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The "field with one element" $\mathbb{F}_1$ is, of course, a very speculative object. Nevertheless, some things about it seem to be generally agreed, even if the theory underpinning them is not; in particular:

  • $\mathbb{F}_1$ seems to have a unique extension of degree $m$, generally written $\mathbb{F}_{1^m}$ (even if this is slightly silly), which is thought to be in some sense generated by the $m$-th roots of unity. See, e.g., here and here and the references therein.

  • If $G$ is a (split, =Chevalley) semisimple linear algebraic group, then $G$ is in fact "defined over $\mathbb{F}_1$", and $G(\mathbb{F}_1)$ should be the Weyl group $\mathcal{W}(G)$ of $G$. For example, $\mathit{SL}_n(\mathbb{F}_1)$ should be the symmetric group $\mathfrak{S}_n$. (See also this recent question.) I think this is, in fact, the sort of analogy which led Tits to suggest the idea of a field with one element in the first place.

These two ideas taken together suggest the following question:

What would be the points of a semisimple (or even reductive) linear algebraic group $G$ over the degree $m$ extension $\mathbb{F}_{1^m}$ of $\mathbb{F}_1$?

My intuition is that $\mathit{GL}_n(\mathbb{F}_{1^m})$ should be the generalized symmetric group $\mu_m\wr\mathfrak{S}_n$ (consisting of generalized permutation matrices whose nonzero entries are in the cyclic group $\mu_m$ of $m$-th roots of unity); and of course, the adjoint $\mathit{PGL}_n(\mathbb{F}_{1^m})$ should be the quotient by the central (diagonal) $\mu_m$; what $\mathit{SL}_n(\mathbb{F}_{1^m})$ should be is already less clear to me (maybe generalized permutation matrices of determinant $\pm1$ when $m$ is odd, and $+1$ when $m$ is even? or do we ignore the signature of the permutation altogether?). But certainly, the answer for general $m$ (contrary to $m=1$) will depend on whether $G$ is adjoint or simply connected (or somewhere in between).

I also expect the order of $G(\mathbb{F}_{1^m})$ to be $m^r$ times the order of $G(\mathbb{F}_1) = \mathcal{W}(G)$, where $r$ is the rank. And there should certainly be natural arrows $G(\mathbb{F}_{1^m}) \to G(\mathbb{F}_{1^{m'}})$ when $m|m'$. (Perhaps the conjugacy classes of the inductive limit can be described using some sort of Kac coordinates?)

Anyway, since the question is rather speculative, I think I should provide guidelines on what I consider an answer should satisfy:

  • The answer need not follow from a general theory of $\mathbb{F}_1$. On the other hand, it should be generally compatible with the various bits of intuition outlined above (or else argue why they're wrong).

  • More importantly, the answer should be "uniform" in $G$: that is, $G(\mathbb{F}_{1^m})$ should be constructed from some combinatorial data representing $G$ (root system, Chevalley basis…), not on a case-by-case basis.

(An even wilder question would be if we can give meaning to ${^2}A_n(\mathbb{F}_{1^m})$ and ${^2}D_n(\mathbb{F}_{1^m})$ and ${^2}E_6(\mathbb{F}_{1^m})$ when $m$ is even, and ${^3}D_4(\mathbb{F}_{1^m})$ when $3|m$.)

Gro-Tsen
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    Imho S_n is GL not SL over F_1. Though there are different opinions. I would think An is analog of SL or PSL or PGL. For example for embeding Sn to GL as permutation matrices goes to GL , moreover representation theory of SL is more complicated than GL so it is not like Sn – Alexander Chervov Jun 21 '17 at 13:58
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    @AlexanderChervov $\mathfrak{S}n$ can be both $\mathit{SL}$ and $\mathit{GL}$ over $\mathbb{F}_1$, which makes perfect sense if the multiplicative group of $\mathbb{F}_1$ is of order $1$, and certainly follows both from the "Weyl groups are points over $\mathbb{F}_1$" principle and from the description I suggest for $\mathit{GL}_n(\mathbb{F}{1^m})$ in general. – Gro-Tsen Jun 21 '17 at 14:24
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    [SL SL] = SL - that holds for every F_q and it is completely algebraic relation, so it would be strange if it is not true for F_1 , is not it ? So it contradics SL(F_1) = S_n, since [S_n S_n] = A_n and quite perfectly agrees with [GL GL] = SL , so for q=1 we have: [S_n S_n] = A_n. That is just my imho. It might be I am wrong. Glad to hear contr-arguments, – Alexander Chervov Jun 21 '17 at 20:15
  • Your suggestion is very interesting - that GL(F_1^n) is set of "permutation matrices" with roots of unity coefficients. Very interesting ! What is Frobenius map ? It is just natural x->x^p ? Can we check Lang-Steiberg(?) like theorem that G = G^{-1} F(G) ? – Alexander Chervov Jun 21 '17 at 20:23
  • @AlexanderChervov I don't find it strange at all: lots of purely group-theoretic facts admit exceptions when they are evaluated over the points of algebraic groups over small finite fields, and $1$ is certainly small. For example, $G_2(\mathbb{F}_q)$ is simple for every $q$ except $2$, which is exactly the sort of phenomenon you mention. I think a pure group-theoretic view is simply not good to approach algebraic groups (which have a richer structure). Also, if you want $G(\mathbb{F}_1)$ to always be simple, you end up with $B_n$, $C_n$ and $D_n$ to have the same (contd.) – Gro-Tsen Jun 21 '17 at 20:26
  • By the way Weyl groups can be realized as "permutation matrices" with +-1 signs and other case as with even number of +1 (if I remember corectly). So in your ideology Weyl group is $GL(F_{1^2})$. Does it have conceptual understanding ? – Alexander Chervov Jun 21 '17 at 20:27
  • (contd.) set of points over $\mathbb{F}_1$, because they will all end up being $\mathfrak{A}_n$. Anyway, I don't know, but I think we should really stick to the principle that "points over $\mathbb{F}_1$ give the Weyl group". For your other questions, I really don't know, but I think $\mathbb{F}_1$ probably doesn't behave too much like the usual finite fields, and I'm not convinced there is anything like a Frobenius (or even a Galois group). Probably best not to speculate too much unless a coherent answer can be made. – Gro-Tsen Jun 21 '17 at 20:30
  • Simplicity is different condition - it does not seems to be given by identity. What I have in mind - that GL is something which has "good reduction" at all points including F_1 . – Alexander Chervov Jun 21 '17 at 20:30
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    See Lorscheid, Theorem 3.14 of his 2nd blueprints paper at https://arxiv.org/pdf/1201.1324.pdf. He describes $G(F_{1^2})$ in his framework, at least. For $n > 2$... why not email Lorscheid? – Marty Jun 22 '17 at 18:39
  • @Marty Thank you, this is indeed interesting. The result in question states that $G(\mathbb{F}_{1^2}$ is the extended Weyl group, which is very nice; unfortunately, I seem to understand that there are conditions on $G$, and also that $G(\mathbb{F}_1)$ is merely a subgroup of the Weyl group, but the author explains why he thinks the latter is unavoidable. – Gro-Tsen Jun 22 '17 at 20:48
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    Perhaps the speculative theory of "spetses" is relevant here: https://link.springer.com/article/10.1007/BF01237357 My rough understanding is that the idea of spetses is to treat all complex reflection groups like the Weyl group of some kind of analog of a reductive algebraic group. – Sam Hopkins Jun 23 '17 at 21:01
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    @SamHopkins I feel that spetses are more or less the answer to the "opposite" question (starting with something which is more or less a Weyl group, we try to construct something like an algebraic group; whereas here I try to go in the opposite direction), but they are indeed worth mentioning. If you know something about them, here would be a place to post a short summary, it would be much appreciated. – Gro-Tsen Jun 23 '17 at 21:45
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    @Gro-Tsen: I see what you mean about "opposite" direction. I should also mention, however, a recent interesting development- the definition of "cyclotomic root system" here: https://arxiv.org/abs/1704.03779. If you could attach such a root system to your generalized algebraic group, then maybe you could extract the generalized Weyl group out of it. – Sam Hopkins Jun 25 '17 at 14:57
  • @AlexanderChervov, re, neither $\operatorname{SL}_2(\mathbb F_2)$ nor $\operatorname{SL}_2(\mathbb F_3)$ is its own commutator subgroup, so the relation probably cannot be said to be "completely algebraic" in your sense. (It is also amusing, though surely irrelevant, that $\operatorname{SL}_2(\mathbb F_2) = \operatorname{GL}_2(\mathbb F_2) = \operatorname{PGL}_2(\mathbb F_2)$ is $\mathrm S_3$.) – LSpice Nov 09 '21 at 22:30

1 Answers1

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Not an answer - just opposite - some additional requirments which good answer may satisfy and further comments.

0) Parabolic subgroups should also be defined. I mean good definition should also come with definition of parabolic subgroups. For example for $S_n$, it seems natural to consider $S_{d_1}\times ... \times S_{d_k}$ as parabolics.

1) Element count should be compatible with "known" (= widely agreed) zeta-functions of P^n (and also Grassmanians, Flags). I mean one can define P^n, Grassmanians, Flags in a standard way just by quotients of G/Parabolic. So one should get number of points for such manifold over $F_{1^l}$, hence one can write a standard Weil's zeta function. For example for S_n for Grassamnian $|S_n/(S_k\times S_{n-k}) | = n!/(k!(n-k)!)$. Requirment: zeta should agree with the "known" one (for example s(s-1)...(s-n+1) for P^n).

2) Structure of representation theory of such groups should be similar to G(F_q). For example for GL(F_1^l) one may expect that similar structure as in Zelevinsky theorem (see MO272686), i.e. all representations (for all "n" GL(n)) organized into Hopf algebra using induction and restriction should be isomorphic to several copies of the same Hopf algebra for S_n, each copy for each cuspidal representation. Interesting question how many cuspidals will one have for fixed GL(n,F_1^l) ?

3) Alvis-Carter duality should work. It is defind as roughly speaking as follows - take character - restrict it to parabolic and than back induction, with summation over all parabolics with appropriate sign. The statement is that: it has order two and isometry on generalized characters. Steinberg representation is dual to trivial. For S_n it is transposition of Young diagram (if I rember correctly).

4) Frobenius action and Lang-Steiberg theorem. One knows that for F_q there is Frobenius and Lang-Steinberg theorem (e.g. G.Hiss page 15) holds that map $G^{-1}F(G)$ is surjection. How to define Frobenius ?

5) "Good" bijection between irreducible representation and conjugacy classes (toy Langldands correspondence). Similar to MO270916 one may expect "good" bijection between conjugacy classes - which might be considered as a toy model for local Langlands correspondence for field with one element. For example for S_n there is well-known "good" bijection between the two sets via Young diagrams.

OP made intersting proposal for an $GL(F_{1^n})$ - monomial matrices with roots of unity entries. It would be very intersting to check whether the properties above holds true for such proposal. What is most unclear for me - what should be Frobenius map ?

Alexander Chervov
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    Definition of GL(F_1^n) same as OP's is contained in Kapranov-Smirnov page 4 (middle) http://www.neverendingbooks.org/DATA/KapranovSmirnov.pdf (Kapranov, Michail; Smirnov, Alexander (1995), Cohomology determinants and reciprocity laws: number field case (PDF)) – Alexander Chervov Jun 25 '17 at 06:55
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    Regarding 2): Zelevinsky showed (section 7 of his LNM book) that his Hopf algebra methods also apply to $G\wr S_n$, for fixed (finite) $G$ and varying $n$. The resulting Hopf algebra has one "primitive" (aka "cuspidal") element for each irreducible representation of $G$. So, if we follow @Gro-Tsen and Kapranov-Smirnov and put $GL_n(\mathbb F_{1^m})=\mu_m\wr S_n$, we might interpret Zelevinsky's result as saying that the cuspidal representations of $GL_n(\mathbb F_{1^m})$ are precisely the characters of the "torus" $GL_1(\mathbb F_{1^m})=\mu_m$. – user85913 Jun 27 '17 at 08:09
  • @t.c. WOW ! WOW ! WOW ! – Alexander Chervov Jun 27 '17 at 08:19
  • @t.c. I am happy with the first sentence , however I do not quite understand the last one – Alexander Chervov Jun 27 '17 at 08:21
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    Sorry, it was indeed quite vague. I am thinking of the following two facts: (1) In Zelevinsky's Hopf algebra $\bigoplus_{n\geq 0} K_0(\mathbb C[GL_n(\mathbb F_q)])$, the primitive representations (i.e. those $\pi$ such that $\delta(\pi)=\pi\otimes 1 + 1\otimes\pi$, where $\delta$ is the coproduct) are precisely the cuspidal ones (in the usual sense of vanishing under all nontrivial parabolic restrictions). (2) In the Hopf algebra $\bigoplus_{n\geq 0} K_0(\mathbb C[G\wr S_n])$, the primitive representations are precisely the irreducible representations of $G\wr S_1$. (to be continued...) – user85913 Jun 27 '17 at 08:38
  • ...(cont) So, if we accept the analogy "$GL_n(\mathbb F_{1^m}):=\mu_m\wr S_n$", and the analogy "cuspidal representation $:=$ irreducible primitive element of the Zelevinsky Hopf algebra" suggested by fact (1), then fact (2) becomes the assertion that the cuspidal representations of $GL_n(\mathbb F_{1^m})$ are precisely the irreducible representations of $GL_1(\mathbb F_{1^m})$. – user85913 Jun 27 '17 at 08:40
  • @t.c. WOW ! ( I cannot reply immeadiately - sorry. But your writing is very intresting!) – Alexander Chervov Jun 27 '17 at 19:44
  • In that sense about GL(Fq) : cuspidals also corresponds to charscters of dome tori... but different ... nonsplit ... – Alexander Chervov Jun 27 '17 at 19:47