Is every convex cone a manifold?

Question

Let $C \subseteq \mathbb{R}^n\setminus \{0 \}$ be a connected convex cone*.

Question: Is $C$ always a topological manifold (perhaps with boundary)? A smooth one? Does anything change if we do not assume the cone is convex? (i.e it's only closed under positive scaling)

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* (I define convex cone to be a subset which is convex and closed under strictly positive scalar multiplication. Equivalently, it's closed under non-zero conic combinations).

Answer 1

Here is a counterexample in dimension $n=3$, i.e. a cone which is not a manifold-with-boundary.

Using $x,y,z$ coordinates, consider the following subset $S$ of the $z=1$ plane: $S$ consists of all points $(x,y,1)$ such that either $x^2 + y^2 < 1$ or $(x,y) = (\cos(2\pi r), \sin(2 \pi r))$ for some rational number $r$. In other words, $S$ is the union (in the $z=1$ plane) of the open unit disc and the points on its boundary whose angle is a rational multiple of $2\pi$. Note that $S$ is a convex subset of the $z=1$ plane.

Now let $C$ be the cone consisting of all rays based at the origin and passing through a point of $S$, i.e. the smallest cone containing $S$.

Note that $C$ is not a manifold with boundary at any of the points $(\cos(2\pi r), \sin(2 \pi r),1)$.

What is going wrong here is that $C$ is not a closed set.

If, on the other hand, you require that the cone $C$ be a closed set, then yes, $C$ is a manifold with boundary. In fact, the same is true for all closed convex subsets of Euclidean space. One still has to be careful because the dimension of $C$ need not be equal to $n$, but if one takes an affine subspace of minimal dimension that contains $C$ then the dimension of $C$ as a manifold with boundary equals the dimension of that affine subspace.

For your other questions,

• Closed cones do not have to be smooth at their boundary points. For an example, take $S$ to be a square in the $z=1$ plane, and take $C$ to be all rays based at the origin passing through $S$. There are still weirder examples where the non-smooth points are dense in the boundary. However, the smooth points themselves are always dense in the boundary.
• If you drop convexity, there is almost nothing that can be said. Take $S$ to be any subset of the $Z=1$ plane whatsever, and take $C$ to be all rays based at the origin passing through $S$.