Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a vague question because I'm not sure what I want to ask.

An ellipsoid has positive curvature everywhere, and bounds a convex subset of $\mathbb R^3$. What I want to say now is "It seems as though a surface with negative curvature can't be convex" but this isn't right, because the ellipsoid itself is not convex, and I could only get at what I was thinking of by observing that it bounds a subset of $\mathbb R^3$.

But this isn't quite what I want, because the ellipsoid is also the boundary of a non-convex subset of $\mathbb R^3$, namely the complement of the convex region. Well, the convex region is bounded, okay, so I can pick out the convex region I want that way, and say that if a surface is the boundary of a bounded subset $S$ of $\mathbb R^3$, and has positive curvature everywhere, then $S$ is convex. But this is not as general as what I want, because a paraboloid also has positive curvature everywhere, and it is the boundary of a convex subset of $\mathbb R^3$, but the claim of the previous sentence doesn't apply to it.

Maybe what I want is something like: let $P$ be a point on a surface, and let $C$ be a small closed curve on the surface that bounds a region $R$ of the surface that contains $P$. Then the subset of $\mathbb R^3$ bounded by $R$ and by (something?) is convex if and only if the Gaussian curvature at $P$ is positive. Or perhaps the condition is that the curvature at every point of $R$ must be non-negative. Or something.

It seems like there are some theorems here, but I don't know quite where to look for them. Are there theorems relating Gaussian curvature of a surface to the convexity of the region of which it is the boundary?

Is it at least true that the boundary surface of a convex subset of $\mathbb R^3$ has non-negative curvature everywhere that its curvature is well-defined? What I am looking for is something like the converse of that.

share|cite|improve this question
I asked a somewhat related question here. – Rahul Feb 13 '13 at 17:04
Seems like there would be many more applicable results if you placed a restriction on the types of surfaces that interest you. Suppose you only considered surfaces of the form $f(x,y,z)=0$. I would guess that it would be fairly easy to relate the properties of this surface with the properties of the set $\{(x,y,z) \in \mathbb R^3 : f(x,y,z) \le 0\}$. This restriction is quite a strong one, and many surfaces would be excluded (I think), but the ones that interest you might still be covered. – bubba Feb 14 '13 at 1:18
up vote 9 down vote accepted

For the final question posed:

Is it at least true that the boundary surface of a convex subset of $\mathbb{R}^3$ has non-negative curvature everywhere that its curvature is well-defined?

The answer is yes. Convexity implies that for every point $p$ on the boundary $\partial \Omega$, we can find a plane $\Pi$ through $p$ such that $\partial\Omega$ lies "on the one side" of $\Pi$. This implies that the second fundamental form is signed (either positive-semi-definite or negative-semi-definite). For surfaces in $\mathbb{R}^3$, the Gaussian curvature is the determinant of the second fundamental form, and hence is the product of the eigenvalues (the principal curvatures), and hence is always positive definite.

Now, as a general comment, convexity is more closely related to the extrinsic curvature of the boundary $\partial\Omega$ rather than the induced intrinsic curvature of it. It just happens that that for embeddings into Euclidean spaces (and especially into $\mathbb{R}^3$), we have easy relations between intrinsic and extrinsic curvatures; this drives what you observe with Gaussian curvature.

In fact, we have the following theorem:

Thm Let $\Omega\subseteq \mathbb{R}^n$ be an open domain, such that $\partial\Omega$ is an orientable smooth codimension 1 Riemannian submanifold. Then $\Omega$ is convex if and only if there exists a choice of orientation on $\partial\Omega$ such that the associated second fundamental form is positive semi-definite.

Sketch of proof:

(1) Essentially by Taylor expansion we see that "the second fundamental form is positive semi-definite everywhere" is equivalent to "locally $\partial\Omega$ is supported by a hyperplane". (The $\Leftarrow$ direction is obvious by taking local parametrization of $\partial\Omega$ as a graph over the hyperplane. As is the $\Rightarrow$ direction if the second fundamental form is positive definite. For the positive semi-definite case, when we encounter a zero eigenvalue, we need to use the fact that the form is positive semi-definite in an entire neighborhood to rule out the case of "bad" higher order Taylor coefficients.)

(2) Once we have the result of the first step, we can appeal to Tietze's theorem which establishes the equivalence between local convexity (existence of local supporting hyperplanes) and global convexity. See the link for a bit more discussion.

In the case of $n = 3$, as mentioned above the positive-semi-definiteness of the second fundamental form is directly tied to the non-negativity of the Gaussian curvature. For higher dimensions, the scalar curvature only controls the second symmetric polynomial of the eigenvalues of the second fundamental form, so is insufficient to control the local convexity of the boundary.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.