**1. Cone types **

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

** 1.1. Definition **

Definition 1Let be a metric space, and , . Theirconeis

Lemma 2If , and , then and

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

*Proof:* Since and ,

Equality implies three equalities

Proposition 3Suppose contains at most pairwise disjoint cones. Then for every admissible graph , .

*Proof:* Suppose edges and belong to different even components of . Then the cones , , , are pairwise disjoint. Indeed, the Lemma says that if , then and are edges, contradiction.

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 4If , and , , . If there exist , so that , , then also , .

*Proof:*

Equality everywhere implies the statement.

** 1.2. Hyperbolic spaces **

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

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

Definition 6For , , , define

Proposition 7 (Cannon 1984)Let be discretely geodesic and -hyperbolic. If , then . In particular, equality implies .

Thus if balls are finite, there are only finitely many cones for .

*Proof:* We show by induction on that every point with is in . It is clear for . Now let with . Choose between and with . Then . By induction hypothesis, . There exists a discrete geodesic such that , and . Suppose . Choose a discrete geodesic from to , put . Then Gromov’s product

So

so . Hence , in particular . Now . So , this shows that , as claimed. This completes the induction proof.

Theorem 8If is non empty, geodesic and -hyperbolic, then is within distance from .

This does not require finiteness of cones. I conjecture that the sharp bound is .

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

*Proof:* Let . Then for all and . Define

Let , choose and such that

- ;
- .

Let be on a geodesic from to such that

(3) .

There exists such that . If , then, using (3) and (1),

If sits at distance from the side of triangle , this contradicts above inequality. So sits at distance from the side . In any case, , hence

Using (2) and (3),

so . Up to now, we have not used extremality. Now we use it. Define and elsewhere. Then, by (1),

thus . By extremality, . This implies that

i.e. .

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

Theorem 9Let be a proper, discretely geodesic and -hyperbolic metric space. Let . Then

- There exists such that , and
- for all , such that is an edge of , there exists between and such that .

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

*Proof:* I skip the first claim and pass to the second. Choose between and such that . Then

Theorem 10Let be a -hyperbolic group in its word metric. Then there exist constants , such that

- Every admissible graph has rank ;
- For every admissible graph of rank , there are at most admissible graphs with rank .

*Proof:* 1. Let . Fix such that . For put

Note that . For every edge of , choose between and , i.e. . Put .

The graph induces a partition

into even and odd components. Let , and choose , such that , are edges. If , the Lemma shows that , are connected by a path of length (passing through ) in , so they belong to the same set in above partition. This shows that .

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