# Amplitude and bigness issues

There are a couple of statements that I have read which are made as though they were trivial, but I am doubtful about them.

1. One is related to an example showing that the s-invariant of an ample line bundle on a projective variety X is an algebraic integer of degree $\leq dim X$. Recall that, given an ideal sheaf $\mathcal{I}\subset \mathcal{O}_X$ on a projective irreducible variety X, the s-invariant of $\mathcal{I}$ with respect to an ample line bundle L is $s_L(\mathcal{I})$ is the minimum $s\in \mathbb{R}$ such that $\mu^{\ast}(sL)-E$ is nef on $X'$, where $$\mu:X'=Bl_{\mathcal{I}}X\rightarrow X$$ is the blow-up along the ideal $\mathcal{I}$, with exceptional divisor E. The author claims that the class $s_L(\mathcal{I})L-E$ is nef (by definition) but not ample and then uses the Campana-Peternell theorem to conclude the result. How does the non-ampleness follow?

2. Now let L be a big and nef divisor on a smooth projective variety X of dimension n and assume that the Seshadri constant of L at some point $x\in X$ is $\epsilon(L;x)>2n$ (we hence have the same inequality at a very general point). Take two general points $x,y\in X$ and consider the blow-up $\mu:X'=Bl_{\{x,y\}}X\rightarrow X$, with corresponding exceptional divisors $E_x$ and $E_y$. The divisor $\mu^{\ast}(\frac{1}{2}L)-nE_x$ is then nef (by definition) and big. How does bigness follow?

Thanks in advance for any insight.

• Regarding 1: the ample cone is open, and its closure is the nef cone. Since the class in question is on the boundary, it cannot be ample. Oct 29, 2012 at 2:07

(1) was handled in the comments. Regarding (2): since $\epsilon(L;x) > 2n$ with strict inequality, in fact $\nu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ is nef for some $c> 0$, where $\nu$ is the blow-up at just $x$. Pulling this back to $X^\prime$, we have $\mu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ nef. But $\mu^\ast \left( \frac{1}{2} L \right) - n E_x$ lies on the segment between $\mu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ and $\mu^\ast \left( \frac{1}{2} L \right)$ (strictly between the endpoints!), with the former nef and the latter big. So $\mu^\ast \left( \frac{1}{2} L \right) - n E_x$ is nef+big = big. Perhaps writing this up just obfuscates things -- just draw a picture of the cones and think about how $\mu^\ast \left( \frac{1}{2} L \right) - t E_x$ moves relative to the nef and pseudoeffective cones, and what $\epsilon$ means in the picture.