4
$\begingroup$

Let $M$ be a smooth manifold and let $G$ be a Lie group smoothly acting on $M$.

Then, under suitable assumptions (if $G$ acts freely and properly on $M$) we have a new smooth manifold $M/G$ corresponding to the orbits of the action.

I would like to know if there is a theorem that states (under suitable assumptions) that the set of fixed points $M^G$ can be equipped with a smooth manifold structure.

I suppose there is such a theorem, because $M^G$ is also the zero set of the infinitesimal generator of the action, which is a smooth vector field, so we have "smooth equations" describing it.

I will greatly appreciate any reference about this topic.

Thanks for your help!

  • 0
    The exponential map in a neighborhood of any point is a diffeomorphism, so if you have a smooth section of linear subscpaces of the tangent bundle, it will determine a submanifold.2012-12-07
  • 0
    @Berci: Thanks. I found a similar idea using the exponential map in the book on heat kernels by Berline-Getzler-Vergne (Proposition 7.12) but I couldn't understand it. Could you elaborate a little bit in an answer below?2012-12-07

1 Answers 1