# Does convex hypersurface necessarily bound a convex domain?

Let $H\in M$ be a convex hypersurface, where $M$ is a complete Riemannian manifold and $H$ is an embedded (complete as a induced metric space) hyper surface without boundary and with positive definite second fundamental form. Is it true that $H$ bound a convex domain $D$ in $M$? i.e. any two point $x, y \in D$ can be connected by a minimal geodesic lies in $D$ and $\partial D=H$.

I suspect this is not true in general, but can anyone provide a countexample or any references?

Counterexample 1: If $D$ is a ball of radius $\frac13$ on the torus $T=\mathbb R^n/\mathbb Z^n$, you can take two points on the opposite sides of the ball such that the minimal geodesic connecting them does not stay in $D$. The boundary $\partial D$ has a strictly positive definite second fundamental form.
Counterexample 2: Let $M$ be your favorite Riemannian manifold and $H$ a convex hypersurface that bounds a domain $D$. Take two points $x,y\in D$ so that $d(x,y)>d(x,H)+d(y,H)$. By rescaling the metric outside $\bar D$, you can force the minimal geodesic joining $x$ and $y$ to exit $\bar D$.
Counterexample 3: It can happen that the hypersurface $H$ does not bound a domain, meaning that $M\setminus H$ has only one connected component. If there are two components in the complement, you can always build a handle between them.