Symmetries vs. Bound in codimension of Nash isometric embedding - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:05:22Z http://mathoverflow.net/feeds/question/87205 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87205/symmetries-vs-bound-in-codimension-of-nash-isometric-embedding Symmetries vs. Bound in codimension of Nash isometric embedding Renato G Bettiol 2012-02-01T04:45:31Z 2012-02-01T16:59:39Z <p>Let $(M^m,g)$ be a compact smooth Riemannian manifold of dimension $m$. From the celebrated <a href="http://en.wikipedia.org/wiki/Nash_embedding_theorem" rel="nofollow">Nash Embedding Theorem</a>, we know there exists a (smooth) isometric embedding $M\hookrightarrow\mathbb R^n$ on the Euclidean space of dimension $n\leq m(3m+11)/2$.</p> <blockquote> <p>My (quite loose) question is whether the <em>amount of symmetry</em> of $(M,g)$ should influence this estimate on the maximal possible codimension for which it embeds isometrically in Euclidean space. In other words, is there any intuition if highly symmetric (vs. generic) metrics on $M$ impose smaller (vs. larger) bounds on the codimension of the isometric embedding?</p> </blockquote> <p>By <em>amount of symmetry</em> I mean any possible ways to measure how symmetric $(M,g)$ is, perhaps the most usual being the <em>symmetry rank</em> (rank of the isometry group); <em>symmetry degree</em> (dimension of the isometry group) and <em>cohomogeneity</em> (dimension of the orbit space $M/Iso(M)$).</p> <blockquote> <p>For example, assume there is an isometric action of the circle $S^1$ on $(M,g)$. Should we (expect to) be able to embed $(M,g)$ isometrically into an Euclidean space with lower (higher?) codimension than if $M$ was endowed with a generic metric?</p> </blockquote> <p>About a possible converse question: if we <em>know</em> a certain metric on $M$ allows it to be embedded with "low codimension", should this indicate anything about how symmetric this metric is? Maybe taking arbitrarily small perturbations of the embedded submanifold one could destroy the symmetries while keeping it embedded (so that there is no relation between those things)?</p> http://mathoverflow.net/questions/87205/symmetries-vs-bound-in-codimension-of-nash-isometric-embedding/87210#87210 Answer by Deane Yang for Symmetries vs. Bound in codimension of Nash isometric embedding Deane Yang 2012-02-01T07:53:18Z 2012-02-01T07:53:18Z <p>Here is a rather incomplete answer:</p> <p>First, it is probably important to distinguish between global and local isometric embeddings. I would say that global embeddings are still quite poorly understood, and symmetry does not seem to help at all. But you might want to look through Gromov's book, Partial Differential Relations, to see if he is able to take advantage of symmetry or not (I don't remember).</p> <p>Local isometric embeddings are only slightly better understood than global. Constant curvature metrics certainly embed in lower dimension. You know this for the spherical and flat metrics. The $n$-dimensional hyperbolic metric can be embedded at least locally into $R^{2n-1}$ using higher dimensional generalization of the sine-Gordon equation (studied by Tenenblat and Terng). I do not know, however, what is known about local isometric embeddings of other symmetric metrics.</p>