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Let $X$ be a smooth projective variety over a field $k$.

It is well-known that the sum of an ample divisor and an effective divisor is big (in fact this can be taken to be the definition of a big divisor). I'm looking for a weakening of this.

Is the sum of a big divisor and a pseudo-effective divisor itself a big divisor?

Recall that a divisor is called pseudo-effective if it lies in the closure of the cone of effective divisors.

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2 Answers 2

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Let's say the big divisor is $B$ and the pseudoeffective one is $F$. Write $B = A + E$ with $A$ ample and $E$ effective (by Kodaira). Then $B+F = A+E+F = A/2 + E + (F + A/2)$. $F+A/2$ is effective since $F$ is psef, and so we've written $B+F$ as ample plus effective, so it's big.

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    $\begingroup$ This looks promising. Though why is $F + A/2$ effective? $\endgroup$
    – Daniel Loughran
    Commented Jun 27, 2017 at 16:01
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    $\begingroup$ @DanielLoughran All classes in some ball around $A/2$ are ample, hence effective. One can look at a ball around $F$ of the same size, and find an effective class in it because $F$ is pseudoeffective. Then $F + A/2$ is that class plus an ample class. $\endgroup$
    – Will Sawin
    Commented Jun 27, 2017 at 20:58
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Theorem 2.2.26 of Lazarsfeld's Positivity in Algebraic Geometry, 2004, is the following:

Theorem: The big cone is the interior of the pseudoeffective cone and the pseudoeffective cone is the closure of the big cone.

These are convex cones. So the sum of a big class and a pseudoeffective class is in the interior of the cone, i.e., once again a big class.

The proof of this theorem is pretty much along the lines of the above answer. So not an independent answer here, just offering a citation.

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