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On a complete smooth Riemannian manifold $(M^n, g)$, define $f(x)= d(p, x)$ where $p, x\in M^n$, and $d$ is the distance determined by the Riemannian metric $g$.

Do we have: Except a set with zero Lebesgue measure, for almost every $t\in \mathbb{R}$, $f^{-1}(t)$ is a $C^{1, 1}, (n- 1)$-dim hypersurface in $M^n$?

If we do not have such general result, under which general assumption on $M^n$, we will have the above result?

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    $\begingroup$ No, just take the cylinder $x^2+y^2=1$ in $xyz$-space. As soon as $t\ge 2\pi$, the level curve $f^{-1}(t)$ is not $C^{1,1}$, as it will have two non-differentiable points: Take $p=(1,0,0)$, then $f^{-1}(t)$ will not be differentiable at $(-1,0,z)$, where $z^2= t^2-4\pi^2$. I think that for most metrics on most manifolds, you'll have similar results. There are a few cases (such as symmetric spaces and simply connected manifolds with nonpositive sectional curvature) where your desired property does hold though. $\endgroup$
    – Robert Bryant
    Commented Jul 17, 2017 at 8:26
  • $\begingroup$ @RobertBryant Nice example! There are actually two questions you asked mmaaatthh, namely smoothness of $f$ and "when is $t$ a regular value?". For the first one, check mathoverflow.net/questions/21295/… There is a nice and complete answer by Ivanov. $\endgroup$
    – M. Dus
    Commented Jul 17, 2017 at 9:06
  • $\begingroup$ @mmaatthh Now, some precisions about Robert Bryant comment on negatively curved manifold. Let $M$ be a Hadamard manifold (simply connected, complete, negatively cuved), then the Cartan-Hadamard theorem ensures that the exponential is everywhere a diffeomorphism. According to Ivanov anwser (see the link in my first comment), this gives you smoothness. Further, the distance function is strictly convex, so that the derivative of $f(x)=d(p,x)$, where $p$ is fixed, never vanishes (except at $p$). So every non-zero $t\in \mathbb{R}$ is a regular value. Sorry for writting two comments. $\endgroup$
    – M. Dus
    Commented Jul 17, 2017 at 9:11
  • $\begingroup$ And to complete the remark about symmetric spaces: If $(M,g)$ is a compact, rank one symmetric space (e.g.,$(M,g)$ is one of $S^n$, $\mathbb{RP}^n$, $\mathbb{CP}^n$, $\mathbb{HP}^n$, or $\mathbb{OP}^2$ with its standard metric), then $f^{-1}(t)$ is a smooth hypersurface (possibly empty) except when $t=0$ or $t=D$, where $D$ is the diameter of $M$. So these are examples as well. Also, for a given $p\in M$, one can modify the metric in a manner rotationally symmetric about $p$ to get nonsymmetric examples with the same property, so 'symmetric space' is not necessary, even in positive curvature. $\endgroup$
    – Robert Bryant
    Commented Jul 17, 2017 at 11:22
  • $\begingroup$ @Robert Bryant, thanks for your detailed answer. And we know if $t$ is not critical value, then $f^{-1}(t)$ is topological hypersurface(from Cheeger) and is in fact Lipschitz, hence is $C^{1}$ almost everywhere. I should ask whether $f^{-1}(t)$ is $C^{1, 1}$ almost everywhere for almost every $t\in \mathbb{R}$, which looks reasonable and helpful to talk about the intrinsic curvature of $f^{-1}(t)$ for most $t$. $\endgroup$
    – mmaatthh
    Commented Jul 17, 2017 at 12:32

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