# The pseudoeffective cone does not contain lines

It seems to be well-known that the pseudoeffective cone $\overline{\text{Eff}}(X)$ of a normal variety $X$ does not contain lines through the origin. How can it be proved? Is there a reference?

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What are you assuming about $X$: is it proper, resp. projective? Also, are you working with numerical equivalence classes of Cartier divisors (or perhaps the associated real vector space)? –  Jason Starr Jun 12 '12 at 15:44
$X$ is projective; I consider the same setting and notation as in the book of Lazarsfeld. A divisor is pseudoeffective if its numerical class is a limit of effective classes in $N^1(X)_{\mathbb R}$. –  fds Jun 12 '12 at 15:59

First, whether a class is pseudoeffective or not depends only on its numerical equivalence class. (The pseudoeffective cone is the closure of the cone of big classes, and $D$ is big if and only if $nD$ is numerically equivalent to $A + E$, where $A$ is an ample divisor on $X$, $E$ is an effective divisor on $X$ and $n > 0$ is an integer.)
Now suppose a class $\beta$ was such that both $\beta$ and $-\beta$ were pseudoeffective. Then $\beta \cdot a_1 \cdots a_k \geq 0$ and $\leq 0$ for any ample classes $a_1, \dots, a_k$. So $\beta \cdot a_1 \cdots a_k = 0$. But any divisor class can be written as a difference of ample classes, so $\beta \cdot \delta_1 \cdots \delta_k = 0$ for all divisor classes $\delta_1, \dots, \delta_k$. Thus, $\beta = 0$.
The standard reference for this material is Lazarsfeld's Positivity in Algebraic Geometry. I don't have a copy handy, but I'm $> 99\%$ certain that this fact is proved somewhere in Volume I.
Edit: I should add that this assumes that $X$ is projective. I don't know of any results without this assumption.
Properness is sufficient. By Chow's Lemma, there exists a projective, birational morphism from a projective variety to the original variety. Under pullback by a birational morphism, effectivity is preserved (hence so is pseudo-effectivity). So for a class $\beta$ as above, the pullback is numerically trivial. In other words, for every curve in the blowup, the image curve in the original variety has $0$ pairing with $\beta$. However, every integral curve in the original variety is the (finite-to-one) image of a curve in the blowup. So $\beta$ is numerically trivial. –  Jason Starr Jun 13 '12 at 15:04