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.

  • 4
    $\begingroup$ 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. $\endgroup$ Commented Oct 29, 2012 at 2:07

1 Answer 1


(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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.