1
vote
0answers
22 views

Extension to rational and real chains

In the paper on stable commutator length, D. Calegari says that generalized $\operatorname{scl}$ function can be extended by linearity to rational group $1$-chains and by continuity to real chains ...
2
votes
1answer
106 views

Cohomology group $H^n(G, \mathbb{R}/\mathbb{Z})$ as a continuous group or a Lie group?

Is there any example of Borel cohomology group $H^n(G, \mathbb{R}/\mathbb{Z})$ for any $G$ such that $H^n(G, \mathbb{R}/\mathbb{Z})$ is a continuous group? Such as a Lie group? Most of the examples ...
1
vote
0answers
77 views

Can group cohomology be used to study fiber bundles?

Is (non-abelian) cohomology used to study vector and principal bundles? Can you give me a text or an article? For example: Consider a vector bundle $E$ with fiber $V$ and base manifold $M$. Consider ...
5
votes
0answers
99 views

Cohomology of covering space

Let $B$ be a base space and $E$ be a covering space of $B$ what is the relation between $H^2(B,\mathbb{Z})$ and $H^2(E,\mathbb{Z})$.?
2
votes
1answer
45 views

What is the map $(\mathbb{C}P^\infty)^n\to G^n(\mathbb{C}^\infty)$ induced by the map $(S^1)^n\to U(n)$?

There is a map $(S^1)^n\to U(n)$, mapping each $n$-tuple to the corresponding diagonal matrix, where $S^1$ is identified with the complex numbers of unit length. There is an induced map ...
1
vote
0answers
54 views

A reference about Dolbeault cohomology

I am looking for a reference about Dolbeault cohomology when the line bundle is not supposed to be positive.
4
votes
1answer
55 views

Topological condition for a group to be of type FL

We say that a group $G$ is of type $FL$ if there exists a resolution $L_\bullet \to \mathbb{Z}$ of finite length of finitely generated, free $\mathbb{Z}G$-modules. Now, un unproven proposition in ...
3
votes
0answers
202 views

group cohomology with coefficient in an induced module

We say that a $G$-module $I$ is induced if $$I\cong L\otimes\mathbb{Z}G$$ where $L$ is an abelian group and the action on $L\otimes\mathbb{Z}G$ is given by the action of $G$ only on the second ...
9
votes
4answers
559 views

Why is the cohomology of a $K(G,1)$ group cohomology?

Let $G$ be a (finite?) group. By definition, the Eilenberg-MacLane space $K(G,1)$ is a CW complex such that $\pi_1(K(G,1)) = G$ while the higher homotopy groups are zero. One can consider the singular ...