## Notes of Urs Lang’s lecture nr 5

1. End of the proof of Theorem 10

Last week, we proved the first half of the following theorem.

Theorem 1 Let ${\Gamma}$ be a hyperbolic group. There exist constants ${M}$, ${N}$ such that

1. Every admissible graph has rank ${\leq M}$.
2. For every admissible graph ${G_0}$ such that rank${(G_0)=0}$, there are at most ${N}$ admissible graphs ${G\subset G_0}$ with rank ${1}$.

Now we prove the second half.

Proof: Let ${G_0}$ have rank ${0}$. Denote by ${\mathcal{G}}$ the set of admissible graphs ${G\subset G_0}$ with rank ${1}$.

We prove that there exists exactly one extremal function ${f\in E'(\Gamma)}$ such that ${G(f)=G_0}$.

Recall that ${\mathcal{C}(z,2\delta +1)}$ is the set of conex ${(v,c(x,v))}$ with ${x\in B(x,2\delta+1)}$. Also, ${C(x,y):=(v_{xy},c(x,_{xy}))}$. We show that there is an injective map ${S:\mathcal{G}}$ to the set of onempty subsets of ${\mathcal{C}(z,2\delta +1)}$. For ${G\in\mathcal{G}}$, there is a unique extremal function ${g\in E'(\Gamma)}$ such that ${\|f-g\|_{\infty}=\frac{1}{4}}$. Then ${X}$ splits into even and odd components, and even components are bipartite, ${X=X_{-1}\cup X_{1}\cup X_{0}}$, and ${X_{a}=\{x\,;\, g(x)=f(x)+\frac{1}{4}a\}}$.

Define ${S(G)}$ be the set of pointed cones ${C(x,y)}$ where ${(x,y)\in X_{-1}\cup X_{1}}$ and ${\{x,y\}}$ is an edge of ${G}$. We claim that the whole graph ${G}$ can be reconstructed from the set ${S(G)}$.

First,

$\displaystyle \begin{array}{rcl} X_{-1}=\{x\,;\, \forall y\textrm{ with }(x,y)\in\mathcal{E}(f),\,C(x,y)\in S(G)\}. \end{array}$

Indeed, if ${x\in X_{-1}}$ and ${\{x,y\}}$ is an edge of both ${G(f)}$ and ${G(g)}$, then ${g(y)\geq f(y)+\frac{1}{4}}$, so ${C(x,y)\in S(G)}$. Conversely, suppose that ${C(x',y')\in S(G)}$ for all ${f}$-edges ${\{x',y'\}}$. Among these, there is an edge of ${g}$. Since ${C(x',y')\in S(G)}$, there is ${(x,y)\in X_{-1}\cup X_{1}}$ which is an edge of ${g}$, and ${C(x,y)=C(x',y')}$. The vertex of these pointed cones coincide. By the ${4}$-point Lemma, ${x}$ and ${x'}$ are connected by a path of length ${\leq 2}$ in ${G(g)}$, so ${x\in X_{-1}}$.

Secondly,

$\displaystyle \begin{array}{rcl} X_{1}=\{y\,;\, \exists x\in X_{-1},\,\textrm{ such that }(x,y)\in\mathcal{E}(f)\}. \end{array}$

One inclusion is easy since ${\mathcal{E}(g)\subset\mathcal{E}(f)}$. Conversely, if ${x\in X_{-1}}$ and ${\{x,y\}\in\mathcal{E}(f)}$, then it follows that ${\{x,y\}\in\mathcal{E}(g)}$, so ${y\in X_1}$.

Clearly, the partition ${X=X_{-1}\cup X_{1}\cup X_{0}}$ determines ${g}$ and hence ${G}$. $\Box$

Remark 1 This gives an exponential bound on the dimension of the injective hull.

We do not know how sharp it can be.

2. Proof of main result for hyperbolic groups

Theorem 2 (Lang, Moezzi 2010) Let ${\Gamma}$ be a hyperbolic group. Then ${E'(\Gamma)}$ is proper and equal to ${E(\Gamma)}$. The collection ${\mathcal{P}=\{P(G)\}}$, ${G}$ admissible graph, is a locally finite polyhedral structure on ${E(\Gamma)}$ with finitely many isometry types of ${\ell_{\infty}^{k}}$ cells, and ${\Gamma}$ acts properly and cocompactly on ${E(\Gamma)}$ by cellular isometries.

Proof: We have already shown that ${\mathcal{P}}$ is finite dimensional and locally finite. Since ${\Gamma}$ is discretely geodesic, ${\mathcal{P}}$ has only finitely many isometry types of cells. This shows that ${E'(\Gamma)}$ is complete, and so agrees with ${E(\Gamma)}$.

Injective hull ${E(X)}$ is unique up to isometries preserving ${e(X)}$ and uniquely determined by their restriction to this subspace, so this gives the ${\Gamma}$-action on ${E(X)}$. Since ${E(\Gamma)}$ is a neighborhood of ${e(\Gamma)}$, ${\Gamma}$ acts cocompactly on ${E(\Gamma)}$. $\Box$

3. Fixed point theorems

3.1. The case of non expansive maps

Theorem 3 (Sine, Soardi 1979) Let ${X}$ be an injective metric space. Let ${L:X\rightarrow X}$ be ${1}$-Lipschitz. Then

1. If ${X}$ is bounded, then ${Fix(L)\not=\emptyset}$.
2. If ${Fix(L)\not=\emptyset}$, then ${Fix(L)}$ is injective and hence contractible.

Question: If ${X}$ is unbounded but ${L}$ has bounded orbits, is ${Fix(L)\not=\emptyset}$ ? Answer turns out to be no.

Example 1 (Prus) Let ${X=\ell_{\infty}}$, let ${L((x_n)_{n\geq 1})=(1+\limsup_{n\rightarrow\infty} x_n ,x_1 ,x_2 ,\ldots)}$. Then ${L}$ is isometric, it has no fixed points, but orbits are bounded.

Note that Prus’s map is not onto.

3.2. The case of isometries

Theorem 4 (Lang 2011) Let ${X}$ be injective. Let ${\Lambda}$ be a group of isometries of ${X}$.

1. If ${\Lambda}$ has bounded orbits, then ${Fix(\Lambda)\not=\infty}$.
2. If ${Fix(\Lambda)\not=\infty}$, then ${Fix(\Lambda)}$ is injective and hence contractible.

Proof: First we show that if ${X}$ is a metric space, ${\Lambda}$ a group of isometries with bounded orbits, then there is ${f\in E(X)}$ that is constant on each orbit.

Take two orbits ${\Lambda x}$, ${\Lambda y}$. Define

$\displaystyle \begin{array}{rcl} D(\Lambda x,\Lambda y):=\sup\{d(x',y')\,;\, x'\in \Lambda x, \, y'\in\Lambda y\}. \end{array}$

This has the properties of a metric except that ${D(\Lambda x,\Lambda x)=diameter(\Lambda x)>0}$ unless ${x}$ is a fixed point. Let ${\Delta(X/\Lambda,D)}$ be the set of all functions ${G:X/\Lambda\rightarrow{\mathbb R}}$ such that

$\displaystyle \begin{array}{rcl} G(\Lambda x)+G(\Lambda y)\geq D(\Lambda x,\Lambda y). \end{array}$

By Zorn’s Lemma, the poset ${(X/\Lambda,\leq)}$ has a minimal element ${F}$. Consider ${F}$ as a function ${f}$ which is constant on orbits, ${f(x)=G(\Lambda x)}$. By construction,

$\displaystyle \begin{array}{rcl} f(x)+f(y)=F(\Lambda x)+F(\Lambda y)\geq D(\Lambda x,\Lambda y), \end{array}$

so ${f\in \Delta(X)}$. By minimality of ${F}$, for all ${x\in X}$ and ${\epsilon>0}$ there exists ${y\in X}$ such that

$\displaystyle \begin{array}{rcl} F(\Lambda x)+F(\Lambda y)\leq D(\Lambda x,\Lambda y)+\epsilon \end{array}$

and

$\displaystyle \begin{array}{rcl} D(\Lambda x,\Lambda y)\leq d(x,y)+\epsilon, \end{array}$

so ${f(x)+f(y)\leq d(x,y)+\epsilon}$, hence ${f\in E(X)}$.

If ${X}$ is injective, ${E(X)=X}$, so ${f=d_z}$ for some ${z\in X}$. ${f}$ constant on orbits implies that ${f=0}$ along ${\Lambda z}$, so ${\Lambda z}$ is a single point, fixed by ${\Lambda}$.

Assuming that ${Fix(\Lambda)\not=\emptyset}$, we show hyperconvexity. Let ${x_i \in Fix(\Lambda)}$ and ${r_i}$ be such that ${d(x_i ,x_j)\leq r_i +r_j}$. Since ${X}$ is hyperconvex, ${Y:=\bigcap_{i}B(x_i ,r_i)}$ is non empty, and it is hyperconvex again (by definition…). For all ${i}$, ${y\in Y}$ and ${L\in\Lambda}$,

$\displaystyle \begin{array}{rcl} d(x_i ,L(y))=d(L(x_i),L(y))=d(x_i ,y)\leq r_i , \end{array}$

so ${L(Y)\subset Y}$. Similarly, ${L^{-1}(Y)\subset Y}$ so ${L(Y)=Y}$. Hence ${\Lambda}$ has a fixed point in ${Y}$, so ${Y\cap Fix(\Lambda)\not=\emptyset}$, which we wanted. $\Box$

3.3. Structure of fixed point sets in ${E(\Gamma)}$, ${\Gamma}$ hyperbolic

Since the cells of ${E(\Gamma)}$ are affine, barycentric subdivision makes sense: For an admissible graph ${G}$, define the center ${b(G)}$ as the barycenter of vertices of ${P(G)}$. Let ${\mathcal{P}'}$ be the collection of simplices corresponding to strictly ascending sequences

$\displaystyle \begin{array}{rcl} P(G_0)\subset P(G_1)\subset \cdots P(G_k). \end{array}$

${\Gamma}$ still acts by simplicial isometries on ${E(\Gamma)'=(E(\Gamma),\mathcal{P}')}$. This is a “${\Gamma}$-CW-complex”, i.e. if an element of ${\Gamma}$ maps a simplex of ${\mathcal{P}'}$ to itself, then it fixes the simplex point wise.

Theorem 5 ${E(\Gamma)'}$ is a finite model for the classifying space ${\underline{E}\Gamma}$ for proper actions.

There are few examples of infinite groups where I can compute ${E(\Gamma)}$. Mainly, groups that act on trees.

Example 2 ${\Gamma={\mathbb Z}_5 \star {\mathbb Z}_4}$ with obvious generating set.

The Cayley graph is a tree where vertices are blown up alternately into pentagons or squares. The injective hull fills in the holes with ${\ell_{\infty}^{2}}$-squares, it is ${2}$-dimensional, whereas Rips’ complex is ${4}$-dimensional.

In fact, in general, cut points of ${X}$ give cut points in ${E(X)}$.

Example 3 ${\Gamma={\mathbb Z}}$ with generating set ${S=\{\pm 1,\pm 2\}}$.

Then ${E(Z)}$ is a diagonal strip in ${\ell_{\infty}^{2}}$ with group elements forming a staircase.

Example 4 ${\Gamma=\langle a,b\,|\, b^3 ,\,abab \rangle}$.

The Cayley graph is a tree of alternating triangles and squares. I suspect that (not yet checked), combinatorially, the injective hull is a ${3}$-dimensional cube complex.

4. Remetrization

I can prove that injective hulls of hyperbolic groups, when they are ${2}$-dimensional, can be remetrized to become ${CAT(0)}$. Call flag a pair ${(G',G)}$ where ${G\subset G'}$ and ${rk(G)=rk(G')+1}$. There are ${2}$ types of ${2}$-dimensional flags, correponding to vertical/horizontal (type 1) or diagonal (type 2) sides of cells.

Proposition 6 Let ${G'}$, ${G_1}$, ${G_2}$ be admissible graphs, such that ${(G',G_i)}$ is a flag of type ${i}$ then ${G=G_1 \cap G_2}$ is also admissible, and ${(G_i ,G)}$ is a flag of opposite type ${3-i}$.

Links are made of sectors of ${\ell_{\infty}^{2}}$ with angles a multiple of ${90^0}$. The figure formed by ${3}$ such sectors contains a tripod but no midpoints, so it cannot be injective. Therefore the link condition for ${CAT(0)}$ is satisfied.

Sisto: Is it true that maximal cells have the same dimension ? Answer: This may depend on the generating set. Look at ${{\mathbb Z}^n}$. There are generating systems which yield ${\ell_{\infty}}$-metric. It seems to me that in general, there is a notion of ${\ell_{\infty}}$-modelled word metric.

Silberman: Your examples have a lot of torsion. What is the injective hull of a torsion free group like ?

Advertisements

## About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri PoincarĂ©, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
This entry was posted in Course and tagged . Bookmark the permalink.