10

Let $M^n \subset \mathbb{R}^{n+1}$ be an isometrically immersed Riemannian hypersurface. The shape operator $s$ is the $(1,1)$ tensor field characterized by $$\langle X, sY \rangle = \langle \text{II}(X,Y), N\rangle,$$ where $X$, $Y$ are vector fields, $\text{II}$ is the second fundamental form, and $N$ is a unit normal vector field. We can then define the mean curvature and Gaussian curvature by $$H = \frac{1}{n}\text{tr}(s), \ \ \ \ \ \ K = \det(s).$$

Gauss' Theorema Egregium is the statement that $K$ is an isometry invariant of $M$ when $\mathbf{n = 2}$. By contrast, $H$ is not. This makes me wonder about the following:

Question: Let $p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \ldots + a_0$ denote the characteristic polynomial of the shape operator $s$. For $n > 2$, are any of the coefficients $a_i$ (local) isometry invariants of $M$? If so, which?

Again, a previous question of mine was meant to get at this, but my thoughts were not quite so clear.

Community
  • 1
Jesse Madnick
  • 32,819
  • 1
    I'm not sure that the gaussian curvature is an isometry invariant for an hypersurface of arbitrary dimension $ n $. Are you sure about it? If $ n=2 $ we know that the gaussian curvature is an isometry invariant because it is equal to the sectional curvature of $ M $ (note in this case the sectional curvature is merely a function on $ M $). –  Mar 02 '13 at 07:56
  • @user55449: Wow, I hadn't realized that at all. Thank you for clarifying; that's a really good point. Hm, I'm not quite sure how to salvage this question. Oh, well. – Jesse Madnick Mar 02 '13 at 08:54
  • OK, I've edited the question accordingly, though it's somewhat more pessimistic. – Jesse Madnick Mar 02 '13 at 09:01

2 Answers2

3

If $n>2$ then all the coefficients $a_m$ are isometry invariants of generic $M$. Say, it sufficient to assume that $M$ has positive curvature, but as you will see much weaker assumptions can be made.

Let $e_i$ be the principle basis and $\kappa_i$ be principle curvatures. The curvature operator of $M$ has eigenvectors $e_i\wedge e_j$ with eigenvalues $K_{ij}=\kappa_i\cdot\kappa_j$. If $K_{ij}>0$ and $n>2$, you can recover $\kappa_i$ from $K_{ij}$, say $$\kappa_i=\sqrt{\frac{K_{ij}\cdot K_{ik}}{K_{jk}}}.$$ Since the values $K_{ij}$ are isometry invariants of $M$ so are all $\kappa_i$ and symmetrizing we get all $a_m$ (up to sign).

About other conditions: say nonzero Gaussian curvature $G=\kappa_1\cdots\kappa_n$ will do as well and the proof is the same. In particular $|G|$ is invariant.

On the other hand for the hyperplane $|H|=\mathrm{const}\cdot |a_1|$ is not invariant, so you should assume something about curvature at the point.

Anton Petrunin
  • 669
  • Interesting. Can you clarify some things for me? Perhaps in an edit? (1) Are you saying that the $\kappa_i$ are intrinsic isometry invariants of generic $M$ when $n > 2$? (2) What specific operator do you mean by "curvature operator of $M$"? (3) I thought negative curvature was in some sense more generic than positive curvature? – Jesse Madnick Mar 02 '13 at 23:07
  • @Jesse: (1) Yes in generic case, say if $G\ne 0$; (2) http://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds (3) Well, if $n>2$, negatively curved mnfld can not be isometric to hypersurface... – Anton Petrunin Mar 02 '13 at 23:52
  • Thank you, that makes sense. Do you have any references for (3)? My differential geometry education has consisted of classical surfaces in $\mathbb{R}^3$, and abstract Riemannian manifolds, with virtually no bridge in between. In particular, I know almost nothing (as you can see) about how the geometry of surfaces in $\mathbb{R}^3$ generalizes (or not) to hypersurfaces in $\mathbb{R}^{n+1}$. – Jesse Madnick Mar 03 '13 at 01:23
  • 1
    @JesseMadnick I don't have a reference for (3), but I think I see the reason: write the 3-dimensional hypersurface locally as$$x_4=\sum_{i,j=1}^3 a_{ij} x_ix_j+o(|x|^2)$$ and observe that whatever signature the form $(a_{ij})$ has, it is either positive semidefinite or negative semidefinite on some two-dimensional subspace. Therefore, a hypersurface cannot have negative sectional curvature. –  Mar 03 '13 at 03:04
2

A reference for a fairly general statement is Spivak's "A Comprehensive Introduction to Differential Geometry," Volume IV, Section 7D. I'll just quote the results here.

Let $M^n \subset \mathbb{R}^{n+1}$ be an isometrically immersed hypersurface. Let $\kappa_1, \ldots, \kappa_n$ be the principal curvatures. Define the symmetric curvatures $K_j$ by $$K_j = \binom{n}{j}^{-1}\sigma_j(\kappa_1, \ldots, \kappa_n),$$ where $\sigma_j$ is the $j$th elementary symmetric polynomial. In particular, \begin{align*} H = K_1 & = \textstyle \frac{1}{n}(\kappa_1 + \cdots + \kappa_n) \\ K = K_n & = \kappa_1 \cdots \kappa_n. \end{align*} are the mean and Gauss-Kronecker curvatures, respectively.

Prop: The set of the $\binom{n}{2}$ numbers $\{\kappa_i \kappa_j \colon i < j\}$ is invariant under isometry. In particular, the even symmetric curvatures $K_{2r}$ are invariant under isometry.

According to Spivak, the "in particular" follows from the algebraic fact that the even powers of the characteristic polynomial of a matrix $A$ can be expressed in terms of the determinants of the $2 \times 2$ submatrices of $A$.

Jesse Madnick
  • 32,819