Which coefficients of the characteristic polynomial of the shape operator are isometric invariants?

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.

Solution 1:

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.

Solution 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 $A$.