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Let $S$ be a simple group of Lie type defining over $\mathbb{F}_q$ where $q$ is a power of a prime $p$ and let $\sigma$ a nontrivial field automorphism of $S$. Set $\mathrm{C}_{S}(\sigma)$ the subgroup of the fixed points of $\sigma$.

Question 1. Is there a formula for $|\mathrm{Irr}(S)|$?

Question 2. Write $S={}^d\Sigma(q)$. Is it true that $\mathrm{C}_{S}(\sigma)={}^d\Sigma(q_0)$ where $q_0$ is also a power of $p$? Is it true that $|\mathrm{Irr}(\mathrm{C}_{S}(\sigma))|<|\mathrm{Irr}(S)|$?

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  • $\begingroup$ $\sigma$ 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. $\endgroup$
    – LSpice
    Commented Apr 10 at 18:16
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    $\begingroup$ 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/… $\endgroup$
    – Sebastien Palcoux
    Commented Apr 11 at 11:35
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    $\begingroup$ 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 $A_1(3)\cong A_4$. $\endgroup$
    – Richard Lyons
    Commented Apr 11 at 17:09
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    $\begingroup$ 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. $\endgroup$
    – Richard Lyons
    Commented Apr 12 at 1:55
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    $\begingroup$ 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 $q-1$. $\endgroup$
    – Richard Lyons
    Commented Apr 12 at 2:13

1 Answer 1

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While Deligne-Lusztig theory gives a method to compute the irreducible characters of finite simple groups of Lie type, it is not always so easy to find a closed form expression for their number. See, for example, the AMM paper by Benson, Feit and Howe https://www.jstor.org/stable/2322289 on the behaviour of the generating function for the number of conjugacy classes of ${\rm GL}(n,q)$. Of course for $q >2$, this is not a simple group, but the formula for the number of conjugacy classes of ${\rm PSL}(n,q)$ will be at least as difficult for large $n$ and $q$.

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  • $\begingroup$ Have now put in a link, but you may need other means of access $\endgroup$
    – Geoff Robinson
    Commented Apr 12 at 15:14
  • $\begingroup$ Thanks for the link. It makes me wonder that are there generating function for the number of conjugacy classes of simple groups of lie type or the $q$-rational points of a simply connected simple algebraic groups? $\endgroup$
    – user44312
    Commented Apr 12 at 15:32
  • $\begingroup$ There probably are in most cases. Maybe look at papers of Fulman, Guralnick and Stanton $\endgroup$
    – Geoff Robinson
    Commented Apr 12 at 15:47
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    $\begingroup$ I think the last problem breaks down in the following way: you need to count/parametrize the semisimple conjugacy classes, and then for each semisimple conjugacy class, you need to count the number of unipotent classes in the centralizer of an element of the chosen semisimple class ( and then sum over all semisimple classes). $\endgroup$
    – Geoff Robinson
    Commented Apr 12 at 15:57
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    $\begingroup$ You have already given me some references on Question 1 and also some hints on part of Question 2. Richard Lyons answered the rest. So, I accept both of your answers. $\endgroup$
    – user44312
    Commented Apr 13 at 16:25

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