Consider the points \begin{align*} a_1 &= (-1, 1, 1, -1) \\ a_2 &= (1, -1, 1, -1) \\ a_3 &= (1, 1, -1, -1) \\ b_1 &= (1, -1, -1, 1) \\ b_2 &= (-1, 1, -1, 1) \\ b_3 &= (-1, -1, 1, 1) \end{align*}
and the convex hull of the points $K = \operatorname{conv} \{ a_1, a_2, a_3, b_1, b_2, b_3 \}$. The book I'm reading claims that $K$ forms a 3-dimensional regular cross-polytope (octahedron). How can the convex hull of 4-dimensional points form a 3-dimensional object? Won't $K$ be a subset of $\mathbb{R}^4$?
Yes, $K$ will indeed be a subset of $\Bbb{R}^4$, but it will be $3$-dimensional, meaning the affine hull of the set is $3$-dimensional. In particular, note that each of the vectors are perpendicular to $(1, 1, 1, 1)$, so they all belong to the $3$-dimensional subspace $$\{x, y, z, w) : x + y + z + w = 0\}.$$ This subspace is convex, hence it must contain the convex hull of the points, i.e. $K$. The subspace is also an affine set, so it must contain the affine hull of $K$, establishing that $K$ is, at most, $3$-dimensional.
On the other hand manual computation reveals that $a_1 - a_2, a_2 - b_1$, and $b_1 - b_3$ are linearly independent, meaning that the affine hull must be at least dimension $3$, confirming that the points form a $3$-dimensional convex set lying in $\Bbb{R}^4$.