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There is well-known heuristic that Weyl groups are reductive algebraic groups over "field with one element".

Probably the best known analogy supporting that heuristic is the limit $q\to1$ for number of elements in $G(F_q)$ - for appropriate "m" it holds:

$$ \lim_{q\to1} \frac { |G(F_q) | } { (q-1)^m} = |Weyl~Group~of~G| $$

For example: $|GL(n,F_q)|= [n]_q! (q-1)^{n}q^{n(n-1)/2} $ so divided by $(q-1)^n$ one gets $[n]_q! q^{n(n-1)/2} $ and at the limit $q\to1$, one gets $n!$ which is the size of $S_n$ (Weyl group for GL(n)). (For other groups see Lorscheid 2009 page 2 formula 1).

Question What are the other analogies supporting heuristics: Weyl groups = algebraic groups over field with one element ?

Subquestion once googling papers on F_1, I have seen quite an interesting analogy from representation theory point of view - it was some fact about induction from diagonal subgroups of symmetric groups $S_{d_1}\times ... \times S_{d_k} \subset S_n$ where $\sum d_i = n$ and similar fact for $GL(n,F_q)$ which was due to Steinberg or Springer or Carter (cannot remember). But I cannot google it again and cannot remember the details :( (Tried quite a lot - I was sure it was on the first or second page of Soule's paper on F_1 - but it is not there, neither many other papers).


Knowing that total element count is okay, we may ask about counting elements with certain properties - like: m-tuples of commuting elements (MO271752), involutions, elements of order $m$, whatever ... From answer MO272059 one knows that there are certain analogies for such counting, however it seems the limits $q\to1$ are not quite clear.

Question 2 Is there any analogy for counting elements with some reasonable conditions ? Hope to see that count for $G(F_q)$ (properly normalized) in the limit $q\to1$ gives answer for Weyl group.

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    I'd be interested to know if one has some kind of limit-expression for the number of real elements in $G(\mathbb{F}_q)$... That seems a good test case for trying to extend this sort of heuristic. – Nick Gill Jun 20 '17 at 09:17
  • @NickGill Well, for GL there are non-real elements, but for S_n all elements are real, (am I correct ?). So there should some limit to ZERO for GL_n. On the other hand for O(n,F_q) all elements are real (am I correct ? ). So we may ask are all elements for corresponding Weyl groups are real ? Is it known ? – Alexander Chervov Jun 20 '17 at 09:24
  • What you say all sounds correct and reasonable -- indeed your comment about $S_n$ suggests that one could also consider rational elements, since all elements of $S_n$ are rational. By the way is that $p$ in the denominator of your limit supposed to be an $n$? – Nick Gill Jun 20 '17 at 09:47
  • @NickGill for GL(n) it is "n" , for other groups I do not remember off-hand... sorry ... – Alexander Chervov Jun 20 '17 at 09:50
  • Oh, sorry, I has presumed that $p$ was the characteristic of the field, but I see now that you have clarified what $p$ is in the previous line. My mistake. – Nick Gill Jun 20 '17 at 09:56
  • @NickGill ohh, yes, use of "p" is misleading notation here, let me change it to say "m" – Alexander Chervov Jun 20 '17 at 10:05
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    @AlexanderChervov: A theorem of Carter says that in a Weyl group every element is "strongly real", i.e. a product $xy$ for some $x, y$ with $x^2 = y^2 = 1$. In particular every element of a Weyl group is real. – Mikko Korhonen Jun 20 '17 at 10:51
  • @MikkoKorhonen Thank you very much ! That I wanted to know for a long time ! Can you give a reference ? So probably Nick Gill is the best person to ask is it true for G(F_q) that all real elements are strongly real ? I know he proved that fact for GL, SL https://arxiv.org/abs/0809.4412 – Alexander Chervov Jun 20 '17 at 11:03
  • @NickGill please look a question in a comment above – Alexander Chervov Jun 20 '17 at 11:04
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    @AlexanderChervov: Conjugacy classes in the Weyl group, Seminar on Algebraic Groups and Related Finite Groups, Lecture Notes in Mathematics 131. – Mikko Korhonen Jun 20 '17 at 11:11
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    No, it's not true that all elements in $G(F_q)$ are strongly real -- the list of which quasisimple groups have all elements (strongly) real was answered by Tiep and Zalesski. This doesn't include all groups of Lie type of course, but it deals with some of them. – Nick Gill Jun 20 '17 at 11:23
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    Sorry, my last comment was misleading -- you asked which groups have all real elements are strongly real, whereas Tiep and Zalesski showed which groups have all elements strongly real. In any case, it is an interesting and relevant paper! And, to address your question,it's not true that all real elements in $G(F_q)$ are strongly real for arbitrary $G$ -- see this paper on the unitary groups https://arxiv.org/pdf/1303.6085.pdf – Nick Gill Jun 20 '17 at 11:28
  • Besides counting elements, other lines of thought might be: thin versus thick buildings, parabolic subgroups and quotients thereof... The analogy between binomial coefficients and Gaussian (i.e., $q$) binomial coefficients can be construed as an analogy between $\mathfrak{S}_n$ and $\mathit{GL}_n(\mathbb{F}_q)$. – Gro-Tsen Jun 20 '17 at 12:01
  • @Gro-Tsen Thank you for your comment. May be you can extend it to an answer. I do not know much about buildings. What about parabolics ? Imho there is NO parabolics in S_n. What about binomial coef... ? – Alexander Chervov Jun 20 '17 at 12:21
  • @MikkoKorhonen by the way, do not you know, is it easy to prove ? (I mean element of Weyl is strongly real ). At least in case of S_n for example ... – Alexander Chervov Jun 20 '17 at 18:29
  • Found that paper on counting analogy : https://fpsac2016.sciencesconf.org/file/245441 GLn(Fq)-analogues of factorization problems in Sn Joel Brewster Lewis1† and Alejandro H. Morales2‡ – Alexander Chervov Jun 22 '17 at 14:34
  • For $p$ not dividing $q$, the $p$-Sylow subgroups of $GL_n(q)$ look a lot like those of $S_n$ (not the same $n$). – j.p. Jun 27 '17 at 16:24
  • @j.p. can you please extend your last comment: " For pp not dividing qq, the pp-Sylow subgroups of GLn(q)GLn(q) look a lot like those of SnSn (not the same nn). " – Alexander Chervov Jul 08 '17 at 18:14
  • Just for myself - http://cage.ugent.be/~kthas/Fun/ site for collective learning blog on F_1 active around 2010 – Alexander Chervov Jul 08 '17 at 18:52
  • Let $k$ be minimal with respect to $p$ dividing $q^k-1$. If $p^2$ does not divide $q^k-1$, then the $p$-Sylow of $GL_n(q)$ is isomorphic to the one of $S_{p\cdot [n/k]}$. Otherwise there is a cyclic group $C_{p^i}$ at the bottom of the iterated wreath product instead of the $C_p$. – j.p. Jul 08 '17 at 22:10
  • @j.p. can you extend your comment to answer, please, what fact do you use about Sylow's subgroup in S_n ? – Alexander Chervov Jul 09 '17 at 18:50

2 Answers2

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This is more an extended comment than a full-fledged answer, but OP encouraged me to write it, and maybe it can encourage someone to post a more precise account along those lines.

A parabolic subgroup of a Coxeter group $W$ with Coxeter generating set $S$ is simply the subgroup (also a Coxeter group) $W_J$ generated by a subset $J\subseteq S$. In case $W$ is a Weyl group, this means we select a subset $J$ of the set $S$ of nodes of the Dynkin diagram. The maximal parabolics are obtained by removing a single node from the Dynkin diagram. In the case of $\mathfrak{S}_n = W(A_{n-1})$ generated by transpositions of adjacent elements, a maximal parabolic is the stabilizer of $\{1,\ldots,k\}$ in $\{1,\ldots,n\}$ where $k$ is the removed node; so the corresponding quotient (I mean, set of cosets) can be viewed as the set of $k$-element subsets of $\{1,\ldots,n\}$, of which there are $\binom{n}{k}$. Now the corresponding parabolic subgroup of $\mathit{GL}_n(\mathbb{F}_q)$ is the stabilizer of a $k$-dimensional subspace, and the quotient is the set of such subspaces (set of points of the Grassmannian), of which there are $\binom{n}{k}_q$ (Gaussian binomial coefficient). The various analogies between ordinary and Gaussian binomial coefficients can then be construed as analogies between the Weyl group and the linear group. Similar things can be said for flag varieties and other Dynkin types, but I don't feel comfortable enough expanding this here.

Along different lines (or maybe not so different), (thick) Tits buildings of spherical type can be seen as a generalization of Coxeter complexes, i.e., "thin" buildings, (of spherical type), and the relation with the algebraic groups on the one hand, and finite Coxeter groups on the other clearly makes the Weyl groups appear similar to algebraic groups over the field with $1$ element. Again, I don't want to expand upon this for fear of saying something wrong, but this should at least suggest a way of looking at things.

One last thing which comes to my mind is about generalized matroids: not only does the set of $k$-element subsets of $\{1,\ldots,n\}$ have a cardinal which has formal similarities with the set of $k$-dimensional subspaces of $\mathbb{F}_q^n$, but their sets also have a structure as a matroid, and again, matroids can be generalized to flag matroids and other Dynkin types.

Gro-Tsen
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  • Thank you for your answer ! So for S_n maximal parabolics seems to be S_{k}\times S_{n-k} and general parabolic S_{k_1} \times ... \times S_{k_l}. So Borel becomes trivial (it is S_1\times ... \times S_1) , but parabolics still exists. There should be analogies with parabolic inductions I guess... – Alexander Chervov Jun 20 '17 at 19:55
  • So for maximal parabolic the size of the factor space S_n / (S_k \times S_{n-k}) is obiously a binomial coefficient n! / k! (n-k)!. While the size of same space for GL(n,F_q) is q-binomial [n]_q! / k_q! (n-k)_q! That is nice.! Would be interesting to see what are properties can be deduced further.. – Alexander Chervov Jun 20 '17 at 20:01
  • By the way definition of parabolic is consistent with naive embedding S_n to GL as permutation matrices and intersecting with standard parabolics in GL. – Alexander Chervov Jun 20 '17 at 20:04
  • It seems Vipul Naik text COMBINATORICS OF THE GENERAL LINEAR GROUP http://www.cmi.ac.in/~vipul/writeupsandpresentations/combinatoricsofglwriteup.pdf is something about that – Alexander Chervov Jun 20 '17 at 20:08
  • It also becomes clearer for me about vector space over F_1 . Vector space over F_q we may think F_q \oplus ... F_q , so F_1 \oplus ... F_1 is over F_1 - this means just a set because: over F_q we can multiply by q each coordinate , but over F_1 we have only one element so multiplcation gives it itself. So what we can say is that preserving the vector space property still works over F_1 . In that way we get analogies with Grassman , binomial, q-binomial . – Alexander Chervov Jun 20 '17 at 20:13
  • May be David Harris preprint http://www.math.harvard.edu/archive/126_fall_98/papers/dgharris.pdf TRANSLATING THE IRREDUCIBLE REPRESENTATIONS OF Sn INTO GLn (Fq) is also about that – Alexander Chervov Jun 21 '17 at 10:50
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[The following comment is too long for the comment box.]

On the subquestion: Zelevinsky's Representations of finite classical groups - a Hopf algebra approach (LNM) may be relevant to what you're thinking of. Zelevinsky builds two Hopf algebras: the first coming from induction and restriction of (complex) representations of the symmetric groups along the inclusion $S_n\times S_m\to S_{n+m}$, the second using parabolic induction and restriction (again, of complex representations) for finite general linear groups along the inclusion $GL_n(\mathbb F_q)\times GL_m(\mathbb F_q)\to GL_{n+m}(\mathbb F_q)$. Zelevinsky shows that the second algebra is a tensor product of copies of the first, with one copy for each pair $(n,\pi)$ where $\pi$ is a cuspidal representation of $GL_n(\mathbb F_q)$.

Here's an attempt, possibly completely misguided, to extract from this an analogy that would be relevant to your question. (I don't know how this relates to the existing literature on $\mathbb{F}_1$. Sorry if I am repeating something that is well known.) If we identify $S_n=GL_n(\mathbb F_1)$, Zelevinsky's result might be interpreted metaphorically as saying that ''the only cuspidal representation of $GL_n(\mathbb F_1)$ is the trivial representation of the trivial group''. Since cuspidal representations of $GL_n(\mathbb F_q)$ are associated to characters of anisotropic tori, and since (I suppose?) the ''group of $\mathbb F_1$-points of a torus over $\mathbb F_1$'' is always the trivial group, this makes some kind of sense.

user85913
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  • Thank you for your answer ! I was just thinking what is "cuspidal" for S_n ? So you suggest it is trivial one. Can we check your suggestion in the following way: inducation from S_a\times S_b \times ... S_l can be seen as parabolic induction. Cuspidal are those NOT obtained by parabolic induction. So your proposal means that all irreps of S_n are somehow obtained by "parabolic induction" - is that true ? – Alexander Chervov Jun 21 '17 at 09:39
  • @AlexanderChervov Yes, it is true for the somewhat silly reason that every irreducible representation of $S_n$ is a subrepresentation of $Ind_{S_1\times\cdots\times S_1}^{S_n} 1$ (the representation induced from the trivial representation of the trivial subgroup---which is a ''Levi subgroup'' according to this analogy). – user85913 Jun 21 '17 at 09:48
  • Remark: it seems to me Levi = Parabolic for S_n , that is natural from Gro-Tsen answer and action on vector spaces over F_q and F_1 – Alexander Chervov Jun 21 '17 at 09:52
  • Okay, but is true for GL that all NOT cuspidals can be parabolic induced from MAXIMAL parabolics ? Is the same true for S_n ? Maximal parabolics are GL(k)\times GL(n-k), and S_k \times S_{n-k} – Alexander Chervov Jun 21 '17 at 09:54
  • Okay, some more thinking - makes me more comfortable with yours example on induction from "Borel". Induction from Borel should give "principal series" representation which is irreducible for generic regular character of torus. So Borel for S_n is trivial = S_1 x ... x S_1 . And so the only induction for it will give regular representation of S_n and so we may think that all irreps of S_n are kind of "principal series", although not "regular". – Alexander Chervov Jun 21 '17 at 10:07
  • @AlexanderChervov : Regarding maximal parabolics, I still think that the answer is yes, just by the transitivity of induction: $Ind_1^{S_n} 1 = Ind_{S_k\times S_{n-k}}^{S_n} (Ind_1^{S_k\times S_{n-k}} 1)$. – user85913 Jun 21 '17 at 10:12
  • Many thanks for your answer ! You also shed light on Zelevinsky result, which I heard a lot but never understood ... Thank you ! – Alexander Chervov Jun 21 '17 at 10:37
  • Question: do not you know - Zelevinsky results holds true for SL(n,F_q) ? Exactly in the same form as for GL ? (What do you think S_n = GL(F_1) or S_n = SL(F_1) ? ) – Alexander Chervov Jun 22 '17 at 11:05
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    @AlexanderChervov : I doubt that it will work in exactly the same way for SL_n, since SL_n x SL_m is not a Levi subgroup of SL_{n+m}, so in the formula for the composition of restriction and induction there will be a sum over S_n-double-cosets (like in GL_n and S_n), but also a sum over characters of the multiplicative group, which will complicate things. On the other hand, Zelevinsky's construction does have analogues for other groups: see e.g. Van Leeuwen (J. Algebra 1991). On the question of GL_n(F_1) vs SL_n(F_1), I'm afraid I don't know enough about F_1 to have an opinion. – user85913 Jun 23 '17 at 08:10
  • @Gro-Tsen the comment above by t.c. somehow supports that GL(F_1) = S_n, not SL. Reprsentation theory of SL, even SL2 is much more tricky than GL - that is not expected for q-S_n = GL – Alexander Chervov Jun 23 '17 at 09:17