Morse Theory, finiteness properties, and Bieri–Stallings groups

In this post we shall discuss finiteness properties of groups and an important class of examples known as the Bieri–Stallings groups. Our main tool will be combinatorial Morse theory.

Eilenberg–MacLane spaces and type $F_n$

In order to define our finiteness properties, we must first define an Eilenberg–MacLane space.

Definition. An Eilenberg–MacLane space for a group $G$ , also known as a $K(G,1)$ , is a connected CW–complex $X$ such that $\pi_1(X)=G$ and the universal cover of $X$ is contractible.

We can construct an Eilenberg–MacLane space for a group $G$ using its presentation ${2}$ –complex $X$ . This is an inductive procedure. In the first step, we attach $3$ –cells to render the second homotopy group trivial. At stage $n$ , we attach ${(n+2)}$ –cells to make the complex ${(n+1)}$ –aspherical. At the end of this procedure we obtain a complex $Y$ which is aspherical with $\pi_1(Y)=G$ and $\pi_n(Y)=0$ for all $n>1$ . Since all the higher homotopy groups are trivial, the universal cover of $Y$ is contractible.

A group is said to be of type $F_n$ if it has an Eilenberg–MacLane space with a finite $n$ –skeleton.

These properties generalize the notions of being finitely generated and of being finitely presented. Any group is of type $F_0$ , since it has an Eilenberg–MacLane space with exactly one vertex. Using our construction of the Eilenberg–MacLane space from the presentation $2$ -complex, one sees that being of type $F_1$ is equivalent to being finitely generated, and being of type $F_2$ is equivalent to being finitely presented.

The Bieri–Stallings groups

Consider the map

$\Phi_n:F(a_1,b_1)\times...\times F(a_n, b_n)\rightarrow \mathbb{Z}$

from the $n$ -fold direct product of rank–two free groups to the integers, defined by mapping all the $a_i$ and $b_i$ to $1$ . The Bieri–Stallings groups $G_n$ are the kernels of the maps $\Phi_n$ .

The Bieri–Stallings groups display a range of finiteness properties.

Theorem. $G_n$ is of type $F_{n-1}$ but not of type $F_n$ .

In our next post we shall discuss a proof of this result in its full generality that uses Brown’s Criterion. In the remainder of this post we shall prove the theorem for two particular cases, namely $G_1$ and $G_2$ , via Morse Theory — a topic we will now introduce.

Combinatorial Morse Theory

An affine cell complex $X$ is a cell complex where, for a fixed $m\geq dim(X)$ , each cell $e$ is equipped with characteristic maps $\chi_{e}:C_e\rightarrow e$ such that $C_e\subset \mathbb{R}^{m}$ is a convex polyhedron and the characteristic map is a continuous function. Additionally $\chi_e$ restricted to any face of $C_e$ is the characteristic function of a cell in $X$ .

A Morse function is a map $f$ from an affine cell complex $X$ to the real numbers with the following properties. The image of the $0$ -skeleton of $X$ under the map is a discrete subset of $\mathbb{R}$ . If the restriction of $f$ on any cell is a constant function then the cell is $0$ -dimensional. And finally, for every cell $e \in X$ , the map $f\chi_e:C_e\rightarrow \mathbb{R}$ can be extended to an affine map from $\mathbb{R}^m$ to $\mathbb{R}$ . Note that the last property implies that the maxima of $f$ restricted to any cell $e$ is achieved at a vertex.

The link of a vertex $v$ is the set of tangent vectors to the space at $v$ which point into cells that contain $v$ . The descending link of a vertex is the subset of the link which consists of tangent vectors that point to cells which have the property that $f$ , when restricted to them, achieves its maximum value at $v$ . The ascending link of a vertex is defined similarly with the word maximum replaced by minimum in the previous sentence.

Now we are ready to state Morse’s Lemma, which will play a central role in much of our reasoning that follows.

Morse’s Lemma. Given closed bounded intervals $J,J^{\prime} \subset \mathbb{R}$ such that $\textup{inf}\{J\}=\textup{inf}\{J^{\prime}\}$ and $J\subsetneq J^{\prime}$ , the only changes in topology that take place (up to homotopy equivalence) while extending the preimage from $f^{-1}(J)$ to $f^{-1}(J^{\prime})$ are the coning off of descending links of vertices in $X$ that are contained in $f^{-1}(J^{\prime}\setminus J)$ . [Note: we define the cone of an empty set to be a vertex, to make room for the special case that some vertices in $X$ may have trivial descending links.]

In our application of Morse’s Lemma below, our Morse function $f$ will be symmetric in the sense that $-f$ will also be a Morse function. So Morse’s Lemma will apply for extending our pre-image downwards with the phrase “descending links” replaced by “ascending links”.

Finiteness properties of $G_1$ and $G_2$ via Morse Theory

This account is based on that of N. Brady in Section 2.4 of Dehn Functions and Non-Positive Curvature, CRM Notes.

First we claim $G_1$ is not finitely generated. Consider a map $m:S^1\vee S^1\rightarrow S^1$ , which takes each circle in the wedge sum homeomorphically to the circle. On the fundamental groups, this induces our map $\Phi_1:F(a,b)\rightarrow \mathbb{Z}$ , the kernel of which is $G_1$ . The lift $f:T\rightarrow \mathbb{R}$ of $m$ to the universal covers, where $T$ is the infinite $4$ -valent tree, is illustrated below and is a Morse function.

Note that $F(a,b)$ acts on $T$ by deck transformations so that for any vertex $x \in T$ , $f(ax) = f(bx) = f(x) + 1$ and $f(a^{-1}x) = f(b^{-1}x) = f(x) - 1$ .

An analysis of $f^{-1}(0)$ will show that $G_1$ is not finitely generated. Note that $f^{-1}(0)$ is a set of vertices of $T$ on which $G_1$ acts freely — indeed, there is a bijective correspondence between $f^{-1}(0)$ and $G_1$ since there are no loops in $T$ and since there is a unique path in $T$ between any two vertices whose edges determine an element of $G_1$ . So $G_1$ would be finitely generated if and only if we could attach finitely many families of $G_1$ –equivariant $1$ -cells to $f^{-1}(0)$ to render it connected.

Suppose this were possible. Then it would be done within the preimage $f^{-1}(I)$ of some bounded interval $I\subset \mathbb{R}$ with integer endpoints since $G_1$ acts horizontally on the tree, and hence on each of the families of $1$ cells mentioned above. Now any two points in $f^{-1}(0)$ can be joined by a path in $f^{-1}(I)$ , and the descending link of each vertex is non empty. So, by Morse’s Lemma, no new path–components are introduced when the pre-image $f^{-1}(0)$ is extended to $f^{-1}(I)$ . So $f^{-1}(I)$ is path connected. But this is false and so a contradiction: at any vertex in $f^{-1}(\textup{sup}(I)+1)$ , the descending link is a $0$ -sphere whose two points are joined by a path within $f^{-1}(I)$ , and this would lead to there being a loop in the tree $T$ .

Next we will show that $G_2$ is finitely generated but not finitely presented. This time we consider a map $l:(S^1\vee S^1) \times (S^1\vee S^1)\rightarrow S^1$ where $l(x,y) = m(x) + m(y)$ and $m : S^1 \vee S^1 \rightarrow S^1$ was defined previously. Lift $l$ to a Morse function $f:T\times T\rightarrow \mathbb R$ on the universal covers, analogous to that in the previous example. Given a point in $T\times T$ the height changes as a sum of the distances traveled along both the individual wedge sums of circles with the prescribed orientation.

We write $\pi_1((S^1\vee S^1) \times (S^1\vee S^1))=F(a,b)\times F(c,d)$ where $a,b,c,d$ represent the paths around each of the four $S^1$ .

We claim that $f^{-1}(0)$ is path–connected. Let $v$ be the lone vertex of $S^1$ . The group $G_2$ acts horizontally and freely on $f^{-1}(0)$ with a compact quotient homeomorphic to the inverse image $l^{-1}(v)$ . But the inverse image $l^{-1}(v)$ is a wedge–sum of four circles, specifically the diagonals $(a(t), c^{-1}(t)), (a(t),d^{-1}(t)), (b(t),c^{-1}(t)), (b(t),d^{-1}(t))$ of the 2–cells arising from the four $S^1 \times S^1$ .

So, fixing a vertex $x$ in $f^{-1}(0)$ , it is enough to show that for every ${g \in G_2}$ there is a path in $f^{-1}(0)$ from $x$ to $gx$ . Let $a,b$ be the two generators on one of the free groups of our product of free groups, and let $c$ be a generator of a different free group in the product. Consider the word $ab^{-1}$ representing some $g \in G_2$ . Well, $ab^{-1}$ freely equals $ac^{-1}cb^{-1}$ and one can travel first along $(a(t),c^{-1}(t))$ and then along $(c(t),b^{-1}(t))$ to reach $gx$ whist staying in $f^{-1}(0)$ throughout. The reader should convince themselves that such an arrangement is possible for any word representing an element of $G_2$ .

As $f^{-1}(0)$ is path connected,

$1\rightarrow \pi_1(f^{-1}(0))\rightarrow \pi_1(l^{-1}(v))\rightarrow G_2\rightarrow 1$

is a short exact sequence. So $G_2$ is finitely generated since $\pi_1(l^{-1}(v)) = F_4$ .

Finally suppose, for a contradiction, $\bf G_2$ is finitely presented. As we saw above, the quotient of $f^{-1}(0)$ by $G_2$ is a wedge–sum of four circles that correspond to a generating set for $G_2$ . Attaching finitely many $2$ -cells to this $1$ -complex gives the presentation ${2}$ –complex ${X}$ of $G_2$ . Now, $f^{-1}(0)$ is the Cayley graph of $G_2$ with generators being the lifts of the $1$ -cells on the presentation 2-complex.

As $T \times T$ is contractible, we can lift the $2$ -cells in ${X}$ to $T\times T$ to obtain a finite set of $G_2$ -equivariant families of $2$ -cells which when pasted on to the $1$ -complex $f^{-1}(0)$ , make it simply connected for reasons similarly associated to the short exact sequence used above. In more detail, since $f^{-1}(0)$ is path connected, and since $G_2$ is the deck group of the covering map $f^{-1}(0)\rightarrow X$ , from the short exact sequence

$1\rightarrow \pi_1(f^{-1}(0))\rightarrow \pi_1(X)\rightarrow G_2\rightarrow 1$

we get that $\pi_1(f^{-1}(0))$ is trivial since $\pi_1(X)\cong G_2$ .

The finite set of $G_2$ -equivariant families of $2$ -cells pasted on to our level set are in the preimage of some interval $J\subset \mathbb{R}$ with integer endpoints that contains $0$ . Since loops in $f^{-1}(0)$ are null-homotopic in $f^{-1}(J)$ , by Morse’s lemma $f^{-1}(J)$ must itself be simply connected. But, as we will explain, this is a contradiction. Consider the descending link of any vertex of height $\textup{sup}(J)+1$ . This is a $1$ -sphere, which must be nullhomotopic inside $f^{-1}(J)$ . But this gives us a nontrivial $2$ -cycle inside the contractible $2$ -complex $T\times T$ .