0
$\begingroup$

If we can construct an Einstein metric for a simply connected Riemannian manifold, why does it necessarily follow it is diffeomorphic to the $3$-sphere?

$\endgroup$
0

1 Answer 1

8
$\begingroup$

Firstly, without the hypothesis that the manifold is compact, this is not true, with hyperbolic space and Euclidean space being the only other complete examples. Moving on, in three dimensions an Einstein metric has constant sectional curvature (this is just linear algebra). Hence, any complete Einstein metric on a simply connected 3-manifold is isometric to the sphere, hyperbolic space or Euclidean space by the classification of space forms which you can find in any text on Riemannian geometry. If the manifold is assumed compact and $1$-connected, then the only option is that the manifold is isometric, hence diffeomorphic, to the $3$-sphere since the others are non-compact.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.