2.11 Convex cones and ampleness
It’s useful to think of ampleness in terms of cones.
Suppose \(V\) is a real vector space and \(S\subseteq V\).
We say that \(S\) is a cone if \(v\in S\), \(t\) positive real \(\Rightarrow t v\in V\).
We say that \(S\) is a convex if \(v,w\in S\), \(t\in [0,1]\Rightarrow tv + (1-t)w\in S\).
The cone over \(S\) is the intersection of all cones in \(V\) containing \(S\).
The convex hull of \(S\) is the intersection of all convex subsets of \(V\) containing \(S\).
It is pretty obvious that an arbitrary intersection of convex subsets of \(V\) is convex and the arbitrary intersection of cones in \(V\) is a cone. So the convex hull of \(S\) is the smallest convex set containing \(S\) and the cone over \(S\) is the smallest cone containing \(S\).
A subset \(S\) of a real vector space \(V\) is a convex cone if and only if the following condition holds: for all \(v,w\in S\) and \(x,y{\gt}0\), \(xv+yw\in S\).
Suppose \(S\) is a convex cone and \(v,w,x,y\) are as above. Then \(\displaystyle xv+yw = (x+y)\left(\frac{x}{x+y}v + \frac{y}{x+y}w\right)\in S\).
For the other direction, suppose the condition holds. Then \(S\) is a cone because, for all \(v\in S\) and \(x{\gt}0\), \(xv = (x/2)v + (x/2)v\in S\). And it is similarly easy to see that \(S\) is convex.
Now, it is easy to see that the convex hull of a set \(S\) is the set of finite sums \(\sum x_i v_i\) with \(v_i\in S\), \(x_i{\gt}0\) and \(\sum x_i = 1\). Similarly, the cone over \(S\) is the set of all expressions \(xv\) with \(x{\gt}0\) and \(v\in S\). Moreover, the cone over the convex hull is the same things as the convex hull of the cone, and this set is the set of all finite sums \(\sum x_i v_i\) with \(v_i\in S\) and \(x_i{\gt}0\). This leads to the following observation.
The set of ample \(\mathbb {R}\)-divisors is the convex hull of the cone over the set of ample integral divisors.
Immediate from the observations above.
Suppose \(A\) and \(B\) are integral divisors with \(A\) ample and \(B\) numerically trivial, and suppose that \(r\in \mathbb {R}\). Then \(A+rB\) is ample.
We can find integers \(r_1\) and \(r_2\) with \(r_1{\lt}r{\lt}r_2\). Then \(A+r_1B\) and \(A+r_2B\) are both ample because ampleness is numerical for integral divisors by Corollary 2.7. So, as
is an ample \(\mathbb {Q}\)-divisor by Corollary 2.26.
If \(A\in \operatorname{\mathrm{Div}_{\mathbb {R}}}X\) is ample. Then any divisor in \(\operatorname{\mathrm{Div}_{\mathbb {R}}}X\) numerically equivalent to \(A\) is also ample.
Suppose \(D\equiv A\). Then \(D = A + B\) where \(B\equiv 0\). We have \(A = \sum _{i=1} a_i A_i\) with \(A_i\) integral and ample and \(a_i\) positive real, and we have \(B = \sum _{j=1}^m b_j B_j\) with \(B_j\) numerically trivial and integral, and \(b_j\) real. We can assume \(m{\gt}0\), since otherwise \(D=A\). By Lemma 2.35, \(A_1 + ma_1^{-1}b_j B_j\) is ample for all \(j\). So \(A_1 + a_1^{-1}B = m^{-1}\sum _{j=1}^m (A_1 + ma_1^{-1}b_j B_j)\) is ample. So \(D= a_1(A_1 + a_1^{-1}B) + \sum _{i\geq 2} a_i A_i\) is ample.
Suppose \(V\) is a vector space and \(S\subseteq V\). The dual cone of \(S\) is the set \(S^{\vee }\subseteq V^*\) consisting of all \(\lambda \in V^*\) such that \(\lambda (v) \geq 0\) for all \(v\in S\).
The dual cone \(S^{\vee }\) is a convex cone in \(V^*\).
Obvious.