I learned from link 1, link 2 link3 that sometimes it is possible to write down all possible group extension. I also know that the group extension is classified by group cohomology.

My questions are:

  1. After we have a result from group cohomology, is it always possible to write down the explicit form of the group extension when the groups are discrete but not necessarily finite and abelian? Is there any algorithm for that?
  2. If the groups involved are topological, is it possible to write down the explicit form of the extended groups? If yes, how to do that? For example, we choose the Borel measurable group cohomology developed by Calvin Moore and we consider the following extension $$1\rightarrow \mathbb{Z} \rightarrow G \rightarrow U(1) \rightarrow 1.$$ One can compute the second group cohomology $\mathcal{H}_{Borel}^2(U(1),\mathbb{Z})=\mathbb{Z}$. Therefore, there should be countably infinitely many ways to extend the $U(1)$ by $\mathbb{Z}$. Obviously, $G=\mathbb{R}$ is one classic example. Is there any way to find the other $G$'s and determine their property?



I don't know what you might mean by "explicit form" but if you have a representative 2-cocycle $f:Q\times Q\to M$ of a cohomology class in $H^2(M,Q)$ with the action of $Q$ on $M$ given by $\theta:Q\to \mathrm{Aut}(M)$, then yes, there is an explicit formula to make a group from $f$. Specifically $G=Q\times M$ as a set and $(q,m)(q',m')=(qq', \theta(q')(m)+m'+f(q,q'))$. See Hall Jr. "Theory of Finite Groups" (and actually all the group theory books I know have this formula, just read the section on cohomology). Algorithms for some of this were given by Derek Holt, see for example Holt, Eick, O`Brien "Handbook of Computational Group Theory"

Perhaps it is obvious, but be cautious that isomorphic groups can be given by inequivalent extensions.

  • $\begingroup$ Normally I calculate the group cohomology of a group from other methods, such as exact sequences. The only thing that I get is an abelian group. I do not know explicitly the function form of the cocycles $f(q,q')$, which element in $M$ do $q$ and $q'$ correspond to. Perhaps, there is a trick to find it in the references you gave. $\endgroup$ – Herman Chu Apr 17 '18 at 22:46
  • $\begingroup$ Well in general I doubt the references will help then. If all you have as an isomorphism $H^2(M,Q)\to \mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_{n}}$ then that identification is anything up to an automorphism, so actual cocycles will have no meaning. However you have $\mathbb{Z}$, which just has 2 automorphisms. So your implicit isomorphisms, if made explicit, would be plausibly concrete up to swapping signs. That said, it sounds like a lot of work to extract explicit isomorphisms from some diagram chase. Good luck. $\endgroup$ – Algeboy Apr 18 '18 at 21:14

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