The continuity of principal coordinate system

Question

$X$ is a $C^k$ hypersurface in $\mathbb R^{n+1}$ and $y$ is a fixed point on $X$. Can we find an orthogonal system $\{e_1(x),e_2(x),\cdots,e_{n+1}(x)\}$ on a neighborhood $U$ of $y$ such that 1. $\{e_1(x),e_2(x),\cdots,e_{n}(x)\}$ consists of all the principal directions of $X$ at $x$. 2. $e_{n+1}(x)$ is the unit normal vector at $x$. 3. $e_i(x) \in C^{k-1}(U)$?

Answer 1

In general, not. Begin by representing $X$ by an implicit equation $F(x)=0$, where $F\in C^k$. Then $\nabla F$ gives you the normal vector $e_{n+1}$ and it's $C^{k-1}$, so that's good. The principal directions come from the Hessian matrix $D^2F$ restricted to the tangent space $(\nabla F)^{\perp}$. Let $H$ be this restriction (it's easier to think of it as a bilinear form on the tangent space). Since it comes from second derivatives, $H$ is only $C^{k-2}$ smooth. You should have expected this, though: if $k=1$, there is no curvature and no principal directions to speak of.

If all eigenvalues of $H$ at $y$ are distinct, then they stay distinct in a neighborhood of $y$. In this neighborhood solving for eigenvectors is straightforward algebra, and they come out to be as smooth as $H$ itself, namely $C^{k-2}$.

If $H(y)$ has repeated eigenvalues, you should not expect even continuity of principal directions in a neighborhood of $y$. I think $z=x^2+y^2+x^2y$ in $\mathbb R^3$ is a counterexample at the point $(0,0,0)$, but did not do all the computations.