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We still need to convince ourselves of some basic facts about the previous lecture, for example is the map injective?
Example: Cut the sphere along the equator. Then the diagram we have isDefinition: If or we say that G splits over C, and we call C the edge group. If or and is not or in the latter case, then we say that splits non-trivially.
Definition: Let be a connected graph (i.e. a 1-dimensional CW-complex). For each vertex (resp. edge ) let (resp. ) be a group. If are vertices adjoining an edge e then let be an injective homomorphism. This data determines a graph of groups .
We say that has:
- underlying graph
- vertex groups
- edge groups
- edge maps
Similarly, we have:
Definition: Let be a connected graph. For each vertex (resp. ) let (resp. ) be a connected CW-complex. If adjoin let be -injective continuous maps. This data determines a graph of spaces . It has underlying graph , vertex spaces , edge spaces , etc. The graph of spaces determines a space as follows: define
where for . We say that is a graph-of-spaces structure (or decomposition) for .
Remark: There is a natural map (by collapsing all the edge and vertex spaces).
Given any graph of groups we can construct a graph of spaces with underlying graph by assigning and realizing the edge maps as continuous maps . We write for . This is well-defined up to homotopy equivalence.
Definition: The fundamental group of is just
- If then
- If then
- Let be an embedded multicurve (disjoint union of circles) inside a surface. Cutting along decomposes into a graph of spaces and into a graph of groups.
Note: The edge maps of are only defined up to free (i.e. unbased) homotopy. Translated to , this means that only the conjugacy class of in matters.
Remark: The map induces a surjection
Here’s a way to construct a graph of groups. Let’s suppose acts on a tree without edge inversions (we can do this by subdividing edges if necessary). Let . The group acts diagonally on The quotient has a structure of a graph of spaces. The underlying graph is and there is a natural map .
Let be a vertex below . The preimage of is just where is the stabilizer of . Similarly, for below , the preimage of is
If adjoins then so the edge map is a covering map and therefore -injective. We have defined a graph of spaces and since is simply connected.
Applying to everything, we have a graph of groups . Its underlying graph is . Its vertex groups are the vertex stabilizers of , its edge groups are the edge stabilizers, and the edge maps are the inclusions. Also, .
Question for next time: Does every graph of groups arise in this way?
Today we will see some methods of constructing groups.
Definition. Let be groups and let and be injective homomorphisms. If the diagram below is a pushout then we say write and we say that is the amalgamated (free) product of and over .
Example. If , we write and say is the free product of and .
As usual, we need to prove existence.
Recall. If is a group, then the Eilenberg-MacLane Space satisfies the following properties:
- is connected;
- for .
- The construction of is functorial;
- is unique, up to homotopy equivalence.
For as above, let and realize as a map and as a map . Now, let , where . By the Seifert-Van Kampen theorem, . Suppose that , and . Then,
In particular, if is finitely generated, then so is , and if are finitely presented and is finitely generated, then is finitely presented.
Example. Let be a connected surface and let be a separating, simple closed curve. Let . Then,
But, what if is non-separating (but still 2-sided)? Then, there are two natural maps representing , where . Associated to , we have a map , , which maps a curve to its signed (algebraic) intersection number with .
Let be a covering map corresponding to . Then,
This has a shift-automorphism . We can now recover :
Defintion. If are injective homomorphisms, then let
Let be the shift automorphism on . Now, is called the HNN (Higman, Neumann, Neumann) Extension of over . We often realize as , where and . It is easy to write down a presentation:. is called a stable letter.
Theorem 12 (Gromov): Let be torsion-free -hyperbolic group. If such that , then for all sufficiently large , .
Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities. For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of .
For the rest of this lecture will be a torsion-free -hyperbolic group, where are primitive (i.e. not proper powers).
Recall that for torsion-free -hyperbolic, primitive implies that .
If and do not commute we can show there is some point on arbitrarily far from .
Hence we have the following lemma.
If and do not commute there is some point on arbitrarily far from .
Proof: Suppose not. That means such that such that . So is in . But the Cayley graph is locally finite so has finitely many elements. By the Pigeonhole Principle such that for some . Then . But then . .
For a moment view and as the horizontal and vertical geodesics in . For two points on and on , we can argue that the geodesic between them curves toward the origin.
And so we have Lemma 14.
Lemma 14: There exists such that , .
Recall that by is a quasi-isometric embedding. So by Theorem 6, and
By Lemma 13 choose such that
. Choose such that . Now, must be -close to so for some point on the geodesic between and , . Then .
For a subgroup , one can choose a closest point projection which is -equivariant. (Write . Choose where and are close and declare to be -equivariant.) is typically not a group homomorphism.
We’re interested in and .
In , there is some such that either or .
Lemma 15: such that , or .
Let . WLOG, is -close to and since is the closest point to (in particular compared to ). So . .
Now we can prove the theorem.
Proof of Theorem 12:
The idea is to use the Ping-Pong Lemma on the Cayley graph.
Let and let , where is provided by Lemma 15. For all we have and likewise for all we have . In particular, .
Let . By -equivariance,
for any . In particular,
by the triangle inequality. Similarly,
for all and all . Because and are quasi-isometrically embedded, it follows that and for .
Therefore, by the Ping-Pong Lemma .