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Firstly, your family $\mathcal{F}$ is not closed under conjugation if $H$ is not normal. Depending on what you want to do, this may not be an issue. There are two references for the construction of $ …

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With …

The toral rank conjecture (or Halperin--Carlsson conjecture) states that if $T^n$ acts with finite isotropy groups on the simply-connected closed manifold $X$, then $$ \sum_i \dim H^i(X;\mathbb{Q})\g …

In fact "$\mathbb{Q}$-acyclic" means "having the same rational homology groups as a point". In particular, the classifying space of any finite group is $\mathbb{Q}$-acyclic, as can be seen by a simple …

No, I don't think so - I think there are more Bredon $1$-cochains than that. The orbit category $\mathcal{O}C_4$ looks like $$ C_4/e \to C_4/C_2 \to C_4/C_4 $$ where the automorphism groups of the ob …

These results follow from the Cartan-Leray spectral sequence, which for a regular covering map $X\to X/W$ and a commutative ring $k$ of coefficients has $$ E_2^{p,q}=H^p(W,H^q(X;k)) $$ (cohomology of …

In what follows I will assume that $G$ is discrete and that $X$ is a simplicial complex with regular $G$-action (see Bredon's "Introduction to compact transformation groups", Chapter III.1). The regul …

The Borel seminar, which is the classic reference for equivariant (Borel) cohomology, containins a wealth of information and is quite readable.

Good question. I think the answer is yes. The unnamed spectral sequence is usually referred to as the isotropy spectral sequence. For a group $G$ acting on $X$ and an abelian group $A$ of coefficient …