quasar987

Well, in general, no, the analogue of the map we created for general group is not going to be a diffeomorphism. For instance, if instead of S^1 and R^1, you take two R^1's (both acting in the same way on R²), then we see fairly easily that the differential will be at every point non surjective. Similarly if we take two S^1's. The key for the above example to work was that the orbits are transverse.