Can free rational curves lift to ramified covers of Fano varieties?

Question

Does there exist $X$ a smooth Fano manifold, $f: Y \to X$ a nontrivial ramified finite cover, $C \subseteq X$ a smooth very free rational curve, such that $f$ is étale over a neighborhood of $C$?

Without the Fano condition, an example is $X = \operatorname{Hilb^2} (\mathbb P^2)$, $Y = \mathbb P^2 \times \mathbb P^2$ blown up at the diagonal, $C$ is the image of a strict transform of a general very free rational curve on $\mathbb P^2 \times \mathbb P^2$, which doesn't intersect the diagonal, so its image doesn't intersect the branch locus and hence the covering is etale over it.

A minimal non-Fano example is the surface $\mathbb P^1 \times \mathbb P^1$ blown up at the four torus fixed points $(0,0),(0,\infty),(\infty,0),(\infty,\infty)$. The ramified cover $\sqrt{xy}$ is ramified at the four divisors at zero and infinity but not at the four exceptional divisors. There is nodal $(2,2)$ curve which intersects each of the exceptional divisors once (and thus the four branched divisors zero times) passing through two general points, so this is an example.

However, there is no Fano surface example. We can fix a point on the branch divisor and a transverse tangent vector. We can degenerate $C$ to a curve going through that point with that tangent vector, which must have an irreducible component positive intersection with the branch divisor. Since $C$ has zero intersection, some other component must have negative intersection with the branch-divisor. Because the branch divisor is a curve, this component must in fact be a component of the branch divisor, with negative self-intersection. So because $X$ is Fano it is a $-1$ curve and we can blow it down, and doing so does not affect the curve class $C$ because $C$ does not intersect the branch divisor.

The motivation is trying to understand Emmanuel Peyre's philosophy in Liberté et accumulation - specifically whether free rational curves, and their associated free rational points, can lie in the exceptional thin subsets in Manin's conjecture.

Answer 1

After some helpful conversations with Johan de Jong, I came up with an example. After finding it I realize I probably should have figured it out earlier.

In fact, $\operatorname{Hilb}^2(\mathbb P^n)$ is an example for any $n>2$.

The blow-up of $\mathbb P^n \times \mathbb P^n$ at the diagonal is a two-to-one cover of $\operatorname{Hilb}^2(\mathbb P^n)$, ramified only at the exceptional divisor. If we take a general rational curve in $\mathbb P^n \times \mathbb P^n$, it will avoid the diagonal, hence lift to the blow-up and descend to $\operatorname{Hilb}^2(\mathbb P^n)$, where it will remain very free, and then lift back to the covering.

It's easier to calculate on this blow-up. Its canonical bundle is $\mathcal O(-n-1,-n-1) \otimes \mathcal O((n-1) E)$. The branch divisor is $E$, so the pullback of the canonical bundle of $\operatorname{Hilb}^2(\mathbb P^n)$ is $\mathcal O(-n-1,-n-1) \otimes \mathcal O((n-2) E)$. It suffices to show that the inverse $\mathcal O(n+1,n+1) \otimes \mathcal O(-(n-2) E)$ is ample. Sections of the line bundle $\mathcal O(1,1) \otimes \mathcal O(- E)$ are given by symplectic bilinear forms. From this, one can see that they define a two-to-one projective embedding and thus are ample (in fact, they are the Plucker coordinates of the Grassmanian of two-dimesnional quotients of $H^0(\mathbb P^n, \mathcal O(1))$ which $\operatorname{Hilb}^2(\mathbb P^n)$ naturally maps to), and then $\mathcal O(3,3)$ is nef, so combining them we get an ample line bundle.