Skip to main content
2 of 2
added 118 characters in body
Michael Joyce
  • 1.8k
  • 1
  • 12
  • 18

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.

Michael Joyce
  • 1.8k
  • 1
  • 12
  • 18