Strictly convex submanifold of $\mathbb{R}^n$ of codimension 1

Question

I am interested in strictly convex submanifolds of $\mathbb{R}^n$. See link for definition.

My question is: suppose $M \subseteq \mathbb{R}^n$ is a strictly convex submanifold without boundary (not necessarily compact) of dimension $n-1$. Does $M$ have the following properties:

1. set complement of $M$ has two connected components, say $A$ and $B$
2. the convex hull of $M$ is equal to either $M\cup A$ or $M\cup B$

Answer 1

Based on the definition proveded in the linked paper, there's a simple topological argument that convex submanifolds partition space.

Let $M\subseteq\mathbb{R}^n$ be a codimension one convex manifold, and define the following subsets of $\mathbb{R}^n$:

• Let $C$ be the convex hull of $M$. By definition, $M=\partial C$. Since $C$ contains its boundary, it is closed.
• Let $A=\text{int}(C)$. By definition, $A$ is open.
• Let $B=\mathbb{R}^n\setminus C$. Since it is the complement of a closed set, $B$ is open.

Thus, we have $\mathbb{R}^n=M\cup A\cup B$, with all three sets disjoint, and $C=M\cup A$. Since $A$ and $B$ are open, they are disconnected in $\mathbb{R}^n\setminus M$.

Note $B$ may not itself be connected. For instance, the submanifold $M\subset\mathbb{R}^2$ given by $x=\pm 1$ is convex, but $B$ has two connected components. $A$ is always connected, since it is convex.

It also may be the case that $A$ is empty (e.g. affine subspaces).