Welcome to MathOverflow

Question

0

hi everybody 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?

edited Aug 17 at 14:27

Spiro Karigiannis
2,282●2●10●25

asked Jun 11 at 13:16

geomtfree
9

geomtfree · Answer 1 · 2012-06-12 07:05:05Z UTC

-1

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?

answered Jun 12 at 7:05

geomtfree
9

1

You should edit your question (and possibly register, so you don't keep creating new accounts). – S. Carnahan♦ Jun 12 at 7:34

Lee Mosher · Answer 2 · 2012-06-11 16:08:04Z UTC

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.

embedding torus

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.