8
$\begingroup$

Let $M$ be a hypersurface (a submanifold of codimension 1) in $\mathbb{R}^{n}$. Is it true that it's Gaussian curvature intrinsic? (when $n>3$).

Reminder:

We focus our attention on a small part of $M$ which is orientable, and choose a smooth unit normal vector field $N$. Then we obtain the shape operator $s$, satisfying: $$sX=-\nabla_xN$$ where $\nabla$ is the Levi-Civita connection on $\mathbb{R}^{n}$. $s$ is self-adjoint, hence there are $n$ real eigenvalues $k_1,...k_n$, and the Gaussian curvature is defined as $\det S=k_1k_2\cdots k_n$


As noted by Ivo Terek, for odd $n$, the answer is positive.

For even $n$, I would still like to know whether the curvature is intrinsic up to sign, i.e if for a given $(M,g)$ only two values are possible? (one is the negative of the other).

Added Clarification

Assume we have an abstract $n−1$ dimensional Riemannian manifold, which can be embedded in $\mathbb{R}^n$. (By abstract I mean we do not have a "preferred" or "canonical" embedding). The Gaussian curvature is defined via an embedding.

The question is whether it's possibe to get different values of the curvature when computing it w.r.t different embeddings. (in the case of even $n$, we consider $k$,$−k$ as the same values, since by changing the direction of the normal, we change the curvature's sign).

For $n=3$ Gauss's Theorema Egregium famously asserts the curvature is an intrinsic invariant, i.e depending only on the metric of $M$, and not on the embedding chosen.

The question is whether this remains true in higher dimensions?

$\endgroup$
  • $\begingroup$ It is proved by Milnor in his unpublished paper. See his collected papers, volume 1. $\endgroup$ – user60933 Apr 27 '17 at 8:09
6
$\begingroup$

To answer the question you asked in the comments: Yes, the Gaussian curvature of a hypersurface is an intrinsic isometry invariant up to sign. One reference for this is Volume 4 of Spivak's Comprehensive Introduction to Differential Geometry. In my second edition it's Corollary 23 in Chapter 7.

$\endgroup$
3
$\begingroup$

It is intrinsic for $n$ odd, and not if $n$ is even. If you pick a normal field $N$, you will compute the Gaussian curvature (Gauss-Kronecker curvature) $K = \det S$, but if you picked the other normal field $-N$ instead, you would have the shape operator $-S$ associated and $\det S = \det(-S)$ only if $n-1$ is even, that is, if $n$ is odd. The curvature shouldn't depend on the choice of normal field if it is to be intrinsic.

$\endgroup$
  • $\begingroup$ Thanks. I agree about the sign ambiguity, in the case of even $n$. But is the absolute value $|K|$ intrinsic? and how do we see $K$ is intrinsic for odd $n$? the fact its sign is independent of the choice of normal is not enough. $\endgroup$ – Asaf Shachar Feb 28 '16 at 16:54
  • $\begingroup$ I'm not sure I follow.. we only have two choices of normal in each point, and if the value of $K$ computed for each choice is the same, then there is no problem (mainly the remark in page 2 here). $\endgroup$ – Ivo Terek Feb 28 '16 at 17:01
  • $\begingroup$ Ok, I see my question perhaps was not phrased accurately. (I meant for other notion of "intrinsic"). Think we have an abstract $n-1$ dimensional Riemannian manifold, which can be embedded in $\mathbb{R}^n$. Now consider different embeddings. The Gaussian curvature is defined via an embedding. The question is whether it's possibe to get different values of the curvature with different embeddings. (in the case of even $n$, we consider $k$,$-k$ as the same values). I am updating the question to make this more explicit. $\endgroup$ – Asaf Shachar Feb 28 '16 at 17:17
  • 1
    $\begingroup$ I think I understood the question now. Theorema Egregium holds for the Gauss-Kronecker curvature if it is well-defined as in my answer.. Take a look in page $227$ of Elementary Topics in Differential Geometry, by Thorpe. I think that Theorem $2$ there is what you're looking for. $\endgroup$ – Ivo Terek Feb 28 '16 at 17:24
  • $\begingroup$ Thanks. This still leaves open the case about even $n$ (up to sign). $\endgroup$ – Asaf Shachar Feb 28 '16 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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