Timeline for Regular semisimple elements in $SL(n,q)$
Current License: CC BY-SA 4.0
18 events
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| Apr 18, 2019 at 8:49 | comment | added | alpoge | Indeed I have overlooked something: in your question you ask for a universal polynomial p_n for which the number of regular semisimple elements in SL_n(q) is p_n(q), and indeed this is quite different from the asymptotic question! I got confused because your above comment seems to suggest you’d like an asymptotic, and I read the question too quickly. Sorry! | |
| Apr 18, 2019 at 8:46 | comment | added | alpoge | The “bad” set is the intersection of disc(charpoly) = 0 and det = 1, it is of codimension 2, thus by e.g. Lang-Weil (actually I prefer just replacing the disc = 0 with its resultant wrt x_{nn} with det = 1, aka eliminating two variables) you get your claim for q >>_n 1. I think people above are trying to give you as precise an answer to your question as possible, on the assumption that you need precise control on the size of the bad set (indeed the above people are exactly who you want to ask to count its size!), but indeed you do get that asymptotic. Let me know if I’ve overlooked something! | |
| Apr 17, 2019 at 23:38 | history | edited | Jim Humphreys | CC BY-SA 4.0 |
added 2 characters in body
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| Apr 17, 2019 at 23:33 | answer | added | Jim Humphreys | timeline score: 1 | |
| Apr 17, 2019 at 15:11 | comment | added | Tree | @peter Mcnamara If we assume $q$ to be odd. what is the appropriate answer to my question? For $GL_{n}$ case, one doesn't need any asymptotics, for $q$ odd, it is always of order $q^{n^2}$, whether $n$ is small or large. So, what happens in $SL_n$? I don't understand: Is my question not valid or is it that results are not known in this case? | |
| Apr 17, 2019 at 4:20 | comment | added | Peter McNamara | @VictorProtsak You are correct and there are 2q^2 non-rss elements in the q odd case. I located my (two, yikes!) errors in my computation. | |
| Apr 17, 2019 at 4:13 | comment | added | Victor Protsak | Clarification: In the last sentence, conjugates must be understood in the matrix sense, i.e. the conjugating element lies in ${\rm GL}(2,q)$. Upon restriction to $G={\rm SL}(2,q)$, it breaks up into two $G$-conjugacy classes. (The $G$-conjugacy class of a non-identity unipotent element of $G$ has $\displaystyle \frac{q^2-1}{2}$ elements, but there are 2 such unipotent conjugacy classes in $G$, with representatives given by $\begin{bmatrix}1&a\\0&1\end{bmatrix}$, where $a$ is, respectively, a square and non-square in $\Bbb{F}_{q}^{\times}$). The total count is not affected, though. | |
| Apr 17, 2019 at 3:23 | comment | added | Victor Protsak | @Peter: I am getting a different count for non-rss elements in the $G={\rm SL}(2,q)$, odd $q$ case. Clearly, the set $G^{\rm rss}$ of regular semisimple elements, and hence its complement, are invariant under the left translation by the center $Z$ of $G$ and $|Z|=2$, so the answer must be even! More precisely, the eigenvalues of a non-rss matrix in $G$ are either both $1$ or both $-1$, and the corresponding sets are related by multiplication by $-I_2$. For the unipotent case, there are $q^2-1$ distinct conjugates of the Jordan block and the identity. So the total count is $2q^2$. | |
| Apr 17, 2019 at 0:58 | comment | added | Peter McNamara | number of regular semisimple elements is always asymptotic to q^{n^2-1} by Deligne's proof of Weil conjectures. | |
| Apr 16, 2019 at 13:24 | comment | added | Tree | @Lspice Ok I see that now,. Yeah we can assume $q$ to be odd leaving power of 2 case! Then any idea if it happens or not! What about the asymptotics though? | |
| Apr 16, 2019 at 12:19 | comment | added | LSpice | The number of elements isn't a polynomial in $q$ in @PeterMcNamara's case, since it is different if $q$ is even versus if it is odd. If you are only concerned about the asymptotics, not the exact count, of the number of rss elements, then I think that is easier. | |
| Apr 16, 2019 at 8:04 | comment | added | Tree | @Peter McNamara In you first comment the number of non regular semisimple element is of the order $q^2$, hence the number of regular semisimple element is of the order $q^3$, which is exactly I am asking for, since $n^2-1=3$ for $n=2$. I don't understand your comment , are you saying that $n=2$ case contradict my statement? | |
| Apr 16, 2019 at 7:59 | comment | added | Tree | @Lspice I have read your references before. These references as you have already said count the number of regular semisimple conjugacy classes, hence doesn't seem to answer my question, which is about enumeration of all such elements. | |
| Apr 16, 2019 at 1:53 | comment | added | Peter McNamara | @LSpice I suspect that for SL_N one will get a different formula for q a power of a prime dividing N, and this phenomenon does not appear for GL_N, where the OP indicates there is a general formula that works for all q. | |
| Apr 16, 2019 at 1:27 | comment | added | LSpice | I can't find a reference on the number of rss elements, but the number of rss classes is well studied. I'm familiar with Steinberg (see, for example, §14 of Steinberg - Endomorphisms of linear algebraic groups); but some Googling also turned up Fleischmann, Janisczczak, and Knörr - The number of regular semisimple classes of special linear and unitary groups. | |
| Apr 16, 2019 at 1:07 | comment | added | LSpice | @PeterMcNamara, I'm not familiar with these counting problems. Does that mean that the statement in the question about the number of rss elements in $\operatorname{GL}(n, q)$ is also true only for large $q$, or does passing to $\operatorname{GL}(n, q)$ magically sort out $\operatorname{SL}(n, q)$ problems? | |
| Apr 16, 2019 at 1:01 | comment | added | Peter McNamara | If n=2 then the number of non-regular-semisimple elements is q^2-q+1 for odd q and q^2 for even q. So what sort of answer are you looking for? Do you care if there is a different formula for a few small characteristics? I expect the general formula is of similar qualitative behaviour to this case. | |
| Apr 15, 2019 at 23:34 | history | asked | Tree | CC BY-SA 4.0 |