Convergence actions and their elements

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

A group of homeomorphisms of a compact space is a convergence group if every infinite sequence of its elements has a subsequence that squeezes the whole space, apart from one point, into smaller and smaller neighbourhoods of another point. Groups of Möbius transformations acting on a sphere at infinity behave like this, and so do hyperbolic groups acting on their boundaries; the definition keeps only this dynamical behaviour.

This lesson proves the first results about such actions on a compact metric space \(M\) with at least three points.

Convergence groups were introduced by Gehring and Martin. The statements of this lesson are those of Section 6 of Bowditch's memoir [Bowditch], which gives the references to the original papers; the proofs here use the definition by sequences throughout.

We assume the core course Point-Set Topology (C90) and the notions of group, subgroup and index.

1. Collapsing sequences

Throughout, \(M\) is a compact metric space and \(\Gamma\) a group acting on \(M\) by homeomorphisms; \(gx\) or \(g(x)\) is the image of \(x\in M\) under \(g\in\Gamma\).

Definition 1.1. A sequence \((g_i)\) in \(\Gamma\) is collapsing with repelling point \(a\) and attracting point \(b\), where \(a,b\in M\), if for every closed set \(K\subseteq M\setminus\{a\}\) and every open set \(U\) containing \(b\), \[ \begin{gathered} g_i(K)\subseteq U\\ \text{for all but finitely many }i . \end{gathered} \] We then say that \((g_i)\) collapses with the pair \((a,b)\). The points \(a\) and \(b\) may coincide.

The action is a convergence action, and \(\Gamma\) a convergence group on \(M\), if every sequence of pairwise distinct elements of \(\Gamma\) has a collapsing subsequence.

Every action of a finite group is a convergence action, since there are no sequences of pairwise distinct elements.

Lemma 1.2. Let \((g_i)\) collapse with the pair \((a,b)\).

  1. The sequence \((g_i^{-1})\) collapses with the pair \((b,a)\).
  2. Every subsequence of \((g_i)\) collapses with the pair \((a,b)\).
  3. If \(x_i\to x\) in \(M\) and \(x\ne a\), then \(g_ix_i\to b\).
  4. For \(h\in\Gamma\), the sequence \((hg_i)\) collapses with \((a,hb)\) and \((g_ih)\) with \((h^{-1}a,b)\).
  5. If \(M\) has at least three points and \((g_i)\) also collapses with \((a',b')\), then \(a'=a\) and \(b'=b\).

Proof. (1) Let \(K'\subseteq M\setminus\{b\}\) be closed and \(U'\) open with \(a\in U'\). Then \(K=M\setminus U'\) is a closed subset of \(M\setminus\{a\}\), and \(U=M\setminus K'\) is an open set containing \(b\). For all but finitely many \(i\) we have \(g_i(K)\subseteq U\), that is, \(g_i(M\setminus U')\) is disjoint from \(K'\), that is, \(g_i^{-1}(K')\subseteq U'\).

(2) is clear.

(3) For large \(N\) the points \(x_i\), \(i\ge N\), stay away from \(a\), so \(K=\{x_i:i\ge N\}\cup\{x\}\) is a compact subset of \(M\setminus\{a\}\). For an open \(U\) containing \(b\), \(g_ix_i\in g_i(K)\subseteq U\) for large \(i\).

(4) Let \(K\subseteq M\setminus\{a\}\) be closed and \(U\) open with \(hb\in U\). Then \(h^{-1}U\) is an open set containing \(b\), so \(g_iK\subseteq h^{-1}U\) and \(hg_iK\subseteq U\) for large \(i\). For the second statement let \(K\subseteq M\setminus\{h^{-1}a\}\) be closed; then \(hK\) is a closed subset of \(M\setminus\{a\}\), and \(g_ihK\subseteq U\) for large \(i\).

(5) Choose \(x\in M\) different from \(a\) and \(a'\). By (3), \(g_ix\to b\) and \(g_ix\to b'\), so \(b=b'\). The same argument for \((g_i^{-1})\), using (1), gives \(a=a'\). \(\blacksquare\)

Example 1.3. Let \(M=\mathbb R\cup\{\infty\}\) be the one-point compactification of the line, a circle, and let \(\mathbb Z\) act through the powers of \(\tau(x)=x+1\), \(\tau(\infty)=\infty\). A closed subset \(K\) of \(M\setminus\{\infty\}\) lies in some interval \([-R,R]\), and \(\tau^n(K)\subseteq[n-R,n+R]\) lies in any given neighbourhood of \(\infty\) once \(|n|\) is large. So every sequence \((\tau^{n_i})\) with \(|n_i|\to\infty\) collapses with the pair \((\infty,\infty)\), and the action is a convergence action. Similarly the powers of \(\sigma(x)=2x\), \(\sigma(\infty)=\infty\), form a convergence group: a closed subset of \(M\setminus\{0\}\) lies in \(\{|x|\ge r\}\cup\{\infty\}\) for some \(r>0\), so \((\sigma^n)_{n\ge1}\) collapses with the pair \((0,\infty)\), and \((\sigma^{-n})_{n\ge1}\) with \((\infty,0)\) by Lemma 1.2(1).

2. The three kinds of elements

Write \(\operatorname{fix}(g)\) for the set of fixed points of \(g\in\Gamma\) in \(M\).

Definition. An element of \(\Gamma\) is elliptic if it has finite order. It is parabolic if it has infinite order and exactly one fixed point, and loxodromic if it has infinite order and exactly two fixed points.

In Example 1.3, \(\tau\) is parabolic and \(\sigma\) is loxodromic.

Theorem 2.1. Let \(\Gamma\) be a convergence group on a compact metric space \(M\) with at least three points, and let \(\gamma\in\Gamma\) have infinite order. Let \((n_i)\) be a sequence of pairwise distinct integers such that \((\gamma^{n_i})\) collapses with a pair \((a,b)\). Then \[ \operatorname{fix}(\gamma)=\{a,b\}. \] Consequently every element of \(\Gamma\) is elliptic, parabolic or loxodromic.

Proof. A fixed point is \(a\) or \(b\). If \(\gamma x=x\) and \(x\ne a\), then \(x=\gamma^{n_i}x\to b\) by Lemma 1.2(3), so \(x=b\).

\(b\) is fixed. Since \(M\) has at least three points, there is \(x\in M\) different from \(a\) and from \(\gamma^{-1}a\). Then \(\gamma^{n_i}x\to b\), so \(\gamma\gamma^{n_i}x\to\gamma b\) by continuity. But \(\gamma\gamma^{n_i}x=\gamma^{n_i}(\gamma x)\) and \(\gamma x\ne a\), so this sequence tends to \(b\). Hence \(\gamma b=b\).

\(a\) is fixed. By Lemma 1.2(1) the sequence \((\gamma^{-n_i})\) of powers of \(\gamma^{-1}\) collapses with \((b,a)\). The previous step, applied to \(\gamma^{-1}\), shows that its attracting point \(a\) is fixed by \(\gamma^{-1}\).

For the last statement let \(\gamma\) have infinite order. The powers \(\gamma^n\), \(n\ge1\), are pairwise distinct, so some subsequence collapses, and \(\gamma\) has one or two fixed points. \(\blacksquare\)

Corollary 2.2. Let \(\gamma\) have infinite order and \(k\ne0\). Then \(\operatorname{fix}(\gamma^k)=\operatorname{fix}(\gamma)\). So powers of parabolic elements are parabolic, powers of loxodromic elements are loxodromic, and an element with at least three fixed points has finite order.

Proof. A sequence of pairwise distinct powers of \(\gamma^k\) has a collapsing subsequence, with a pair \((a,b)\). It consists of powers of \(\gamma^k\) and of powers of \(\gamma\), so by Theorem 2.1, applied to both elements, \(\operatorname{fix}(\gamma^k)=\{a,b\}=\operatorname{fix}(\gamma)\). \(\blacksquare\)

Proposition 2.3 (parabolic elements). Let \(\gamma\) be parabolic with fixed point \(p\). Then every sequence \((\gamma^{n_i})\) with \(|n_i|\to\infty\) collapses with the pair \((p,p)\).

Proof. Suppose not. Then there are a closed \(K\subseteq M\setminus\{p\}\), an open \(U\) containing \(p\), and infinitely many \(i\) with \(\gamma^{n_i}(K)\not\subseteq U\). Among these, choose pairwise distinct exponents and then a collapsing subsequence. By Theorem 2.1 its pair \((a,b)\) has \(\{a,b\}=\operatorname{fix}(\gamma)=\{p\}\), so it collapses with \((p,p)\), and \(\gamma^{n_i}(K)\subseteq U\) for large \(i\) in the subsequence. This is a contradiction. \(\blacksquare\)

3. Loxodromic elements

Proposition 3.1. Let \(\gamma\) be loxodromic. Its two fixed points can be named \(a\) and \(b\) so that the sequence \((\gamma^n)_{n\ge1}\) collapses with the pair \((a,b)\) and \((\gamma^{-n})_{n\ge1}\) with \((b,a)\).

The point \(b\) is the attracting and \(a\) the repelling fixed point of \(\gamma\); for \(\gamma^{-1}\) the roles are exchanged.

Proof. Some subsequence \((\gamma^{n_i})\), with \(n_i\ge1\), collapses with a pair \((a,b)\). By Theorem 2.1, \(\{a,b\}\) is the set of the two fixed points, so \(a\ne b\).

Choose an open \(U\) with \(b\in U\) and \(a\notin\overline U\). Since \(\overline U\) is a closed subset of \(M\setminus\{a\}\), there is \(m=n_i\ge1\) with \(\gamma^m(\overline U)\subseteq U\). Put \[ W=U\cap\gamma(U)\cap\dots\cap\gamma^{m-1}(U). \] This is an open set containing \(b\), because \(b\) is fixed. It satisfies \(\gamma(W)\subseteq W\): indeed \(\gamma(W)=\gamma(U)\cap\dots\cap\gamma^m(U)\) and \(\gamma^m(U)\subseteq U\). Hence \(\gamma^j(W)\subseteq W\) for all \(j\ge0\). Also \(\overline W\subseteq\overline U\subseteq M\setminus\{a\}\).

Now let \(K\subseteq M\setminus\{a\}\) be closed and \(V\) open with \(b\in V\). There are \(p=n_i\) with \(\gamma^p(K)\subseteq W\) and \(q=n_j\) with \(\gamma^q(\overline W)\subseteq V\). For \(n\ge p+q\), \[ \begin{aligned} \gamma^n(K)&=\gamma^q\gamma^{n-p-q}\gamma^p(K)\\ &\subseteq\gamma^q\gamma^{n-p-q}(W)\\ &\subseteq\gamma^q(W)\subseteq V . \end{aligned} \] So \((\gamma^n)_{n\ge1}\) collapses with \((a,b)\), and \((\gamma^{-n})_{n\ge1}\) with \((b,a)\) by Lemma 1.2(1). \(\blacksquare\)

Proposition 3.2 (a fundamental domain). Let \(\gamma\) be loxodromic with repelling point \(a\) and attracting point \(b\). There are \(m\ge1\) and a compact set \(K\subseteq M\setminus\{a,b\}\) with \[ \bigcup_{k\in\mathbb Z}\gamma^{mk}(K)=M\setminus\{a,b\}. \]

Proof. Choose an open \(N\) with \(b\in N\) and \(a\notin\overline N\). By Proposition 3.1 there is \(m\ge1\) with \(\gamma^m(\overline N)\subseteq N\). Let \[ K=\overline N\setminus\gamma^m(N). \] It is compact; it does not contain \(b\), because \(b\in\gamma^m(N)\); and it does not contain \(a\).

Let \(x\in M\setminus\{a,b\}\) and \(S=\{k\in\mathbb Z:\gamma^{mk}x\in\overline N\}\). Since \(x\ne a\), \(\gamma^{mk}x\to b\) as \(k\to\infty\), so \(S\) contains all large \(k\). Since \(x\ne b\), \(\gamma^{-mk}x\to a\) as \(k\to\infty\), and \(a\) has a neighbourhood disjoint from \(\overline N\); so \(S\) is bounded below. Let \(k_0=\min S\) and \(y=\gamma^{mk_0}x\in\overline N\). If \(y\) were in \(\gamma^m(N)\), then \(\gamma^{m(k_0-1)}x\in N\), contradicting the minimality of \(k_0\). So \(y\in K\) and \(x\in\gamma^{-mk_0}(K)\). Conversely the images of \(K\) avoid \(a\) and \(b\). \(\blacksquare\)

Proposition 3.3. Let \(\gamma\) be loxodromic with fixed points \(a,b\), and let \[ G_{a,b}=\{g\in\Gamma:\ ga=a,\ gb=b\}. \] Then the cyclic group \(\langle\gamma\rangle\) has finite index in \(G_{a,b}\).

Proof. Take \(m\) and \(K\) as in Proposition 3.2 and fix \(x\in M\setminus\{a,b\}\). For \(g\in G_{a,b}\) the point \(gx\) is not \(a\) or \(b\), so there is \(k(g)\in\mathbb Z\) with \(\gamma^{mk(g)}gx\in K\). We show that the set \[ F=\{\gamma^{mk(g)}g:\ g\in G_{a,b}\} \] is finite. Then every \(g\) lies in \(\langle\gamma\rangle f\) for some \(f\in F\), and the index is at most the number of elements of \(F\).

Suppose \(F\) is infinite and let \((h_i)\) be a sequence of pairwise distinct elements of \(F\). Each \(h_i\) fixes \(a\) and \(b\) and has \(h_ix\in K\). Pass to a subsequence collapsing with a pair \((a',b')\). If \(a\ne a'\), then \(a=h_ia\to b'\), so \(a=b'\); thus \(a\in\{a',b'\}\), and in the same way \(b\in\{a',b'\}\). Since \(a\ne b\), \(\{a',b'\}=\{a,b\}\). In particular \(x\ne a'\), so \(h_ix\to b'\), and \(b'\in K\) because \(K\) is closed. This contradicts \(K\cap\{a,b\}=\varnothing\). \(\blacksquare\)

Proposition 3.4. Let \(\gamma\) be loxodromic with fixed points \(a,b\). An element \(h\in\Gamma\) that fixes one of the points \(a,b\) fixes both.

Proof. Replacing \(\gamma\) by \(\gamma^{-1}\) if necessary, let \(a\) be the repelling and \(b\) the attracting point, and \(ha=a\). Suppose \(hb\ne b\), and put \(z=h^{-1}b\). Then \(z\ne b\), and \(z\ne a\) because \(ha=a\ne b\). By Corollary 2.2 no power \(\gamma^p\) with \(p\ne0\) fixes \(z\). In particular the points \(\gamma^nz\), \(n\in\mathbb Z\), are pairwise distinct, and \(w=\gamma^{-1}z\) is not \(a\) or \(b\).

Take \(m\) and \(K\) as in Proposition 3.2. For \(n\ge1\) consider the point \(u_n=h\gamma^{-n}w=h\gamma^{-n-1}z\). It is not \(a\), because \(\gamma^{-n-1}z\ne a\) and \(h\) fixes \(a\); it is not \(b\), because \(\gamma^{-n-1}z\ne z=h^{-1}b\). So there is \(k_n\in\mathbb Z\) with \(\gamma^{mk_n}u_n\in K\). Put \[ g_n=\gamma^{mk_n}h\gamma^{-n}. \] Then \[ \begin{gathered} g_na=a,\qquad g_nw\in K,\\ g_n(\gamma^nz)=\gamma^{mk_n}hz=b . \end{gathered} \]

The set \(\{g_n:n\ge1\}\) is finite. Otherwise some subsequence consists of pairwise distinct elements and collapses with a pair \((a',b')\). The set \[ C=\{\gamma^nz:n\ge1\}\cup\{b\} \] is compact, because \(\gamma^nz\to b\); it does not contain \(a\), and it does not contain \(w=\gamma^{-1}z\). As in the proof of Proposition 3.3, \(a\in\{a',b'\}\), because all \(g_n\) fix \(a\).

If \(w\ne a'\), then \(g_nw\to b'\) along the subsequence, so \(b'\in K\). Then \(a\ne b'\), so \(a=a'\). Hence \(C\) is a closed subset of \(M\setminus\{a'\}\), and \(g_n(C)\) lies in any given neighbourhood of \(b'\) for large \(n\) in the subsequence. But \(b=g_n(\gamma^nz)\in g_n(C)\), so \(b=b'\in K\), which is false.

If \(w=a'\), then \(a\ne a'\), so \(a=b'\). Again \(C\) is a closed subset of \(M\setminus\{a'\}\), so \(b=g_n(\gamma^nz)\) lies in any given neighbourhood of \(b'=a\) for large \(n\), which is false because \(a\ne b\).

Conclusion. There are \(n<n'\) with \(g_n=g_{n'}\). With \(p=n'-n\ge1\) and \(q=m(k_{n'}-k_n)\) this reads \(h\gamma^p=\gamma^qh\). Then \[ h(\gamma^pz)=\gamma^q(hz)=\gamma^qb=b=h(z), \] so \(\gamma^pz=z\). This contradicts the choice of \(z\). \(\blacksquare\)

Corollary 3.5. Let \(\gamma\) be loxodromic with fixed points \(a,b\).

  1. The stabilizers of \(a\) and of \(b\) in \(\Gamma\) both equal \(G_{a,b}\), and \(\langle\gamma\rangle\) has finite index in it. So these stabilizers are virtually infinite cyclic.
  2. Every element of infinite order in \(G_{a,b}\) is loxodromic with fixed points \(a,b\). No parabolic element of \(\Gamma\) fixes \(a\) or \(b\).
  3. The fixed point sets of two loxodromic elements of \(\Gamma\) are equal or disjoint.

Proof. (1) is Propositions 3.4 and 3.3. (2) An element of infinite order that fixes \(a\) or \(b\) fixes both, so it has two fixed points and is loxodromic by Theorem 2.1, with fixed point set \(\{a,b\}\). (3) If a loxodromic \(\gamma'\) fixes \(a\) or \(b\), it fixes both by Proposition 3.4, and it has only two fixed points. \(\blacksquare\)

Corollary 3.6. Let \(\Gamma\) be a convergence group on a compact metric space \(M\) with at least three points, and \(p\in M\). The stabilizer of \(p\) in \(\Gamma\) either is virtually infinite cyclic and contains a loxodromic element, or consists of elliptic and parabolic elements.

Proof. If a loxodromic element fixes \(p\), Corollary 3.5(1) applies. Otherwise every element of the stabilizer is elliptic or parabolic by Theorem 2.1. \(\blacksquare\)

4. Exercises

4.1. Let \(\gamma\) be loxodromic with repelling point \(a\) and attracting point \(b\), and \(h\in\Gamma\). Show that \(h\gamma h^{-1}\) is loxodromic with repelling point \(ha\) and attracting point \(hb\).

4.2. Let \(\gamma\) be loxodromic with fixed points \(a,b\) and \(C\subseteq M\setminus\{a,b\}\) compact. Show that \(\gamma^n(C)\cap C\ne\varnothing\) for only finitely many \(n\in\mathbb Z\).

4.3. Let \(\Gamma\) be a convergence group on \(M\), with at least three points, and suppose that \(\Gamma\) fixes a point \(p\) and contains a loxodromic element. Show that \(\Gamma\) is virtually infinite cyclic.

4.4. Let \(\Gamma\) be a convergence group on \(M\) and \(M'\subseteq M\) a nonempty closed \(\Gamma\)-invariant subset. Show that \(\Gamma\) is a convergence group on \(M'\). Deduce that if \(M'\) has at least three points, only finitely many elements of \(\Gamma\) fix every point of \(M'\).

4.5. Let \(\psi(x)=2x+1\), \(\psi(\infty)=\infty\), on the circle \(M=\mathbb R\cup\{\infty\}\) of Example 1.3. Show that the powers of \(\psi\) form a convergence group on \(M\) and that \(\psi\) is loxodromic, and find its repelling and its attracting point.

5. Solutions

4.1. The element \(h\gamma h^{-1}\) has infinite order, and \((h\gamma h^{-1})^n=h\gamma^nh^{-1}\). By Lemma 1.2(4), applied twice, \((h\gamma^nh^{-1})_{n\ge1}\) collapses with the pair \((ha,hb)\). By Theorem 2.1 its fixed point set is \(\{ha,hb\}\), which has two points.

4.2. Let \(U\) be an open set containing \(b\) whose closure is disjoint from \(C\). Since \(C\) is a closed subset of \(M\setminus\{a\}\), Proposition 3.1 gives \(\gamma^n(C)\subseteq U\) for all large \(n\), so \(\gamma^n(C)\cap C=\varnothing\). The same argument for \(\gamma^{-1}\), with a neighbourhood of \(a\), covers large negative \(n\).

4.3. A loxodromic element \(\gamma\in\Gamma\) fixes \(p\), so \(p\) is one of its two fixed points, and \(\Gamma\) is the stabilizer of \(p\). Apply Corollary 3.5(1).

4.4. Let \((g_i)\) be pairwise distinct elements of \(\Gamma\) and pass to a subsequence that collapses on \(M\) with a pair \((a,b)\). If \(M'\) is a single point \(p\), the subsequence collapses on \(M'\) with the pair \((p,p)\), because \(M'\setminus\{p\}\) is empty. Otherwise choose \(x\in M'\) with \(x\ne a\). Then \(g_ix\to b\) and \(g_ix\in M'\), so \(b\in M'\) because \(M'\) is closed; the same argument for the inverses gives \(a\in M'\). A closed subset \(K\) of \(M'\setminus\{a\}\) is closed in \(M\), and a relatively open subset of \(M'\) containing \(b\) has the form \(U\cap M'\) with \(U\) open in \(M\); for large \(i\), \(g_i(K)\subseteq U\cap M'\). So the subsequence collapses on \(M'\) with the pair \((a,b)\).

For the second statement suppose infinitely many elements fix every point of \(M'\), and take a sequence of pairwise distinct ones. A subsequence collapses on \(M'\) with a pair \((a,b)\). Choose \(x\in M'\) different from \(a\) and \(b\). Then \(x=g_ix\to b\), so \(x=b\), a contradiction.

4.5. For \(n\in\mathbb Z\), \(\psi^n(x)=2^n(x+1)-1\). A closed subset \(K\) of \(M\setminus\{-1\}\) lies in \(\{x:|x+1|\ge r\}\cup\{\infty\}\) for some \(r>0\), and \(\psi^n\) maps this set into \(\{x:|x+1|\ge2^nr\}\cup\{\infty\}\), which lies in any given neighbourhood of \(\infty\) for large \(n\). So \((\psi^n)_{n\ge1}\) collapses with the pair \((-1,\infty)\) and \((\psi^{-n})_{n\ge1}\) with \((\infty,-1)\). A sequence of pairwise distinct powers has a subsequence with exponents tending to \(+\infty\) or to \(-\infty\), so the powers of \(\psi\) form a convergence group. The element \(\psi\) has infinite order and its fixed points are \(-1\) and \(\infty\), so it is loxodromic, with repelling point \(-1\) and attracting point \(\infty\).

References