Proposition 21.9.19FD0EEED0DBF

Proof. Let $Q$ be a face, choose a vertex $v\in Q$, and use the notation of lemma 21.8.2(4). Since $u_1,\ldots ,u_ n$ is the basis dual to $e_1,\ldots ,e_ n$, every $x\in N_{\mathbb R}$ can be written as $x=\sum _ i\alpha _ iu_ i$ with $\alpha _ i=\langle e_ i,x\rangle $, and $\langle m-v,x\rangle =\sum _ it_ i\alpha _ i$ for $m=v+\sum _ it_ ie_ i\in P$. If $v$ minimizes $\langle \cdot ,x\rangle $ on $P$, then $\alpha _ i=\langle (v+e_ i)-v,x\rangle \geq 0$, because $v+e_ i\in P$. If moreover every point of $Q$ minimizes it, and the $i$th facet at $v$ does not contain $Q$, choose $m\in Q$ with $t_ i=\ell _ i(m)>0$; then $\sum _ jt_ j\alpha _ j=0$ with nonnegative terms, so $\alpha _ i=0$. Hence $x$ is a nonnegative combination of the normals of the facets containing $Q$, all of which pass through $v$, that is, $x\in \sigma _ Q$. Conversely, if $x=\sum _\rho c_\rho u_\rho $ with $c_\rho \geq 0$ and all $\rho $ containing $Q$, then $\langle m,x\rangle =\sum _\rho c_\rho (\ell _\rho (m)-a_\rho )$, and each $\ell _\rho $ attains its minimum $0$ on $P$ at every point of $Q$; so does $\langle \cdot ,x\rangle $. This proves the characterization, and it shows that $\sigma _ Q$ is generated by part of the basis $u_1,\ldots ,u_ n$ of $N$.

Every face $Q$ is the intersection of $P$ with the facets containing it. Indeed, if $Q$ is the set where $\langle \cdot ,x\rangle $ is minimal, then $x\in \sigma _ Q$, and writing $x=\sum _\rho c_\rho u_\rho $ as above we see that the minimum is attained exactly where $\ell _\rho =0$ for all $\rho $ with $c_\rho >0$; this set contains the intersection of the facets containing $Q$, which in turn contains $Q$.

The faces of $\sigma _ Q$ are the cones generated by subsets of its generators. Indeed, if $m\in M_{\mathbb R}$ is nonnegative on $\sigma _ Q$ and $x$ is a nonnegative combination of the generators, then $\langle m,x\rangle $ is a sum of nonnegative terms, so the face $\{ x\in \sigma _ Q:\langle m,x\rangle =0\} $ is generated by the generators annihilated by $m$. Conversely, since the generators are part of a basis of $N_{\mathbb R}$, for every subset of them there is an $m$ that vanishes on this subset and equals $1$ on the remaining generators, and $m$ is nonnegative on $\sigma _ Q$. If $J$ is a set of facets containing $v$, the set $Q'=\{ m\in P:\ell _\rho (m)=0\text{ for }\rho \in J\} $ is a face, namely the set where $\sum _{\rho \in J}\langle \cdot ,u_\rho \rangle $ is minimal. It contains $v+se_ i$ for small $s>0$ whenever the $i$th facet at $v$ is not in $J$, so the facets containing $Q'$ are exactly those in $J$, and $\sigma _{Q'}=\operatorname {cone}\{ u_\rho :\rho \in J\} $. Thus every face of $\sigma _ Q$ is in the family.

Every $x$ attains its minimum on $P$ on some face $Q$, and then $x\in \sigma _ Q$ by the characterization; hence the cones cover $N_{\mathbb R}$. Let $Q_1,Q_2$ be faces, and let $Q$ be the intersection of $P$ with all facets containing both. It is a face (if no facet contains both, $Q=P$; otherwise it is the set where the sum of the corresponding functionals is minimal), and every face containing $Q_1\cup Q_2$ contains $Q$, since it is the intersection of $P$ with the facets containing it. Since $Q\supset Q_1\cup Q_2$, it is the smallest face containing $Q_1\cup Q_2$. By the characterization, $x\in \sigma _{Q_1}\cap \sigma _{Q_2}$ if and only if the face where $\langle \cdot ,x\rangle $ is minimal contains $Q_1\cup Q_2$, that is, if and only if it contains $Q$; so $\sigma _{Q_1}\cap \sigma _{Q_2}=\sigma _ Q$. The generators of $\sigma _ Q$, the normals of the facets containing $Q_1\cup Q_2$, form a subset of the generators of $\sigma _{Q_ k}$, so $\sigma _ Q$ is a face of $\sigma _{Q_1}$ and of $\sigma _{Q_2}$.

For the last assertion, let $|x|=1$ and choose a vertex $v$ minimizing $\langle \cdot ,x\rangle $ on $P$. The coefficients $\alpha _ i=\langle e_ i,x\rangle $ above are nonnegative and satisfy $\alpha _ i\leq |e_ i|$. Their sum $\lambda $ is positive, because $x\neq 0$, and at most $\sum _ i|e_ i|$. Let $\Lambda $ be the maximum of these finite sums over all vertices. Dividing by $\lambda $ and omitting the zero coefficients gives the required expression, with all $u_{\rho _ i}$ in the cone $\sigma _{\{ v\} }$. In dimension zero the fan consists of the zero cone, and there is no unit vector. $\square$