12
votes
2answers
286 views

Group cohomology without G-modules (a.k.a. what does this bar construction compute?)

Without any prior exposure to the cohomology of groups, one might naively proceed by replacing a group by a sort of resolution. For instance, let's take $G = \mathbb{Z}^2$, and "resolve": $$ 0 \to ...
6
votes
1answer
615 views

Double coset formulas for Orthogonal groups [Solved]

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition ...
5
votes
1answer
216 views

Truncation of BG?

Let $G$ be a topological group. In some cases, e.g. when $G$ is discrete or when the spaces $G^n$ are locally contractible and the coefficients are discrete, the cohomology of the classifying space ...
5
votes
2answers
527 views

Subobject-poset (co-)homology

Given a group, there is another way to define its "(co-)homology" using a classifying space. Specifically, one takes the partially ordered set of its proper non-trivial subgroups (if they exist), and ...
8
votes
1answer
946 views

Why isn't the orbifold cohomology of $pt/G$ equal to the cohomology of $BG$?

The classifying space of a group $G$ is given by taking a contractible space $E$ equipped with a free $G$-action, and looking at the quotient, which we dub $BG$. The homotopy type of this space (and ...