What is the shape of all convex combinations of $\geq$ five vectors in $\mathbb{R}^3$?

Question

The convex combinations of two linearly independent vectors in $\mathbb{R}^3$ span a line. The convex combinations of three linearly independent vectors in $\mathbb{R}^3$ span a solid triangle. The convex combinations of four vectors in general position in $\mathbb{R}^3$ span something a lopsided tetrahedron, I think. How can we describe the convex combinations of five vectors in $\mathbb{R}^3$? Can we do better than saying it is "the convex hull of the union of the lopsided tetrahedra which are the combinations of any four of them taken at a time"? I.e.

$$C(x_1, \dotsc, x_5) = \operatorname{Hull}\left( \cup_{1\leq i \leq 5} T_i \right),\, \text{for } T_i = C(x_1, \dotsc, \hat{x}_i, \dotsc, x_5)?$$

(Here $C$ is a function mapping vectors to the set of all their convex combinations.)

More generally, what about $n+2$ vectors in $\mathbb{R}^n$?

Answer 1

Thanks to Carathéodory's theorem, you don't need to take convex hull after union: $$C(x_1, \dotsc, x_5) = \bigcup_{1\leq i \leq 5} T_i\quad \text{for } T_i = C(x_1, \dotsc, \hat{x}_i, \dotsc, x_5)$$ (Similarly in higher dimensions: every point of convex hull is a convex combination of some $(n+1)$ vertices).

However, this representation is somewhat redundant: the tetrahedra overlap a lot. As Rahul said in a comment, the set is actually the union of 1 or 2 tetrahedra (apart from degenerate cases). Here's a geometrical explanation: imagine you are at one of these five points. The other four form a tetrahedron $T$.

If you are inside of $T$, then $C = T$. If you are outside, then when looking at $T$, you see either $1$, $2$, or $3$ of its faces.

• If you see $1$ face of $T$, then $C$ is the union of $T$ and the tetrahedron built on this face, using your point as a vertex.
• If you see $3$ faces of $T$, then $C$ is the tetrahedron built on the face you don't see, using your point as a vertex.
• If you see $2$ faces of $T$, they are separated by an edge. Call this edge $PQ$. The other two vertices of $T$ form an edge $RS$ which you don't see (except for the vertices $R,S$ themselves). Consider the triangle $YRS$ (where $Y$ is you), which divides $T$. Build two tetrahedra with the base $YRS$: one with vertex $P$, the other with vertex $Q$. Their union is $C$.

One could translate the above argument into inequalities and cases with subcases, but who would want to read that?