A subset $B$ of $\mathbb{R}^{3}$ is convex if, for every two points $p,q \in B$, the straight-line segment connecting $p$ and $q$ is entirely contained in $B$. A smooth surface $S$ in $\mathbb{R}^{3}$ is convex if it lies on the boundary of a convex set.
It is well-known that a characteristic property of convex surfaces is the following:
Property 1: For each point $p \in S$, the surface $S$ lies on one side of the tangent space $T_{p}S$, i.e., $S$ is contained in one of the two half-spaces defined by $T_{p}S$.
The notion of convexity can be extended to the case where the ambient manifold is an arbitrary Riemannian $3$-manifold $M$ as follows:
A subset $B$ of $M$ is geodesically convex if, for every two points $p,q \in B$, there is a unique minimizing geodesic which connects $p$ and $q$ and is entirely contained in $B$. A smooth surface $S$ in $M$ is convex if it lies on the boundary of a convex set.
My question is: Does a geodesically convex surface in $M$ satisfy an analogue of Property 1? A meaningful analogue of Property 1 could be that $\exp(T_{p}S) \cap
S =\{p\}$ for all $p \in S$.
Best Answer
Let's look one dimension lower; we can then take a cartesian product with $\Bbb R$ to handle the 3-dimensional case.
On the standard torus in 3-space (an embedding of $[0, 1] \times [0, 1]$ with the usual identifications) with the product metric, draw a small diamond spanning the outermost "equator" like the orange diamond $D$ in the diagram below:
If $D$ is small enough, something very like it will be geodesically convex. Now in $E = \partial D$ consider the green point $P$. The tangent space to $E$ at $P$ exponentiates to the red curve, which is part of some torus-knot-like curve. If the slope of the diamond-side is irrational (back in $[0, 1] \times [0, 1]$), then this curve's extension (the result of the $\exp$ operation you described) ends up being dense in the torus, hence intersects $D$.
So the answer to your query, at least as written, is "no."
There might be some good case to be made for a local claim, i.e., that there's some disk at the origin in $T_p(M)$ whose exponential doesn't meet the interior of $B$, but that'd be much harder to prove (for me, at least!) -- you'd have to use some actual differential geometry!