## Notes of Jean-François Lafont’s Cambridge lecture 22-06-2017

Hyperbolic groups whose boundary is a Sierpinski ${n}$-space

Joint work with Bena Tshishiku.

1. Sierpinski ${n}$-space

Start with an ${(n+1)}$-dimensional sphere. Remove a dense family of balls with disjoint interiors. Get ${S_n}$. Up to homeo, balls need not be round. One merely needs that their diameters tend to 0.

Any homeo of ${S_n}$ permutes the distinguished peripheral spheres.

Examples.

1. Free groups have ideal boundary ${S_0}$.
2. If ${M}$ is a compact negatively curved ${n}$-manifold with nonempty totally geodesic boundary, then ${\partial\tilde M=S_{n-2}}$.
3. Let ${\Gamma}$ be a nonuniform lattice of isometries of ${H^n}$. Then ${\Gamma}$ is cocompact on the complement ${A}$ of a union of horospheres, hence ${\partial A=S_{n-2}}$.

1.1. Cannon conjecture

What properties of the group follow from specifying the topology of the boundary ? This is what Cannon’s conjecture is about: if ${\partial\Gamma=S^2}$, must ${\Gamma}$ be a cocompact lattice in ${H^3}$?

Here is a topological variant of Cannonc’s conjecture.

Theorem 1 (Bartels-Lueck-Weinberger) If ${\Gamma}$ is torsion-free hyperbolic, and ${\partial \Gamma=S^{n-1}}$, and ${n\geq 6}$, then there exists a unique closed aspherical ${n}$-manifold ${M}$ with ${\Gamma=\pi_1(M)}$.

2. Result

Theorem 2 If ${\Gamma}$ is torsion-free hyperbolic, and ${\partial \Gamma=S_{n-2}}$, and ${n\geq 7}$, then there exists a unique aspherical ${n}$-manifold ${M}$ with nonempty boundary with ${\Gamma=\pi_1(M)}$. Moreover, every boundary component of ${M}$ corresponds to a quasi-convex subgroup of ${\Gamma}$.

3. Proof

3.1. Step 1

Kapovitch-Kleiner: ${\Gamma}$ is a relative ${PD(n)}$-group, relative to the collection of stablizers of peripheral spheres.

3.2. Step 2

Realize ${\Gamma}$ as ${\pi_1(X)}$, where ${X}$ is a finite relative ${PD}$ complex, relative to a finite subcomplex ${Y\subset X}$. We use the Rips complex for ${B\Gamma}$ but the Bartles-Lueck-Weinberger complexes for parabolic subgroups ${\Lambda_i}$.

3.3. Surgery theory

Browder-Novikov-Sullivan-Wall surgery theory provides obstructions to finding a manifold homotopy equivalent to ${X}$. They belong to the space ${S(X)}$ that appears in the algebraic surgery exact sequence

$\displaystyle \begin{array}{rcl} \cdots\rightarrow H_n(X,L_\cdot)\rightarrow H_n({\mathbb Z}\Gamma)\rightarrow S(X)\rightarrow H_{n-1}(X,L_\cdot)\rightarrow\cdots. \end{array}$

A similar exact sequence appears in 4-periodic surgery exact sequence, with ${L}$ replaced with a very similar ${\bar L}$ (and ${S(X)}$ with ${\bar S(X)}$). They have the same homotopy groups and differ only in their 0-spaces

$\displaystyle \begin{array}{rcl} \textrm{for }L_\cdot, ~G/TOP ; \quad \textrm{for }\bar L_\cdot, ~G/TOP\times L_0({\mathbb Z}). \end{array}$

There is a long exact sequence

$\displaystyle \begin{array}{rcl} \cdots\rightarrow H_n(X,L_0({\mathbb Z}))\rightarrow S_n(X)\rightarrow\bar S_n(X)\rightarrow H_{n-1}(X,L_0({\mathbb Z}))\rightarrow\cdots. \end{array}$

It turns out that ${H_n(X,L_0({\mathbb Z}))=H_n(X,{\mathbb Z})=0}$. Furthermore, thanks to the (L-theoretic) Farrell-Jones isomorphism conjecture (which holds for hyperbolic groups, Bartels-Lueck-Reich), ${\bar S_n(X)=0}$. Hence ${S_n(X)=0}$, the obstruction vanishes, so there exists a homology manifold model for ${B\Gamma}$.

3.4. ${CAT(0)}$ groups

Bartels-Lueck-Reich cover ${CAT(0)}$ groups. In the relative case (replace spheres with Sierpinski spaces), much of the argument carries over, but the first step.