embedding torus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:27:35Z http://mathoverflow.net/feeds/question/99296 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99296/embedding-torus embedding torus geomtfree 2012-06-11T13:16:54Z 2012-08-17T14:27:17Z <p>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?</p> http://mathoverflow.net/questions/99296/embedding-torus/99306#99306 Answer by Lee Mosher for embedding torus Lee Mosher 2012-06-11T16:08:04Z 2012-06-11T16:08:04Z <p>I suppose that by "usual" you mean \$\mathbb{R}^3\$ with the semi-Riemannian metric \$dx^2 + dy^2 - dz^2\$?</p> <p>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.</p> http://mathoverflow.net/questions/99296/embedding-torus/99339#99339 Answer by geomtfree for embedding torus geomtfree 2012-06-12T07:05:05Z 2012-06-12T07:05:05Z <p>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?</p>