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Let $G$ and $H$ be two topological groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of abstract groups.

Is there a way to introduce a topology on $E$ such that $\mathcal{E}$ becomes an extension of topological groups? If there is a way, is it unique?

Similarly, let $G$ and $H$ be Lie groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of topological groups.

Is there a way to introduce a smooth structure on $E$ such that $\mathcal{E}$ becomes an extension of Lie groups? If there is a way, is it unique?

Thank you all in advance.

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2 Answers

The answer to the first question is no. In general, the automorphism group of $G$ as an abstract group will be bigger than its continuous automorphism group. For instance, if we take $G$ to be the additive group of $\mathbb{R}$, we can pick a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space and permute basis elements to get many discontinuous automorphisms of $G$. Then the semidirect product of $\operatorname{Aut}(G)$ by $G$ cannot be made into a topological extension, as conjugation would not be continuous.

I think the answer will still be no even if we restrict to central extensions, although I do not know a counterexample off the top of my head.

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@Evan: Thanks for your comment.. –  jap Jul 13 '11 at 5:13
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You can describe abstract group extensions of H with G by 2-cocycles of group cohomology. If you have an extension E, you get an induced H-operation on G, by conjugating in E (take any set-theoretic section of $E\to G$). The extensions of G with H with this H-operation on G are classified up to isomorphism, by the second group cohomology. You get a 2-cocycle corresponding to E by taking any set-theoretic section $s : G\to E$ of $E\to G$ that maps 1 to 1 and write down the 2-cocycle $c : H \times H \to G$ by $c(h,h'):=s(h)s(h')s(hh')^{-1}$. Then equip the set $G\times H$ with the multiplication $(g,h)(g',h') := (g+h.g'+c(h,h'),hh')$. More explicitly, this is $(g,h)(g',h') = (g+s(h)g's(h)^{-1}+s(h)s(h')s(hh')^{-1},hh')$. This is again a group extension and it is isomorphic to E.

One can show that all extensions are of the type I just constructed for a given cocycle, up to isomorphism. The extensions in the same isomorphism class differ only by a coboundary. A good reference would be Weibel's homological algebra book.

To have an extension with a topological group structure implies that the corresponding cocycle is continuous and in general, this can not be expected. Observe that continuity doesn't follow from the axioms for 2-cocycles and depends on the topological structures of G and H, which you don't want to change.

So I think, it's wrong in general as well as for central extensions, which would be the case of a trivial H-action on G, where you still don't get any continuity for free. At the same time, I don't know of any trivial counter-example. You might take any non-continuous map from the real numbers times real numbers to the real numbers and form the corresponding "twisted" semi-direct product as sketched above. Then you can not get a topological group structure on the extension such that the extension is in the category of topological groups.

As for the smooth case, the same idea applies, where one would need the cocycle to be smooth as well.

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Is it evident that "to have an extension with a topological group structure implies that the corresponding cocycle is continuous"? After all, the section $s$ used to produce the cocycle might be discontinuous. I can see that the cohomology class will contain some cocycles that are continuous in certain places and other cocycles that are continuous in other places, but I don't see how to get one cocycle that's continuous globally. (Am I just being dense?) –  Andreas Blass Jul 13 '11 at 1:19
    
Thanks all for your comments. Could someone provide some example. –  jap Jul 13 '11 at 5:14
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