Let $S$ be a smooth closed submanifold of $M$. Let $U$ be a tubular neighborhood such that for any $x\in U\setminus S$ there is a unique minimizing geodesics. We now consider the distance squared to $S$, $d_S^2(x):=dist^2(x,S)$, $x\in U$. Can we write down the taylor expansion of $d_S^2(\exp_p(t\nu))$, for any $t$ as long as $\exp_p(t\nu)\in U$? Here $\nu=\nabla d_S(p)$ or in general $\nu$ is any vector in $T_pM$). When we consider the distance function from a point $p$, $d(p,.)$, and write the Teylor expansion of $d_p(\exp_p(tv),\exp_p(tw))$, we see the Ric(v,w) in the expansion.

When instead of a point, we have a distance function from a submanifold $S$, I think that the Hessian of $\nabla^2 d_S$ on $S$ should be equal to the second fundamental form of $S$ and its eigenvalues give the principal curvatures. I wondering if we can choose some direction and calculate the Taylor expansion and see other curvature terms in the expansion. What would other curvature terms that might appear?

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    $\begingroup$ If I understand your question correctly, isn't the Taylor expansion just $t^2$? $\endgroup$ – Ryan Budney Nov 30 '16 at 20:53
  • $\begingroup$ Ryan, that's right if $\nu = \nabla d_S$. If not, then it's less straightforward. $\endgroup$ – Deane Yang Dec 1 '16 at 3:55
  • $\begingroup$ Thanks for your comments. I have edited the question. I would be happy to know your thoughts. $\endgroup$ – Math101 Dec 1 '16 at 9:15

Have a look at (where your $S$ is a mostly curve):

  • MR2024928 Reviewed Gray, Alfred Tubes. Second edition. With a preface by Vicente Miquel. Progress in Mathematics, 221. Birkhäuser Verlag, Basel, 2004. xiv+280 pp.

A remark: If $S$ is a point, the second derivative (with respect to Riemannian normal coordinates) of the squared distance from the point is the Riemannian metric. The Taylor expansion of the Riemannian metric (with respect to Riemannian normal coordinates centered at $S$) has vanishing linear term (since the Christoffels vanish at $S$ in the exponential chart), and the second order term is exactly Riemannian curvature at $S$: It is a form of the numerator of sectional curvature. This is how Riemann found his curvature (sketched in the Habilitations-Lecture, and in his paper submitted to French Academy in 1861).

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