Is it possible to extend this homomorphism?

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?

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

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.
• 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. Commented Sep 14, 2019 at 8:51
• @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. Commented Sep 14, 2019 at 10:42
• @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 $\alpha\beta=0$. Commented Sep 16, 2019 at 11:06