1
$\begingroup$

Let $N$ be a compact Riemanian manifold and $G$ be its isometry group. Let $M=\chi^{\infty}(N)$ be the space of smooth vector fields on $N$. There is a natural right action of $G$ on $M$ with $X.g=g^*(X),\; g\in G, X\in M$, the push forward of $X$ under the isometry $g$. So $M$ is a $G$ module.

How can one express the group cohomologies $H^n(G,M)$ explicitely?Is there a reference which contain such computations? What can be said about a riemannian manifold whose all cohomology groups $H^n(G,M)$ vanish?

Edit: according to the comment of Neal I understand the following part of the previous version is a trivial question:

Does this sequence of cohomologies determine the geometry of $N$? Namely is it true to say that two nonisometric metrics on $N$ give two different cohomology sequence?

$\endgroup$
  • 4
    $\begingroup$ Most metrics should have trivial isometry groups. $\endgroup$ – Neal Apr 20 '18 at 19:22
  • $\begingroup$ @Neal thanks for your comment. Is there a reference who computed this cohomology in the non rigid case, e.g for constant curvature? $\endgroup$ – Ali Taghavi Apr 21 '18 at 3:54
  • $\begingroup$ I don't have a reference handy, but I think even for constant curvature one often has trivial isometry group? (Euclidean tori do not.) I would start exploring with hyperbolic surfaces, c.f. mathoverflow.net/questions/222154/… $\endgroup$ – Neal Apr 23 '18 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.