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Last time we proved that if \mathcal{G} is a graph of groups, then G = \pi_1 \mathcal{G} acts on a tree T, namely the underlying graph of the universal cover of X_\mathcal{G}.  Lemmas 16, 17, and 18 describe the vertex and edge sets of T and the action of G.  The following is immediate.

Theorem 14 (Serre): Every graph of groups is developable.

Corollary: If \mathcal{G} is a graph of groups and v \in V(\Gamma), then the map G_v \rightarrow G = \pi_1 \mathcal{G} is injective.

The tree T is called the Bass-Serre Tree of \mathcal{G}.  We will usually equip T with a length metric in which each edge has length 1.  The approach we’ve taken is due to Scott-Wall.  Here’s a sample application.

Lemma 19: Let G = A \ast_C B.  If g \in G is torsion, then g is conjugate into A or B.

Proof. Let T be the Bass-Serre Tree.  Fix a vertex v \in T amd consider \{g^n v \mid n \in \mathbb{Z} \}.  This is a finite set.

Exercise 19: There is an x \in T that minimizes \mathrm{max}_{n \in \mathbb{Z}} d(x, g^n v) (when g is torsion).

Such an x is fixed by g.  Therefore, g stabilizes a vertex.  The stabilizers of the vertices are precisely the conjugates of A and B\Box

Our next goal is to understand the elements of G = \pi_1 \mathcal{G} — which are trivial and which not?  We’ll start with a presentation.  If \Gamma is a tree, then G can be thought of as a sequence of amalgamated free products.  If \Gamma is a rose, then G can be thought of as a sequence of HNN-extensions.  In general, fix a maximal tree \Lambda \subseteq \Gamma.  This determines a way of decomposing G as a sequence of amalgamated products followed by a sequence of HNN-extensions.  This process is sometimes called “Excision.”  It follows that G = \pi_1 \mathcal{G} has a “presentation” like this:

G = \langle \{ G_v \mid v \in V(\Gamma) \}, \{t_e \mid e \in E(\Gamma)\} \mid \\ \{t_e \partial_+^e(g)t^{-1}_e = \partial_-^e(g) \mid e \overline{t}(\Gamma), g \in G_e\}, \{t_e = 1 \mid e \in E(\Lambda)\} \rangle

We can write any g \in G in the following form:

g = g_0 t^{\epsilon_1}_{e_1} g_1 t^{\epsilon_2}_{e_2} \cdots t^{\epsilon_{n-1}}_{e_{n-1}} g_{n-1}t^{\epsilon_n}_{e_n} g_n

where \epsilon_i = \pm 1, e_1^{\epsilon_1} \cdots e_n^{\epsilon_n} is a loop in \Gamma, and g_i \in G_{v_i} where v_i is the terminus of e_i^{\epsilon_i} and the origin of e_{i+1}^{\epsilon_{i+1}}.  A priori, we just know that g = g_0 t^{\epsilon_1}_{e_1} g_1 \cdots t^{\epsilon_n}_{e_n} g_n.  Suppose, say, that e_1 doesn’t originate from v_0.  Then insert stable letters corresponding to edges in \Lambda that form a path from v_0 to the origin of e_1.  Now repeat.

Ping-Pong Lemma

Question. Let G be a group and a, b \in G.  When is \langle a, b \rangle \cong F_2?

Ping-Pong Lemma: Let G be a group acting on a set X and a, b \in G. Assume:

  1. a and b have infinite orders.
  2. There exist X_1, X_2 \subseteq X such that X_2 \nsubseteq X_1 and a^m X_2\subseteq X_1, b^m X_1 \subseteq X_2 for all m \neq 0.

Then \langle a, b \rangle \cong F_2.

Proof. Consider \varphi : F_2 = \langle x, y \rangle \twoheadrightarrow \langle a,b \rangle \leq G such that \varphi (x) = a and \varphi (y) = b.  Choose a reduced word w for a nontrivial element in F_2, i.e., either w = x^{m_1} y^{n_1} x^{m_2} \cdots or w = y^{m_1} x^{n_1} y^{m_2} \cdots and m_i, n_i \neq 0.

Case 1: w = x^{m_1} y^{n_1} \cdots x^{m_k}.  Then \varphi (w) X_2 = a^{m_1} b^{n_1} \cdots a^{m_k} X_2 and a^{m_k} X_2 \subseteq X_1, so \varphi (w) X_2 \subseteq a^{m_1} b^{n_1} \cdots b^{n_{k-1}} X_1 and b^{n_{k-1}} X_1 \subseteq X_2, and so on until \varphi (w) X_2 \subseteq a^{m_1} X_2 \subseteq X_1, so \varphi (w) X_2 \neq X_2.  Therefore \varphi (w) \neq 1.

Case 2: w = y^{m_1} x^{n_1} \cdots y^{m_k}.  Then, by Case 1, \varphi (xwx^{-1}) \neq 1, so \varphi (x) \varphi (w) \varphi (x^{-1}) \neq 1.  Therefore \varphi (w) \neq 1.

Case 3: w = x^{m_1} y^{n_1} \cdots y^{n_k}.  (similar to above)

Case 4: w = y^{m_1} x^{n_1} \cdots x^{n_k}.  (similar to above)

Free Groups Are Linear

Theorem 3. F_2 is linear.

Proof. \mathrm{SL}_2\mathbb{R} acts on \mathbb{R}^2 by linear transformations.  Let a = \left(\begin{array}{cc} 1&2 \\ 0&1 \\ \end{array}\right) and b = \left(\begin{array}{cc} 1&0 \\ 2&1 \\ \end{array}\right).  Then a^m = \left(\begin{array}{cc} 1&2m \\ 0&1 \\ \end{array}\right) and b^m = \left(\begin{array}{cc} 1&0 \\ 2m&1 \\ \end{array}\right). Let X_1 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| > |y|\right\} and X_2 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| < |y|\right\}.  Then a^m X_2 \subseteq X_1, b^m X_1 \subseteq X_2 for all m \neq 0, so \langle a, b \rangle \cong F_2 by the Ping-Pong Lemma.

Corollary 1. Finitely generated free groups are linear, hence residually finite.

Proof. The case of \mathbb{Z} is obvious; otherwise, this follows from F_n \hookrightarrow F_2 as proved in Exercise 1.

Separability

Definition. Let G be a group.  The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.

Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.

Exercise 3. Let G be a group. X \subseteq G is separable if and only if for all g \in G \smallsetminus X, there exists a homomorphism to a  finite group q : G \rightarrow Q such that q(g) \not \in q(X).  Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.

Hint. For the “if” direction, let g \in G and consider q^{-1} \circ q(g).  For the other direction, use the definition of subbase and that p : G \rightarrow P, q: G \rightarrow Q \Rightarrow \text{ker}(p \times q) = \text{ker}(p) \cap \text{ker}(q).

Definition. Let G be a group.

  1. G is Extended RF (ERF) if any subgroup of G is separable.
  2. G is Locally ERF (LERF, subgroup separable) if any finitely generated subgroup is separable.

Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all g \in G \smallsetminus H, there exists a finite-index subgroup K \leq G such that H \leq K \leq G and g \not \in K.

Proof. In the “only if” direction, by the previous exercise, for all g \in G \smallsetminus H, there exists a homomorphism to a  finite group q : G \rightarrow Q such that q(g) \not \in q(H). Then g \not \in  q^{-1} \circ q(H) = K.  Conversely, let g \in G \smallsetminus H. By hypothesis, there exists a finite-index subgroup K \leq G such that H \leq K \leq G and g \not \in K.  Let L = \cap_{h \in G} K^h. Note that this is a finite number of intersections (|G/N_G (K)|, to be precise). There exists a finite quotient q : G \rightarrow G/L.  Then  q^{-1} \circ q(H) = H \cdot L.  Therefore, g \not \in  q^{-1} \circ q(H), i.e., q(g) \not \in q(H), and the lemma follows by the previous exercise.

Scott’s Criterion (1978). Let X be a Hausdorff topological space and G = \pi_1 (X,x).  Let X' \rightarrow X be a covering and H = \pi_1 (X',x'). Then H is separable in G if and only if for any compact \Delta \subseteq X', there exists and intermediate finite-sheeted cover \hat{X} \rightarrow X such that X' \rightarrow \hat{X} embeds \Delta into \hat{X}.

Exercise 4. Let H \leq G be a separable subgroup.

  1. If G' \leq G, then G' \cap H is separable in G’.
  2. If G \leq G' has finite index, then H is separable in G’.
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