Timeline for Is it possible to extend this homomorphism?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2019 at 11:06 | comment | added | MSMalekan | @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 $\alpha\beta=0$. | |
| Sep 16, 2019 at 11:00 | vote | accept | MSMalekan | ||
| Sep 14, 2019 at 13:41 | history | edited | Sean Eberhard | CC BY-SA 4.0 |
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| Sep 14, 2019 at 10:42 | comment | added | Uri Bader | @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. | |
| Sep 14, 2019 at 8:51 | comment | added | Sean Eberhard | 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. | |
| Sep 14, 2019 at 8:48 | comment | added | Sean Eberhard | 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. | |
| Sep 14, 2019 at 7:24 | history | answered | Uri Bader | CC BY-SA 4.0 |