional representation of the cyclic group generated by $g$. If $G$ has no abelian quotient then you're doomed.

$\begingroup$ I was wondering if we might implicitly be assuming the Hilbert space to be infinitedimensional, or (equivalently?) allowing the extended hom to take values in a larger space. $\endgroup$ Commented Sep 14, 2019 at 8:48

$\begingroup$ As another example, if $G$ is the discrete Heisenberg group and $\alpha$ is in the center and maps to an element with nonunit determinant then any extension must be infinitedimensional. $\endgroup$ Commented Sep 14, 2019 at 8:51

1$\begingroup$ @SeanEberhard Let $G$ be the discrete Heizenberg group and $Z<G$ its center. A unirep of $Z$ extends to $G$ iff every irrep appears in it with high enough multiplicity (typically infinite) as could be seen by Stonevon Neumann (irreps of $G$ have constant central character). It follows that many natural unireps of $Z$, eg the regular rep, cannot be extended. Your example follows as well from this. $\endgroup$ Commented Sep 14, 2019 at 10:42

$\begingroup$ @UriBader Thanks for your answer. What is the answer if we assume that $\alpha$ has more than two elements in its support? Indeed, in my problem, I suppose that $\alpha$ is a zerodivisor, that's mean there is a nonzero $\beta$ in $\mathbb CG$ such that $\alpha\beta=0$. $\endgroup$ Commented Sep 16, 2019 at 11:06