In our previous note Amalgamated products and HNN extensions (IV): Markov properties, we saw that for every $n \geq 4$ and for every finitely-presented group $G$, there exists a $n$-dimensional closed manifold whose fundamental group is $G$. However, such a result does not hold for $n\leq 3$; for the case $n=2$, see for example the note On subgroups of surface groups. The present note is devoted to the 3-dimensional case $n=3$.

In particular, we will prove that the groups $\mathbb{Z}^n$ for $n \geq 4$, $\mathbb{F}_m \times \mathbb{F}_n$ for $m \geq 3, \ n \geq 2$, and $\mathbb{F}_m \times \mathbb{Z}^n$ for $n ,m \geq 2$ are not 3-manifold groups.

Definitions: A handlebody of genus $g$ is an orientable 3-manifold (with boundary) obtained by attaching $g$ “handles” to a 3-ball. Roughly speaking, it is a full closed surface of genus $g$.

A way to construct 3-manifolds (without boundary) is to glue two handlebodies $H_1$ and $H_2$ of some genus $g$ along their boundaries by a homeomorphism $h : \partial H_1 \to \partial H_2$. We say that the triple $(H_1,H_2,h)$ is a Heegaard splitting of genus $g$ of the resulting manifold.

Theorem 1: Every closed 3-manifold admits a Heegaard splitting.

Sketch of proof. Let $M$ be a closed 3-manifold, which will be thought of as a three-dimensional simplicial complex. Let $M_1$ denote a small neighborhood of the 1-skeleton $\Gamma_1$ of $M$ and $M_2 = M \backslash M_1$ its complement. If $\Delta$ is a 3-simplex, then $M_1 \cap \Delta$ and $M_2 \cap \Delta$ are illustrated by the following picture:

Notice that, if $\Gamma_2 \subset M$ is the graph whose vertices are the centers of the 3-simplices of $M$ and whose edges link two vertices corresponding to adjacent 3-simplices, then $M_2$ may be thought of as a neighborhood of $\Gamma_2$.

For $i=1,2$, let $T_i$ denote a maximal subtree of $\Gamma_i$. Then, $M_i$ may be thought of as a neighborhood of $T_i$, which is homeomorphic to a 3-ball, with one handle for each edge of the complement $\Gamma_i \backslash T_i$. Therefore, $M_i$ is a handlebody of genus $1- \chi(\Gamma_i)$, since, if $v( \cdot)$ and $e(\cdot)$ denotes respectively the number of vertices and edges of a graph,

$\chi( \Gamma_i)= v(\Gamma_i)- e(\Gamma_i)$

and

$1= \chi(T_i)=v(T_i)-e(T_i)=v(\Gamma_i)-e(T_i)$

imply that

$e(\Gamma_i) - e(T_i) = 1- \chi(\Gamma_i)$.

Therefore, we have proved that $M$ is the union of two handlebodies which intersect along their boundaries. To conclude, we need to argue that $M_1$ and $M_2$ have the same genus, or equivalently, that $\chi(\Gamma_1)= \chi (\Gamma_2)$.

Let $v(M), \ e(M), \ f(M), \ c(M)$ denote respectively the number of vertices, edges, faces and 3-cells of $M$. By construction,

$\chi (\Gamma_1)= v( \Gamma_1)- e(\Gamma_1) = v( M) - e(M)$

and

$\chi( \Gamma_2)= v( \Gamma_2) - e( \Gamma_2) = c(M) - f(M)$.

However, by Poincaré’s duality, $\chi(M)=0$, hence

$\chi ( \Gamma_1) - \chi( \Gamma_2)= v(M)-e(M) + f(M)- c(M) = \chi (M)=0$.

The proof is complete. $\square$

From a Heegaard splitting $(H_1,H_2,h)$ of genus $g$, it is possible to find a presentation of the fundamental group of the associated 3-manifold $M$:

We apply van Kampen’s theorem to the decomposition $M = H_1 \cup H_2$. The handlebodies $H_1$ and $H_2$ have a free group of rank $g$ as fundamental groups, and they meet in $M$ along a closed surface $\Sigma$ of genus $g$. Let $\{ c_1, \ldots, c_{2g} \}$ be a generating set of $\pi_1(\Sigma) \leq \pi_1(H_1)$. Then,

$\pi_1(M)= \langle a_1,\ldots, a_g, b_1, \ldots, b_g \mid h_*(c_i)=c_i, \ 1 \leq i \leq 2g \rangle$,

where $\{ a_1, \ldots, a_g \}$ (resp. $\{ b_1, \ldots, b_g \}$) is a generating set of $\pi_1(H_1)$ (resp. $\pi_1(H_2)$).

It is worth noting that such a presentation has the same number of generators and relations; we say that the presentation is balanced. Therefore, we have proved:

Corollary 2: Every 3-manifold group admits a balanced presentation.

So, a natural question is: which groups admit a balanced presentation? In fact, this question is linked to (co)homology of groups, which we define now.

A space $X$ is aspherical whenever $\pi_n(X)= 0$ for all $n \geq 2$; according to Whitehead’s theorem, it amounts to require the universal covering $\tilde{X}$ to be contractible. An interesting property of such spaces is that they are uniquely determined, up to homotopy, by their fundamental groups, that is to say, two aspherical spaces are homotopy equivalent if and only if their fundamental groups are isomorphic. Given a group $G$, a $K(G,1)$ space or a classifying space for $G$ is an aspherical space whose fundamental group is isomorphic to $G$; according to our previous remark, such a space is uniquely determined by $G$, up to homotopy. In particular, this allows us to define the $n$-th (co)homology group of $G$ as the $n$-th (co)homology group of such a classifying space.

In order to justify that (co)homology groups are always defined, we need to prove that every group $G$ has a classifying space. First, to a given presentation $\langle X \mid \mathcal{R} \rangle$ of $G$ we associate a two-dimensional CW-complex $Y$ as follow: Define $Y^{(1)}$ as a bouquet of $| X |$ circles, each labelled by a generator of $X$. Then, for every relation $w \in \mathcal{R}$, glue a $2$-cell along the path labelling by $w$. Using van Kampen’s theorem, it is not difficult to notice that the fundamental group of our complex $Y$ is isomorphic to $G$. However, it may not be aspherical. Now, let us define an increasing sequence of CW-complexes $Y= Y_1 \subset Y_2 \subset \cdots$ such that $\pi_1(Y_n) \simeq G$ and $\pi_k(Y_n)= \{ 1 \}$ for all $k \leq n$ (such a sequence will define a Postnikov tower). The spaces $(Y_{n})$ are defined inductively in the following way: Let $(\varphi_{\alpha} : \mathbb{S}^n \to Y_{n-1})$ be a generating set for $\pi_n(Y_{n-1})$; then, define $Y_n$ by gluing $n$-cells to $Y_{n-1}$ using the $\varphi_{\alpha}$. By construction, $\pi_n(Y_n) = \{1 \}$, and we claim that the embedding $Y_{n-1} \hookrightarrow Y_n$ is $\pi_i$-injective for all $i< n$ by cellular approximation. Finally, let $Y_{\infty}= \bigcup\limits_{n \geq 1} Y_n$. By construction, $Y_{\infty}$ is aspherical, and because $Y_{\infty}$ have been constructed just by gluing $n$-cells to $Y$ with $n \geq 3$, we deduce that $\pi_1(Y_{\infty}) \simeq \pi_1(Y) \simeq G$ (since the fundamental group depends only on the $2$-skeleton of a CW-complex). Therefore, $Y_{\infty}$ is a classifying space of $G$.

For more information on Postnikov towers, see section 4.3 of Hatcher’s book, Algebraic Topology.

Now, we are ready to linked the deficiency of a presentation, that is the number of its generators minus the number of its relations, with the first and second homology groups of the underlying group. The result below is mentioned by Brown in the fifth exercice of chapter II.5 of his book Cohomology of groups. Here is the topological approach he suggests:

Theorem 3: Let $\langle X \mid R \rangle$ be a presentation of a group $G$ with $|X|=n$ and $|R|=m$. Let also $r$ denote $\mathrm{rk}_{\mathbb{Z}} G^{ab}$. Then $\mathrm{rk}_{\mathbb{Z}} H_2(G) \leq r-n+m$.

Proof. Let $Y$ be the CW-complex associated to the given presentation. According to what we have said above, we may add $n$-cells to $Y$, with $n \geq 3$, in order to “kill” higher homotopy groups, ie., our new CW-complex $Y_{\infty}$ is aspherical. Moreover, because we did not modify the $2$-squeleton of $Y$, $\pi_1(Y)$ is isomorphic to $\pi_1( Y_{\infty})$. Therefore, $Y_{\infty}$ is a $K(G,1)$, hence $H_{\ast}(Y_{\infty}) \simeq H_{\ast} (G)$.

Notice that $1-n+m = \chi(Y)= 1- \mathrm{rk}_{\mathbb{Z}} H_1(Y) + \mathrm{rk}_{\mathbb{Z}} H_2(Y)$, hence

$\mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m$.

On the other hand, we clearly have an epimorphism $H_2(Y) \twoheadrightarrow H_2(Y_{\infty})$, therefore:

$\mathrm{rk}_{\mathbb{Z}} H_2(G)= \mathrm{rk}_{\mathbb{Z}} H_2( Y_{\infty}) \leq \mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m.$ $\square$

Corollary 4: If a group $G$ admits a balanced presentation, then $\mathrm{rk}_{\mathbb{Z}} H_2(G) \leq \mathrm{rk}_{\mathbb{Z}} G^{ab}$.

Thus, now we have a necessary criterion to determine whether or not a group admits a balanced presentation. The Proposition below applies this criterion to the right-angled Artin groups (we defined them in our previous note Some SQ-universal groups).

Proposition 5: A right-angled Artin group $A(\Gamma)$ admits a balanced presentation if and only if $\Gamma$ has more vertices than edges.

Sketch of proof. To the canonical presentation of $A(\Gamma)$ is associated a cube complex $Y$, called Salvetti complex: $Y$ has only one point, one edge for each generator of $A(\Gamma)$, and $n+1$ generators span an $n$-cube if and only if they pairwise commute. Now, it can be proved that the Salvetti complex is nonpositively curved so that its universal cover is CAT(0); because a CAT(0) space is contractible, we deduce that $Y$ is a classifying space of $A(\Gamma)$. Moreover, it is not difficult to prove that the $n$-th homology group of $Y$ is free on the set of $n$-cubes. Therefore, $\mathrm{rk}_{\mathbb{Z}} H_1(A(\Gamma))$ corresponds to the number of vertices of $\Gamma$, and $\mathrm{rk}_{\mathbb{Z}} H_2(A(\Gamma))$ to its number of edges.

Thus, if $A(\Gamma)$ admits a balanced presentation, according to Theorem 3, $\Gamma$ must have more vertices than edges. The converse is obvious. $\square$

For more information on CAT(0) geometry, see Bridson and Haefliger’s book, Metric spaces of nonpositive curvature.

Now, we are ready to prove that the groups mentionned at the beginning of this note do not admit a balanced presentation, and thus are not the fundamental groups of closed 3-manifolds.

Corollary 6: Let $n \geq 1$. Then $\mathbb{Z}^n$ admits a balanced presentation if and only if $n \leq 3$.

Proof. $\mathbb{Z}^n$ is the right-angled Artin group associated to the complete graph $K_n$. Because $K_n$ has $n$ vertices and $(n-1)!$ edges, it is sufficient to apply Proposition 5 to conclude. $\square$

Corollary 7: Let $m,n \geq 1$. Then $\mathbb{F}_n \times \mathbb{F}_m$ admits a balanced presentation if and only if $m \leq 2$ and $n=1$ or $m=1$ and $n \leq 2$.

Proof. $\mathbb{F}_n \times \mathbb{F}_m$ is the right-angled Artin group associated to the bipartite complete graph $K_{m,n}$. Because $K_{m,n}$ has $m+n$ vertices and $mn$ edges, it is sufficient to apply Proposition 5 to conclude. $\square$

Corollary 8: Let $m,n \geq 1$. Then $\mathbb{F}_m \times \mathbb{Z}^n$ admits a balanced presentation if and only if $m=1$ and $n \leq 2$ or $n \geq 2$ and $m \geq 2$.

Proof. Let $D_m$ denote the graph with $m$ vertices and no edges. Then $\mathbb{F}_m \times \mathbb{Z}^n$ is the right-angled Artin group associated to the join $K_n \ast D_m$. Because $D_n \ast D_m$ has $n+m$ vertices and $mn+ (n-1)!$ edges (here, we suppose $0!=0$), it is sufficient to apply Proposition 5 to conclude. $\square$