Vasily Rogov Vasily Rogov This user account status is Approved About About Posts Posts Comments Comments Ah, you are right, this happens because $A$ might be not isometric to $B$, but isometric to it's connected component… On Isometries of a product of Riemannian manifolds Actually, here is an answer by Ivan Solonenko on a similar (but more particular) question. In fact he proves that… On Isometries of a product of Riemannian manifolds >Notation question Yes, thank you, that's a typo; >Silly nitpick I think, we don't need the connectedness. >Elementary question Hmm...… On Isometries of a product of Riemannian manifolds
