Timeline for The number of irreducible characters of simple groups of Lie type
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13 at 3:31 | vote | accept | user44312 | ||
| Apr 12 at 14:44 | comment | added | user44312 | Thanks very much for the detailed answer. It is very helpful. | |
| Apr 12 at 4:34 | answer | added | Geoff Robinson | timeline score: 4 | |
| Apr 12 at 2:13 | comment | added | Richard Lyons | Here are many more examples: If $n$ and $q$ are odd and $\sigma$ has order $2$, and $S=A_n(q)$ (finite simple), then $C_S(\sigma)$ is not simple, although its unique noncyclic composition factor is the simple group $A_n(q_0)$. Same if $\sigma$ has order $p$ and $p$ divides both $n+1$ and $q-1$. | |
| Apr 12 at 1:55 | comment | added | Richard Lyons | No, there are examples with rank and q arbitrarily large. For simplicity let's just consider the case $d=1$. If you drop the assumption that $S$ is simple and replace it by the assumption that $S$ is the q-rational points of a {\bf simply connected simple algebraic} group, then $S$ is quasisimple (with few small exceptions) and so is $C_S(\sigma)$. The nonabelian simple quotients of $S$ and $C_S(\sigma)$ are $L(q)$ and $L(q_0)$, respectively. The situation is more complicated when $S={^d}L(q)$ with $d>1$. It is not even universally agreed what "field automorphism" should mean in that case. | |
| Apr 12 at 1:29 | comment | added | user44312 | Thanks for the example. I guess this situation only happens when the rank and $q$ are small. Am I right? | |
| Apr 11 at 17:09 | comment | added | Richard Lyons | It is not true in general that if $S={^d}L(q)$ is simple, then $C_S(\sigma)\cong {^d}L(q_0)$, in your notation. For example if $S$ is the simple group $A_1(9)\cong A_6$ and $\sigma$ has order $2$, then $C_S(\sigma)\cong\Sigma_4\cong PGL_2(3)$ whereas $A_1(3)\cong A_4$. | |
| Apr 11 at 15:05 | comment | added | user44312 | Thanks. The data from Luebeck's website suggest that $|\mathrm{Irr}(S)|=f(q)$ is increasing when $q$ grows. | |
| Apr 11 at 11:35 | comment | added | Sebastien Palcoux | There are formulas for $|\mathrm{Irr}(S)|$ if you specify the family. For example, $|\mathrm{Irr}({\rm PSL}(2,q))| = q+1$ if $q$ even, and $(q+5)/2$ if $q$ odd. You can find formulas for other families in Frank Luebeck website: math.rwth-aachen.de/~Frank.Luebeck/chev/nrclasses/… | |
| Apr 11 at 1:58 | comment | added | user44312 | For a field automorphism of $S$, I mean an $\mathrm{Aut}(S)$-conjugate of an element in $\mathrm{Aut}(S)$ which is induced by a field automorphism of $\mathbb{F}_q$. | |
| Apr 11 at 1:29 | comment | added | LSpice | Now what is a field automorphism of a group? If it means an automorphism of the group of $\mathbb F_q$-rational points of an algebraic group coming from an $\mathbb F_{q_0}$-automorphism of $\mathbb F_q$ and an $\mathbb F_{q_0}$-structure on the algebraic group, then isn't the first part of your question 2 automatic? | |
| Apr 11 at 1:24 | history | edited | user44312 | CC BY-SA 4.0 |
added 55 characters in body
|
| Apr 10 at 18:16 | comment | added | LSpice | $\sigma$ is an automorphism of $\mathbb F_q$ or of $S$? If the former, then we can deduce from it an automorphism of $S$; but, if the latter, then I don't know what "the fixed field of $\sigma$" means. | |
| Apr 10 at 16:19 | history | asked | user44312 | CC BY-SA 4.0 |