Recall that a group $G$ is of type F if there exists a compact $K(G,1)$.

There are many examples of groups which are not of type F but which are virtually of type F, that is, they have finite-index subgroups of type F. For example, $SL(n,\mathbb{Z})$, mapping class groups, automorphism groups of free groups, etc.

In all the examples I am aware of, the problem comes down to torsion (a group of type F must be torsion-free), and the required finite-index subgroup is just one that has no torsion.

Question: Does anyone know of a torsion-free group that is not of type F but is virtually of type F?


It is conjectured that no such example exists. It is conjectured that $G$ is of type $F$, but it is hard to prove a group is type $F$ without explicitly exhibiting a classifying space. The class of groups of type $FP$ is a well-behaved proxy. Moreover, it is conjectured that every finitely presented group of type $FP$ is actually of type $F$. It is known that a finitely presented group of type $FP$ has a classifying space which is "finitely dominated" meaning that it is a retract in the homotopy category of a finite complex, which is just as good for most homotopical purposes. I will prove that if a torsion-free group $G$ has a finite index subgroup $H$ of type $FP$, then $G$ is itself of type $FP$. Applying this to your situation, since $H$ is finitely presented, so is $G$, so the conjecture implies that $G$ is of type $F$. It is possible that the hypothesis that $H$ is of type $F$ allow an unconditional proof, but I do not expect it.

$G$ has type $FP$ if the trivial $G$-module $\mathbb Z$ has a finite length resolution by finitely generated projective $G$-modules. Type $F$ is equivalent to a finite length resolution by finitely generated free $G$-modules, plus finite presentation. The advantage of type $FP$ is that it is the intersection of two properties that can be treated separately, namely finite dimension and finite generation of cohomology $FP_\infty$. These can be interpreted as the existence of two resolutions, one of finite length and other by finite modules. The two properties can be combined to produce a resolution that is simultaneously finitely generated and finite length, but only if one allows projective modules and not just free modules, so type $F$ does not decompose this way.

The easy part is finite generation of homology. If $G$ contains a finite index subgroup $H$, then the intersection of the conjugates of $H$ is a finite index normal subgroup $N$. The Hochschild-Serre spectral sequence for the group extension $N\to G\to G/N$ allows us to combine the $FP_\infty$ properties of $N$ and $G/N$ to conclude that $G$ is $FP_\infty$.

The other part is finite homological dimension. By assumption $H$ has a finite dimensional classifying space with universal cover $EH$. Consider the space of $H$-equivariant maps from $G$ into $EH$. That is, intertwining the left action of $H$ on $G$ and the only action of $H$ on $EH$. This leaves the right action of $G$ on $G$, which makes this a $G$-space, a candidate for $EG$. For the quotient to be a model of $BG$, we need the space to be contractible and the action to be free. This space is isomorphic to $EH^{[G:H]}$, hence contractible and finite dimensional. The size of the isotropy is bounded by the index. So the isotropy is a finite subgroup of the torsion-free group $G$, so actually $G$ acts freely on this space. This shows that $G$ has a finite dimensional classifying space. In fact, the dimension of $G$ is the same as the dimension of $H$, although that does not come out of this construction.

Reference: Ken Brown's book, Chapter VIII or his notes.

  • $\begingroup$ Bestvina and Brady disproved the conjecture $F=FP$ (Inventiones Math. 1997). Their examples are certain normal subgroups of 2-dimensional RAAGs: The subgroups they construct are $FP_2$ (and, hence, $FP$ since the groups are 2-dimensional) but are not finitely-presented. $\endgroup$ – Misha Aug 23 '14 at 4:28
  • $\begingroup$ Still, in Sarah's question finite presentability is not an issue, and thus from FP plus finite presentability we can deduce F, answering negatively the question (torsion-free + virtually F implies F). $\endgroup$ – YCor Aug 23 '14 at 12:56
  • $\begingroup$ I edited to include finite presentation. As YCor says, the BB examples cannot be used for this question and the current conjecture FP+fp=F would apply. $\endgroup$ – Ben Wieland Aug 23 '14 at 16:14
  • 1
    $\begingroup$ @Misha - I don't think the Bestvina--Brady examples come from 2-dimensional RAAGs. Indeed, they come from RAAGs defined by the 1-skeleton of a flag triangulation of a homology sphere. $\endgroup$ – HJRW Aug 24 '14 at 19:36
  • $\begingroup$ I asserted some conjectures, but I didn't quite isolate them. The obstruction for a finitely presented finite dimensional $FP_\infty$ group to have a finite classifying space is for the $G$-module $\mathbb Z$ to be trivial in the reduced $K$-group $\bar{K}_0(\mathbb ZG)$. The $K$-theory generalization of one of the Kaplansky conjectures says that this $K$-group is zero (even if the group is just torsion-free). I don't know if this conjecture has a name, other than the $K$-theory (variant) Kaplansky conjecture. This is subsumed by the monster Farrell-Jones conjecture. $\endgroup$ – Ben Wieland Mar 22 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.