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Let $G$ be a torsion free group and $\alpha$ be a non-zero element in its complex group algebra. Assume that $\mathfrak A$ is the Banach sub-algebra of $\ell^1(G)$ generated by $\alpha$. Is it possible to extend a non-zero 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?

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  • $\begingroup$ What is the situation for the Heisenberg group and $\alpha$ a generator of the center? $\endgroup$ Commented Sep 14, 2019 at 7:23

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No. Take $\alpha=g$, a group element, and consider a non-trivial one dimensional representation of the cyclic group generated by $g$. If $G$ has no abelian quotient then you're doomed.

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  • $\begingroup$ I was wondering if we might implicitly be assuming the Hilbert space to be infinite-dimensional, 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 infinite-dimensional. $\endgroup$ Commented Sep 14, 2019 at 8:51
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    $\begingroup$ @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 Stone-von 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. $\endgroup$
    – Uri Bader
    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 zero-divisor, that's mean there is a nonzero $\beta$ in $\mathbb CG$ such that $\alpha\beta=0$. $\endgroup$ Commented Sep 16, 2019 at 11:06

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