I repeatedly heard that K(F_1) is the sphere spectrum. Does anyone know about the proof and what that means?

$\begingroup$ In case you're not aware of this feature, you can change which answer you accepted, shall you decide to. $\endgroup$– Ilya NikokoshevOct 24, 2009 at 16:50

2$\begingroup$ Let's be serious. Anything may arise from the heuristics on the field with one element. It's numerology. I don't think anyone has real hope to prove the Riemann hypothesis from these heuristics. $\endgroup$– Fernando MuroJun 23, 2011 at 17:25
3 Answers
I understand that this is because GL_{n}(F_{1}) is supposed to be Sigma_{n}, the symmetric group on n letters. Thus K(F_{1}) = K(finite sets) which is the sphere spectrum by the BarrattPriddyQuillenSegal theorem.
But I have no idea why GL_{n}(F_{1}) should be Sigma_{n}...
Yes, taking GL _n(F _1) to be Sigma _n one can make sense both of the Q and the +construction and both yield the same answer as shown by Deitmar in http://arxiv.org/abs/math/0605429.
GL _n(F _1)=Sigma _n is suggested by several observations. One is that counting formulas for subspaces of n dimensional vector spaces over F _q turn into counting formulas for subsets of nelement sets, if one sets q=1. So one could say that an ndimensional vector space over F_1 is an nelement set and GL _n(F _1)=Aut(F _1^n)=Sigma _n. See Cohn's very nicely written http://arxiv.org/abs/math/0407093 for this.
One gets another hint by looking at the Tits building for GL _n(F _q) (that is a simplicial complex where the group acts). There is a natural limit for q going to one 1  what then remains is the socalled chamber of the building and the symmetry group of that is Sigma _n.
Further hints that one should just drop addition (in comparison to the usual notion of module) come from arithmetic geometry, but that is maybe less convincing and a longer story...

2$\begingroup$ I don't have a direct reference to this, but if you compare representation theory and the theory of group actions on finite sets, it seems to me that you can see more evidence that the correct way of thinking about F_1 involves forgetting about addition. $\endgroup$ Oct 21, 2009 at 17:58

$\begingroup$ Is GLn(F1) = Sigma_n right? The cardinality doesn't work out when I take the q>1 limit of GLn(Fq) (off by a factor of (q1)^n). $\endgroup$ Oct 21, 2009 at 18:10

1$\begingroup$ @harrison: Yes, burnside ring and representation ring, marks and characters, etc. behave very similarly. Even some proofs carry over, e.g. Steinberg adapted a proof of Frobenius about the irreducibility of some permutation representation to the linear representation analogon (I saw this in a lecture by C. Soule, Nashville 2009) $\endgroup$ Oct 24, 2009 at 13:41

1$\begingroup$ @Reid:What one usually does to count F1points, when the number of points of a variety is given by a polynomial (which is not unusual), is to "multiply away" the zeros at q=1 and then insert 1. This then is consistent with the symmetry group of the chamber view point  for any Chevalley group not only GLn. $\endgroup$ Oct 24, 2009 at 13:42
Here is another heuristic, related to what RandalWilliams said above. The sphere spectrum is the unit object in nice categories of spectra. That is, ring spectra are algebras over the sphere spectrum. Now, to every scheme X you can associate a Ktheory ring spectrum K(X), and this is contravariant. So, in the usual theory there is a morphism K(Z)>K(X) for all schemes X. So, finding F_1 also means finding something (its Ktheory spectrum) that maps to the (homotopy) limit of all Ktheory spectra. That this should be the unit object of the category of spectra doesn't seem very surprising.

$\begingroup$ Thanks. I just looked where the idea comes from: Manin mentions it in "Lectures on zeta functions and motives (according to Deninger and Kurokawa)" shortly and refers to Priddy "Transfer, symmetric groups, and stable homotopy theory", fitting to the first answer. $\endgroup$ Oct 25, 2009 at 10:25