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Lemma 27 Revisited. Suppose is a covering map. Then there is a covering map such that is the fibre product of and .
Proof. Let be the fibre product of and . There is a map given by . Let be the fibre product of and ; i.e.
There is a map given by . This is a covering map and injective, so it is a homeomorphism.
Let be continuous, be a covering map and , choices of basepoint. We have already seen that a choice of such that determines an elevation of to at . Fix such a . The pre-image of in is in bijection with the set of cosets
This raises the question, when do two cosets determine the same elevation?
Exercise 24. and determine the same elevation if and only if
that is, the set of elevations of to is in bijection with .
Let and be graphs of spaces and suppose we have the following data.
(a) A combinatorial map given by and .
(b) Covering maps for each .
(c) Covering maps for each , such that whenever adjoins .
This determines a continuous map . When is really a covering map?
Theorem 17. is a covering map if
(i) for all adjoining , the edge map is an elevation of ; and
(ii) wherever adjoining and , every elevation to arises as an edge map of .
Proof (sketch). It’s enough to consider our local model : and . and be covering maps defining and a map . By Lemma 27, is a covering map if and only if is a fibre product with respect to some covering:
Every map in the diagram is injective, so (for each component)
and it follows that is the fibre product of and . The result follows.
Question (Gromov). Classify groups up to quasi-isometry.
1) Ends. Roughly, if is a metric space, is the number of components of the boundary at of . If , then captures algebraic information.
Definition. For functions , , we say if there exists such that . If and then .
2) Growth. If is a group and is a finite generating set.
where is the set of elements such that . This is a quasi-isometric invariant of .
Example. is exponential.
3) If is finitely presented and is quasi-isometric to then is also finitely presented.
4) Let is finitely presented; so . Let . Then
where , , and . The question: how hard is it to write in such a product? Define to be minimum in any such expression of in (1). Let
This function is the Dehn function of , which measures how hard the word problem is to solve in . The -class of is a quasi-isometric invariant.
Remark. Having a solvable word problem is equivalent to having a computable Dehn function.
Hyperbolic Metric Spaces
We want a notion of metric spaces (and hence for groups) that captures hyperbolicity (that is, for one, that triangles are thin).
In what follows, is always a geodesic metric space. We’ll write for a geodesic between and (not necessarily unique).
Definition. Let , and let . We say that is -slim if
where , and the same for both and (that is, for each geodesic “side” of the triangle, it is contained in a neighborhood of the other two geodesic sides of the triangle).
Definition. is Gromov hyperbolic (or -hyperbolic, or just hyperbolic) if every geodesic triangle, , is uniformly -slim; that is, there exists such that every is -slim.
Example (a). Any tree is -hyperbolic. Every geodesic triangle is a “tripod”.
Example (b). is not -hyperbolic for any .
Example (c). (and hence ) is hyperbolic (and indeed, any space of principal negative sectional curvature bounded away from zero).
Given a geodesic triangle and let . We ask how far from the other sides is ? Well, inscribe a semi-circle centered at inside of ; pick the largest such inscribed semi-circle, and call its radius . So is -slim, where is the largest ; that is, is the radius of the largest semi-circle that can be inscribed in .
So to find , we look at semi-circles; for this, we need a fact about .
Fact. For any , , where are angles of the triangle.
This leads to a uniform bound on the area, and hence the radius of semi-circles inscribed in .
To define hyperbolic groups, we want to prove hyperbolicity is a quasi-isometric invariant of geodesic metric spaces. We need to “quasi-fy” the definition of -hyperbolic.
Definition. A quasi-geodesic is a quasi-isometric embedding of a closed interval.
Exercise 13. Let by in polar coordinates. Show that is a quasi-isometric embedding.
We will prove this behavior does not happen in hyperbolic metric spaces.