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Let $X$ be a smooth Fano variety of Picard number one. Let $Y \subset X$ be a closed subvariety. Let $d$ be an integer such that there is a dominating component $K_d$ of rational curves of degree $d$ in $X$. In other words, the family $K_d$ of curves covers $X$.

Question: For all such positive integer $d$, does there exist a rational curve $l \in K_d$ passing through a general point $p \in Y$ which lie in $Y$?

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  • $\begingroup$ Who lies in $Y$ --- point $P$ (or $p$?), or the curve? $\endgroup$
    – Sasha
    Commented May 19, 2023 at 4:28
  • $\begingroup$ The curve. The point $p$ is on $Y$. There was some typo. I have corrected. $\endgroup$
    – LAPRAS
    Commented May 19, 2023 at 4:31
  • $\begingroup$ If the curve lies in $Y$ then $X$ is irrelevant for the problem --- you are just asking if there is a curve in $Y$ passing through $p$. Of course, this depends very much on the properties of $Y$. For instance, if $Y$ is a smooth K3 surface the answer is negative. $\endgroup$
    – Sasha
    Commented May 19, 2023 at 4:45
  • $\begingroup$ There's a typo in the title $\endgroup$
    – YCor
    Commented May 19, 2023 at 5:02
  • $\begingroup$ Assume that $Y$ is unirulled. If $K^Y_d$ is a component of rational curves of degree $d$, then my question is equivalent to asking whether $K_d^Y \cap K_d$ has dimension at least $\text{dim}(Y)$. Is that always true ? $\endgroup$
    – LAPRAS
    Commented May 19, 2023 at 5:03

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No, just take $Y$ the product of $\mathbb P^1$ with a variety with no rational curves where the $\mathbb P^1$ is embedded in $X$ as something of degree $2$, and take $d$ odd. The only rational curves in $Y$ will be covers of the $\mathbb P^1$ which have even degree.

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