A uniruled variety is the one which admits a dominant map from $X \times \mathbb P^1$. I think it is true that uniruled varieties are rationally connected. Is the converse true? What about low dimensional cases, i.e. surfaces and threefolds?
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1$\begingroup$ For surfaces, rational connectivity coincides with rationality. In general, it is even an open problem whether there are any rationality connected varieties which are not unirational. This is called an "embarrassing question" in notes by J. Kock on lectures by Joe Harris: mat.uab.es/~kock/RLN/rcv.pdf . For a recent confirmation that this is still an open problem, see Jason Starr's answer here: mathoverflow.net/questions/131086/… . $\endgroup$– Vesselin DimitrovCommented Aug 24, 2013 at 9:31
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4$\begingroup$ PS: I realized I had misread your question. Certainly uniruled varieties need not be rationally connected: e.g., a ruled surface $C \times \mathbb{P}^1$, where $C$ is a curve of positive genus, is not rationally connected as it has non-zero holomorphic $1$-forms (coming from $C$). $\endgroup$– Vesselin DimitrovCommented Aug 24, 2013 at 10:19
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$\begingroup$ ... And yes, rationally connected varieties are uniruled. This should be obvious: the standard definition of rational connectivity is that there is a dominant map from $T \times \mathbb{P}^1 \times \mathbb{P^1}$ induced diagonally from a morphism from $T \times \mathbb{P}^1$. $\endgroup$– Vesselin DimitrovCommented Aug 24, 2013 at 10:52
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2$\begingroup$ @VesselinDimitrov: Thanks for the answer and for clearing my confusion. $\endgroup$– AdamCommented Aug 24, 2013 at 11:17
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2 Answers
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Yes, rationally connected varieties are uniruled (a fact which follows essentially by definition).
In characteristic zero, one can think about this also by exploiting the following characterization of uniruled and rationally connected varieties, that can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4.
Let $X$ be a smooth projective variety of dimension $d$, defined over a field $\mathbb{K}$ of characteristic zero. Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$ with $a_i \geq r$ for every $i$. Now one has the following
Proposition. Assume $\textrm{char}(\mathbb{K})=0$. Then $X$ is uniruled if and only through a general point $x \in X$ there is a $0$-free rational curve. Moreover $X$ is rationally connected if and only if through a general point $x \in X$ there is a $1$-free rational curve.
Remark. The condition $\textrm{char}(\mathbb{K})=0$ is an essential one. For instance, if $\textrm{char}(\mathbb{K})=p>0$ then the Fermat hypersurface of degree $p^r+1$ in $\mathbb{P}^N$, with $N \geq 4$ and $r \geq 1$, is uniruled by lines, none of which are free. Moreover, if $p^r >N$ any such a hypersurface has ample canonical class, then it contains no free rational curves at all.
answered Aug 25, 2013 at 12:08
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$\begingroup$ But why do you write "in characteristic zero?" Over any uncountable field, uniruledness simply means that there is a rational curve through any general point; rational connectivity is in turn the stronger property that there is a rational curve through any two general points. $\endgroup$ Commented Sep 24, 2013 at 14:52
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$\begingroup$ You are right. It is the characterization by using free rational curves that only holds in characteristic zero. I corrected the answer and added a remark, thank you for the observation. $\endgroup$ Commented Sep 24, 2013 at 15:11
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$\begingroup$ Yes, this is the difference between inseparable and separable rational connectivity (which only arises in positive characteristic, of course). The proposition remains true upon adding the qualifier "separably" in front of "uniruled" and "rationally connected"; this is Theorems IV 1.9 and IV 3.7 in Kollar's book. The example you give is not separably uniruled. $\endgroup$ Commented Sep 24, 2013 at 16:04
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Let $k$ be an algebraically closed uncountable field of characteristic zero. If $X$ is rationally connected then through two general points of $X$ there is a rational curve. In particular there is a rational curve through a general point of $X$. This means that the evaluation map
$$ev:\mathbb{P}^1\times Mor(\mathbb{P}^1,X)\rightarrow X$$
is dominant. Since $Mor(\mathbb{P}^1,X)$ has at most countably many components and $X$ is irreducible, there exists a component $Y$ of $Mor(\mathbb{P}^1,X)$ such that
$$ev:\mathbb{P}^1\times Y\rightarrow X$$
is dominant. Therefore $X$ is uniruled.
