Notes of Kevin Wortman’s lecture

Cohomology of arithmetic groups over function fields

I show that {SL(3,F[t])}, {F} a prime field, virtually has infinite {b_2}.

1. The problem

More generally, we deal with finite extensions {K} of {F(t)} (functions on {\mathbb{P}^1}), like {F(t,\sqrt{t^3 -t})} (functions on an elliptic curve). And rings of {S}-integers, like {F[t]} (poles only at {\infty}), {F[t,t^{-1}]} (poles only at 0 and {\infty}).

Let {G} be an absolutely almost simple {K}-group with {\mathrm{rank}_K(G)>0}, like {SL_n}, {Sp_2n}, {SO_\Phi}, for {\Phi} a quadratic form in at least 5 variables…. {G(O_S)} embeds in {\prod_{v \in S}G(K_v)}, {K_v =} completion of {K} at place {v}, as a non-uniform lattice. Let

\displaystyle  \begin{array}{rcl}  k=\sum_{v\in S}\mathrm{rank}_K(G(K_v)). \end{array}

It is the dimension of the Bruhat-Tits building of the product.

Theorem 1 (Bux-Wortman 2005, Bux-Köhl-Witzel 2011) {G(O_S)} is of type {F_{k-1}} but not {F_k}.

So, potentially, cohomology in degree {k} might be infinite dimensional. It is indeed the case.

Theorem 2 (Stuhler 1980) If {G=SL_2}, {H^k(\Gamma,F)} is infinite dimensional for some finite index subgroup {\Gamma} of {G(O_S)}.

I have a generalization.

Theorem 3 If the {K}-type of {G} is not {E_8}, {F_4}, {G_2} or {BC_n}, then {H^k(\Gamma,F)} is infinite dimensional for some finite index subgroup of {G(O_S)}.

2. A simple example

Let us treat the abelian case: If {U=F[t]}, then {H^1(U,F)} is infinite. It is obvious, but I will give a fancy proof that is an introduction to general ideas. Take {F=\mathbb{F}_2}.

{U} acts on a tree {H}. Vertices are elements of {U}. Put a common parent at height {n} for two polynomials {f} and {g} if {degree(f-g)=n}. Let {U_n=\{}polynomials of degree {\leq n\}}. It is a finite subgroup. Let {U^n=\{}polynomials of valuation {>n} at {0\}}, so that {U=U_n \times U^n}. The quotient {H/U^n} is a finite tree with an infinite ray attached to its root. Let {\{y_0=0,y_1,...\}} be the vertices along the infinite ray issuing from the 0 polynomial. {U_n} acts on {H/U^n}. Let {\Theta_n : H \rightarrow H/U^n}. It is {U_n}-equivariant. Stuhler used equivariant cohomology to compute cohomology of {Sl_2(F[t])}, but the method gets unwieldy for other groups. Let us proceed differently, and introduce a different space {X_1}, the full simplex with vertex set {U}. It is simply connected with a free action of {U} (so {H^1 (U)=H^1(X_1/U)}), it comes with an equivariant map {\psi:X_1 \rightarrow H}.

I describe explicit cocycles {\phi_n}: {\phi_n(\mathrm{edge} (y_{n-1} y_n))=1}, {\phi_n=0} elsewhere. {\Phi_n=\phi_n\circ\theta_n\circ\psi} is {U}-invariant, so it descends to {X_1/U}. It is a cocycle. Let {B_m} be an arc from 0 to {t^m} in {X_1}. Its image in {X_1/U} is a cycle. On computes

\displaystyle  \begin{array}{rcl}  \Phi(\pi(B_m))=1\textrm{ if }m=n, \quad =0\textrm{ if }n>m, \end{array}

so cohomology is infinite.

3. Now let us pass to {\Gamma=Sl_2(F[t])}

Proposition 4 {H^1(Sl_2(F[t]),F)} is infinite.

Proof.

View {\Gamma} as a subgroup of {Sl_2(F((t^{-1})))}. Let {T} be the Bruhat-Tits tree of that group. {H} is a subtree there, it is a horoball. {\Gamma} acts cocompactly on {T\setminus \Gamma H}. View {U} as the unipotent subgroup in {\Gamma}. It acts on {T}, and the induced action on {H} is the same as before.

Let {Y_1} be the full simplex with vertex set {\Gamma}. Average translates of cochain {\Phi_n} over cosets in {\Gamma/U} to get a {\Gamma}-invariant cochain {\Lambda_n}. It descends to a 1-cocycle on {Y_1/\Gamma}. Again, its evaluation on {B_m} shows cohomology is infinite.

Why does the theorem have exceptions ? The proof applies when (up to finite index), unipotent subgroups are abelian. This fails for {F_4}, {G_2}, …

About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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