# Equivariant and basic cohomology

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).

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.
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. – Taladris Jun 18 '12 at 1:25