(of corresponding dimensions). how can I prove this? I think my main stumbling block is my general ignorance of group cohomology.
integral cohomology groups of an aspherical manifold are isomorphic to the integral cohomology groups of it's fundemental group
-
2To answer this, we need to know what is your definition of group cohomology. – 2012-01-05
-
0I have only learned the definition contained in Dummit and Foote wherein it is defined to be $Ext_{\mathbb{Z}G}(\mathbb{Z},\mathbb{Z})$. Are there non-equivalent definitions? – 2012-01-05
-
1All definitions are equivalent, but to prove something is equivalent to one of them you need to actually pick one of them! – 2012-01-05
-
2By the way, it is usually best if the body of the question is self contained and, in particular, does not require the title to make sense (your titiel/question does not satisfy this condition): have you ever seen a book whose first sentence starts in the cover? – 2012-01-05
-
0agreed! though I'm willing to work out the equivalence between your definition and mine if yours lends itself to a proof of the statement. – 2012-01-05
2 Answers
Suppose $X$ is an aspherical CW-complex (for example, an anspherical smooth manifold), so that the universal covering space $\tilde X$ is contractible (an allow me to be sloppy with basepoints...)
The group $G=\pi_1(X)$ acts properly discontinuously on $\tilde X$, and this action induces an action of $G$ on the singular complex $C_\bullet(\tilde X)$, so that this complex can be seen as a complex of $G$-modules. Since $\tilde X$ is contractible, the homology of the complex $C_\bullet(\tilde X)$ is zero in all non-zero degrees, and isomorphic to $\mathbb Z$ in degree zero—a tiny bit of care will show that the homology in degree zero is in fact a trivial $G$-module.
Now, a little fiddling with definitions will show that
Lemma. for each $p$, the group of $p$-chains $C_p(\tilde X)$ is a free $G$-module.
In fact, that fiddling will show
Porism. a basis of $C_p(\tilde X)$ as a $G$-module can be identified with the set of singular $p$-simplices in $X$.
It follows that $C_\bullet(\tilde X)$ is in fact a free resolution of the trivial $G$-module $\mathbb Z$. The general nonsense that goes under the name of homological algebra tells us now that we can compute $H^\bullet(G,\mathbb Z)=\mathrm{Ext}_{\mathbb ZG}^\bullet(\mathbb Z,\mathbb Z)$ as the cohomology of the complex $\hom_{\mathbb ZG}(C_\bullet(\tilde X),\mathbb Z)$. Now, the porism above (and a similar observation regarding differentials...) allows us to identify the complex $\hom_{\mathbb ZG}(C_\bullet(\tilde X),\mathbb Z)$ with the slightly simpler $\hom_{\mathbb Z}(C_\bullet(X),\mathbb Z)$.
Finally, the cohomology of the complex $\hom_{\mathbb Z}(C_\bullet(X),\mathbb Z)$ is, by definition, the integral singular cohomology of the space $X$.
More details can be found in MacLane's Homology book, if I recall correctly.
-
0Thanks, Mariano. Much appreciated – 2012-01-05
One has in general $H^*(BG;\mathbb Z)\cong H^*(G;\mathbb Z)$ where the left hand side is singular cohomology, and the right hand side is group cohomology. In the case at hand, $H^*(\pi_1(X);\mathbb Z)\cong H^*(B\pi_1(X);\mathbb Z)$. However, $B\pi_1(X)$ is a $K(\pi_1(X),1)$ space, and so is $X$ since it's aspherical. Since $X$ is a CW complex (it's a manifold) and CW Eilenberg-Maclane spaces are unique up to weak homotopy equivalence, this means that $X$ is weak homotopy equivalent to $K(\pi_1(X),1)=B\pi_1(X)$. Since cohomology is weak homotopy invariant, this gives us the result.
-
0thank you. what is meant by "BG"? – 2012-01-05
-
0The space $BG$ is the [classifying space](http://en.wikipedia.org/wiki/Classifying_space) of $G$. For a discrete group (i.e., not a topological group), $BG$ will be a $K(G,1)$ space. – 2012-01-05
-
0great. thanks very much. – 2012-01-05