Are there some intrinsic invariants of surfaces other than Gaussian curvature?

Question

The principal curvatures of a surface is denoted by $\kappa_{1}, \kappa_{2}$.

Let $P(x,y)$ be a polynomial with real coefficients. Assume that $P(\kappa_{1}, \kappa_{2})$ is an intrinsically invariant quantity of all surfaces in $\mathbb{R}^{3}$(It is invariant under isometries of surfaces).

Is it true to say that $P(x,y)$ is in the form $P(x,y)=F(xy)$ for some one variable polynomial $F$?

In fact this question, which is motivated by "Gauss theorema egregium", asks:

Are there some "Theorema Egregiums" other than "Gauss theorema Egregium"?

Answer 1

As others have pointed out, it's not hard to show that any function $F(\kappa_1,\kappa_2)$ that is intrinsic to the surface metric must be a function of $K = \kappa_1\kappa_2$, so that settles what one might call the 'lowest-order' case. However, there are certainly higher-order versions. For example, the expression $|\nabla K|^2$ is an intrinsic invariant, and it can be expressed as a polynomial in the second fundamental form and its first covariant derivative (a good exercise in a curves and surfaces course). One might want to think of this as a 'higher-order' version of Gauss' theorem, but it's not very exciting because, in some sense, it's a derivative of Gauss' theorem.

A natural question that arises (and the one that I thought the OP wanted to ask, based on the title of the question) is whether there is any higher-order theorem of this kind that is not just a derivative (of some order) of Gauss' theorem. The answer to this question is 'no', in the following more precise sense:

Suppose given a surface described locally as a graph $z = f(x,y)$ where $f(0,0) = f_x(0,0) = f_y(0,0) = 0$, so that $f$ has a Taylor series expansion of the form $$ f = \tfrac12 c_{20} x^2 + c_{11} xy + \tfrac12 c_{01} y^2 + \tfrac16 c_{30} x^3 + \cdots = \sum_{i+j\ge2} \tfrac1{i!j!} c_{ij}\, x^iy^j. $$ Then Gauss' theorem says that $K(0,0) = c_{20}c_{02}-{c_{11}}^2$. In fact, as is not difficult to show, if one takes the Taylor series of $K$ to be of the form $$ K = \sum_{i+j\ge0} \tfrac1{i!j!} b_{ij}\, x^iy^j, $$ that there exist formulae of the form $b_{ij} = B_{ij}(c)$ where $B_{ij}$ is a universal polynomial in the $c_{kl}$ for which $k+l\le i+j+2$. In fact, for each order $d$, one can collect these to define polynomial mappings $$ B_d: \oplus_{k=2}^{d+2} S^k(\mathbb{R})\longrightarrow \oplus_{k=0}^{d}S^k(\mathbb{R}) $$ that represent the formula giving the Taylor series of $K$ to order $d$ in terms of the Taylor series of $f$ to order $d{+}2$.

The version of the question that I have in mind is whether every formula expressing an intrinsic invariant of the induced metric of finite order in terms of the second fundamental form and its covariant derivatives must factor through some $B_d$ at the series level. (Gauss' Theorem and the above arguments show that the answer is 'yes' for intrinsic invariants of order $0$.) The answer is that, indeed, every finite order intrinsic invariant function on the domain of $B_d$ must factor through $B_d$. This is a consequence of the usual proofs of the isometric embedding theorem for real-analytic surfaces.

Answer 2

A surfaces of constant curvature $K$ admit number of local embedding into $\mathbb{E}^3$ as the surfaces of revolution. Direct calculations show that any pair $k_1$ and $k_2$ such that $K=k_1\cdot k_2$ appear this way.

So, "yes", any $P(x,y)=F(x\cdot y)$ for some $F$.