Let $S$ be a uniruled surface, ie admits a dominant map $ f:X \times \mathbb{P}^1$. Why then it's canonical divisor $\omega_X$ cannot be trivial? Motivation: I want to understand why $K3$ surfaces cannot be uniruled?
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2$\begingroup$ In positive characteristic it can be uniruled. In characteristic zero, a general rational curve of a uniruling has semi positive normal bundle. Since also the tangent bundle of the curve is positive, by adjunction the first Chern class of the ambient variety has positive degree on the rational curve. $\endgroup$– Jason StarrCommented Apr 5, 2023 at 21:14
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2$\begingroup$ No GAGA is not needed at all. You can read more about this in the textbooks by Debarre and by Koll'ar. $\endgroup$– Jason StarrCommented Apr 5, 2023 at 22:12
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1$\begingroup$ In characteristic zero, it is perhaps easier to simply say that a nonzero global section of $\omega_S$ would give a nonzero section of $\omega_{X \times \mathbb{P}^1}$ (by pullback), but such a section cannot exist. (This uses that $f$ is separable, so the argument does not work in positive characteristic.) $\endgroup$– nafCommented Apr 6, 2023 at 9:14
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1$\begingroup$ @naf: is there quick argument to see that a nonzero section of $\omega_{X \times \mathbb{P}^1}$ cannot exist? one idea comes into my mind to check it is to construct a contradiction on intersection count of associated divisors by using the adjunction formula to restrict the canonical bundle of $X \times \mathbb{P}^1$ to a general member $\{x\} \times \mathbb{P}^1$. Is this the "standard" argument to see that $\omega_{X \times \mathbb{P}^1}$ cannot have nonzero sections, or is there another "elementary" way to see it? $\endgroup$– user267839Commented Apr 6, 2023 at 16:56
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1$\begingroup$ This follows from the fact that $\omega_{\mathbb{P}^1}$ has no nonzero section since $\omega_{X \times Y} = \omega_X \boxtimes \omega_Y$ for any smooth varieties $X$ and $Y$. $\endgroup$– nafCommented Apr 7, 2023 at 10:00
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1 Answer
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The comment by user @naf is completely correct: there is a simpler argument than the argument I sketched. However, the argument I sketched gives a stronger result, which Mumford conjectured is sharp.
Because of the semipositivity, for every smooth, projective variety $S$ over a characteristic zero field $k$, for a geometric generic point $x$ of $S$, the tangent direction at $x$ of any rational curve contained in $X$ and containing $x$ is contained in the stalk of the following coherent sheaf, $$ \mathcal{F}:=\bigcap \text{Ker}\left( T_{S/k} \xrightarrow{\phi} \bigotimes^r \Omega_{S/k}\right), $$ where the intersection is over all integers $r\geq 0$ and all $\mathcal{O}_S$-module homomorphisms $\phi:T_{S/k} = \Omega^\vee_{S/k} \to \bigotimes^r \Omega_{S/k}$. In particular, for a nonzero, skew-symmetric section $\phi:T_{S/k} \to \Omega_{S/k}$ in dimension $2$, the kernel is the zero sheaf. Thus, there is no rational curve containing a geometric generic point on a surface with positive $h^0(S,\omega_{S/k})$. The same argument proves that $S$ is non-uniruled (for $S$ of arbitrary dimension) if there exists an integer $m>1$ such that $h^0(S,\omega^{\otimes m}_{S/k})$ is nonzero, i.e., if the $m^{\text{th}}$ plurigenus is nonzero
Mumford conjectured that the sheaf above, $\mathcal{F}$, is (generically) the kernel of the derivative map for the "rational quotient", i.e., the quotient of $S$ by all free rational curves. This quotient is only regular on a dense Zariski open subset of $S$, which we may as well take to be an open on which the quotient morphism is smooth to its image and on which $\mathcal{F}$ is locally free. Mumford conjectures that the restriction of $\mathcal{F}$ to this open equals the kernel of the derivative map of this smooth morphism.
Mumford's Conjecture is wide open; it is not even clear that the sheaf $\mathcal{F}$ is involutive. By the Rationally Connected Fibration Theorem, we do know that Mumford's conjecture is equivalent to the "Uniruledness Conjecture" (sometimes also attributed to Mumford): $S$ is uniruled if and only if the $m^{\text{th}}$ plurigenus equals zero for every $m>0$.