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could anyone please help me? why is it impossible to embed a torus in R^3 with index 1 ( usual euclidean space with index 1 as a semi-riemannian manifold) as a semi-riemannian submanifold?


thanx. but i didn't understand. let me ask another question. why with this metric R^3 doesn't have any compact semi-riemannian submanifold?

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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Daniel Moskovich, Andres Caicedo, BS. Sep 22 '13 at 7:24

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1 Answer 1

I suppose that by "usual" you mean $\mathbb{R}^3$ with the semi-Riemannian metric $dx^2 + dy^2 - dz^2$?

If so, for any compact surface $S$ embedded in $\mathbb{R}^3$, since the metric is invariant under translation on $\mathbb{R}^3$ you are allowed to translate $S$ so that it is tangent to the light cone $x^2 + y^2 - z^2 = 0$. At any point of tangency the restricted metric is indefinite. To do this translation, first translate $S$ so that it is strictly inside the light cone, then let $S$ gently descend until the first moment that it touches the light cone.

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