Gale diagram

Remarks

The Gale transform is determined up to an invertible linear map of $\R^{n-d-1}$, the change of basis of $N$. Many authors call the configuration $g_1, \dots, g_n$ a Gale diagram, and some first rescale each nonzero $g_i$ by a positive factor; this does not affect the property below.

Key property: let $T \subseteq \set{1, \dots, n}$ be nonempty and $S$ its complement. Consider an affine function $\ell$ on $\R^d$ with $\ell(v_i) = 0$ for $i \in S$ and $\ell(v_i) > 0$ for $i \in T$. Such an $\ell$ exists if and only if $0$ lies in the relative interior of the convex hull of $\set{g_i : i \in T}$. Indeed, the vectors $(\ell(v_1), \dots, \ell(v_n))$ are exactly the vectors $y$ with $\sum_i y_i g_i = 0$. So when the $v_i$ are the vertices of a convex polytope $P$, the sets $\set{v_i : i \in S}$ that are vertex sets of faces of $P$ are read off from the $g_i$.

Example: for the $4$ vertices $0, e_1, e_2, e_1 + e_2$ of a square in $\R^2$, $n - d - 1 = 1$, and $N$ is spanned by $(1, -1, -1, 1)$. So the Gale transform is $1, -1, -1, 1$ in $\R$. For $T = \set{1, 4}$, the hull of $g_1 = g_4 = 1$ is $\set{1}$, which misses $0$, so $\set{e_1, e_2}$ is not a face: it is a diagonal. For $T = \set{1, 2}$, the hull of $1$ and $-1$ contains $0$ in its interior, so $\set{e_2, e_1 + e_2}$ is an edge.

Non-example: the points $0, e_1, 2e_1$ in $\R^2$ do not span $\R^2$ affinely, so the construction does not apply.