Schwarzschild metric as induced metric

[paweld]

According to Nash theorem http://en.wikipedia.org/wiki/Nash_embedding_theorem" every Riemannian manifold can be isometrically embedded
into some Euclidean space. I wonder if it's true also
in case of pseudoremanninan manifolds. In particular is it possible to find
a submanifold in pseudoeuclidean space that, the metric induced on it will be
Schwarzschild metric? How many dimensions we need?

 

[George Jones]

Chris Clarke* showed that every 4-dimensional spacetime can be embedded isometically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

* Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428

 

[Passionflower]

You need 6 dimensions to embed a Schwarzschild solution. I think that all GR solutions can be (locally) embedded in 10 dimensions.

 

[paweld]

You need 6 dimensions to embed a Schwarzschild solution.

How do you know it?