F_1, buildings, the braid group, GL_n(F_1[t,1/t])

It used to be you had to talk about “the field with one element” very quietly, and only among people whose opinion of you was secure. The reason, of course, is that there is no field with one element. Which doesn’t stop people from saying “But if there _were_ a field with one element, what would it be like?”

Nowadays all kinds of people are musing about this odd question in the bright light of day, and no one finds it kooky. John Baez covered the basics in a 2007 issue of This Week’s Finds. And as of a few weeks ago the field with one element has its own blog, “Ceci N’est Pas Un Corps.”

From a recent post on CNPUC, I learned the interesting fact that the braid group on n strands can be thought of as GL_n(F_1[t]).

So here’s a question: what is GL_n(F_1[t,1/t])?

Proposed answer below the fold.

(Usual disclaimers — blogging is a non-professional activity, which means that what I write below hasn’t been carefully checked, or really checked at all. In particular, it’s quite possible that what I write below is either completely wrong, or already known to experts on F_1. Or, I suppose, both! )

Let’s start with GL_2. Suppose q is an honest prime power; then GL_2(F_q[t,1/t]) acts by left multiplication on the coset space

GL_2(F_q((t))) / GL_2(F_q[[t]]) (*)

(here F_q[[t]] denotes the ring of formal power series and F_q((t)) = F_q[[t]][1/t] the field of Laurent series.)

The cosets (*) are naturally identified with the set of homothety classes of F_q[[t]]-lattices in (F_q((t)))^2. This description provides a natural graph structure on (*): we say two lattices L_1 and L_2 are adjacent when one is contained in the other with index q, and we say that two elements of (*) are adjacent if the corresponding homothety classes [L_1] and [L_2] have representatives which are adjacent.

The graph thus constructed turns out to be an infinite regular tree T of degree q+1, called the Bruhat-Tits tree of GL_2(F_q((t))); it carries an action of GL_2(F_q((t)), in which the stabilizer of every vertex is isomorphic to GL_2(F_q[[t]]). Serre’s book Trees provides a beautiful account of everything you might want to know about groups acting on trees in general, and on Bruhat-Tits trees in particular.

What does this have to do with our original question? Write G for GL_2(F_q[t,1/t]). Then G acts on T, and one can check that the action of G on edges of T is transitive, while the action on vertices has two orbits. Now we know from topology that if you want to understand a G, what you ought to do is find a contractible space EG on which G acts — then the quotient EG/G is the classifying space BG, whose topology often helps us understand the structure of the original group G.

So T is clearly our candidate for EG. And the quotient of T/G has two vertices and a single edge. It looks like this:

* —– *.

But wait — that’s a closed interval, which is again contractible! Since pi_1(BG) = G, not the identity, we did something wrong. What’s missing is that G doesn’t act freely on T. That means that to get the right answer, we have to think of the two vertices above, not as actual points, but as what you might call “stacky points” — each one should be thought of as “a point modulo G_v,” where G_v is a point stabilizer in G. Similarly, the edge above should be thought of as “a line segment modulo G_e,” where G_e is an edge stabilizer.

Fortunately, we know how to compute fundamental groups of funny gluings-together of stacky points of this kind — what the above argument ends up showing is that G is an amalgam:

G = G_v *_{G_e} G_v (**)

Note that G_v and G_e can be described quite explicitly: G_v is the intersection of G with GL_2(F_q[[t]]), which is precisely GL_2(F_q[t]), while G_e turns out to be the congruence subgroup Gamma_0(t) of GL_2(F_q[t]); that is, it is the preimage of a Borel subgroup of GL_2(F_q) under the projection GL_2(F_q[t]) -> GL_2(F_q) which sends t to 0.

Now what does all this say when F_q becomes F_1?

Well, GL_n(F_1[t]) is supposed to be the braid group on n strands, and GL_n(F_1) is supposed to be the symmetric group S_n. When n=2, the “Borel subgroup” of S_2 is surely a point stabilizer — in other words, the trivial group. And the braid group on 2 strands is just a Z, with the projection to GL_2(F_1) = S_2 being the usual projection Z -> Z/2Z. To sum up, we have

G_v = Z

G_e = 2Z.

So (**) suggests that

GL_2(F_1[t,1/t]) = Z *_(2Z) Z

which is a nice nonabelian group; the quotient by the central 2Z is an infinite dihedral group.

What happens for larger n? Now there’s still a nice contractible space on which GL_n(F_q[t,1/t]) acts, and the quotient, instead of being a line segment with stackiness, is an n-simplex with stackiness. And you can use this set-up to compute GL_n(F_q[t,1/t]) in terms of GL_n(F_q[t]) and various congruence subgroups intermediate between GL_n(F_q[t]) and the kernel of reduction mod t.

This suggests that we might also formally construct GL_n(F_1[t,1/t]) in a similar way. But I was too lazy to do this (come on, the field doesn’t even exist!) so let me explain what I think is a pretty good guess as to the answer.

Note that we never said what the Bruhat-Tits tree of GL_2(F_1((t))) is. But it ought to be a regular tree of degree 1+1 = 2, which is to say, an infinite line. And GL_2(F_1[t]), aka the braid group, ought to act on this line in a way that fixes a vertex. Now trees of degree 2 are a lot more rigid than trees of degree bigger than 2; the only automorphism fixing a vertex is inversion! So the braid group acts through its S_2-quotient.

What is the Tits building of GL_3(F_1((t)))? Well, if I worked this out right, the Tits building of GL_3(F_q((t))) is a 2-complex in which every vertex has 2(q^2 + q + 1) incident edges, and (q+1)(q^2+q+1) incident 2-simplices. This suggests that the Tits building of GL_3(F_1((t))) should have 6 incident edges and 6 incident triangles to every vertex — and so what can it be but the tiling of the plane arising from the root lattice of SL_3? Again, this is much more rigid than the honest building obtained for q>1; the natural group of automorphisms of this lattice is the affine Weyl group, which in this case is a semidirect product of Z^2 by S_3. (When n=2, the affine Weyl group is the infinite dihedral group we encountered above.) So GL_3(F_1((t))), whatever it is, should map to this affine Weyl group.

Indeed, on this point we are not merely guessing — Lisa Carbone has worked out the complex that deserves to be called the building of GL_n(F_1((t))) (.pdf slides of her talk on the subject) and indeed it carries the geometry of the affine Weyl group. If you’re in Madison, please note — Carbone will talk in our algebraic geometry seminar about this stuff on Friday, December 5.

We are now ready to propose an answer to the question we started with:

Write W_n for the affine Weyl group of SL_n, and Br_n for the braid group on n strands. Both W_n and Br_n carry natural projections to S_n. Then GL_n(F_1[t,1/t]) is the fiber product of W_n and Br_n over S_n. In other words,

GL_n(F_1[t,1/t]) is the subgroup of W_n x Br_n consisting of pairs (w,b) projecting to the same element of S_n.

(If you wanted to ask a question with some actual content, you could ask whether this would follow from applying to F_1 the description of GL_n(F_q[t,1/t]) as an amalgam of congruence subgroups of GL_n(F_q[t]).)

In particular, GL_n(F_1[t,1/t]) fits into an exact sequence

1 -> Z^(n-1) -> GL_n(F_1[t,1/t]) -> Br_n -> 1

in which Br_n acts on Z^(n-1) via the reduced permutation representation of its quotient S_n.

So I suppose my claim is that GL_n(F_1[t,1/t]) is something like the mapping class group of a disc with n boundary components instead of n punctures.

Anybody believe this?