Theorem 19.9.9CC2BF9294FA3

Proof. Choose $g_0=(b,e)\in G$ and an integral basis $\lambda _1,\ldots ,\lambda _ d$ of $\Lambda $. Every element of $G$ has a unique expression

These coordinates identify $G$ with $\mathbb Z^{d+1}$. Let $S'$ be the image of $S$ under this identification. It generates $\mathbb Z^{d+1}$, and the height of the image of $(x,m)\in S$ is $q=m/e\geq 0$.

The points $(m^{-1}x,1)$, for $(x,m)\in S$ with $m>0$, span the $(d+1)$-dimensional real vector space $\operatorname {span}_{\mathbb R}G$. Since they all have height one, their affine span has dimension $d$. Taking the convex hull and the closure does not change this affine span, which proves the dimension assertion.

Intersect the cone containing $S$ with $\operatorname {span}_{\mathbb R}G$. The linear change of coordinates carries this intersection to a closed cone with compact height-one slice whose only point at height zero is the origin. Indeed, a vector $(x,m)$ with $|x|\leq Cm$ has horizontal coordinates bounded by a constant times $m$, hence by a constant times $q$. The height-one body of $S'$ is

where the right-hand side is expressed in the basis $\lambda _1,\ldots ,\lambda _ d$. It is full-dimensional in $\mathbb R^ d$, and therefore

Apply the semigroup counting theorem of subsection 19.9.5 to $S'$, which generates its full lattice. The points of $S'$ at height $q$ correspond bijectively to the valuations of the nonzero elements of $W_{eq}$, and by lemma 19.9.2 there are $\dim W_{eq}$ of them. Hence

Dividing by $e^ d$ gives the asserted limit, and its positivity gives eventual nonvanishing. For $d=0$ the body is a point of volume one, so the integer dimensions converge to one and are eventually equal to one. Finally, every degree with a nonzero section lies in the projection $e\mathbb Z$ of $G$, which proves the vanishing outside these multiples. $\square$