Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A group $G$ is cohomologically $p$- complete if the canonical map from $G$ to it's pro$-p$ completion $\hat G^p$ induces an isomorphism on cohomology $H^\ast_{cont}(\hat G^p, \mathbb{Z}_p) \rightarrow H^\ast(G, \mathbb{Z}_p).$

In the paper "3-manifold groups are virtually residually-p", http://arxiv.org/abs/1004.3619 Aschenbrenner and Friedl define a $p$-efficient graph of groups $G$ to have the following properties:

1) $\pi_1(G)$ is residually-$p$ finite.

2) for all vertices v and edges e, the subgroups $G_v$ and $G_e$ of G are closed in the pro-p topology on $G$;

3) for all vertices v and edges e, the pro-p topology on $G$ induces the pro-$p$ topology on $G_v$ and on $G_e$.

They then have the following lemma: Suppose $G$ is a $p$-efficient graph of finitely generated groups such that the vertex and edge groups are cohomologically $p$-complete. Then $\pi_1(G)$ is cohomologically $p$-complete.

On the other hand, an earlier paper "Groups with the same cohomology as their pro-$p$ completions" http://arxiv.org/pdf/0809.3046v3.pdf by Lorensen said the following for amalgamated products and HNN extensions:

Let $G = G_1 \ast_{G_0} G_2$ and suppose $G_i$ is topologically p-embedded in G for each i, and $G_0$ is topologically p-embedded in $G_1$ and $G_2$. If $G_i$ are all cohomologically p-complete, then $G$ is cohomologically p-complete.(He has a similar statement for HNN extensions.)

If I am reading this correctly, topologically p-embedded is exactly condition (3) above. So, my question is, are the other conditions necessary- particularly residually p-finiteness? Is there a difference between amalgamated products and general graphs of groups that requires conditions (1) and (2)?

Thanks very much for your time,

Kevin

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.