Topology on extensions of topological groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:24:15Z http://mathoverflow.net/feeds/question/70139 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70139/topology-on-extensions-of-topological-groups Topology on extensions of topological groups jap 2011-07-12T16:19:50Z 2011-07-12T23:55:54Z <p>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.</p> <p>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?</p> <p>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.</p> <p>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?</p> <p>Thank you all in advance.</p> http://mathoverflow.net/questions/70139/topology-on-extensions-of-topological-groups/70147#70147 Answer by Evan Jenkins for Topology on extensions of topological groups Evan Jenkins 2011-07-12T16:54:18Z 2011-07-12T16:54:18Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/70139/topology-on-extensions-of-topological-groups/70191#70191 Answer by Konrad Voelkel for Topology on extensions of topological groups Konrad Voelkel 2011-07-12T23:55:54Z 2011-07-12T23:55:54Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>As for the smooth case, the same idea applies, where one would need the cocycle to be smooth as well.</p>