## 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 ${F_4}$, ${G_2}$, …