Math-1267 · mathematics · Group theory

Math-1267

Kourovka Notebook Problem 21.116

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Statement

Definitions and mathematical statement

Let $T$ be an infinite rooted, locally finite, spherically homogeneous tree: every vertex at level $n$ has the same number $m_n\geq2$ of children. Suppose $G$ acts faithfully by root-preserving automorphisms, transitively on each level. The rigid stabilizer $\operatorname{rist}_G(v)$ consists of elements acting trivially outside the subtree below $v$. Put $\operatorname{rist}_G(n)=\prod_{v\text{ at level }n}\operatorname{rist}_G(v)$; the factors commute because their subtrees are disjoint. The group is branch if it has such an action and $[G:\operatorname{rist}_G(n)]<\infty$ for every $n$.

Treat $G$ as a discrete group. Let $C_b^k(G;\mathbb R)$ be the vector space of bounded functions $f:G^{k+1}\to\mathbb R$ invariant under simultaneous left translation of all arguments. The homogeneous coboundary is \[ (df)(g_0,\ldots,g_{k+1})= \sum_{i=0}^{k+1}(-1)^i f(g_0,\ldots,\widehat{g_i},\ldots,g_{k+1}). \] Its cohomology $H_b^k(G;\mathbb R)=\ker d/\operatorname{im}d$ is bounded cohomology with trivial real coefficients; the image is not replaced by its closure. The group is boundedly acyclic if this vanishes for all $k\geq1$.

Is every branch group boundedly acyclic? In cocycle language, every bounded invariant cocycle of positive degree would have to be the coboundary of a bounded invariant cochain. No amenability assumption is imposed.

Short English statement

Does every branch group have vanishing bounded cohomology with trivial real coefficients in all positive degrees?

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