4
$\begingroup$

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 :)).

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?

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.

Hope my question is not too trivial. If it is, please give me a good reference on the subject (with a differential geomety point of view if possible).

Thanks in advance.

Edit: To answer a comment, here is the definition of basic cohomology Let $(M,F)$ be a foliated manifold. A basic form on M is a form $\omega$ satisfying $\iota_X\omega=0$ and ${\mathcal L}_X\omega=0$ for every vector field tangent to the foliation at every point. The exterior derivative d sends a basic form to a basic form, so the space $\Omega_{\textrm{bas}}(M)$ of basic forms is a differentiel complex. The basic cohomology is its cohomology ring. So, its coefficient ring is $\mathbb R$ and I am considering equivariant cohomology with the same ring.

Improve this question
$\endgroup$
6
  • $\begingroup$ Can you please recall for me what basic cohomology is, and maybe say which coefficients you consider. $\endgroup$
    – Konrad Waldorf
    Jun 15, 2012 at 15:23
  • $\begingroup$ Sorry for late answer (week-end...) and thank you for your interest in my question. Let $(M,{\mathcal F})$ be a foliated manifold. A basic form on $M$ is a form $\omega$ satisfying $i_X\omega=0$ and $L_X\omega=0$ for every vector field tangent to the foliation at every point. The exterior derivative $d$ sends a basic form to a basic form, so the space $\Omega_{bas}(M)$ of basic forms is a differentiel complex. The basic cohomology is its cohomology ring. So, its coefficient ring is $\mathbb R$ and I am considering equivariant cohomology with the same ring. $\endgroup$
    – Taladris
    Jun 18, 2012 at 1:25
  • 1
    $\begingroup$ The Paper arxiv.org/pdf/1004.1043v1.pdf might be of interest to you, even though it does not exactly adress your question. $\endgroup$
    – ThiKu
    Oct 15, 2013 at 15:57
  • $\begingroup$ @user39082: Thank you very much. Unfortunately, I already know this example. It deals with a mix of basic and equivariant cohomologies: a group action transverse to a foliation. But it is too complicated for me. $\endgroup$
    – Taladris
    Oct 16, 2013 at 15:42
  • 3
    $\begingroup$ I would really really like it if people would not post links to the PDF of an arXiv e-print but to the abstract. $\endgroup$
    – José Figueroa-O'Farrill
    Mar 14, 2015 at 17:13

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.