Use topology to study your group

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Use topology to study your group

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[QUICK DESCRIPTION]

If you want to study a group $G$, try to realize $G$ as the fundamental group of a topological space.  This works best when $G$ is infinite and discrete, especially if $G$ is finitely presented and torsion-free.

[PREREQUISITES]

A basic knowledge of group theory and topology, especially the [[w:fundamental group | fundamental group]] and [[w:covering space |covering spaces]].

[GENERAL DISCUSSION]

Actions are a good way of studying a group $G$.  The actions of $G$ by linear transformations on a vector space are the subject of [[Representation theory front page|representation theory]], for instance.  Topological spaces are another particularly fruitful object for actions.  Of course, there are many different sorts of topological space and many different sorts of action that one might want to study, depending on the group in question.  One common hypothesis is that the group G should act freely and properly discontinuously on the topological space $X$---for brevity, for the remainder of this article such actions shall be referred to as ''geometric''.

If $G$ is the fundamental group of a reasonably nice topological space $Y$ then $G$ acts geometrically on the universal cover, and conversely if $G$ admits such an action on a space $X$ then the quotient space $Y$ has fundamental group $G$.  So the study of geometric actions is equivalent to the study of fundamental groups.  Indeed, the construction of an [[w:Eilenberg-MacLane_space|Eilenberg--Mac Lane Space]] provides such a space for any group, and better still the construction is functorial.

Using this idea, one aims to study a group $G$ by finding a particularly nice topological space $X$ on which $G$ acts, or equivalently by exhibiting $G$ as the fundamental group of a nice space $Y$.  Conversely, attractive hypotheses on $Y$ impose restrictions on $G$.  For instance, if $Y$ is a CW-complex with finite one-skeleton then $G$ is finitely generated (in particular countable), and if $Y$ has finite two-skeleton then $G$ is finitely presented.  (This is an equivalent way of looking at [[Presentations of groups|group presentations]].  Any presentation for $G$ describes a two-complex with fundamental group $G$---the generators determine the one-skeleton and the relations the two-skeleton.)  It is easy to prove these facts using the [[w:Seifert–van_Kampen_theorem|Seifert–-van Kampen Theorem]].

If $Y$ is [[w:Aspherical_space|aspherical]] then the homology and cohomology of $Y$ are equal to the group homology and cohomology of $G$, so if $Y$ is also compact then a variety of other conditions are imposed including that $G$ is torsion-free.  Therefore, if we want $Y$ to be very nice---compact and aspherical---then $G$ will have to be finitely presented and torsion-free.  (If the torsion-free hypothesis is too onerous, one approach is to remove the requirement that the action of $G$ on $X$ be free.  In this case the resulting quotient $Y$ is best not thought of as just a space, but rather as a space with some extra structure.)  So we have come to the following precept.

[frame]If you are interested in a group $G$, try to find a nice space $Y$ with fundamental group $G$.  This is likely to work particularly well if $G$ is finitely presented and torsion-free.[/frame]

These ideas apply very nicely to free groups.

[EXAMPLE NSThm]

In this example we will give a very simple proof of the Nielsen--Schreier Theorem, which asserts that every subroup of a free group is free, by exhibiting free groups as the fundamental groups of graphs.

By a graph we mean a connected, 1-dimensional CW-complex.  In particular, we allow multiple edges (1-cells) between pairs of vertices (0-cells) and also loops---edges that adjoin only one vertex (although such phenomena can be removed by subdividing).

A graph with just one vertex and $n$ edges is called a rose with $n$ petals.  (Here $n$ need not be finite.)  The Seifert--van Kampen Theorem implies that the fundamental group of a rose with $n$ petals is precisely the free group on $n$ generators.   More generally, let $Y$ be an arbitrary graph and let $T$ be a maximal tree in $Y$.  Then $Y/T$ is a rose and the quotient map $Y\to Y/T$ is a homotopy equivalence.  This proves the following.

[theorem freegraph] A group G is free if and only if G is the fundamental group of a graph.[/theorem]

The Nielsen--Schreier Theorem follows immediately from this and elementary covering-space theory.

[theorem ns|Nielsen--Schreier Theorem]  Every subgroup of a free group is free.[/theorem]

Let $F$ be a non-abelian free group and let $H$ be a subgroup.  Let $Y$ be a rose such that $F$ is the fundamental group of $Y$.  By standard covering space theory, there is a covering space $Y'$ of $Y$ with fundamental group $H$.  But a covering space of a graph is a graph, so $H$ is free. This completes the proof.

[/EXAMPLE]

Indeed, these techniques work so well for free groups that a large proportion of the modern study of free groups is conducted in terms of graphs.  So one has the following, rather more specific, precept.

[frame]If your group G is free, try to rephrase your question in terms of the topology of graphs.<comment thread="72" />[/frame]

A lot of modern group theory can be seen as an attempt to generalize these techniques to larger classes of groups and spaces: [[w:Δ-hyperbolic_space|hyperbolic metric spaces]] and [[w:CAT(0)|CAT(0) metric spaces]], for instance, can be seen in this light.

[EXAMPLE finind]

Here's another example of a fact about free groups that is very simple to prove using topology.

[proposition freefinind]
A finitely generated free group only has finitely many subgroups of given finite index $k$.
[/proposition]
[proof]
Let $F$ be a free group generated by $n$ elements and as above let $Y$ be a rose with $n$ petals, so $F\cong \pi_1(Y,v)$ where $v$ is the unique vertex of $Y$.

A subgroup $H$ of index $k$ corresponds to a covering map $Y'\to Y$ of degree $k$ together with a choice of base vertex in $Y'$.  In particular, $Y'$ is a graph with precisely $k$ vertices and $nk$ edges.  Clearly, there are only finitely many such graphs $Y'$.  Furthermore, for each such $Y'$ there are precisely $k$ choices of base vertex and only finitely many choices of covering map $Y'\to Y$.

We have seen that $H$ can be described by a finite amount of data.  This proves the proposition.
[/proof]

[ref Proposition #freefinind] is particularly useful because, via the universal property of free groups, it follows that the same holds for ''every'' finitely generated group.
[/EXAMPLE]

The Schreier Index Formula is a third nice example.

[EXAMPLE]
Suppose $F$ is free on $n$ generators and $H$ is a subgroup of finite index $k$.  By [ref Theorem #ns] above, $H$ is free.  But what can we say about the rank $m$ of $H$?  There is a very nice answer to this question, using Euler characteristic.

As before, think of $F$ as the fundamental group of a rose $Y$ with $n$ petals.  The Euler characteristic of $Y$ is equal to the number of vertices minus the number of edges, so $\chi(Y)=1-n$.  What's more, Euler characteristic is a homotopy invariant, so it follows that the Euler characteristic of ''any'' graph is equal to 1 minus the rank of the fundamental group.

As before, we let $Y'$ be the covering space of $Y$ corresponding to $H$.  We can now compute $m$, the rank of $H$, by [[double counting]].  On the one hand, we have seen that $\chi(Y')=1-m$. On the other, the Euler characteristic of a covering space is precisely the degree of the covering map multiplied by the Euler characteristic of the base space, so $\chi(Y')=k(1-n)$.  Equating these and rearranging, we have proved the following theorem.

[theorem sif|Schreier Index Formula]Let $F$ be a free group of rank $n$ and let $H$ be a subgroup of finite index $k$.  Then
[math]
\mathrm{rank}(H)=1-k(1-n).
[/math]
[/theorem]
[/EXAMPLE]

[EXAMPLE] An example involving trees and amalgamated products, as in Serre's
book, would be good here.

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