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Let $X$ be $\mathbb{P}^2$ blowing up a point, and $E$ be the exceptional divisor. Its Picard group module numerically equivalence is generated by $\mathcal{O}_X(E)$ and $H:=\pi^*\mathcal{O}_{\mathbb{P}^2}(1)$. Hence $H_1(X)$ is of $2$ dimensional. Let $\overline{NE}(X)$ be the Mori cone of $X$ (i.e. the closure of the cone generated by effective $1$-cycles quotient out by numerical equivalence). Then what is the shape of this cone?

I initially thought it is just $\mathbb R_+ E \oplus \mathbb R_+ H$. But suppose $C$ is a general hypersurface on $X$ and $C\cdot E=a$. If $\pi_*(C) = b\mathcal{O}_{\mathbb{P}^2}(1)$, then $C = bH-aE$ in $N_1(X)$. So it is a problem to determine allowable combinations of $a$ and $b$.

Besides, I was wondering if there is a way to determine Mori cones for $\mathbb P^n$ blowing up $m$ points (maybe for smaller $n,m$?)? In my case, $X$ is toric, so should be computable. But when $m$ is large, it is no longer toric...

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  • $\begingroup$ You should not say "numerically effective 1-cycles", because that is sometimes used as a synonym for "nef". Instead say "numerical classes of effective 1-cycles". $\endgroup$
    – Nefertiti
    Commented Apr 11, 2017 at 8:31

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In the specific case you ask about, the answer is that $\overline{NE}(X)$ is spanned by $E$ and $H-E$. One way to see this is that every irreducible curve $C$ other than $E$ itself must satisfy $E \cdot C \geq 0$, and every irreducible curve $C$ must satisfy $(H-E) \cdot C \geq 0$ since the class $H-E$ is basepoint-free.

As you suggest, whenever we blow up at most $n+1$ general points in $\mathbf P^n$, the result is a toric variety $X$, and in this situation there is a straightforward way to compute $\overline{NE}(X)$.

Once we start to blow up more points, things get much trickier. Starting from $\mathbf P^2$ we can blow up at most 8 points and still end up with a del Pezzo surface; the Cone Theorem then guarantees that $\overline{NE}(X)$ is a finite rational polyhedral cone, and we can actually compute it in all these cases. (This is a good exercise so I won't write the answer; it is written in many places.) For 9 points we no longer have a del Pezzo surface, and in fact the cone no longer has finitely many extremal rays, but we can still give a good description.

(Similar results are available for blowups of $\mathbf P^3$ in at most 8 points, and in higher dimensions.)

That's about as good as it gets, though. Remarkably, as far as I am aware even for the blowup of $\mathbf P^2$ in 10 points the complete description of $\overline{NE}(X)$ is still not known, despite much work.

For more information about the limit of what is known, you can search for recent work on the so-called SHGH conjecture and Nagata's conjectire.

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  • $\begingroup$ Thank you for your answer! However, I still did not see why $E$ and $H-E$ form the boundary of $\overline{NE}(X)$, could you give more details? $\endgroup$
    – Li Yutong
    Commented Apr 14, 2017 at 14:17
  • $\begingroup$ @LiYutong: it's really better to work these things out for yourself. Take an irreducible curve $C$ different from $E$. Can it have $E \cdot C <0$? Can it have $(H-E) \cdot C <0$? $\endgroup$
    – Nefertiti
    Commented Apr 19, 2017 at 8:31
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Just to clarify Nefertiti's answer:

First, $|H-E|$ is base point free because one can take any line $l$ passing the point $x$ which is blown up. Then $H \sim \pi^*\mathcal{O}(l) = \mathcal{O}(\pi^{-1}(L))$, hence $\pi^{-1}(L)-E \sim H-E$. Thus $|H-E|$ is base point free, in particular, it is nef.

Suppose $xE+yH$ is a class of effective curve, then $(xE+yH)\cdot (H-E)\geq 0$. By $E \cdot E=-1, E \cdot H=0$, one has $x+y \geq 0$. As $-E+H$ is effective, we determine one edge (i.e. spanned by $-E+H$). If $E+bH$ is also effective for some $b<0$, then $(E+bH)\cdot H = b$, a contradiction. Hence $\overline{NE}(X) = \mathbb{R}_{+}(H-E)+\mathbb{R}_{+}(E)$.

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