In Birational Geometry of Algebraic Varieties, Kollar and Mori write that for a line bundle "being big is essentially the birational version of being ample" (page 67). Recall that a line bundle $L$ on a projective variety $X$ of dimension $d$ is big if

$$ \limsup_{n \to \infty } \dfrac{H^0(X,L^n)}{n^d} \neq 0.$$

In other words, the rate of growth of the spaces of global sections is as big as possible. Big line bundles tend to exhibit behavior analogous to ample line bundles. I will give a couple of examples. In what follows, let $X$ be a variety over the complex numbers and let $L$ be a line bundle on $X$.

  1. Suppose $X$ is normal. If $L$ is ample, some power of $L$ defines an embedding in a projective space. Analogously, if $L$ is big, some power of $L$ defines a map

$$ \varphi_m: X \dashrightarrow H^0(X,L^m)$$

that is birational onto its image (Positivity in Algebraic Geometry I, page 139).

  1. If $L$ is ample, some power of $L$ is globally generated. On the other hand, if $L$ is big, some positive power of $L$ is generically globally generated; that is, the natural map

$$ H^0(X,L^m) \otimes \mathcal{O}_{X} \rightarrow L^m$$

is generically surjective (Positivity in Algebraic Geometry I, page 141).

Now, to get to my question, recall that if $L$ is ample, there exists a natural number $m$ such that the multiplication maps

$$ H^0(X,L^a) \otimes H^0(X,L^b) \rightarrow H^0(X,L^{a+b}) $$

are surjective for $a,b \geq m$ (Positivity in Algebraic Geometry I, page 32).

Question: Do big line bundles have a property analogous to the surjectivity of multiplications maps?

It is not clear to me what this property should be, but I would hope that these multiplication maps have eventually high rank in some suitable sense.


1 Answer 1


If $R(L)=\oplus H^0(mL)$ is not finitely generated, the above surjectivity will fail, however it will hold "asymptotically" for any big line bundle $L$. In fact, by Fujita's approximation of big classes (see eg. Lazarsfeld's Positivity book Theorem 11.4.4), for any $\epsilon >0$ there is a birational modification $f:X'\to X$ such that $f^*L=A+E$ where $A$ is an ample $\mathbb Q$-divisor and $E$ is an effective $\mathbb Q$-divisor such that ${\rm vol}(A)>{\rm vol}(L)-\epsilon$. Thus, by the ample case, there is an $m>0$ such that $H^0(aA)\otimes H^0(bA)\to H^0((a+b)A)$ is surjective for all $a,b\geq m$ sufficiently divisible (so that $aA$ and $bA$ are Cartier). Since $f_*\mathcal O _{X'}(aA)\subset \mathcal O _X(aL)$, we see that if $$V_{a,b}={\rm Im} \left( H^0(aL)\otimes H^0(bL)\to H^0((a+b)L))\right),$$ then $\dim V_{a,b}/h^0((a+b)L)>(1-\epsilon)$ for $m\gg 0$.

  • $\begingroup$ Thank you. This is exactly the sort of statement I had hoped would be true. I have a couple of quick questions. 1. Since $a$ and $b$ should clear the denominators of $A$ for the argument to work, we could still expect the dimension of $V_{a,b}$ to be quite small even for arbitrarily big $a$ and $b$. Is that correct? 2. Does your argument use the fact that the volume is a limit rather than just a limsup? Thanks again. $\endgroup$ Commented Aug 24, 2020 at 23:39
  • 1
    $\begingroup$ The statement is true for all $a,b$ sufficiently big, but more annoying to state/prove. Essentially, we have proven the statement for ll $a,b$ divisible by $m$. Since $L$ is big, we can assume that $H^0(cL)>0$ for all $c\geq m$. So in general we write $a=km+a'$ with $m\leq a'<2m$, $b=jm+b'$ for $m\leq b'<2m$ and then $V_{a,b}\supset V_{km,jm}$ and we still have $\dim V_{km,jm}/h^0((a+b)L)>(1-\epsilon)$ for $j,m\gg 0$. $\endgroup$
    – Hacon
    Commented Aug 25, 2020 at 4:27
  • $\begingroup$ And, yes, I used that the limit and lim sup agree (but this is easy to prove) $\endgroup$
    – Hacon
    Commented Aug 25, 2020 at 4:28

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