Is the higher order mean curvature Hr ( for higher dimension) of an hypersurface is instrinsic or extrinsic? For the mean and gaussian curvature there is no problem. I think that if r is odd then Hr is extrinsic (Like H1) and if r is even then Hr is instrinsic.

$\begingroup$ It is known that the only intrinsic local differential invariants of a Riemannian manifold are the metric, the curvature tensor, and its covariant derivatives. For a hypersurface, the Gauss equations say that the latter are all quadratic functions of the second fundamental form and its covariant derivatives, obtained by differentiating the Gauss equations. More invariants can be obtained by contracting indices using the metric tensor. I do not know what you mean by "higher order mean curvature", but it would have to fit within my explanation above to be intrinsic. $\endgroup$– Deane YangOct 28, 2017 at 2:23

$\begingroup$ From the clues of mean and Gaussian curvature I'm guessing $H_r$ is the $r^\text{th}$ elementary symmetric polynomial of the principal curvatures? $\endgroup$– Anthony CarapetisOct 28, 2017 at 5:05

$\begingroup$ When r is even, if we change the direction of the unit normal vector, the Hr change the sign, this is because we have an even product of principal curvature in their definition but in the cas of r odd Hr don't change the sign. this is true I know that instrinsic quantities depend only of the first fundamental form but I see that Hr depend on principal curvatures so the second fundamental form. this is what make aa doubt to my conjecture $\endgroup$– Mohammed MohammedOct 28, 2017 at 13:06
1 Answer
If we let $S: TM \to TM$ denote the shape operator of a hypersurface $M \to \mathbb R^{n+1}$ and $\mathfrak R:\Lambda^2TM\to\Lambda^2TM$ the curvature operator, then the Gauss equation can be written $\mathfrak{R} = \Lambda^2 S.$ Here $\Lambda^2S$ is the obvious induced map on bivectors $(\Lambda^2 S)(v \wedge w) = Sv \wedge Sw.$
Thus the (pointwise) intrinsic Riemannian invariants of $S$ are exactly those that can be written in terms of $\Lambda^2 S$.
Since $S$ is selfadjoint, we can choose an orthonormal basis $e_i$ for $T_p M$ so that $S = \mathrm{diag}(\kappa_1,\ldots,\kappa_n)$. Just as we can get the standard mean curvature by taking the trace of $S$, we can get the higher invariants of $S$ by taking traces of $\Lambda^k S \colon$ we compute
$$\begin{align} \mathrm{tr}(\Lambda^k S) &= \sum_{i_1<\ldots<i_k} \langle e_{i_1}\wedge\cdots\wedge e_{i_k}, (\Lambda^k S)(e_{i_1}\wedge\cdots\wedge e_{i_k}) \rangle \\ &= \sum_{i_1<\ldots<i_k}\langle e_{i_1}, S e_{i_1} \rangle\cdots\langle e_{i_k}, S e_{i_k} \rangle\\ &= \sum_{i_1<\ldots<i_k}\kappa_{i_1}\cdots \kappa_{i_k} \end{align}$$ which is (perhaps up to a constant) the "higher order mean curvature" $H_k$. Thus we can confirm your conjecture that this is intrinsic for even $k$: in this case we can write $$(\Lambda^k S)(v_1\wedge\ldots\wedge v_k) = (\Lambda^2 S)(v_1 \wedge v_2)\wedge \cdots \wedge(\Lambda^2 S)(v_{k1}\wedge v_k),$$ so $\Lambda^2 S$ determines all the even invariants. On the other hand, when $k$ is odd, the transformation $S \to S$ leaves $\Lambda^2 S$ invariant but changes the sign of $\Lambda^k S$, so the odd invariants of $S$ are not invariant in this strict sense.
This is perhaps a bit of a copout, though, since we can change the sign of $S$ just by flipping our unit normal while leaving all the geometry the same. Thus, I think we should really ask whether $H_k$ is intrinsic. Since any rank1 $S$ induces $\Lambda^2 S = 0$, this is clearly false for $k=1$; so let's assume $k \ge 3$. Noting that the unknown $a = \mathrm{rank}(S)$ and the known $b = \mathrm{rank}(\mathfrak R)$ are related by $b = \binom a 2$ (just look at their diagonalizations), we can handle two cases separately:
 If $b < 3$, then $\mathrm{rank}(S) < 3 \le k$ and thus $\Lambda^k S = 0$, so $H_k = 0$ is determined by $\mathfrak R.$
 If $b \ge 3$, then there's a nice rigidity theorem that tells us $\mathfrak R$ determines $S$ up to sign  see the first theorem in Chapter 12 of Spivak vol 5, or this MSE answer of mine.
Either way, we can determine $H_k$ from $\mathfrak R$; so all the higher odd invariants (excluding the actual mean curvature) are intrinsic up to their sign.