Equariant and basic cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:58:27Z http://mathoverflow.net/feeds/question/99687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99687/equariant-and-basic-cohomology Equariant and basic cohomology Taladris 2012-06-15T09:21:25Z 2012-06-15T09:21:25Z <p>Hello everyone, </p> <p>I have difficulties to understand the connection between equivariant and basic cohomology. I understand the definition of them but not how they are related (the Weil algebra killed me :)). </p> <p>For example, consider a smooth manifold $M$ with a smooth action of a Lie group $G$. If we assume all orbits have the same dimension, this induces a foliated structure on M whose leaves are the orbites of the action. So, we have two cohomologies to look at: the basic and the equivariant ones. How are they related? </p> <p>I have no special assumptions on $G$ and $M$. $G$ has no reason to be compact, and the action is not even proper (for example, the foliation of $S^1\times S^1$ induced by the foliation of $[0,1]\times[0,1]$ by parallel lines of non rational slope is a case I would like to understand). In the special case I am looking, $G$ is commutative but I am also interested in what happens when $G$ is not.</p> <p>Hope my question is not too trivial. If it is, please give me a good reference on the subject (with a differential geometer point of view if possible).</p> <p>Thanks in advance. </p>