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Let $Y$ be a rationally connected variety over an algebraically closed field, and let $$\phi:X\dashrightarrow Y$$ be a rational fibration such that the general fiber of $\phi$ is rationally chain connected. Is it true that $X$ is rationally chain connected?

If we assume that the general fiber of $\phi$ is smooth and rationally connected can we conclude that $X$ is rationally connected?

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  • $\begingroup$ I think you edited the question. The original hypothesis was that the general fiber be rationally chain connected, not smooth and rationally connected. With the new hypothesis, of course Francesco Polizzi's answer is correct. $\endgroup$
    – Jason Starr
    Commented Oct 17, 2013 at 11:30

2 Answers 2

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Over $\mathbb{C}$, the answer to the second question is yes.

In fact, since rational connectedness is a birational property, one can solve the indeterminacy of the rational map $\phi$ in order to obtain a dominant morphism $f \colon Z \to Y$ (whose general fiber is birational to the general fiber of $\phi$) and then apply the following result, due to Graber, Harris and Starr:

Theorem. Let $f \colon Z \to Y$ be any dominant morphism of complex varieties. If $Y$ and the general fiber of $f$ are rationally connected, then $Z$ is rationally connected.

See the paper Families of rationally connected varieties, J. Amer. Math. Soc. 16 (2003), no. 1, 57–67, in particular Corollary 1.3.

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    $\begingroup$ Yes, I knew this theorem. However I was wondering if it is still true when we $f$ is just a rational map and not a morphism and the general fiber of $f$ is just rationally chain connected. $\endgroup$
    – Puzzled
    Commented Oct 16, 2013 at 14:26
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    $\begingroup$ Ok, maybe you should have added more background :-) Anyway, the answer to your second question is yes because rational connectedness is a birational property, so just solve the indeterminacy of the rational map $\phi$ and apply Graber-Harris-Starr result. I edited the answer. $\endgroup$
    – Francesco Polizzi
    Commented Oct 16, 2013 at 14:36
  • $\begingroup$ Thank you very much Francesco. Your argument is clear. Can we say something when the general fiber of $\phi$ is singular and rationally chain connected? For instance can we conclude that $X$ is rationally chain connected? $\endgroup$
    – Puzzled
    Commented Oct 16, 2013 at 14:51
  • $\begingroup$ Rational chain connectedness is not a birational property (think of a cone over an elliptic curve, which is rationally 2-connected but whose resolution is only unirational), so my previous argument does not work. At the moment, I do not know. Maybe someone else, more expert on the topic, could answer. $\endgroup$
    – Francesco Polizzi
    Commented Oct 16, 2013 at 15:08
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What are you assuming about $X$? Consider the closed subscheme $X$ of $\mathbb{P}^2 \times \mathbb{P^2}$ consisting of those points $([s,t],[X,Y,Z])$ satisfying the equation $$s^3(X+uY + u^2Z)(X+\omega uY+\omega^2 u^2 Z)(X+\omega^2 u Y +\omega^2 u^2 Z) = 0,$$ where $u$ is a coordinate on some irrational cyclic $3$-sheeted cover of $\mathbb{P}^1$, e.g., $u$ is a cube root of $(t/s)^3 + 1$. Of course this $X$ is very much not normal. But it fibers over the base $\mathbb{P}^1$, the fibers are rationally chain connected (just triangles of lines), yet every rational curve in $X$ is contained in a fiber of the projection to $\mathbb{P}^1$.

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  • $\begingroup$ So Corollary 1.3 of Families of rationally connected varieties, J. Amer. Math. Soc. 16 (2003), requires $X$ to be smooth? $\endgroup$
    – Puzzled
    Commented Oct 16, 2013 at 16:24
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    $\begingroup$ "So Corollary 1.3 ... requires $X$ to be smooth?" No, that corollary does not have a smoothness hypothesis. However, the hypothesis is that a general fiber is rationally connected. You asked about the case where the fiber is rationally chain connected, and that is weaker than rationally connected. The counterexample above shows that one cannot replace the hypothesis on the fiber by rationally chain connected. $\endgroup$
    – Jason Starr
    Commented Oct 16, 2013 at 16:33
  • $\begingroup$ Ok, now everything is clear. Thank you very much. $\endgroup$
    – Puzzled
    Commented Oct 16, 2013 at 16:36

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