Timeline for Dimensions of irreducible representations of $GL(n,F_q)$ are polynoms in q having roots ONLY at roots of unity and zero?
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7 events
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| Oct 18, 2021 at 22:27 | comment | added | Dr. Evil | Right...Thanks for the clarification. | |
| Oct 18, 2021 at 21:29 | comment | added | David E Speyer | I don't believe so. For $GL_1$, there is no canonical bijection between $\mathbb{F}_q^{\times}$ and $\mathrm{Hom}(\mathbb{F}_q^{\times}, \mathbb{C}^{\times})$, so I don't see why it would be better for $n$ larger, and all of the sources I know make this distinction. PS: Thanks for the useful answer! | |
| Oct 18, 2021 at 21:22 | comment | added | Dr. Evil | Are the two not canonically isomorphic? For instance, for S_n we know that the set of conjugacy classes is in canonical bijection with the set of irreducible characters (because both are in canonical bijections with partitions)...I had always assumed something similar holds for GL_n(F_q). Are you saying this is not the case? | |
| Oct 18, 2021 at 13:07 | comment | added | David E Speyer | A bit of a correction: Functions $\Phi \to \mathcal{P}$ canonically index conjugacy classes in $GL_n(\mathbb{F}_q)$. In order to canonically parametrize irreps, one should replace $\Phi$ with $\Gamma$, which is an inductive limit of $\mathrm{Hom}(\mathbb{F}_{q^n}^{\times}, \mathbb{C}^{\times})$ along norm maps; see Section 3.1 of arxiv.org/abs/math/0612668 . | |
| Aug 14, 2018 at 7:03 | history | edited | Dr. Evil | CC BY-SA 4.0 |
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| Aug 14, 2018 at 6:54 | history | edited | Dr. Evil | CC BY-SA 4.0 |
added 2553 characters in body
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| Aug 14, 2018 at 6:17 | history | answered | Dr. Evil | CC BY-SA 4.0 |