# Quantitative upper bound on mean curvature of an isometric embedding

By Nash embedding theorem, any complete Riemannian manifold $$M$$ can be isometrically embedded in $$\mathbb{R}^N$$, for sufficiently large $$N$$.

The proof of the theorem is quite involved, and it is not clear to me how the mean curvature of the produced embedding depends on $$M$$.

My question is the following. Is there a way to build an isometric embedding with mean curvature bounded from above in terms of appropriate intrinsic geometric data of the original manifold (e.g. curvature bounds, injectivity radius, etc)?

An example of the result that I would hope is the following statement: There exists a constant $$c_n>0$$, depending only on $$n$$, such that for any complete $$n$$-dimensional Riemannian manifold $$M$$ there exists $$N>0$$ and an isometric embedding of $$M$$ in $$R^N$$ such that the mean curvature of the embedding is not greater than $$c_n \max\{K,1/\mathrm{i}\}$$, where $$K$$ is the upper bound on the absolute value of the sectional curvatures of $$M$$ and $$\mathrm{i}$$ is a lower bound on the injectivity radius of $$M$$.

I am in particular interested in the case when $$M = \mathbb{R} \times_f Z$$ is the warped product of the euclidean $$\mathbb{R}$$ and a compact Riemannian manifold $$(Z,h)$$ with warping factor $$f : I \to Z$$.

I found plenty of lower bounds to the mean curvature of an isometric embedding in terms for so-called $$\delta$$-invariants defined by Chen (see e.g. here), but none of this results implies that an isometric immersion satisfying such a bound actually exists.

This question is somewhat related with this one.

EDIT: In the above, by "mean curvature" I mean the norm of the trace of the second fundamental form of the embedding. The second fondamental form is a symmetric tensor with values in the normal bundle.

• Could you say why a bound on the second fundamental form isn’t sufficient for your needs? – Deane Yang Jan 4 at 19:16
• The second fundamental form of the embedding is exactly what I would like to control (more precisely, the norm of its trace). I tried to clarify a bit the question in light of your comment. – Raziel Jan 4 at 20:00
• This is not my area of expertise, but it seems that the fact that you can isometrically embedded $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ in a way that has arbitrarily large mean curvature suggests you are going to have to essentially reproduce Nash's proof, but with quantitative estimates. – RBega2 Jan 4 at 20:06
• Have you looked at Gromov’s book Partial Differential Relations? It’s possible that it addresses this or similar questions. – Deane Yang Jan 4 at 21:26
• I think the answer is already provided in previous MathOverflow discussion. What's not mentioned there is that the second deformation step is probably best addressed using Gunther's proof, which is far simpler than previously known proofs. Tao (terrytao.wordpress.com/2016/05/11/…) has a nice description of the entire proof. You just need to verify that at each step, you can bound the second fundamental form in terms of the curvature, injectivity radius, and any prior bound on the second fundamental form. This is messy but should be straightforward. – Deane Yang Jan 5 at 0:05

## 1 Answer

I think (with one caveat) that your question has a negative answer for families of very squashed spheres.

Let $$D_a^\pm=\{(x,y,\pm a^{-1}): x^2+y^2\leq a\}$$ be the disk of radius $$a$$ at height $$\pm a^{-1}$$ and let $$T_a= \{(x,y,z):(x-a \cos(\theta))^2+(y-a\sin (\theta))^2+z^2=a^{-2}, \theta\in [0,2\pi]\}$$ be the the surface obtained by rotating the semicircle of radius $$a^{-1}$$ centered at $$(0,0, a)$$ around the $$z$$-axis.

Set $$\Gamma_a=D_a^+\cup T_a \cup D_a^-.$$ This is a $$C^{1,1}$$ convex surface. Clearly as long as $$a\geq 1$$:

• The Gauss curvature is bounded between $$0$$ and $$2$$.
• The maximum of the mean curvature is at least $$a$$.
• The injectivity radius is bounded from below by $$a$$.
• If desired these can be smoothed without changing the relevant properties.

The convexity ensures this surface is isometrically rigid in $$\mathbb{R}^3$$ so there is no other isometric immersion into $$\mathbb{R}^3$$. Letting $$a\to \infty$$ shows that one can't have your desired estimate.

The caveat: I don't know whether one has (or should expect) this rigidity when one embeds into higher codimension.

• Yes, the question has a positive answer only if the codimension is sufficiently high. – Deane Yang Jan 4 at 23:56
• @DeaneYang That doesn't surprise me. That being said, Is there a (straightforward) way to embed the families in my answer into $\mathbb{R}^N$ so that the mean curvature is bounded? – RBega2 Jan 5 at 0:13
• I don't know how to do it explicitly. – Deane Yang Jan 5 at 0:32
• @RBega2, in my question the dimension of the ambient can be arbitrarily large. Hence your nice example (for which I thank you) does not provide an answer. – Raziel Jan 5 at 8:32