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A group $G$ is of type $FP_2$ if it admits a partial projective resolution of $\mathbb{Z}G$-modules $$ P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow \mathbb{Z} \rightarrow 0$$ with each $P_i$ being finitely generated and projective.

In the event $G$ acts freely and cocompactly on a space $X$ with trivial first homology, then $G$ is of type $FP_2$. The converse holds if $G$ has cohomological dimension $2$, but does it hold in general?

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    $\begingroup$ This is not true. Bestvina-Brady group show FP_n is not F_n. perhaps you are thinking of geometric dimension vs cohomological dimension. There either geometric dimension 2 is not cd 2 or whitehead conjecture is false. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 20, 2014 at 23:44
  • $\begingroup$ Thanks for the reply, however I don't see how $FP_n$ not being equivalent to $F_n$ implies this is false. Perhaps you could explain more? Yes, Bestvina and Brady construct a group $G$ which has $cd(G) = 2$, is of type $FP_2$, but not finitely presented. However, by the Eilenberg-Ganea Theorem, $G$ acts on freely on a $2$-dimensional acyclic (not necessarily simply connected) space $X$. Using Schanuel's Lemma and the fact that $G$ is of type $FP_2$ one gets that $X$ is $G$-cocompact. $\endgroup$
    – Gaelan Hanlon
    Commented Jan 20, 2014 at 23:59
  • $\begingroup$ Eilenberg-Ganea conjecture is open for n=2. It is known either this is false or Whitehead conjecture is false. Any group acting freely on a 2-dimensional acyclic complex has cd 2. The augmented cellular chain complex would give a free resolution of the trivial module. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 21, 2014 at 0:06
  • $\begingroup$ Eilenberg-Ganea shows cd=geometric dimension in dimension >2. I think you confuse FP with geometric dimension. Thompson's group F has FP_2 but no finite dimensional classifying space. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 21, 2014 at 0:08
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    $\begingroup$ I think the question has a positive answer: indeed it's shown in Brown's book that a FP2 group $F/R$ can be written as quotient of a fp group $F/R_1$ ($R_1$ finite) by a perfect normal subgroup $N/R_1$. Thus adding 2-gons corresponding to $F/R_1$ to the Cayley graph of $F/R$ works. $\endgroup$
    – YCor
    Commented Jan 21, 2014 at 7:10

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It is true that a group $G$ is $FP_2$ if and only if $G$ acts freely cellularly on a connected CW-complex with trivial first homology group. I see from the comments that you are worried about finite generation of $G$: this is not a problem, because a group $G$ is $FP_1$ over any (non-trivial) ring if and only if $G$ is finitely generated. In fact, the elements $g_1,...,g_n$ generate $G$ if and only if the elements $1-g_1,...,1-g_n$ generate the augmentation ideal of $\mathbb{Z}G$.

Now suppose that $G$ is $FP_2$, and take an infinite presentation of $G$ with finitely many generators. The Cayley graph for $G$ with respect to this generating set is a connected 1-dimensional complex with a free cocompact $G$-action. Since $G$ is $FP_2$, the first homology group of this graph is finitely generated as a $\mathbb{Z}G$-module. Now attach free orbits of 2-cells corresponding to the relators, one relator at a time. If you attached all the relators you would get a 2-complex with trivial fundamental group. Since the first homology of the graph you started with is finitely generated, you only need to attach finitely many of the orbits of cells to get a complex with trivial first homology. This proves something slightly stronger: if you take the Cayley 2-complex for any presentation for $G$ with finitely many generators, then it has a $G$-invariant subcomplex that is cocompact and has trivial first homology.

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