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 ?

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/
This entry was posted in Course and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s