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)<d(x,v)}. 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<r<s<t<s+\epsilon} and {z\in X} such that

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

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)<d(w,z)-r}. 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<r+2\epsilon <\delta +3\epsilon}. 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.

 

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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/
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