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Let we have a complex of abelian topological or lie groups $$\ldots \to G_{n}\to G_{n+1}\to \ldots$$ such that the image of $G_{n}$ is a closed subgroup of $G_{n+1}$. Then we have a complex of fundamental groups $$\ldots \to \pi_{1}(G_{n})\to \pi_{1}(G_{n+1})\to \ldots$$

Are there some standard theorems or results about a comparison betwee the cohomology of $\pi_{1}(G_{n})$ and the $\pi_{1}$ of cohomolgy of $G_{n}$?

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If you consider an exact sequence of groups, it's a fibration of topological spaces, so you get a homotopy long exact sequence. So that makes me think $\pi_n$ behaves like a derived functor from topological abelian groups to abelian groups. In particular this is the derived functor of $\pi_0$.

So there is the spectral sequence for applying a derived functor to a complex whose first page whose second page is $\pi_m$ of cohomology of $G_n$.

However, I don't think the first page is $\pi_m(G_n)$. One can see this from an exact sequence, like $\mathbb Z/2 \to S^1 \to S^1$. All the homotopy groups of the cohomology are trivial, because it is exact. But $\pi_m(G_n)$ has many nonzero terms and you can't make them all vanish just by applying a differential.

I don't know enough about spectral sequences to say exactly what the relationship between these two objects is, but I believe it is a question primarily of homological algebra.

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  • $\begingroup$ thanks for the answer. I should concentrate on it. But another question: In this question, to what extent I need to consider "abelian" group. This is a motivation to ask:Let $H$ be a closed (but not necessarily normal) subgroup of a topological group $G$. Then is it true that the topological space $G/H$, left cosets, has an abelian fundamental group? $\endgroup$ – Ali Taghavi Nov 13 '14 at 22:54
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    $\begingroup$ @AliTaghavi No, take a finite non-abelian subgroup of a connected topological group - say the quaternions inside $SU_2$. $\endgroup$ – Will Sawin Nov 13 '14 at 23:01
  • $\begingroup$ So the fundamental group of the quotient is order 8 group quternions. As the free action defines a universal covering space. Now what is the shape of the quotient space?Another question: could you please expand your answer with a possible revision or some reference? Or at least email me some more information? Any way, thanks again for your interesting answer and comment. $\endgroup$ – Ali Taghavi Nov 15 '14 at 7:46

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