Is it possible to calculate the degree of a toric divisor directly from the fan of the toric variety? If so, how is this done? Or is there some alternative way to calculate the degree of these divisors?
Degree of Toric Divisors
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algebraic-geometry
toric-geometry
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3The set of Cartier T-divisors on a fan is exactly the set of piecewise linear integral functions on the fan. Given such a function, one can recover the divisor by finding the values of the function at the first lattice elements which are the coefficients of the irreducible divisors associated to rays ( 1- dimensional cones) – 2012-07-22
1 Answers
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Let the toric variety be $X$; let $d = \mathrm{dim} \ X$. For any Cartier $T$-divisor $D$, we can describe $\dim H^0(X, \mathcal{O}(D))$ as the number of lattice points in a certain polytope $P_D$. See Section 5.3 of Fulton's Introduction to Toric Varieties. The degree of $D$ is $d! \mathrm{Vol}(P_D)$.
Proof: Let $h(n)$ be the Hilbert polynomial $h(n) = \dim H^0(X, \mathcal{O}(nD))$. We know that the leading term of $h(n)$ is $\mathrm{deg}(\mathcal{O}(D)) n^d/d!$. But also, the number of lattice points in $n \cdot P_D$ is $\mathrm{Vol}(P_D) n^d + O(n^{d-1})$. Equating these two, $\mathrm{deg}(\mathcal{O}(D)) = d! \mathrm{Vol}(P_D)$. $\square$
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0I'm curious, wouldn't this imply that $\mathrm{deg}(\mathcal{O}(D))=D^d$ according to Fulton's book (p. 111)? (This lead to my own question http://math.stackexchange.com/questions/174656/degree-of-a-divisor-self-intersection) – 2012-07-24