Tube formula for a hypersurface in a Riemannian manifold

Question

Let $(M,g)$ be a complete Riemannian manifold and $N$ a closed (orientable) hypersuface of $M$. Let $d$ be the signed distance from $N$ and $N_r=\{x\in M: 0<d(x,N)<r\}$. For $r$ small enough $$\Phi:N\times[0,r)\to N_r$$ $$(x,t)\mapsto (x,\exp_x(t\,\textbf{n}(x)))$$ is a diffeomorphism. Here $\textbf{n}$ is the inward unit normal. The following formula always holds: $$|\det D\Phi(x,t)|= 1+O(t),$$ by the Taylor expansion of $\det D\Phi(x,t)$ and the fact that $\det D\Phi(x,0)$=1. I wonder if there is an explicit formula for the Jacobian of $\Phi$, $|\det D\Phi(x,t)|$ in general. I found some formulas under some curvature conditions in Gray's book Tubes. What about the general case?

Answer 1

Let $e_1, \dots, e_{n-1}$ be an orthonormal basis of $T_xN$ parallel translated along the normal geodesic $t \mapsto \Phi(x,t)$. Let $J_1, \cdots, J_{n-1}$ be Jacobi fields along the normal geodesic such that $J_i(0) = e_i$ and $J'_i(0) = 0$. Let $J_n = e_n$ be the unit velocity vector of the normal geodesic. Then given $t \ge 0$, $$ \det D\Phi(x,t) = J_1\wedge\cdots\wedge J_n = \det [J^j_i], $$ where $J_i = J_i^je_j$. Since $J_i'(0) = 0$, for each $i$, the first order term vanishes. The second order term at $t=0$ can be found using the Jacobi equations satisfied by $J_i$. It's essentially the Ricci curvature. This can be found in any presentation of the Bishop-Gromov inequality, which by now appears in many textbooks on Riemannian geometry. It also appears in a classic paper by Heintze and Karcher.

Answer 2

I very strongly recommend reading Alfred Gray's Tubes:

Gray, Alfred, Tubes, Progress in Mathematics (Boston, Mass.) 221. Basel: Birkhäuser (ISBN 3-7643-6907-8/hbk). xiii, 280~p. (2003). ZBL1048.53040.

(for bonus points, how do I get Birkhäuser to typeset properly?)