Hello, I'm a newbie to mathoverflow. I reading a paper about Fano varieties (over C) and there is an assumption that uniruled varieties with picard number 1 are Fano...why is this true? Sorry if this is obvious thing. I am just learning the definitions still.
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3 Answers
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Since $X$ is a projective variety with $\textrm{Num}(X) \cong \mathbb{Z}$, the canonical divisor $K_X$ is either ample, or anti-ample or numerically trivial.
On the other hand we know that $X$ is uniruled, so it contains a free rational curve, namely the image of a map $$f \colon \mathbb{P}^1 \longrightarrow X$$ such that $f^*T_X$ is generated by global sections. In this case,
$$f^*T_X=\mathcal{O}_{P^1}(a_1) \oplus \cdots \oplus \mathcal{O}_{P^1}(a_n),$$
with $a_1 \geq a_2 \geq \cdots \geq a_n \geq 0.$
Moreover, since $T_{P^1}$ embeds into $f^*T_X$, it follows $a_1 \geq 2$. Therefore
$$K_X \cdot f_{*} \mathbb{P}^1= - \sum_{i=1}^n a_i \leq -2,$$
hence $K_X$ is not nef. This implies that $K_X$ is anti-ample, i.e. $X$ is Fano.
answered May 31, 2011 at 16:25
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1$\begingroup$ The canonical divisor could also be trivial (though of course this doesn't affect the argument). $\endgroup$– nafCommented May 31, 2011 at 16:30
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1$\begingroup$ Francesco, just to do a little bit of nitpicking, let me point out that the Picard number being $1$ does not imply that $\mathrm{Pic}X\simeq \mathbb Z$. In general there might be some torsion classes. Of course, this does not matter in your argument, and at the end $X$ is simply connected so there are actually no torsion classes, but I don't think this is clear a priori. Cheers. $\endgroup$ Commented Jun 1, 2011 at 6:30
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$\begingroup$ Sandor, of course you are right. I was thinking only about those divisor classes which are not numerically trivial. I will edit the answer, thank you. $\endgroup$ Commented Jun 1, 2011 at 6:50
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1$\begingroup$ Yeah, I know. You can just replace $\mathrm{Pic}$ with $\mathrm{Num}$. $\endgroup$ Commented Jun 1, 2011 at 7:07
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Here is another way of seeing this: If $X$ us uniruled, then as in Polizzi's answer we find a rational curve $f:\mathbb{P}^1\to X$ through a general point on $X$. Since $f^*K_X$ has negative degree on $\mathbb{P}^1$, any global section in $H^0(X,mK_X)$ must vanish on the image of $f$, hence on a dense subset of $X$, and hence on $X$. In particular, $K_X$ cannot be ample, and so $X$ is Fano.
In fact, the above argument shows that if $X$ is uniruled, then all the plurigenera $p_m=h^0(X,mK_X)$ vanish for all $m\ge 0$. In terms of Kodaira dimension, this means that $\kappa(X)=-\infty$.
An open problem in birational geometry is whether the converse also holds (see Debarre's book). For curves, this is obvious, since $p_1$ is just the genus of $X$. In dimension 2, Castelnuovo’s criterion asserts that a surface $X$ is rational if and only if $q(X) = p_{12}(X) = 0$, where $q(X)$ is the irregularity $h^0(X, \Omega_X^1)$. In dimension 3, this true by a deep result by Miyaoka and the MMP. For higher dimensions, only a few partial results are known.
answered May 31, 2011 at 22:43
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1$\begingroup$ JC, why is it clear that "$f^*K_X$ has negative degree on $\mathbb P^1$"? I suppose this follows from Mori-Mukai or their argument and it is not a deep fact, but I don't see how it is a "since". Cheers. $\endgroup$ Commented Jun 1, 2011 at 6:35
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1$\begingroup$ I don't think one needs to use Mori-Mukai: If X is uniruled, there is a smooth variety $Y$ of dimension $dim(X) - 1$ and a dominant rational map $f: Y \times \mathbb{P}^1 \to X$. Since $X$ is proper, $f$ is defined outside a subset of codimension $2$ so one can pullback pluricanonical forms via $f$. Moreover,
$K_{Y \times \mathbb{P}^1} = p_1^*K_Y \otimes p_2^*K_{\mathbb{P}^1}$so no positive tensor power of $K_{Y \times \mathbb{P}^1}$ has a non-zero section. Since we are in characteristic zero, $f$ is separable, so this gives a contradiction. $\endgroup$– nafCommented Jun 1, 2011 at 14:23 -
$\begingroup$ @ulrich: I did not say that one needs Mori-Mukai! I just said that one needs to use something. $\endgroup$ Commented Jun 1, 2011 at 14:54
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$\begingroup$ @Sandor: OK. The only reason for my comment was to point out that one doesn't need to use anything non-elementary and to indicate where the characteristic zero assumption is used. $\endgroup$– nafCommented Jun 1, 2011 at 16:38
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$\begingroup$ @ulrich: Seriously? Did you read my comment? $\endgroup$ Commented Jun 1, 2011 at 21:51
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Just to give yet another argument using more or less the definition of uniruledness.
$X$ is uniruled if there exists a dominant morphism from a variety $Y\times \mathbb P^1$ where $\dim Y=\dim X-1$. Therefore the image of $\{y\}\times \mathbb P^1$ for a general $y\in Y$ moves in a family of dimension $\dim Y$ in $X$ and hence has a semi-positive normal bundle and in particular the determinant of its normal bundle has to be a non-negative line bundle. Then the adjunction formula shows that $-K_X$ is positive on this curve.
If the Picard number is $1$, then $\mathrm{Pic}X\otimes \mathbb Q\simeq \mathbb Q$. If $L$ is a fixed ample divisor, then for any divisor $D$ there exists $a,b\in \mathbb N$ such that $aD\sim bL$. Since there exists at least one curve on which $-K_X$ is positive, it follows that $-K_X\sim \alpha L$ for some $\alpha\in \mathbb Q_+$ and hence it is ample itself. It follows that $X$ is Fano.
answered Jun 1, 2011 at 6:26