## The Tits alternative and non-positive curvature

The following is largely based on the paper The Tits Alternative for CAT(0) Cubical Complexes by Sageev and Wise.

1.1. The Tits alternative

A group ${G}$ satisfies the Tits alternative if for all subgroups ${H < G}$, either ${H}$ is virtually solvable or it contains a non-abelian free group. One way to think of this is that a group satisfying the Tits alternative has either “small” subgroups (subgroups and quotients of solvable groups are solvable), or “large” subgroups (extensions by groups containing a free group contain a free group). The term originates from the work of Jacques Tits in which he showed that finitely generated linear groups have this property (later named the Tits alternative).

1.2. Known results

The following groups are known to satisfy the Tits alternative:

1. Finitely generated linear groups (Tits)
2. hyperbolic groups (Gromov, Ghys and de la Harpe)
3. ${\mathrm{Out}(F_n)}$ (Bestvina-Feighn-Handel)
4. mapping class groups of compact surfaces (Ivanov, McCarthy)
5. some large subclasses of CAT(0) groups (Ballmann-Swiatkowski, Sageev-Wise, Xie)

Thompson’s group F is an example of a group not satisfying the Tits alternative as it contains no free subgroup of rank greater than one and isn’t virtually solvable (Bleak, for instance, constructs non-virtually solvable subgroups). Another is the Grigorchuk group. In fact any group of intermediate growth can’t satisfy the Tits alternative: it can’t have a non-abelian free subgroup (otherwise its growth would in fact be exponential), and by the MilnorWolf theorem, if it were virtually solvable, it would have to have a finite index nilpotent subgroup (implying polynomial growth by Gromov’s theorem) or an exponentially growing solvable subgroup (again contradicting intermediate growth).

It’s still an open question whether the Tits alternative holds for all ${\mathrm{CAT}(0)}$ groups, i.e. for groups which act properly and cocompactly by isometries on some ${\mathrm{CAT}(0)}$ space.

1.3. Hyperbolic groups

In case of hyperbolic groups, a stronger version of the Tits alternative holds:

Theorem 1. A subgroup of a hyperbolic group either

1. is finite,
2. is virtually infinite cyclic, or
3. contains a nonabelian free group.

An elementary proof of the above in the case where the subgroup is torsion free is presented on Henry Wilton’s geometric group theory blog (a precursor and inspiration for the blog you’re now reading). The general case requires additional results about hyperbolic elements and is presented (in French) in Sur les groupes hyperboliques d’après Mikhael Gromov (Theorem 37, page 157).

1.4. Groups acting on ${\mathrm{CAT}(0)}$ cube complexes

Theorem 2 (Sageev-Wise). Suppose ${G}$ acts properly on a finite dimensional ${\mathrm{CAT}(0)}$ cube complex ${X}$ and has a bound on the order of finite subgroups. Then any subgroup of ${G}$ either contains ${F_2}$ or is virtually abelian.

In particular, if the group acts on such an ${X}$ freely (and thus has no non-trivial finite subgroups), its subgroups contain ${F_2}$ or are virtually abelian.

I’ll sketch the proof of this theorem following the exposition in the original paper and add a few explanatory details.

We first prove the theorem assuming that ${G}$ is finitely generated.

Since ${X}$ is assumed to be finite dimensional, we proceed, as you may have already guessed, by induction on the dimension. If the dimension is 0, then the group ${G}$ must be finite (since the action is proper), so we’re done. Now we assume the theorem holds for ${G}$ acting on all such cube complexes of dimension less than ${n}$. Assume ${\mathrm{dim}(X) = n}$. To apply the induction step, we will identify a proper action of a subgroup of ${G}$ on a lower dimensional subcomplex of ${X}$. This subcomplex will be a “hyperplane” (defined next) and the subgroup acting will be its stabilizer.

A ${\mathrm{CAT}(0)}$ cube complex ${X}$ is a cubical cell complex which is simply connected and in which every link is flag. Two edges of ${X}$ are square equivalent if they are opposite edges of a square of ${X}$. Let ${\sigma}$ be an ${n}$-cube (identified with the standard unit cube in ${\mathbb{R}^n}$). A midcube of ${\sigma}$ is an intersection of ${\sigma}$ with an ${\mathbb{R}^n}$ hyperplane parallel to a face of ${\sigma}$ and containing its center. For instance, there is only one midcube in an edge, two midcubes in a square, etc.

A hyperplane in ${X}$ is the union of all midcubes meeting the equivalence class of an edge (under the equivalence relation generated by square equivalence). A few facts about hyperplanes:

1. A hyperplane meets each cube in at most one midcube.
2. A hyperplane separates ${X}$ into two components.
3. The ${\mathrm{CAT}(0)}$ cube structure on ${X}$ induces a ${\mathrm{CAT}(0)}$ cube structure on any hyperplane in ${X}$.

1.5. Tangent: ends of groups

Let ${G}$ be any finitely generated group, with ${S}$ being the set of generators. Then the number of ends of ${G}$, denoted ${e(G)}$, is the number of components of the boundary $\partial G$. It’s a classic fact that the number of ends can only equal 0 (when ${G}$ is finite), 1 (e.g. ${\mathbb{Z}^2}$), 2 (e.g. ${\mathbb{Z}}$), or infinity (e.g. ${F_2}$).

The number of ends of a finitely generated group ${G}$ relative to a subgroup ${H}$, denoted ${e(G,H)}$, is the number of ends of the graph ${\mathrm{Cay}_S(G) / H}$, i.e. the quotient of the Cayley graph of ${G}$ by the action of ${H}$. Equivalently, it is the number of ends of the graph in which vertices are cosets of ${H}$ and vertices ${Hx}$ and ${Hxg}$ are connected if and only if ${g \in S}$.

It is a famous result of Stallings that ${e(G) > 1}$ if and only if ${G}$ splits as an amalgamated product or an HNN-extension over a finite group. In the language of Bass-Serre theory, we have:

For relative ends, the picture is more complicated:

The one of the key theorems which embody the (${\Leftarrow}$) implication above is the Algebraic Torus Theorem (Dunwoody-Swenson), given below. First, however, we must establish the subgroup with respect to which ${G}$ has multiple ends.

Theorem 3 (Sageev). Suppose ${G \curvearrowright X}$, a finite dimensional ${\mathrm{CAT}(0)}$ cube complex, without a global fixed point. Then there is a hyperplane ${J \subset X}$ such that ${e(G, \mathrm{stab}(J)) > 1}$.

Theorem 4 (Algebraic Torus). Suppose ${G}$ is finitely generated, and ${H}$, a subgroup of ${G}$, is virtually polycyclic with ${e(G,H) > 1}$. Then one of the following holds.

1. ${G}$ is virtually polycyclic.
2. There exists a short exact sequence

$\displaystyle 1 \rightarrow P \rightarrow G \rightarrow G/P \rightarrow 1$

in which ${P}$ is virtually polycyclic and ${G/P}$ is non-elementary fuchsian (see below).

3. ${G}$ splits over a virtually polycyclic group.

1.6. Tangent: fuchsian groups

A fuchsian group is a subgroup of ${\mathrm{PSL}_2(\mathbb{R})}$. Thus, a fuchsian group ${G}$ acts discretely on the hyperbolic plane ${\mathbb{H}}$. Consider the set ${Gz}$, for some ${z \in \mathbb{H}}$. With respect to the hyperbolic metric on ${\mathbb{H}}$, the set ${Gz}$ has no limit points: either in ${\mathbb{H}}$, since the action is discrete, or in ${\partial\mathbb{H}}$, since every point in ${\mathbb{H}}$ is infinitely far from the boundary. However, considering the Euclidean distance ${\mathbb{H}}$ inherits by being embedded as the Poincaré disk in ${\mathbb{R}^2}$, ${Gz}$ may have limit points in ${\partial\mathbb{H}}$. Basic hyperbolic geometry implies that:

1. the limit points occur only in ${\partial\mathbb{H}}$,
2. for any other ${z' \in \mathbb{H}}$, the sets of limit points of ${Gz'}$ and ${Gz}$ are identical, and
3. the cardinality of the set of limit points is 0, 1, 2 or uncountable.

A fuchsian group is non-elementary when the set of limit points is uncountable. Such a group contains ${F_2}$.

The non-elementary fuchsian groups are really the only interesting ones: the elementary ones are either finite or virtually cyclic.

Let’s recap the proof so far. We proceed by induction, and assume the theorem true for ${\mathrm{dim}(X) = n-1}$. For ${X}$ of dimension ${n}$, since ${G}$ is infinite and acts properly on ${X}$, it has no global fixed points. Hence we apply Sageev’s theorem and conclude that there is a hyperplane ${J \subset X}$ s.t. ${e(G, \mathrm{stab}(J)) > 1}$. Let ${H = \mathrm{stab}(J)}$. Then ${H}$ acts on ${J}$ properly. By induction, either ${H}$ has an ${F_2}$ subgroup, and then so does ${G}$ so we’re done, or ${H}$ is virtually f.g. abelian. In this case, we apply the Algebraic Torus Theorem to ${G}$ and ${H}$, and conclude that (1) ${G}$ is virtually polycyclic, or (2) it has a non-elementary fuchsian quotient, or (3) it splits over a virtually polycyclic group. We deal with these possibilities one by one.

If ${G}$ is virtually polycyclic, then we can apply the following two results of Bridson to conclude that it’s in fact virtually abelian.

Lemma 5.

1. If ${G \curvearrowright X}$, a ${\mathrm{CAT}(0)}$ complex with finitely many shapes, then ${G}$ acts semi-simply with a discrete set of translation lengths.
2. If ${G}$ acts semi-simply and properly on a ${\mathrm{CAT}(0)}$ space, then any virtually solvable subgroup ${H < G}$ is virtually abelian.

(The number of shapes of a metric cell complex is the number distinct isometry classes of cells. In our case, ${X}$ has finitely many shapes since every cube of a given dimension is isometric and ${X}$ is has finite dimension. An action is semi-simple if there are points in the space realizing the translation lengths of each element. I.e., if for every $g \in G$ there’s a point realizing $\mathrm{inf}_{x \in X}d(x, gx)$).

If ${G}$ has a non-elementary fuchsian quotient, which must contain an ${F_2}$, then ${G}$ itself must contain an ${F_2}$.

Finally, suppose ${G}$ splits over a virtually polycyclic group ${P}$. Then by the first part of Bridson’s lemma above, ${G}$ acts semi-simply, and by the second part, ${P}$ must be virtually abelian. We analyze amalgamated products and HNN-extensions separately.

Suppose ${G = A \ast_P B}$. If ${[A:P] > 2}$ or ${[B:P]>2}$, then ${G}$ contains an ${F_2}$, by the normal form theorem for amalgamated products. If ${[A:P] = [B:P] = 2}$, then the Bass-Serre tree, on which ${G}$ acts, is homeomorphic to a line in which every edge is stabilized by a conjugate of ${P}$. Thus there’s an index 2 subgroup ${G' < G}$ acting by translations and ${\mathrm{ker}(G' \rightarrow \mathbb{Z}) = P}$. Hence ${G' \cong P \rtimes \mathbb{Z}}$, and so ${G}$ is virtually polycyclic and hence virtually abelian by Bridson’s lemma. This takes care of the amalgamated product case. To analyze an HNN-extension by ${P}$, as well as to prove this theorem for non-finitely generated ${G}$, another theorem is required.

Theorem 6 (Bridson-Haefliger). Suppose ${G \curvearrowright X}$ by semi-simple isometries and ${X}$ is ${\mathrm{CAT}(0)}$. Suppose also that

1. there’s a bound on the dimension of isometrically embedded flats,
2. the set of translation lengths of ${G}$ is discrete at 0, and
3. there is a bound on the order of finite subgroups.

Then any ascending sequence of virtually abelian subgroups stabilizes.

Suppose ${G = C \ast_P}$, with ${i_1}$ and ${i_2}$ being the two injections of ${P}$ into ${C}$. If ${[C:i_1(P)] > 1}$ and ${[C:i_2(P)] > 1}$, then by the normal form theorem for HNN-extensions, ${G}$ has an ${F_2}$ subgroup. If ${[C:i_1(P)] = 1}$, then ${[C:i_2(P)] = 1}$. [Otherwise, if ${i_2(P) = ti_1(P)t^{-1} < i_1(P)}$, consider the following subgroups of ${G}$:

$\displaystyle i_1(P) < t^{-1}i_1(P)t < t^{-2}i_1(P)t^2 < \cdots$

By the Bridson-Haefliger theorem above, this sequence must stabilize, hence ${i_1(P) = ti_1(P)t^{-1} = i_2(P)}$.] Then ${G \cong P \rtimes \mathbb{Z}}$, and by Bridson’s lemma, ${G}$ is virtually abelian.

This finishes the proof for finitely generated ${G}$. If ${G}$ is not finitely generated, it contains an ascending, non-stabilizing sequence of finitely generated subgroups

$\displaystyle G_1 < G_2 < G_3 < \cdots$

If any ${G_i}$ contains an ${F_2}$, we’re done. If not, then by the theorem applied to finitely generated groups, each is virtually abelian. Then conditions of the Bridson-Haefliger theorem above are satisfied, however, so a non-stabilizing ascending sequence is impossible. Hence if ${G}$ is not finitely generated, it contains an ${F_2}$.