Timeline for the character tables of irreducible representations of $SL(3,Z_q)$
Current License: CC BY-SA 3.0
9 events
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| Mar 12, 2017 at 9:19 | comment | added | Uri Bader | Qiaochu Yuan, you are definitley right saying that applying Clifford Theory is the first step in attacking this problem. However, a glance look at the Avni-Klopsch-Onn-Voll paper mentioned in my answer shows that this is only the tip of the iceberg, and there is a way to go from there. | |
| Mar 10, 2017 at 5:21 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Mar 9, 2017 at 23:16 | comment | added | Qiaochu Yuan | In the general case of an extension $1 \to N \to G \to H \to 1$, it's still the case that $H$ acts on the set of irreducible representations of $N$, and the next step is still to look at the stabilizers of the action, but now one has to study the projective representation theory of the stabilizers with respect to certain $2$-cocycles. I don't know how easy these computations are to do and I don't know a reference for them. | |
| Mar 9, 2017 at 23:15 | comment | added | Qiaochu Yuan | @Xiao-Gang: it doesn't split in that sense, but you can still use Clifford theory to answer the question (in principle), although the answer is complicated. The easiest special case of Clifford theory to understand is that of a semidirect product $G \cong A \rtimes H$ where $A$ is abelian; in this case $H$ acts on the group of characters $\hat{A}$ of $A$ and the representation theory of $G$ breaks up into the representation theories of the stabilizers of this action. This should be familiar to physicists from the case that $G$ is the Poincare group; the stabilizers are the "little groups." | |
| Mar 9, 2017 at 23:10 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Mar 9, 2017 at 23:09 | comment | added | Qiaochu Yuan | @Andrei: yes, you're right. In fact the condition is trace zero for all values of $d$. | |
| Mar 9, 2017 at 12:50 | comment | added | Xiao-Gang Wen | @Qiaochu Yuan Thank you very much for the answer. I wonder if the extension $1 \to N \to SL_d(\mathbb{Z}/p^n \mathbb{Z}) \to SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) \to 1 $ split or not. If it splits, then $SL_d(\mathbb{Z}/p^n \mathbb{Z}) =N\times SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $, and we can use the character tables of $N$ and $SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $ to construct that of $SL_d(\mathbb{Z}/p^n\mathbb{Z}) $. But if it does not spit, can we still use the character table of $N$ and $SL_d(\mathbb{Z}/p^{n-1}\mathbb{Z}) $ to construct that of $SL_d(\mathbb{Z}/p^n\mathbb{Z}) $. | |
| Mar 9, 2017 at 9:05 | comment | added | Andrei Smolensky | The matrix $M$ is not arbitrary, consider the simplest case of $2{\times}2$ matrices, then $\det(1+p^{n-1}M)\equiv1\pmod{p^n}$ means that $\operatorname{tr}(M)\equiv0\pmod{p}$. | |
| Mar 9, 2017 at 3:44 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |