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Can we have some (new) examples of group extensions $G/N=Q$ with continuous (e.g. Lie groups) $G$ and $Q$, but a finite discrete $N$? Note that $1 \to N \to G \to Q \to 1$.

What I know already contains:

$$SU(2)/\mathbb{Z}_2=SO(3).$$

$$\frac{\mathbb{R}/{\mathbb{Z}}}{\mathbb{Z}_n}={\mathbb{R}}/{(n\mathbb{Z})}.$$

What else are the examples that you can provide?

A systematic answer to obtain new examples (a few or even a list) is of course most welcome. ; )

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  • $\begingroup$ What do you mean by continuous groups more precisely? Do you just mean connected topological groups? $\endgroup$ – Tobias Kildetoft Mar 30 '17 at 19:49
  • $\begingroup$ Yes. And Lie groups. $\endgroup$ – annie heart Mar 30 '17 at 19:50
  • $\begingroup$ But being just a Lie group does not mean much if you for example allow it to be zero-dimensional. I am still not entirely sure what you mean by continuous here, as that word usually only applies to maps, not groups. $\endgroup$ – Tobias Kildetoft Mar 30 '17 at 19:59
  • $\begingroup$ OK, I guess I am just a applied math or an engineer here. Dont get confused on the term continuous, I mean something like Lie groups like SU(N). $\endgroup$ – annie heart Mar 30 '17 at 20:01
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In general, there is a way to classify extensions of group $1 \to N \to G \to Q \to 1$ for given $Q$ and $N$. You need

  • a group morphism $\phi$ from $Q$ to $Out(N)$ (the outer automorphism group of $N$),

  • and the cohomological class of a $2$-cocycle in $H^2(Q,Z(N)_{\phi})$ ($Z(N)_{\phi}$ means that $Z(N)$ is considered as a $Q$-module with the action given by $\phi$).

You can read about the link between extensions of groups and second cohomology groups in Adem and Milgram's $\textit{Cohomology of Finite Groups}$.

I assume that by $G$ and $Q$ continuous groups, you mean topological groups. As far as I can tell, once you are given such extension, where $N$ is finite and $Q$ is a topological group there is only one way to extend the topology, the same is true for a Lie group. This comes from the fact that whenever we have $1 \to N \to G \to Q \to 1$ where $N$ is finite and $G$ and $Q$ are topological groups, the projection map $G\to Q$ is necessarily a topological cover.

Edit : as written in the comment below, you also need to add the condition that $\phi$ is continuous to ensure that $G$ is a topological group (here $N$ has the discrete topology).

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    $\begingroup$ Seems to me that if $Q$ is connected there are very few options for $\phi$. $\endgroup$ – Jyrki Lahtonen Mar 30 '17 at 8:26
  • $\begingroup$ Thanks! +1, could you provide us some new examples? $\endgroup$ – annie heart Mar 30 '17 at 15:26
  • $\begingroup$ Can you comment the difference between the 2nd cohomology group $H^2(Q,Z(N)_{\phi})$ and $H^2(Q,N)$? $\endgroup$ – annie heart Mar 30 '17 at 15:31
  • $\begingroup$ @Clément Guérin, Nice, what role does the Torsion play here? $\endgroup$ – wonderich Mar 30 '17 at 15:47
  • $\begingroup$ @annieheart, $H^2(Q,N)$ refers twisted cohomology with non-abelian coefficients, it is very hard to even define the second cohomology group and the higher cohomology groups. Whereas $H^2(Q,(Z(N)_{\phi})$ can be nicely define with the non-homogenous cochain definition. As for the new examples, you can take $G$ to be the Heisenberg group and $N$ to be its center. It may be interesting. Just bear in mind that $\phi$ needs to be trivial if $Q$ is connected (see Jyrki's Lahtonen's comment). $\endgroup$ – Clément Guérin Apr 2 '17 at 8:23
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Let $G = Q = \{z; z \in \Bbb C, \left| z \right| = 1\}$ and $N = \{1,-1\}$.

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  • $\begingroup$ That is nice +1, but I had it in $\frac{\mathbb{R}/{\mathbb{Z}}}{\mathbb{Z}_n}={\mathbb{R}}/{(n\mathbb{Z})}.$ with $n=2$ is your example. $\endgroup$ – annie heart Mar 30 '17 at 18:57
  • $\begingroup$ Let us try some more nontrivial examples. $\endgroup$ – annie heart Mar 30 '17 at 18:58
  • $\begingroup$ Same construction on the quaternions with norm $1$ and $N$ the quaternion group. $\endgroup$ – Marc Bogaerts Mar 30 '17 at 18:59
  • $\begingroup$ Thanks please update more examples on the answer so we can see explicitly. ; ) $\endgroup$ – annie heart Mar 30 '17 at 19:01
  • $\begingroup$ My previous example is wrong since $N$ is not a notmal subgroup. The good news is that $N = \{\1,-1}$ still works. Moreover for any "classic" group one can take for $N$ any finite subgtoup of the center (scalar matrices) such as the $n$ -th roots of unity. $\endgroup$ – Marc Bogaerts Mar 31 '17 at 7:42

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