Let $(M,g)$ be a Riemannian manifold, $\Omega\subset M$ be a closed set with smooth boundary $\partial\Omega$ and $\nu$ be the unit normal of $\partial\Omega$ pointing into $\Omega$.

$\Omega$ is said to be geodesically convex iff $\forall x_0, x_1\in\Omega$ $\exists c:[0,1]\stackrel{\text{geodesic}}\to(M,g)$ s.t. $c(0)=x_0, c(1)=x_1$, $c([0,1])\subset\Omega$, $\mathrm{Length}[c]=d_g(x_0,x_1)$.

Suppose $\Omega$ is geodesically convex. Then...

[Q.1] Does it hold that the 2nd fundamental form of $\partial\Omega$ toward $\nu$ is nonnegative definite at each point on $\partial\Omega$?

[Q.2] Let $\psi_r(x):=\mathrm{exp}^g_x [r\nu(x)]\in N$ $(x\in\partial\Omega)$. Then for small $|r|$, $\psi_r$ is an embedding. Here, does it hold that the inner 2nd fundamental form of $\psi_r$ is nonnegative definite at each point on $\partial\Omega$ when $r>0$ and is sufficinetly small?

Thank you.


I will address your first question. First define, for a fixed point $p\in\partial \Omega$, three conditions on $\partial \Omega$.

a) There is an open subset $U\subset M$ with $p\in U$, such that any two points in $U\cap \Omega$ can be joined by length minimising geodesic $c:[0,1]\rightarrow M$ with $c[0,1]\subset U\cap \Omega$.

b) Any geodesic $c:(-\epsilon,0]\rightarrow M$ with $c(-\epsilon,0)\subset \mathrm{int(\Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(\dot c(0),\nu(p))\neq 0$.

c) The second fundamental form $l_\nu(\cdot,\cdot)=g(\nabla_\cdot \cdot, \nu)$ is non-negative at $p$.

We will prove that a) $\Rightarrow$ b) $\Rightarrow$ c), which answers your first question.

Step 1 Take geodesic normal coordinates $x^1,\dots,x^n$ , centred at $p$, with $\partial_n\vert_p=\nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:\mathbb{R}^{n-1}\rightarrow \mathbb{R}$ such that $$ \Omega \cap V =\{x^n\ge f(x^1,\dots,x^{n-1})\}, $$ where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,\dots,X_{n-1}$ on $V$ by $$ X_j(q)=\partial_j\vert_q + \partial_jf(x^1(q),\dots,x^{n-1}(q))\partial_n\vert_q. $$ If $q\in V \cap \partial \Omega$, then $X_j(q) \in T_q\partial \Omega$ and hence for $1\le j\le n-1$ we have $$ \partial_jf(0) = X_j^k(p) g_{kl}(p) \nu^l(p)=g(X_j(p),\nu(p)) = 0 \tag{1}. $$

Step 2 We claim that $$ \text{b)} \quad \Leftrightarrow \quad f \text{ has a local minimum at 0.} \tag{2} $$ Note that in the coordinates fixed above, b) means that $\{x^n=0\}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(\Omega)\cap V =\{x^n> f(x^1,\dots,x^{n-1})\}$ near $p$, or in other words that $f(x^1,\dots,x^{n-1})\ge 0$ in a neighbourhood of $p$. This proves (2).

Step 3 We're now in a position to prove a) $\Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $q\in \partial \Omega \cap U$ such that $$x^n(q)=f(x^1(q),\dots,x^{n-1}(q))<0. \tag{3}$$ Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L =\{x^j=tx^j(q): 0\le t \le 1\}$. Assuming a), we must have $L\subset \Omega \cap U$, which implies $$ tx^n(q) \ge f(tx^1(q),\dots,tx^{n-1}(q)). $$ Divide by $t$ and take the limit $t\rightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)\ge0$, which is a contradiction to $(3)$.

Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1\le j \le n-1$ and $1\le k \le n$ we have $$ \nabla_kX_j (p)= (\Gamma_{kj}^l \partial_l + \partial_k \partial_j f \partial_n + \Gamma_{kn}^r \partial_r)(p) = \partial_j\partial_kf(p) \cdot \nu(p) $$ and since $f$ is independent of $x^n$, we further obtain $$ \nabla_{X_k}X_j(p) = \partial_j\partial_kf(p) \cdot \nu(p), $$ which implies $l_{\nu}(X_k,X_j)\vert _p = \partial_j\partial_kf(p)$. Since the $X_j$ form a basis of $T_p\partial \Omega$ we have $$ \text{c)}\quad \Leftrightarrow \quad f \text{ has non-negative Hessian at $0$}. \tag{4} $$ From $(2)$ and $(4)$ it is evident that b) implies c).

Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $\partial \Omega$ is strictly positive, all geodesics coming from $\Omega$ hit $\partial \Omega$ transversally.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.