My understanding of Poincare duality (via wikipedia) is that for an $n$-dimensional orientable closed manifold $X$, for any coefficient ring $R$, there is an isomorphism $$H^k(X,R)\cong H_{n-k}(X,R)$$ Here, the wiki article says that this isomorphism depends on making a "choice of orientation with respect to the coefficient ring $R$", which here I'm interpreting as - The isomorphism always exists, but the choice of isomorphism depends on a choice of orientation.

Is this correct so far?

Assuming the above is correct, then if $Z$ is a contractible orientable space, and $G$ is a group acting freely on $Z$, then $Z/G$ is the classifying space for $G$.

Is it true that $H^i(Z/G,R)\cong H^i(G,R)$, and $H_i(Z/G,R)\cong H_i(G,R)$? (where $H^i(G,R),H_i(G,R)$ is group cohomology, with $R$ viewed as its underlying abelian additive group with trivial $G$-action).

If things are still correct (please correct me if I'm making any mistakes!), then Poincare duality should give some nontrivial relations between group cohomology and group homology.

For example, if $G$ acts freely on an orientable contractible space of dimension 2, then for any ring $R$ (identified with its underlying additive group), the central extensions of $G$ by $R$ (classified by elements of $H^2(G,R)$) should be in bijection with elements of $H_0(G,R)\cong R$ - but this is definitely false, since if $G$ has trivial Schur multiplier, one can compute that $H^2(G,R) = 0$ for any ring $R$.

Where have I gone wrong?

I suspect the comparison between $H_i(Z/G,R)$ and $H_i(G,R)$ and $H^i(Z/G,R)$ and $H^i(G,R)$ must be subtler than I expect. Even so, are there examples of theorems which use Poincare duality (or other properties of singular (co)homology) to produce nontrivial results in group (co)homology?

Is there a good reference which discusses the comparison between group (co)homology and the singular (co)homology of classifying spaces in detail?

EDIT: In response to some of the comments, I suppose part of the question is - For which groups $G$ (I'm mostly interested in finite groups and the fundamental groups of Riemann surfaces) can one find a contractible manifold $Z$ on which $G$ acts freely and cocompactly?

  • $\begingroup$ "Is this correct so far" Yes, as far as I can see. That's what I would interpret it to mean as well. In the simplest case, for instance, there are two isomorphisms from $\Bbb Z$ to $\Bbb Z$, so it's not too difficult to imagine that if $H_0(X,\Bbb Z)\cong H^n(X,\Bbb Z)\cong \Bbb Z$, then changing between the two orientations of the manifold changes between the two isomorphisms of the (co)homology groups. I assume that this is indeed the case, although I haven't checked. $\endgroup$
    – Arthur
    Dec 29, 2016 at 23:36
  • $\begingroup$ I don't know the group theory side of the story, but if $G$ acts freely on an orientable contractible space of dimension 2 then $G$ is either free or the fundamental group of a surface. In those cases respectively one either has that $H^2(G,R) = 0$ or that $G$ has nontrivial Schur multiplier. I don't see how you've come to the conclusion that central extensions of $G$ by $R$ should be in bijection with $R$. That's... not true. $\endgroup$
    – user98602
    Dec 30, 2016 at 0:18
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    $\begingroup$ Infinite dimensional Poincaré duality is not going to be very useful ;-) $\endgroup$ Dec 30, 2016 at 2:35
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    $\begingroup$ It's straightforward from the definition of group (co)homology. Your free resolution of $R$ as a $\Bbb Z[G]$-module is as the cochain algebra of the universal cover, $Z$. $\endgroup$
    – user98602
    Dec 30, 2016 at 3:26
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    $\begingroup$ Hatcher's 'Algebraic Topology' handles the equality of the group (co)homology and the (co)homology of the classifying space if I remember correctly. Search the chapter about local coefficients. $\endgroup$
    – cjackal
    Dec 30, 2016 at 4:21

1 Answer 1


I will address your "edited" question: "For which groups $G$ can one find a contractible manifold $Z$ on which $G$ acts freely and cocompactly?" (I assume you also want to assume "properly discontinuously" to get Hausdorff quotient spaces.)

A very good (even if dated) reference for this question is Chapter 8 of Ken Brown's book "Cohomology of groups". A necessary condition for existence of such an $n$-dimensional manifold is that $G$ is an $n$-dimensional Poincare duality group (a $PD(n)$ group) of type $F$. Equivalently, there exists a finite $K(G,1)$ and $H^i(G, {\mathbb Z}G)\cong {\mathbb Z}$ for $i=n$ and zero otherwise. It is a famous open problem (due to C.T.C. Wall) that the converse holds as well. (To be more precise, Wall asked only for $G$ to be $PD(n)$, he did not assume that $G$ admits a finite $K(G,1)$; Mike Davis constructed examples for all $n\ge 4$ showing that the latter condition is needed.) This conjecture is known to be true for $n=1$ (that's easy), for $n=2$ (very hard: Eckmann, Linnell and Muller, 1982), and is wide-open for $n\ge 3$. Rob Kirby discusses this conjecture in great detail in his list of problems in topology. For $n\ge 5$ the conjecture is a part of the one of the central problems in higher dimensional manifold topology, namely the Borel Conjecture. The deepest (in my mind) result so far is that Wall's conjecture holds for hyperbolic groups with boundary homeomorphic to $S^{n-1}$, $n\ge 6$. This is due to Barthels, Lueck and Weinberger, see here. If I had to guess, Wall's conjecture is true for $n=3$ and is false for $n\ge 5$. No idea what happens in between ($n=4$).

Now, as for your paranthetical question "I'm mostly interested in finite groups and the fundamental groups of Riemann surfaces", that part is easy: Finite groups are never $PD(n)$, unless trivial, in which case, $\{1\}$ acts freely and cocompactly on the singleton. As for the fundamental groups of Riemann surfaces, every such group is either free (and nontrivial), in which case it is not $PD(n)$ group for any $n$ or acts properly discontinuously freely and cocompactly on $R^2$, which is contractible, of course.


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