User gurs - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T03:51:20Zhttp://mathoverflow.net/feeds/user/11932http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groupssmooth cohomology of Lie groupsgurs2011-01-24T07:06:35Z2011-01-24T10:35:12Z
<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&version=1.0&verb=Display&handle=euclid.mmj/1028989726" rel="nofollow">Hu</a> and <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0K-45H34FP-21&_user=4584237&_coverDate=09%252F30%252F1973&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000063492&_version=1&_urlVersion=0&_userid=4584237&md5=cde7528bc489e42bf4ee877df0afc525&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#52158Answer by gurs for cohomology theory for algebraic groupsgurs2011-01-15T10:30:47Z2011-01-15T10:37:28Z<p>Thanks all for the valuable information.</p>
http://mathoverflow.net/questions/50277/sections-of-topological-group-extensionsSections of topological group extensionsgurs2010-12-24T08:14:32Z2011-01-01T09:44:59Z
<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#53034Comment by gursgurs2011-01-24T12:48:22Z2011-01-24T12:48:22ZThanks 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#53034Comment by gursgurs2011-01-24T10:56:21Z2011-01-24T10:56:21ZThanks 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-groupsComment by gursgurs2011-01-24T09:28:16Z2011-01-24T09:28:16ZMany thanks. But, what kind of Lie group extensions does $H^2(G,A)$ characterize.http://mathoverflow.net/questions/50277/sections-of-topological-group-extensionsComment by gursgurs2010-12-24T09:13:28Z2010-12-24T09:13:28ZLet 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-extensionsComment by gursgurs2010-12-24T09:04:25Z2010-12-24T09:04:25ZThanks. But I want a non-trivial extension.