Nef cone of Hilbert scheme of $n$ points

Question

Suppose $\operatorname{Nef}(X)$ is a rational polyhedron with extremal rays $\{F_i\}_i$. Now, consider the Hilbert scheme of $n$ points $X^{[n]}$ and the embedding $\operatorname{Nef}(X)\subset \operatorname{Nef}(X^{[n]})$. Let $\operatorname{Nef}(X^{[n]})\subset\Lambda\subset N^1(X^{[n]})$, such that $\Lambda$ is sitting inside the cone generated by $\operatorname{Nef}(X)$ and a divisor $aL^{[n]}-\frac{1}{2}B\in N^1(X^{[n]}),$ where $L\in \operatorname{Nef}(X).$

Then, I don't understand how one proves the following statement:

If $D\in\Lambda$ spans an extremal ray, then there is an extremal ray $F_i$ of $\operatorname{Nef}(X)$ such that $D$ is the unique $L_{[n]}$-orthogonal ray in the plane spanned by $F_i^{[n]}$ and $aL^{[n]}-\frac{1}{2}B$. Here $B$ is the exceptional divisor of the Hilber-Chow morphism.

My question: Why are all the extremal rays of $\Lambda$, $L_{[n]}$-orthogonal?

Reference: Page 5 of the paper "The nef cone of the Hilbert scheme of points on rational elliptic surfaces and the cone conjecture" by John Kopper.

Answer 1

$\DeclareMathOperator{\Nef}{Nef}$There is more to the situation in the paper than you have written here. I will use the notation of the paper instead of your notation.

In the paper, $X$ is a rational elliptic surface. We write $F$ for the class of a fiber, and $E$ signifies a $(-1)$-curve. The nef cone $\Nef X$ is spanned by these classes (as $E$ varies over all the $(-1)$-curves). These induce curve classes $F_{[n]}$ and $E_{[n]}$ on $X^{[n]}$. There is also the class $C_0$ of a curve contracted by the Hilbert-Chow morphism.

The cone $\Lambda \subset N^1(X^{[n]})$ is the cone of divisors which are positive on the curve classes $F_{[n]}$, $C_0$, and $E_{[n]}$. It contains the nef cone $\Nef(X)$ embedded in $\Nef(X^{[n]})$.

Kopper then considers the cone $\Lambda' \subset N^1(X^{[n]})$ which is spanned by $\Nef(X)$ and $(n-1)F^{[n]}-\frac{1}{2}B$, and proves that $\Lambda \subset \Lambda'$.

Crucially, Kopper computes at the end of the proof of Lemma 3.1 that the curve classes $E_{[n]}$ and $C_0$ are nonnegative on $(n-1)F^{[n]}-\frac{1}{2}B$, and therefore nonnegative everywhere on $\Lambda'$. (This remark at the end of Lemma 3.1 perhaps seems slightly out of place, and perhaps should have been discussed in the paragraph following the proof.) On the other hand, the new vertex $(n-1)F^{[n]}-\frac{1}{2}B$ is negative on $F_{[n]}$. Thus, $\Lambda\subset \Lambda'$ can be cut out by the single additional restriction that the intersection with the curve class $F_{[n]}$ has to be nonnegative.

The picture is that $\Lambda$ is just the intersection of the cone $\Lambda'$ with the half-space determined by the inequality $D.F_{[n]}\leq 0$. The extremal rays of $\Lambda'$ are all either in $\Nef(X)$ or the single ray $(n-1)F^{[n]}-\frac{1}{2}B$. The only places $\Lambda$ can have new extremal rays is along the plane $D.F_{[n]}=0$, since the cone doesn't change anywhere else. And the hyperplane section of $\Lambda'$ gotten by intersecting with $D.F_{[n]}=0$ basically looks like a copy of $\Nef(X)$, since $\Lambda'$ is a cone over $\Nef(X)$. The extremal rays all lie on the lines between extremal rays of $\Nef(X)$ and the divisor $(n-1)F^{[n]}-\frac{1}{2}B$.

Also see the paper https://arxiv.org/abs/1509.04722 where a slightly easier situation is considered by similar methods in Section 5.