Let's first state the situation. I'm interested in Hadamard manifolds $M$ (i.e. complete, simply connected manifolds with sectional curvature $K \leq 0$, which are homeomorphic to $\mathbb{R}^n$ by Cartan-Hadamard (we may also assume $-1 \leq K$ if that helps)), but even the case of $\mathbb{R}^n$ would be a nice first step.

As usual, a set $A \subseteq M$ is called convex if for every $p,q \in A$ there is a unique geodesic segment $c$ in $M$ connecting $p$ and $q$ which is entirely contained in $A$. Since connecting geodesics always exist and are unique in Hadamard manifolds, this only amounts to requiring this geodesic segment to lie in $A$.

One can readily show that a bounded open (closed) convex set is homeomorphic to an open (closed) ball $D^n$ in $\mathbb{R}^n$ (take the exponential map). Furthermore, intersections of convex sets are convex.

Now let $A, B \subseteq M$ be bounded convex sets with $A\cap B \neq \emptyset$. By the above statements $A \cong D^n \cong B$ and $A \cap B \cong D^n$, but how the ball $A \cap B$ sits inside $A$ might be complicated: for instance think of $A$ as being the unit ball in the Euclidean plane and $B$ as a filled triangle around the origin intersecting the boundary of the ball near the three corners, but not near the middle of the three edges. You can generalize this by replacing the triangle with a convex polygon with as many corners as you like, creating an even more complicated situation (where complexity is measured e.g. by the homology of $A \setminus A \cap B$).

My question is: under which circumstances is the pair $(A, A\cap B)$ homeomorphic to $(D^n, D^n_+)$ (where $D^n_+$ is a half ball inside the ball $D^n$, i.e. the intersection of $D^n$ with a half space of $\mathbb{R}^n$)?

This question is quite general and it might even be too general to answer. So let's rather look at the following scenario (which is my real motivation).

Let $A$ be a ball $B_r(p) \subseteq M$ of radius $r > 0$ around $p \in M$ (this is convex). Let $C$ be a convex set in $M$ of distance $d(p,C) > r'$ (with $r' \geq r$) from $p$ and let $B$ be the $r'$-neighborhood $B_{r'}(C)$ of $C$ (i.e. all points in $M$ having distance $< r'$ to some point of $C$). By convexity of the distance function, $B_{r'}(C)$ is also convex. In this situation, is $(A, A\cap B) \cong (D^n, D^n_+)$?

If you draw a picture the situation looks like the statement is so trivial it wouldn't require an argument, but somehow I wasn't able to produce one. Some of my thoughts:

  • It's crucial for $p$ to be farther away from $C$ than the radius $r'$ in the construction of $B$ and the radius $r'$ to be at least as big as $r$. Else we could take $A$ to be an $r$-ball around the origin in $\mathbb{R}^2$ and $C$ to be the $x$-axis. By taking $r' < r$, $A \setminus B$ would be nonempty and consist of two connected components. Also by just requiring $p$ to be of distance $> r'$ from $C$ (but not requiring $r' \geq r$) we could obtain the same situation by just taking $C$ to be a suitable parallel to the $x$-axis, again cutting $A$ into two pieces.
  • One can see that it's necessary for $(\partial A) \cap B$ to be an $(n-1)$-dimensional disc inside the boundary $S^{n-1} \cong \partial A$. By the generalized Schoenflies theorem (see https://en.wikipedia.org/wiki/Schoenflies_problem#Generalizations) this might even turn out to be sufficient.
  • Perhaps the exponential map might help to translate the problem into Euclidean space, but then again convexity is a metric property (and the exponential map is only a radial isometry).

Thank you for reading this wall of text.


Your Answer

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

Browse other questions tagged or ask your own question.