Theorem of Bryant in higher dimensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:41:09Z http://mathoverflow.net/feeds/question/81440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81440/theorem-of-bryant-in-higher-dimensions Theorem of Bryant in higher dimensions gary 2011-11-20T17:48:19Z 2011-11-20T19:23:23Z <p>hallo,</p> <p>i have the following question. i read about Bryant's theorem which sais that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? Tanks in advance. </p> <p>Mira</p> http://mathoverflow.net/questions/81440/theorem-of-bryant-in-higher-dimensions/81444#81444 Answer by Robert Bryant for Theorem of Bryant in higher dimensions Robert Bryant 2011-11-20T19:23:23Z 2011-11-20T19:23:23Z <p>First, the hypotheses of the theorem I proved require that $Y$ be compact and oriented, in addition to requiring that $g$ be real-analytic.</p> <p>Second, the method I used (the Cartan-Kähler Theorem) extends, essentially without modification, to higher dimensions as long as $Y$ is compact and parallelizable and $g$ is real-analytic.</p> <p>Real-analyticity is certainly necessary, since a minimal submanifold of a real-analytic Riemannian manifold (such as a Calabi-Yau manifold in any dimension) is necessarily real-analytic itself.</p> <p>By contrast, not all special Lagrangian submanifolds of a Calabi-Yau are parallelizable. Thus, parallelizability is not necessary in general, but I don't know how to remove that hypothesis in the existence proof. For example, I do not know whether every real-analytic metric on $S^4$ is obtainable by embedding it as a special Lagrangian in some $4$-dimensional Calabi-Yau. </p>