Let $(S,g)$ be a complete Riemannian manifold and $M\subset S$ an embedded submanifold (of the same dimension) with non-empty boundary $\partial M$. I am interested in understanding the relation between different notions of convexity for $M$:
A) Global notions of convexity:
(A1) For every $x,y\in \mathrm{int} M$ there exists a geodesic $c:[0,1]\rightarrow \mathrm{int} M$ with $c(0)=x$, $c(1)=y$ and $\mathrm{Length}(c)=d(x,y)$.
(A2) For every $x,y\in M$ there exists a geodesic $c:[0,1]\rightarrow M$ with $c(0)=x$, $c(1)=y$ and $\mathrm{Length}(c)=d(x,y)$.
(A3) For every $x,y\in M$ there exists a geodesic $c:[0,1]\rightarrow M$ with $c(0)=x$, $c(1)=y$, $c((0,1))\subset \mathrm{int}M$ and $\mathrm{Length}(c)=d(x,y)$.
Further one could consider (A1'), (A2') and (A3') by requiring the respective geodesics to be unique.
B) Notions of convexity defined via the boundary.
Denote with $l_\nu(x):T_x\partial M \times T_x\partial M \rightarrow \mathbb{R}$ the second fundamental form of $\partial M$ with respect to the inward pointing unit normal $\nu$, i.e. $l_\nu(x)[X,Y]=\langle \nabla_X Y\vert_x,\nu(x)\rangle_{g(x)}$. Then we can also consider the following convexity conditions:
(B1) For every $x\in \partial M$ we have $l_\nu(x) \ge 0$ (positive semi-definite).
(B2) For every $x\in \partial M$ we have $l_\nu(x) > 0$ (positive definite).
Question: What is the relationship between notions in group A and the ones in group B? For example, does (B2) imply (A1)?
Here is what I know so far: Here I prove that $(A1)\Rightarrow (B1)$ and that $(B2)$ implies that all geodesics leave $M$ transversally.
If $M\subset \mathbb{R}^2$ is a compact, then it's not hard to see that $(B2)\Rightarrow (A1)$: Assuming that (A1) fails, we can pick two points $x,y\in \mathrm{int} M$ for which the connecting line leaves $\mathrm{int} M$. Say $z \in \partial M$ is an intersection point of this line with $\partial M$ such that the line $\overline{xz}$ lies in $M$. Now move $z$ along the boundary until $\overline{xz}\cap M^c\neq \emptyset$, then $\overline{xz}$ hits $\partial M$ tangentially, which contradicts (B2).
Approach for the general case (Edit 24th April): We want to show that (B2) implies (A1): Fix $x\in \mathrm{int}(M)$. We have to show that there is an open set $\Omega \subset T_xM$ such that $\exp_x(\Omega)=\mathrm{int} M$. Denote $S_xM$ the unit-sphere in $T_xM$ and let $\tau: S_xM\rightarrow [0,\infty]$ be the exit time defined by $$\tau(v)=\sup\{t\ge 0: \exp_x(tv)\in M \quad \text{ for all } 0\le s \le t\}.$$ Now (B2) implies that all geodesics starting in $x$ hit $\partial M$ transversally and, further using the implicit function theorem, we obtain that the subset $U\subset S_xM$ on which $\tau$ is finite, is open and that $\tau$ is smooth on $U$. Then $\Omega = \{sv\in T_xM: v\in S_xM, 0\le s<\tau(v)\}$ is an open subset of $T_xM$ with smooth boundary that satisfies $\exp_x(\Omega)\subset \mathrm{int}M$.
If we additionally assume that $x$ does not have any conjugate points in a neighbourhood of $M$, then $\exp_x$ is a local diffeomorphism in a neighbourhood of $\overline{\Omega}$ and thus restricts to an immersion $(\overline{\Omega},\partial \Omega)\rightarrow (M,\partial M)$ of two manifolds with boundary, sending boundary into boundary. I suspect that some topology argument should give surjectiveness for free, yielding $\exp_x(\Omega)=\mathrm{int}(M)$.
However, I don't see at the moment how this argument can be saved if one leaves away the extra assumption.