You are currently browsing the tag archive for the ‘Graphs of groups’ tag.
Definition. A group splits freely if acts on a tree without global fixed point and such that every edge stailizer is trivial. If does not split freely, then is called freely indecomposable.
Example. . Equivalently, acts on without global fixed points. So splits freely.
If but splits freely, then for .
Definition. The rank of is the minimal such that surjects .
It is clear that .
Grushko’s Lemma. Suppose is surjective and is minimal. If , then such that for .
Pf. Let be simplicial and let be a graph of spaces with vertex spaces and edge space a point. So where .
Let be a graph so that and realize as a simplicial map . Let . Because is minimal, is a forest, contained in . The goal is to modify by a homotopy to reduce the number of connected components of .
Let be the component that contains . Let be some other component. Let a path in from to .
Look at . Because is surjective, there exists such that . Therefore if , then is null-homotopic in and gives a path from to .
We can write as a concaternation as such that for each , . By the Normal Form Theorem, there exists such that is null-homotopic in .
We can now modify by a homotopy so that . Therefore and the number of components of has gone down. By induction, we can choose so that is a tree. Now factors through . Then and there is a unique vertex of that maps to . So every simple loop in is either contained in or as required.
An immediate consequence is that .
Grushko’s Theorem. Let be finitely generated. Then where each is freely indecomposable and is free. Furthermore, the integers and are unique and the are unique up to conjugation and reordering.
Pf. Existence is an immediate corollary of the fact that rank is additive.
Suppose . Let be the graph of groups. Let be the Bass-Serre tree of .
Consider the action of on . Because is freely indecomposable, stabilize a vertex of . Therefore is conjugate into some .
Now consider the action of on . is a graph of groups with underlying graph , say, and is a free factor in . But there is a covering map that induces a surjection . Therefore, . The other inequality can be obtained by switching and .
As before, are and are graphs of spaces equipped with maps , , , and such that
Lemma 28: Suppose that every edge map of is an elevation. Then the map is -injective.
Proof: The idea is to add extra vertex spaces to so that satisfies Stalling’s condition. As before, we have inclusions:
If does not satisfy Stalling’s condition then one of these maps is not surjective. Without loss of generality, suppose that
does not arise as an edge map of . Suppose . Then is a subgroup of . Let be the corresponding covering space of . Since contains , there is a lift of . Now replace by and repeat. After infinitely many repetitions, the result satisfies the hypothesis of Stalling’s condition, and is contained in a subgraph of spaces.
Theorem 18: If and are residually finite groups then is also residually finite.
Proof: Let , let be a point, and fix maps . This defines a graph of spaces . By the Seifert-van Kampen Theorem, . Let be the graph of spaces structure on the universal cover of , and let be a compact subset. We may assume that is connected; we may also assume that if (Bass-Serre tree) and , then . Let be the map to the underlying map of the Bass-Serre tree. Then is a finite connected subgraph, . For each , let , a compact subspace of . Because are residually finite, we have a diagram
where is a finite-sheeted covering map and embeds in . Let be defined as follows. Set ; for a vertex , the vertex space is the corresponding to . For each corresponding to , define so that the diagram
commutes. Now sum.
Exercise 26: If and are residually finite and is finite, prove that is residually finite.
Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map can be extended to a covering map .
Let be graphs of spaces equipped with maps , and as before. Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges. We will aim to do the same thing with graphs of spaces.
Definition: Let be a graph of spaces, and let be the map to the underlying graph. If is a subgraph, then has a graph-of-spaces structure with underlying graph . Call a subgraph of spaces of .
We’re seeking a condition on such that is realized as a subgraph of spaces of some with a covering map such that the following diagram commutes:
Definition: For each edge map of , and each a vertex of , let
For each possible degree , let be the set of elevations of degree . We will say satisfies Stallings’ condition if and only if the following two things hold:
(a) Every edge map of is an elevation of the appropriate edge map of .
(b) For each and , there is a bijection .
So in Figure 1, the graph of spaces is something you might be able to turn into a covering. In the picture, is represented by the blue circles, and is represented by the green circles. Observe that the blue circles are in bijection with the green circles.
Corollary: satisfies Stallings’ condition if and only if can be realized as a subgraph of spaces of some such that
(a) , and
(b) there is a covering map such that the following diagram commutes:
Proof of Corollary. First we’ll show that if can be extended to a covering map as described above, then satisfies Stallings’ condition. By Theorem 17, every edge map of is an elevation. So there are inclusions
Furthermore, these maps are surjective, and clearly degree-preserving.
Now assume that satisfies Stallings’ condition. Then we build as follows. Let . As above, we have degree-preserving inclusions (this time, not surjections)
Extend these inclusions to bijections . Now we set . Each of these is an elevation
This defines an edge space and an edge map . Consider the corresponding elevation in :
Because and are of the same degree, we have a covering transformation . So we can identify them, and use as the other edge map. By construction, satisfies the conditions of Theorem 17, so there is a suitable covering map .
Exercise 25: (This will be easier later, but we have the tools necessary to do this now.) Prove that if and are LERF groups, then so is .