User gurs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:51:20Z http://mathoverflow.net/feeds/user/11932 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groups smooth cohomology of Lie groups gurs 2011-01-24T07:06:35Z 2011-01-24T10:35:12Z <p>Let $G$ be a Lie group and $A$ a smooth $G$-module. Define $C^n(G,A)=\{ f: G^n \to A|~f~\text{is smooth}\}$ and $\partial^n: C^n \to C^{n+1}$ by the standard formula as used in the cohomology of abstract groups. I think this cohomology must be well studied. Can somebody provide me some references for this cohomology. A similar cohomology for topological groups has been studied by <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.mmj/1028989726" rel="nofollow">Hu</a> and <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&amp;_udi=B6V0K-45H34FP-21&amp;_user=4584237&amp;_coverDate=09%252F30%252F1973&amp;_rdoc=1&amp;_fmt=high&amp;_orig=search&amp;_origin=search&amp;_sort=d&amp;_docanchor=&amp;view=c&amp;_acct=C000063492&amp;_version=1&amp;_urlVersion=0&amp;_userid=4584237&amp;md5=cde7528bc489e42bf4ee877df0afc525&amp;searchtype=a" rel="nofollow">Heller</a> using continuous cochains.</p> <p>Also, can somebody tell me what kind of Lie group extensions $H^2(G,A)$ correspoind to.</p> http://mathoverflow.net/questions/52099/cohomology-theory-for-algebraic-groups/52158#52158 Answer by gurs for cohomology theory for algebraic groups gurs 2011-01-15T10:30:47Z 2011-01-15T10:37:28Z <p>Thanks all for the valuable information.</p> http://mathoverflow.net/questions/50277/sections-of-topological-group-extensions Sections of topological group extensions gurs 2010-12-24T08:14:32Z 2011-01-01T09:44:59Z <p>Does there exists an example of an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or continuous in a neighbourhood of identity) and satisfy the propery that $s(x^{-1}) =s(x)^{-1}$. One can see that the extension $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{S}^1 \to 1$ does not admit such a section.</p> <p>For discrete groups there are many extensions admitting such a section. For example, consider the extension $0 \to \mathbb{Z} \stackrel{i}{\to} \mathbb{Z} \times \mathbb{Z}_2 \stackrel{p}{\to} \mathbb{Z}_4 \to 0$ with $i(x)= (2x, [x])$ and $p(x, [y])=[x + 2y]$. Clearly, the section $s$ defined by $s([0])=(0,[0])$, $s([1])=(-1,[1])$, $s([2])=(0,[1])$ and $s([3])=(1,[1])$ satisfy the property that $s(x^{-1})= s(x)^{-1}$. I am looking for an example in the setting of compact connected topological groups.</p> http://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groups/53034#53034 Comment by gurs gurs 2011-01-24T12:48:22Z 2011-01-24T12:48:22Z Thanks for your comments. If $E$ is $A \times G$ as a smooth manifold, then the extension has an obvious smooth section and the proof works as per your suggestion. I am also feeeling that it should be true. Thanks again. http://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groups/53034#53034 Comment by gurs gurs 2011-01-24T10:56:21Z 2011-01-24T10:56:21Z Thanks for your comments and providing the references. I would like to add that, in the continuous cohomology for topological groups due to Hu and Heller, they showed that the second cohomology group classifies topologically split group extensions. More precisely, if $G$ is a topological group and $A$ is a topological $G$-module, then $H_{cont}^2(G,A)$ classies all extensions $1 \to A \to E \to G \to 1$ such that as a space $E$ is $A \times G$. I am guessing that the similar result is true for Lie groups when we work with smooth cohains as mentioned in my question. Is is true? http://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groups Comment by gurs gurs 2011-01-24T09:28:16Z 2011-01-24T09:28:16Z Many thanks. But, what kind of Lie group extensions does $H^2(G,A)$ characterize. http://mathoverflow.net/questions/50277/sections-of-topological-group-extensions Comment by gurs gurs 2010-12-24T09:13:28Z 2010-12-24T09:13:28Z Let me write my question again. Does there exist an extension of topological groups $1 \to N \to E \to G \to 1$ admitting a section $s:G \to E$ which is continuous (or only continous in a neighbourhood of identity) and satisfy the property that $s(x^{-1})=s(x)^{-1}$ for all $x \in G$ but is not a homomorphism. http://mathoverflow.net/questions/50277/sections-of-topological-group-extensions Comment by gurs gurs 2010-12-24T09:04:25Z 2010-12-24T09:04:25Z Thanks. But I want a non-trivial extension.