Timeline for Characterization of the family of simple groups PSL(2,q) by tensor multiplicity
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2021 at 15:34 | comment | added | Sebastien Palcoux | @GeoffRobinson: the point is that PSL2 is the only (usual) infinite family of non-abelian finite simple groups on which $m(G)$ is bounded above. And it turns out that this above bound is $3$. | |
| Jan 31, 2021 at 22:17 | answer | added | Frank Lübeck | timeline score: 5 | |
| Jan 31, 2021 at 12:17 | vote | accept | Sebastien Palcoux | ||
| Jan 31, 2021 at 17:14 | |||||
| Jan 30, 2021 at 17:48 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
update of classification2 up to 10000000
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| Jan 30, 2021 at 13:02 | answer | added | Denis Chaperon de Lauzières | timeline score: 4 | |
| Jan 30, 2021 at 11:25 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
Addition of the way of Geoff Robinson
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| Jan 30, 2021 at 11:24 | comment | added | Sebastien Palcoux | @GeoffRobinson: yes, in fact, let $G$ be a non-abelian finite simple group with $|G| \le 10^6$, let $\chi$ be an irreducible character of maximal degree and let $k$ be the class number. Then $\chi(1) \le 3k$ if and only if $G$ is isomorphic to $\mathrm{PSL(2,q)}$ for some $q$ or $\mathrm{PSU(3,3)}$, see the added list in the post. I guess it is also true without retriction on $|G|$. | |
| Jan 29, 2021 at 22:45 | comment | added | Geoff Robinson | As an even cruder bound, it is (more than) enough to find an irreducible character with $\chi(1) > 3k$, where the simple group $G$ has $k$ conjugacy classes. It is a conjecture (originally formulated in a more general form by D. Gluck, mainly motivated by solvable groups) that a simple group $G$ should have an irreducible character $\chi$ with $\chi(1) > |G|^{\frac{1}{3}}.$ There are probably only finitely many families of simple groups of Lie type with $k = k(G) > \frac{|G|^{1/3}}{3}$ ( which include ${\rm PSL}(2,q)$), and finitely many alternating groups satisfy that inequality. | |
| Jan 29, 2021 at 15:08 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
Improvement of the checking method as suggested by Mikko Korhonen in comment + an interesting classification for |G|<10^7
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| Jan 29, 2021 at 8:07 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
addition of a remark
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| Jan 29, 2021 at 7:53 | comment | added | Sebastien Palcoux | @MikkoKorhonen: The answer to your question is yes for $|G| \le 10^6$, where the smallest possible $\chi(1)^2/\Sigma(G)$ is $256/63 \simeq 4.06$ (given by $G = \mathrm{PSU}(3,3)$ only), so that $4 \Sigma(G) < \chi(1)^2$ is also true. | |
| Jan 29, 2021 at 6:16 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
checking update: now up to 10^7
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| Jan 28, 2021 at 6:23 | comment | added | Mikko Korhonen | Let $\Sigma(G)$ be the sum of the degrees of complex irreducible characters of $G$. Here is another question. For $G$ finite simple and not isomorphic to $\operatorname{PSL}_2(q)$, does there exist an irreducible character $\chi$ such that $3 \Sigma(G) < \chi(1)^2$? Then in the tensor product $\chi \otimes \chi$ some irreducible occurs with multiplicity $> 3$. There are upper bounds for $\Sigma(G)$ in the literature, maybe for $G$ of Lie type taking $\chi$ to be the Steinberg character would suffice. | |
| Jan 28, 2021 at 4:39 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
made the code readable
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| Jan 28, 2021 at 3:54 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
made the post a bit clearer
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| Jan 27, 2021 at 22:24 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
minor edit
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| Jan 27, 2021 at 21:34 | history | asked | Sebastien Palcoux | CC BY-SA 4.0 |