In our previous note Amalgamated products and HNN extensions (IV): Markov properties, we saw that for every and for every finitely-presented group
, there exists a
-dimensional closed manifold whose fundamental group is
. However, such a result does not hold for
; for the case
, see for example the note On subgroups of surface groups. The present note is devoted to the 3-dimensional case
.
In particular, we will prove that the groups for
,
for
, and
for
are not 3-manifold groups.
Definitions: A handlebody of genus is an orientable 3-manifold (with boundary) obtained by attaching
“handles” to a 3-ball. Roughly speaking, it is a full closed surface of genus
.
A way to construct 3-manifolds (without boundary) is to glue two handlebodies and
of some genus
along their boundaries by a homeomorphism
. We say that the triple
is a Heegaard splitting of genus
of the resulting manifold.
Theorem 1: Every closed 3-manifold admits a Heegaard splitting.
Sketch of proof. Let be a closed 3-manifold, which will be thought of as a three-dimensional simplicial complex. Let
denote a small neighborhood of the 1-skeleton
of
and
its complement. If
is a 3-simplex, then
and
are illustrated by the following picture:
Notice that, if is the graph whose vertices are the centers of the 3-simplices of
and whose edges link two vertices corresponding to adjacent 3-simplices, then
may be thought of as a neighborhood of
.
For , let
denote a maximal subtree of
. Then,
may be thought of as a neighborhood of
, which is homeomorphic to a 3-ball, with one handle for each edge of the complement
. Therefore,
is a handlebody of genus
, since, if
and
denotes respectively the number of vertices and edges of a graph,
and
imply that
.
Therefore, we have proved that is the union of two handlebodies which intersect along their boundaries. To conclude, we need to argue that
and
have the same genus, or equivalently, that
.
Let denote respectively the number of vertices, edges, faces and 3-cells of
. By construction,
and
.
However, by Poincaré’s duality, , hence
.
The proof is complete.
From a Heegaard splitting of genus
, it is possible to find a presentation of the fundamental group of the associated 3-manifold
:
We apply van Kampen’s theorem to the decomposition . The handlebodies
and
have a free group of rank
as fundamental groups, and they meet in
along a closed surface
of genus
. Let
be a generating set of
. Then,
,
where (resp.
) is a generating set of
(resp.
).
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 is aspherical whenever
for all
; according to Whitehead’s theorem, it amounts to require the universal covering
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
, a
space or a classifying space for
is an aspherical space whose fundamental group is isomorphic to
; according to our previous remark, such a space is uniquely determined by
, up to homotopy. In particular, this allows us to define the
-th (co)homology group of
as the
-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 has a classifying space. First, to a given presentation
of
we associate a two-dimensional CW-complex
as follow: Define
as a bouquet of
circles, each labelled by a generator of
. Then, for every relation
, glue a
-cell along the path labelling by
. Using van Kampen’s theorem, it is not difficult to notice that the fundamental group of our complex
is isomorphic to
. However, it may not be aspherical. Now, let us define an increasing sequence of CW-complexes
such that
and
for all
(such a sequence will define a Postnikov tower). The spaces
are defined inductively in the following way: Let
be a generating set for
; then, define
by gluing
-cells to
using the
. By construction,
, and we claim that the embedding
is
-injective for all
by cellular approximation. Finally, let
. By construction,
is aspherical, and because
have been constructed just by gluing
-cells to
with
, we deduce that
(since the fundamental group depends only on the
-skeleton of a CW-complex). Therefore,
is a classifying space of
.
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 be a presentation of a group
with
and
. Let also
denote
. Then
.
Proof. Let be the CW-complex associated to the given presentation. According to what we have said above, we may add
-cells to
, with
, in order to “kill” higher homotopy groups, ie., our new CW-complex
is aspherical. Moreover, because we did not modify the
-squeleton of
,
is isomorphic to
. Therefore,
is a
, hence
.
Notice that , hence
.
On the other hand, we clearly have an epimorphism , therefore:
Corollary 4: If a group admits a balanced presentation, then
.
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 admits a balanced presentation if and only if
has more vertices than edges.
Sketch of proof. To the canonical presentation of is associated a cube complex
, called Salvetti complex:
has only one point, one edge for each generator of
, and
generators span an
-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
is a classifying space of
. Moreover, it is not difficult to prove that the
-th homology group of
is free on the set of
-cubes. Therefore,
corresponds to the number of vertices of
, and
to its number of edges.
Thus, if admits a balanced presentation, according to Theorem 3,
must have more vertices than edges. The converse is obvious.
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 . Then
admits a balanced presentation if and only if
.
Proof. is the right-angled Artin group associated to the complete graph
. Because
has
vertices and
edges, it is sufficient to apply Proposition 5 to conclude.
Corollary 7: Let . Then
admits a balanced presentation if and only if
and
or
and
.
Proof. is the right-angled Artin group associated to the bipartite complete graph
. Because
has
vertices and
edges, it is sufficient to apply Proposition 5 to conclude.
Corollary 8: Let . Then
admits a balanced presentation if and only if
and
or
and
.
Proof. Let denote the graph with
vertices and no edges. Then
is the right-angled Artin group associated to the join
. Because
has
vertices and
edges (here, we suppose
), it is sufficient to apply Proposition 5 to conclude.