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A basic reference for the background to this post is my monograph.

Let $G$ be a group, and let $[G,G]$ denote the commutator subgroup. Every element of $[G,G]$ can be expressed as a product of commutators; the commutator length of an element $g$ is the minimum number of commutators necessary, and is denoted $\text{cl}(g)$. The stable commutator length is the growth rate of the commutator lengths of powers of an element; i.e. $\text{scl}(g) = \lim_{n \to \infty} \text{cl}(g^n)/n$. Recall that a group $G$ is said to satisfy a law if there is a nontrivial word $w$ in a free group $F$ for which every homomorphism from $F$ to $G$ sends $w$ to $\text{id}$.

The purpose of this post is to give a very short proof of the following proposition (modulo some background that I wanted to talk about anyway):

Proposition: Suppose $G$ obeys a law. Then the stable commutator length vanishes identically on $[G,G]$.

The proof depends on a duality between stable commutator length and a certain class of functions, called homogeneous quasimorphisms

Definition: A function $\phi:G \to \mathbb{R}$ is a quasimorphism if there is some least number $D(\phi)\ge 0$ (called the defect) so that for any pair of elements $g,h \in G$ there is an inequality $|\phi(x) + \phi(y) - \phi(xy)| \le D(\phi)$. A quasimorphism is homogeneous if it satisfies $\phi(g^n) = n\phi(g)$ for all integers $n$.

Note that a homogeneous quasimorphism with defect zero is a homomorphism (to $\mathbb{R}$). The defect satisfies the following formula:

Lemma: Let $f$ be a homogeneous quasimorphism. Then $D(\phi) = \sup_{g,h} \phi([g,h])$.

A fundamental theorem, due to Bavard, is the following:

Theorem: (Bavard duality) There is an equality $\text{scl}(g) = \sup_\phi \frac {\phi(g)} {2D(\phi)}$ where the supremum is taken over all homogeneous quasimorphisms with nonzero defect.

In particular, $\text{scl}$ vanishes identically on $[G,G]$ if and only if every homogeneous quasimorphism on $G$ is a homomorphism.

One final ingredient is another geometric definition of $\text{scl}$ in terms of Euler characteristic. Let $X$ be a space with $\pi_1(X) = G$, and let $\gamma:S^1 \to X$ be a free homotopy class representing a given conjugacy class $g$. If $S$ is a compact, oriented surface without sphere or disk components, a map $f:S \to X$ is admissible if the map on $\partial S$ factors through $\partial f:\partial S \to S^1 \to X$, where the second map is $\gamma$. For an admissible map, define $n(S)$ by the equality $[\partial S] \to n(S) [S^1]$ in $H_1(S^1;\mathbb{Z})$ (i.e. $n(S)$ is the degree with which $\partial S$ wraps around $\gamma$). With this notation, one has the following:

Lemma: There is an equality $\text{scl}(g) = \inf_S \frac {-\chi^-(S)} {2n(S)}$.

Note: the function $-\chi^-$ is the sum of $-\chi$ over non-disk and non-sphere components of $S$. By hypothesis, there are none, so we could just write $-\chi$. However, it is worth writing $-\chi^-$ and observing that for more general (orientable) surfaces, this function is equal to the function $\rho$ defined in a previous post.

We now give the proof of the Proposition.

Proof. Suppose to the contrary that stable commutator length does not vanish on $[G,G]$. By Bavard duality, there is a homogeneous quasimorphism $\phi$ with nonzero defect. Rescale $\phi$ to have defect $1$. Then for any $\epsilon$ there are elements $g,h$ with $\phi([g,h]) \ge 1-\epsilon$, and consequently $\text{scl}([g,h]) \ge 1/2 - \epsilon/2$ by Bavard duality. On the other hand, if $X$ is a space with $\pi_1(X)=G$, and $\gamma:S^1 \to X$ is a loop representing the conjugacy class of $[g,h]$, there is a map $f:S \to X$ from a once-punctured torus $S$ to $X$ whose boundary represents $\gamma$. The fundamental group of $S$ is free on two generators $x,y$ which map to the class of $g,h$ respectively. If $w$ is a word in $x,y$ mapping to the identity in $G$, there is an essential loop $\alpha$ in $S$ that maps inessentially to $X$. There is a finite cover $\widetilde{S}$ of $S$, of degree $d$ depending on the word length of $w$, for which $\alpha$ lifts to an embedded loop. This can be compressed to give a surface $S'$ with $-\chi^-(S') \le -\chi^-(\widetilde{S})-2$. However, Euler characteristic is multiplicative under coverings, so $-\chi^-(\widetilde{S}) = -\chi^-(S)\cdot d$. On the other hand, $n(S') = n(\widetilde{S})=d$ so $\text{scl}([g,h]) \le 1/2 - 1/d$. If $G$ obeys a law, then $d$ is fixed, but $\epsilon$ can be made arbitrarily small. So $G$ does not obey a law. qed.

More ambitious than simply showing that a group is infinite is to show that it contains an infinite subgroup of a certain kind. One of the most important kinds of subgroup to study are free groups. Hence, one is interested in the question:

Question: When does a group contain a (nonabelian) free subgroup?

Again, one can (and does) ask this question both about a specific group, and about certain classes of groups, or for a typical (in some sense) group from some given family.

Example: If $\mathcal{P}$ is a property of groups that is inherited by subgroups, then if no free group satisfies $\mathcal{P}$, no group that satisfies $\mathcal{P}$ can contain a free subgroup. An important property of this kind is amenability. A (discrete) group $G$ is amenable if it admits an invariant mean; that is, if there is a linear map $m: L^\infty(G) \to \mathbb{R}$ (i.e. a way to define the average of a bounded function over $G$) satisfying three basic properties:

1. $m(f) \ge 0$ if $f\ge 0$ (i.e. the average of a non-negative function is non-negative)
2. $m(\chi_G)=1$ where $\chi_G$ is the constant function taking the value $1$ everywhere on $G$ (i.e. the average of the constant function $1$ is normalized to be $1$)
3. $m(g\cdot f) = m(f)$ for every ${}g \in G$ and $f \in L^\infty(G)$, where $(g\cdot f)(x) = f(g^{-1}x)$ (i.e. the mean is invariant under the obvious action of $G$ on $L^\infty(G)$)

If $H$ is a subgroup of $G$, there are (many) $H$-invariant homomorphisms $j: L^\infty(H) \to L^\infty(G)$ taking non-negative functions to non-negative functions, and $\chi_H$ to $\chi_G$; for example, the (left) action of $H$ on $G$ breaks up into a collection of copies of $H$ acting on itself, right-multiplied by a collection of right coset representatives. After choosing such a choice of representatives $\lbrace g_\alpha \rbrace$, one for each coset $Hg_\alpha$, we can define $j(f)(hg_\alpha) = f(h)$. Composing with $m$ shows that every subgroup of an amenable group is amenable (this is harder to see in the “geometric” definition of amenable groups in terms of Folner sets). On the other hand, as is well-known, a nonabelian free group is not amenable. Hence, amenable groups do not contain nonabelian free subgroups.

The usual way to see that a nonabelian free group is not amenable is to observe that it contains enough disjoint “copies” of big subsets. For concreteness, let $F$ denote the free group on two generators $a,b$, and write their inverses as $A,B$. Let $W_a, W_A$ denote the set of reduced words that start with either $a$ or $A$, and let $\chi_a,\chi_A$ denote the indicator functions of $W_a,W_A$ respectively. We suppose that $F$ is amenable, and derive a contradiction. Note that $F = W_a \cup aW_A$, so $m(\chi_a) + m(\chi_A) \ge 1$. Let $V$ denote the set of reduced words that start with one of the strings $a,A,ba,bA$, and let $\chi_V$ denote the indicator function of $V$. Notice that $V$ is made of two disjoint copies of each of $W_a,W_A$. So on the one hand, $m(\chi_V) \le m(\chi_F) = 1$, but on the other hand, $m(\chi_V) = 2 (m(\chi_a)+m(\chi_A)) \ge 2$.

Conversely, the usual way to show that a group $G$ is amenable is to use the Folner condition. Suppose that $G$ is finitely generated by some subset $S$, and let $C$ denote the Cayley graph of $G$ (so that $C$ is a homogeneous locally finite graph). Suppose one can find finite subsets $U_i$ of vertices so that $|\partial U_i|/|U_i| \to 0$ (here $|U_i|$ means the number of vertices in $U_i$, and  $|\partial U_i|$ means the number of vertices in $U_i$ that share an edge with $C - U_i$). Since the “boundary” of $U_i$ is small compared to $U_i$, averaging a bounded function over $U_i$ is an “almost invariant” mean; a weak limit (in the dual space to $L^\infty(G)$) is an invariant mean. Examples of amenable groups include

1. Finite groups
2. Abelian groups
3. Unions and extensions of amenable groups
4. Groups of subexponential growth

and many others. For instance, virtually solvable groups (i.e. groups containing a solvable subgroup with finite index) are amenable.

Example: No amenable group can contain a nonabelian free subgroup. The von Neumann conjecture asked whether the converse was true. This conjecture was disproved by Olshanskii. Subsequently, Adyan showed that the infinite free Burnside groups are not amenable. These are groups $B(m,n)$ with $m\ge 2$ generators, and subject only to the relations that the $n$th power of every element is trivial. When $n$ is odd and at least $665$, these groups are infinite and nonamenable. Since they are torsion groups, they do not even contain a copy of $\mathbb{Z}$, let alone a nonabelian free group!

Example: The Burnside groups are examples of groups that obey a law; i.e. there is a word $w(x_1,x_2,\cdots,x_n)$ in finitely many free variables, such that $w(g_1,g_2,\cdots,g_n)=\text{id}$ for every choice of $g_1,\cdots,g_n \in G$. For example, an abelian group satisfies the law $x_1x_2x_1^{-1}x_2^{-1}=\text{id}$. Evidently, a group that obeys a law does not contain a nonabelian free subgroup. However, there are examples of groups which do not obey a law, but which also do not contain any nonabelian free subgroup. An example is the classical Thompson’s group $F$, which is the group of orientation-preserving piecewise-linear homeomorphisms of $[0,1]$ with finitely many breakpoints at dyadic rationals (i.e. points of the form $p/2^q$ for integers $p,q$) and with slopes integral powers of $2$. To see that this group does not obey a law, one can show (quite easily) that in fact $F$ is dense (in the $C^0$ topology) in the group $\text{Homeo}^+([0,1])$ of all orientation-preserving homeomorphisms of the interval. This latter group contains nonabelian free groups; by approximating the generators of such a group arbitrarily closely, one obtains pairs of elements in $F$ that do not satisfy any identity of length shorter than any given constant. On the other hand, a famous theorem of Brin-Squier says that $F$ does not contain any nonabelian free subgroup. In fact, the entire group $\text{PL}^+([0,1])$ does not contain any nonabelian free subgroup. A short proof of this fact can be found in my paper as a corollary of the fact that every subgroup $G$ of $\text{PL}^+([0,1])$ has vanishing stable commutator length; since stable commutator length is nonvanishing in nonabelian free groups, this shows that there are no such subgroups of $\text{PL}^+([0,1])$. (Incidentally, and complementarily, there is a very short proof that stable commutator length vanishes on any group that obeys a law; we will give this proof in a subsequent post).

Example: If $G$ surjects onto $H$, and $H$ contains a free subgroup $F$, then there is a section from $F$ to $G$ (by freeness), and therefore $G$ contains a free subgroup.

Example: The most useful way to show that $G$ contains a nonabelian free subgroup is to find a suitable action of $G$ on some space $X$. The following is known as Klein’s ping-pong lemma. Suppose one can find disjoint subsets $U^\pm$ and $V^\pm$ of $X$, and elements $g,h \in G$ so that $g(U^+ \cup V^\pm) \subset U^+$$g^{-1}(U^- \cup V^\pm) \subset U^-$, and similarly interchanging the roles of $U^\pm, V^\pm$ and $g,h$. If $w$ is a reduced word in $g^{\pm 1},h^{\pm 1}$, one can follow the trajectory of a point under the orbit of subwords of $w$ to verify that $w$ is nontrivial. The most common way to apply this in practice is when $g,h$ act on $X$ with source-sink dynamics; i.e. the element $g$ has two fixed points $u^\pm$ so that every other point converges to $u^+$ under positive powers of $g$, and to $u^-$ under negative powers of $g$. Similarly, $h$ has two fixed points $v^\pm$ with similar dynamics. If the points $u^\pm,v^\pm$ are disjoint, and $X$ is compact, one can take any small open neighborhoods $U^\pm,V^\pm$ of $u^\pm,v^\pm$, and then sufficiently large powers of $g$ and $h$ will satisfy the hypotheses of ping-pong.

Example: Every hyperbolic group $G$ acts on its Gromov boundary $\partial_\infty G$. This boundary is the set of equivalence classes of quasigeodesic rays in (the Cayley graph of) $G$, where two rays are equivalent if they are a finite Hausdorff distance apart. Non-torsion elements act on the boundary with source-sink dynamics. Consequently, every pair of non-torsion elements in a hyperbolic group either generate a virtually cyclic group, or have powers that generate a nonabelian free group.

It is striking to see how easy it is to construct nonabelian free subgroups of a hyperbolic group, and how difficult to construct closed surface subgroups. We will return to the example of hyperbolic groups in a future post.

Example: The Tits alternative says that any linear group $G$ (i.e. any subgroup of $\text{GL}(n,\mathbb{R})$ for some $n$) either contains a nonabelian free subgroup, or is virtually solvable (and therefore amenable). This can be derived from ping-pong, where $G$ is made to act on certain spaces derived from the linear action (e.g. locally symmetric spaces compactified in certain ways, and buildings associated to discrete valuations on the ring of entries of matrix elements of $G$).

Example: There is a Tits alternative for subgroups of other kinds of groups, for example mapping class groups, as shown by Ivanov and McCarthy. The mapping class group (of a surface) acts on the Thurston boundary of Teichmuller space. Every subgroup of the mapping class group either contains a nonabelian free subgroup, or is virtually abelian. Roughly speaking, either elements move points in the boundary with enough dynamics to be able to do ping-pong, or else the action is “localized” in a train-track chart, and one obtains a linear representation of the group (enough to apply the ordinary Tits alternative). Virtually solvable subgroups of mapping class groups are virtually abelian.

Example: A similar Tits alternative holds for $\text{Out}(F_n)$. This was shown by Bestvina-Feighn-Handel in these three papers (the third paper shows that solvable subgroups are virtually abelian, thus emphasizing the parallels with mapping class groups).

Example: If $G$ is a finitely generated group of homeomorphisms of $S^1$, then there is a kind of Tits alternative, first proposed by Ghys, and proved by Margulis: either $G$ preserves a probability measure on $S^1$ (which might be singular), or it contains a nonabelian free subgroup. To see this, first note that either $G$ has a finite orbit (which supports an invariant probability measure) or the action is semi-conjugate to a minimal action (one with all orbits dense). In the second case, the proof depends on understanding the centralizer of the group action: either the centralizer is infinite, in which case the group is conjugate to a group of rotations, or it is finite cyclic, and one obtains an action of $G$ on a “smaller” circle, by quotienting out by the centralizer. So one may assume the action is minimal with trivial centralizer. In this case, one shows that the action has the property that for any nonempty intervals $I,J$ in $S^1$, there is some ${}g \in G$ with $g(I) \subset J$; i.e. any interval may be put inside any other interval by some element of the group. For such an action, it is very easy to do ping-pong. Incidentally, a minor variation on this result, and with essentially this argument, was established by Thurston in the context of uniform foliations of $3$-manifolds before Ghys proposed his question.

Example: If $\rho_t$ is an (algebraic) family of representations of a (countable) free group $F$ into an algebraic group, then either some element $g \in F$ is in the kernel of every $\rho_t$, or the set of faithful representations is “generic”, i.e. the intersection of countably many open dense sets. This is because the set of representations for which a given element is in the kernel is Zariski closed, and therefore its complement is open and either empty or dense (one must add suitable hypotheses or conditions to the above to make it rigorous).