# Symmetries vs. Bound in codimension of Nash isometric embedding

Let $(M^m,g)$ be a compact smooth Riemannian manifold of dimension $m$. From the celebrated Nash Embedding Theorem, we know there exists a (smooth) isometric embedding $M\hookrightarrow\mathbb R^n$ on the Euclidean space of dimension $n\leq m(3m+11)/2$.

My (quite loose) question is whether the amount of symmetry 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?

By amount of symmetry I mean any possible ways to measure how symmetric $(M,g)$ is, perhaps the most usual being the symmetry rank (rank of the isometry group); symmetry degree (dimension of the isometry group) and cohomogeneity (dimension of the orbit space $M/Iso(M)$).

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?

About a possible converse question: if we know 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)?

• You probably already know this, but it is worth remarking that while there is an equivariant Nash embedding theorem, for all $n \geq 4$ there is a left-invariant metrics on $S^3$ (the unit quaternions) such that with this metric $S^3$ cannot be equivariantly embedding in $\mathbb{R}^k$ for any $k \leq n$. See MR0583381 (81h:53059) Moore, John Douglas; Schlafly, Roger On equivariant isometric embeddings. Math. Z. 173 (1980), no. 2, 119–133. – Andy Putman Feb 1 '12 at 5:11
• (I know the above doesn't answer your question, but it is somewhat related) – Andy Putman Feb 1 '12 at 5:12
• @Andy may be "k>= 4" ? otherwise I am not clear... Any way very interesting comment – Alexander Chervov Feb 1 '12 at 8:05
• @Alexander Chernov : Yes, $k \geq 4$ is correct, sorry for the typo. – Andy Putman Feb 1 '12 at 17:57

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.