Counterexample showing that G-invariant de Rham cohomoloy different from cohomology of G-invariant sub-complex? – mathoverflow.net

Counterexample showing that G-invariant de Rham cohomoloy different from cohomology of G-invariant sub-complex? – mathoverflow.net

10:03 No Comments
If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex ...

from Hot Questions - Stack Exchange OnStackOverflow
via Blogspot

Share this

Unknown

0 Comment to "Counterexample showing that G-invariant de Rham cohomoloy different from cohomology of G-invariant sub-complex? – mathoverflow.net"

Post a Comment

Subscribe to: Post Comments (Atom)