# Notion of curvature for a volume embedded in $R^3$

This question might sound slightly vague, but please bear with me.

If I have an orientable, closed, sufficiently smooth surface in $R^3$, I can define its principal curvatures, mean curvature as well as the implicit (Gauss) curvature and plenty else using the machinery of Riemannian geometry. In particular, notions of curvature are perfectly well defined and understood for such surfaces.

Now suppose the object of my interest is not such a surface per se, but a certain volume (well-behaved subset of $R^3$) that includes this surface as a sort of a "mean". One example would be a real life hollow sphere, whose mean surface is a 2-sphere, but whose wall has non-zero "thickness".

What is the correct notion of curvature (intrinsic as well as extrinsic) for such entities?

From what (little) I know, such an object is a "3-manifold with boundary" so does one need to apply the definition of curvatures that apply to 3 manifolds, making the curvature a tensor? In that case, would the Gauss (scalar) curvature of such an object be 0?

Pretty much everything you wrote is technically correct - any open subset of $\mathbb R^3$ is a 3-manifold, and if you include the boundary then it is a 3-manifold with boundary; and it inherits the Riemannian metric from $\mathbb R^3$. The geometry you get is not very interesting, though - since curvature is a local notion, the fact that your set is open means that it is locally isometric to the (flat!) space it sits inside. Thus you can define the Riemannian/Ricci/Scalar curvatures of such a manifold, but they are all zero; so they're not going to be useful in quantifying anything about the set. From the pure DG perspective the only non-trivial geometry here is that of the boundary surface, which you seem familiar with.

If you're interested in a theoretical approach, the only thing I could think of would be to study the geometry of an appropriate topological skeleton - this is in some sense attempting to recover the "mean" surface you talk about. In general skeletons are not submanifolds, but if you choose the correct definition they may be almost everywhere regular enough to define the curvature - I'm not really sure. Hopefully it's at least a lead for some reading.

If you're planning on applying this, perhaps look into surface meshing and normal estimation for point clouds. I'm not sure how well these techniques address the "thickness" issue - just another lead for you.

• Thank you for your reply and the link to "topological skeleton" - I'll check it out. Jun 10, 2015 at 17:04

A $3$-manifold, seen inside $\Bbb R^4$ is nothing more than a hypersurface. We have this generalization of the Gaussian curvature, called the sectional curvature, which for $2$-manifolds reduces to the Gaussian curvature that we already know. This indeed uses the Riemann curvature tensor. From this you can compute the scalar curvature using the Ricci tensor, but I see no straightforward reason to see why the scalar curvature should be zero, since for $2$-manifolds, it also reduces to the Gaussian curvature (modulo a factor of $2$).

Parametrizations are of the form: $${\bf x}(u^1,u^2,u^3) = (x(u^1,u^2,u^3),y(u^1,u^2,u^3),z(u^1,u^2,u^3)),$$ so we have the first fundamental form given by $g_{ij} = \langle {\bf x}_i,{\bf x}_j \rangle,$ with $1 \leq i,j \leq 3$. I would treat the problem as follows: consider your volume in $\Bbb R^4$, using the inclusion: $$(x,y,z)\hookrightarrow (0,x,y,z)$$ or something similar. Then you can consider ${\bf N} = {\bf x}_i \wedge {\bf x}_j \wedge {\bf x}_k$, a normal vector, and the shape operator associated. The Gauss-Kronecker curvature is the product of the eigenvalues of the shape operator. For $2$-manifolds this also reduces to the Gaussian-Curvature.

So you have a lot of ways to generalize the Gaussian curvature - it depends on which kind of analysis you want to make. Meaning that there are several results which are stated in terms of these different curvatures.

As far as I know, studying submanifolds $M^k \subset \Bbb R^n$ is not so easy, but the theory of hypersurfaces $M^{n-1}\subseteq \Bbb R^n$, such as the one in our discussion here, has a lot of analogies with the classical theory of surfaces in $\Bbb R^3$. This book, from Montiel and Ros, as far as I recall, do some stuff for hypersurfaces, and talk a bit about the Gauss-Kronecker curvature I said above. I'm not sure that this explanation was what you wanted, but I hope it helps.

• Ivo, maybe I missed something, but the OP seemed to be talking about domains with boundary in $\Bbb R^3$. You're thinking about a hypersurface without boundary in $\Bbb R^4$, sitting inside a hyperplane, which really doesn't help things, IMHO. Jun 10, 2015 at 13:32
• Welp, more likely I missed something, than you. Makes sense, though. I'll leave my answer there.. at least I think I addressed the "What is the correct notion of curvature (intrinsic as well as extrinsic) for such entities?" properly - meaning that there is no "correct" notion.. Jun 10, 2015 at 15:30