## Notes of Urs Lang’s lecture nr 4

1. Cone types

This is a tool to get upper bounds on ranks, and therefore dimensions of injective hulls.

1.1. Definition

Definition 1 Let ${X}$ be a metric space, and ${x}$, ${y\in X}$. Their cone is

$\displaystyle \begin{array}{rcl} c(x,y)=\{z\in X\,;\,d(x,y)+d(y,z)=d(x,z)\}. \end{array}$

Lemma 2 If ${f\in \Delta_1 (X)}$, and ${\{x,y\}\in\mathcal{E}(f)}$, then ${\{x,z\}\in\mathcal{E}(f)}$ and ${f(x)=f(y)+d(y,z)}$

This adds a lot of edges to the graph, and hopefully decreases the rank.

Proof: Since ${f\in \Delta(X)}$ and ${\{x,y\}\in\mathcal{E}(f)}$,

$\displaystyle \begin{array}{rcl} f(z)&\geq& d(x,z)-f(z)=d(x,y)+d(y,z)-f(x)\\ &=&f(y)+d(y,z)\geq f(z). \end{array}$

Equality implies three equalities $\Box$

Proposition 3 Suppose ${X}$ contains at most ${k}$ pairwise disjoint cones. Then for every admissible graph ${G}$, ${rk(G)\leq\frac{k}{2}}$.

Proof: Suppose edges ${\{x,y\}}$ and ${\{x',y'\}}$ belong to different even components of ${G(f)}$. Then the cones ${c(x,y)}$, ${c(y,x)}$, ${c(x',y')}$, ${c(y',x')}$ are pairwise disjoint. Indeed, the Lemma says that if ${z\in c(x,y)\cap c(x',y')}$, then ${\{x',z\}}$ and ${\{x,z\}}$ are edges, contradiction. $\Box$

This is not very efficient. For finitely generated groups, for instance, translations map cones to cones, and there may be infinitely many disjoint cones. Here is a better criterion.

Lemma 4 If ${f\in \Delta (X)}$, and ${x}$, ${x'}$, ${v\in X}$. If there exist ${y\in c(x',v)}$, ${y'\in c(x,v)}$ so that ${\{x,y\}}$, ${\{x',y'\}\in\mathcal{E}(f)}$, then also ${\{x,y'\}}$, ${\{x',y\}\in\mathcal{E}(f)}$.

Proof:

$\displaystyle \begin{array}{rcl} f(x)+f(y)+f(x')+f(y')&=&d(x,y)+d(x',y')\\ &\leq&d(x,v)+d(v,y)+d(x',v)+d(v,y')\\ &=& d(x,y')+d(x',y)\\ &\leq&f(x)+f(y)+f(x')+f(y'). \end{array}$

Equality everywhere implies the statement. $\Box$

1.2. Hyperbolic spaces

Definition 5 (Gromov 1987) Let ${X}$ be discretely geodesic. Say that ${X}$ is ${\delta}$-hyperbolic if for every triangle, there is a map to a tripod which is isometric up to an additive error at most ${\delta}$.

We shall use Cannon’s theorem that hyperbolic groups have finitely many cones. The group structure is not needed.

Definition 6 For ${x}$, ${v\in X}$, ${\rho\geq 1}$, define

$\displaystyle \begin{array}{rcl} L_{\rho}(x,v):=\{u\in X\,;\, d(u,v)\leq\rho \textrm{ and } d(x,u)< d(x,v)\}. \end{array}$

Proposition 7 (Cannon 1984) Let ${X}$ be discretely geodesic and ${\delta}$-hyperbolic. If ${L_{\delta+1}(x',v)\subset L_{\delta+1}(x,v)}$, then ${c(x,v)\subset c(x',v)}$. In particular, equality ${L_{\delta+1}(x',v)= L_{\delta+1}(x,v)}$ implies ${c(x,v)= c(x',v)}$.

Thus if balls are finite, there are only finitely many cones ${c(x,v)}$ for ${x\in X}$.

Proof: We show by induction on ${\ell}$ that every point ${y\in c(x,v)}$ with ${d(x,y)=\ell}$ is in ${c(x',v)}$. It is clear for ${\ell=0}$. Now let ${y\in c(x,v)}$ with ${d(v,y)=\ell+1}$. Choose ${y'}$ between ${v}$ and ${y}$ with ${d(v,y')=\ell}$. Then ${y'\in c(x,v)}$. By induction hypothesis, ${y'\in c(x',v)}$. There exists a discrete geodesic ${\gamma'}$ such that ${\gamma'(0)=x'}$, ${\gamma'(k)=v}$ and ${\gamma'(k+\ell)=y'}$. Suppose ${k\geq 1}$. Choose a discrete geodesic from ${x'}$ to ${y}$, put ${u:=\gamma(k-1)}$. Then Gromov’s product

$\displaystyle \begin{array}{rcl} (v\cdot y)_{x'}=\frac{1}{2}(d(x',v)+d(x',y)-d(v,y))\geq k-1. \end{array}$

So

$\displaystyle \begin{array}{rcl} d(u,v)&\leq& d(\gamma(k-1),\gamma'(k-1))+d(\gamma(k-1),\gamma(t))\\ &\leq&\delta+1, \end{array}$

so ${u\in L_{\delta+1}(x',v)}$. Hence ${u\in L_{\delta+1}(x,v)}$, in particular ${d(x,u). Now ${d(u,y)\geq d(x,y)-d(x,u)>d(x,y)-d(x,v)=\ell+1}$. So ${d(x',y)=d(x',u)+d(u,y)>k-1+\ell+1=k-\ell}$, this shows that ${y\in c(x',v)}$, as claimed. This completes the induction proof. $\Box$

Theorem 8 If ${X}$ is non empty, geodesic and ${\delta}$-hyperbolic, then ${E(X)}$ is within distance ${\leq\delta}$ from ${e(X)}$.

This does not require finiteness of cones. I conjecture that the sharp bound is ${\delta/2}$.

Idea of the proof. For the real line, extremal functions ${f}$ look like distance functions outside a bounded interval. Use the largest function of the form ${d_x -}$const which is less that ${f}$. Since ${f=d_x -}$const at at least one point on each side of ${x}$, const has to be ${0}$. On a ${\delta}$-hyperbolic space, it goes the same way with an additive error.

Proof: Let ${f\in E(X)}$. Then ${f(x)\geq d_z (x)-f(z)}$ for all ${x}$ and ${z}$. Define

$\displaystyle \begin{array}{rcl} s:=\inf\{t\geq 0\,;\,\exists z\textrm{ such that } f(x)\geq d_z (x)-t\}. \end{array}$

Let ${\epsilon>0}$, choose ${s-\epsilon and ${z\in X}$ such that

1. ${f(x)\geq d_z(x)-t}$;
2. ${f(y).

Let ${w}$ be on a geodesic from ${z}$ to ${y}$ such that

(3) ${d_z (w)=\min\{d_z (y),\delta+2\epsilon\}}$.

There exists ${x}$ such that ${f(x). If ${d_z (w)=\delta+2\epsilon}$, then, using (3) and (1),

$\displaystyle \begin{array}{rcl} d(w,x)>f(x)+r \geq d_z (x)+(t-r)=d_z (x)-d_z (w)+\delta. \end{array}$

If ${w}$ sits at distance ${\leq\delta}$ from the side ${zx}$ of triangle ${xyz}$, this contradicts above inequality. So ${w}$ sits at distance ${\leq\delta}$ from the side ${yx}$. In any case, ${d(w,y)\leq (x\cdot z)_y}$, hence

$\displaystyle \begin{array}{rcl} d(w,x)\leq d(x,y)-d(w,y)+\delta. \end{array}$

Using (2) and (3),

$\displaystyle \begin{array}{rcl} 2r&<& d_z (y)-f(y)+d(w,x)-f(x)\\ &\leq& d_z (y)+d(w,x)-d(x,y)\\ &\leq&d_z (y)-d(w,y)+\delta\\ &=&d_z (w)+\delta, \end{array}$

so ${t. Up to now, we have not used extremality. Now we use it. Define ${g(z):=t}$ and ${g(x)=f(x)}$ elsewhere. Then, by (1),

$\displaystyle \begin{array}{rcl} g(x)+g(z)=f(x)+t \geq d_z (x), \end{array}$

thus ${g\in\Delta(X)}$. By extremality, ${f(z)\leq g(z)=t<\delta+3\epsilon}$. This implies that

$\displaystyle \begin{array}{rcl} \|f-d_z \|_{\infty}=f(z)=\delta, \end{array}$

i.e. ${d(f,e(z))\leq\delta}$. $\Box$

Lee: Is the proof natural for trees, and then can be adapted to hyperbolic spaces, including additive errors of ${\delta}$ ? Answer: right, this is the way I found it.

Theorem 9 Let ${X}$ be a proper, discretely geodesic and ${\delta}$-hyperbolic metric space. Let ${f\in E(X)}$. Then

1. There exists ${z\in X}$ such that ${f(z)\leq\delta+\frac{1}{2}}$, and
2. for all ${x}$, ${y}$ such that ${\{x,y\}}$ is an edge of ${G(f)}$, there exists ${v}$ between ${x}$ and ${y}$ such that ${d_z (v)\leq 2\delta +1}$.

In other words, an extremal function has a kind of center ${z}$, and edges (viewd as geodesics) cannot pass far from it.

Proof: I skip the first claim and pass to the second. Choose ${v}$ between ${x}$ and ${y}$ such that ${d_x (v)\rfloor (z\cdot y)_x \rfloor}$. Then

$\displaystyle \begin{array}{rcl} d_z (v)&\leq& d_x (z) -d_x (v)+\delta\\ &=&\lceil (x\cdot y)_z \rceil\\ &\leq& (x\cdot y)_z +\frac{1}{2}. \end{array}$

$\displaystyle \begin{array}{rcl} (x\cdot y)_z &=&\frac{1}{2}(d_z (x)+d_z (y)-d(x,y))\\ &\leq&\frac{1}{2}(f(x)+f(z)+f(y)-f(z)-d(x,y))\\ &\leq& f(z)\leq\delta +\frac{1}{2}. \end{array}$

$\Box$

Theorem 10 Let ${\Gamma}$ be a ${\delta}$-hyperbolic group in its word metric. Then there exist constants ${M}$, ${N}$ such that

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

Proof: 1. Let ${f\in E'(\Gamma)}$. Fix ${z}$ such that ${f(z)\leq\delta+\frac{1}{2}}$. For ${r\geq 0}$ put

$\displaystyle \begin{array}{rcl} \mathcal{C}(z,r):=\{(v,c(x,v))\,;\, v\in B(z,r),\, x\in\Gamma\}. \end{array}$

Note that ${|\mathcal{C}(z,r)|=|B(e,r)||\{c(x,e)\,x\in\Gamma\}|<\infty}$. For every edge ${\{x,y\}}$ of ${G(f)}$, choose ${v_{xy}\in B(z,2\delta +1)}$ between ${x}$ and ${y}$, i.e. ${y\in c(x,v_{xy})}$. Put ${C(x,y):=(v_{xy},c(x,v_{xy}))\in\mathcal{C}(z,2\delta +1)}$.

The graph ${G(f)}$ induces a partition

$\displaystyle \begin{array}{rcl} X=\bigcup([x_k]_1 \cup [x_k]_{-1}) \cup \tilde{X} \end{array}$

into even and odd components. Let ${x}$, ${x'\in\Gamma}$ and choose ${y}$, ${y'}$ such that ${\{x,y\}}$, ${\{x',y'\}}$ are edges. If ${C(x,y)=C(x',y')}$, the Lemma shows that ${x}$, ${x'}$ are connected by a path of length ${2}$ (passing through ${y}$) in ${G(f)}$, so they belong to the same set in above partition. This shows that ${rk(G(f))\leq\frac{1}{2}|\mathcal{C}(z,r)|}$. $\Box$

That’s all for today, I will complete the proof next time.