embedding torus - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T14:27:35Zhttp://mathoverflow.net/feeds/question/99296http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/99296/embedding-torusembedding torusgeomtfree2012-06-11T13:16:54Z2012-08-17T14:27:17Z
<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#99306Answer by Lee Mosher for embedding torusLee Mosher2012-06-11T16:08:04Z2012-06-11T16:08:04Z<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#99339Answer by geomtfree for embedding torusgeomtfree2012-06-12T07:05:05Z2012-06-12T07:05:05Z<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>