differential-geometrymanifoldsriemannian-geometry
What are some examples of a Riemannian manifold which is both complete and geodesically convex? An obvious example is $R^n$. An open hemisphere of $S^n$ is geodesically convex but not complete. I don't believe hyperbolic space is geodesically convex, although I might be wrong.
Edit 1: We say Riemannian manifold $M$ is geodesically convex if given any two points in $M$, there is a unique minimizing geodesic contained within $M$ that joins those two points.
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!
Best Answer
Any complete simply connected manifold with negative curvature is geodesically connected. The hyperbolic space is geodesically connected. The reason is that such manifolds have neither conjugate points nor focal points and consequently is simply geodesically convex.