26
$\begingroup$

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 $3|m$.)

$\endgroup$
  • 5
    $\begingroup$ 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 $\endgroup$ – Alexander Chervov Jun 21 '17 at 13:58
  • 1
    $\begingroup$ @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. $\endgroup$ – Gro-Tsen Jun 21 '17 at 14:24
  • 1
    $\begingroup$ [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, $\endgroup$ – Alexander Chervov Jun 21 '17 at 20:15
  • $\begingroup$ 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) ? $\endgroup$ – Alexander Chervov Jun 21 '17 at 20:23
  • 2
    $\begingroup$ See Lorscheid, Theorem 3.14 of his 2nd blueprints paper at arxiv.org/pdf/1201.1324.pdf. He describes $G(F_{1^2})$ in his framework, at least. For $n > 2$... why not email Lorscheid? $\endgroup$ – Marty Jun 22 '17 at 18:39
7
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ Definition of GL(F_1^n) same as OP's is contained in Kapranov-Smirnov page 4 (middle) neverendingbooks.org/DATA/KapranovSmirnov.pdf (Kapranov, Michail; Smirnov, Alexander (1995), Cohomology determinants and reciprocity laws: number field case (PDF)) $\endgroup$ – Alexander Chervov Jun 25 '17 at 6:55
  • $\begingroup$ 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" $GL_1(\mathbb F_{1^m})=\mu_m$. $\endgroup$ – t.c. Jun 27 '17 at 8:09
  • $\begingroup$ @t.c. WOW ! WOW ! WOW ! $\endgroup$ – Alexander Chervov Jun 27 '17 at 8:19
  • $\begingroup$ @t.c. I am happy with the first sentence , however I do not quite understand the last one $\endgroup$ – Alexander Chervov Jun 27 '17 at 8:21
  • 1
    $\begingroup$ 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...) $\endgroup$ – t.c. Jun 27 '17 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.