Let $X$ be a smooth projective rationally connected variety over $\mathbb C$. Let $C$ be a curve class in $X$ such that rational curves equivalent to $C$ connect every two points.


Let $f: Y \to X$ be a finite morphism from a normal variety $Y$ to $X$, generically of degree $d$. Let $D$ be the branch divisor of $f$. (This is the pushforward of the ramification divisor, which is well-defined because the ramification divisor depends only on the codimension 1 points, which are regular.)



> 
> Do we have the inequality
>  $$ C \cdot D \geq 2 d - o(d) ?$$
> 
> 
> 


Here the $o(1)$ goes to $0$ as $d$ goes to $\infty$ for any fixed $X, C$. 


Of course we know that rationally connected varieties are simply connected, so $D \neq 0$, and $C$ is ample, so $C \cdot D >0$. This question asks whether we can show a much stronger inequality, so that $X$ isn't even close to being simply connected.


I can show a positive answer in some special cases, like projective space: Pull back any cover to the space of pairs of a line and a point on that line. It remains connected because the space of pairs of a line through a point is simply connected. Stein factorize the map from the pullback to the Grassmanian of lines in $\mathbb P^n$. We obtain a ramified cover of the Grassmanian, each point of which corresponds to a connected cover of a line. But the ramified cover of the Grassmanian is ramified only at the lines contained in $D$, which is a codimension $2$ subset, so it is unramified by purity, so it is a trivial cover as the Grassmanian is simply connected. Thus a general line has a connected inverse image is $Y$, so by Riemann-Hurwitz its intersection with $D$ must be at least $2d-2$.


But this argument uses in a couple places special facts about the Grassmanian of lines that I don't know how to generalize to an arbitrary rationally connected variety.


Of course for projective spaces with the line class, this lower bound is optimal, as can be seen by the cover adjoining the $d$th root of the ratio of two coordinates.



Are rationally connected varieties robustly simply connected?
Denote the curve class by $\beta$. Let $M$ be the normalization of a closed subvariety of $\overline{\mathcal{M}}\_{0,2}(X,\beta)$ such that the restricted evaluation morphism, $$\text{ev}|\_M = (\epsilon\_1,\epsilon\_2):M\to X\times X,$$ is surjective and generically finite. Denote by $m$ the degree of $\text{ev}|\_M$. For every morphism $f:Y\to X$ as above, the claim is that $$\langle [D\_f], \beta \rangle \geq 2d - 2(m!).$$


Fix a rational point, $$x\_0:\text{Spec}(\mathbb{C}) \to X\setminus D\_f,$$ (so I am assuming the field is $\mathbb{C}$, and I am about to start arguing geometrically, but there should be a better argument that avoids this). Denote by $M\_{x\_0}$ the fiber product with its remaining evaluation morphism to $S$, $$M\_{x\_0} = M\times\_{\epsilon\_2,X,x\_0} \text{Spec}(\mathbb{C}) \xrightarrow{e\_1} X. $$ This morphism is generically finite with a branch divisor $E\_{x\_0}$. Let $F$ denote $D+E\_{x\_0}$. Denote $X^o = X\setminus F$, and similarly $Y^o = f^{-1}(X^o)$ and $M^o\_{x\_0} = e\_1^{-1}(X^o)$. Then the morphisms, $$f^o : Y^o \to X^o, \ \ e\_1^o : M^o \to X^o, $$ are both finite and étale. Fix another rational point, $$ x\_1:\text{Spec}(\mathbb{C}) \to X^o,$$ and fix points above it, $$y\_1:\text{Spec}(\mathbb{C}) \to Y^o,\ \ m\_1:\text{Spec}(\mathbb{C}) \to M^o.$$ There are pushforward homomorphisms, $$\pi\_1(f^o): \pi\_1(Y^o,y\_1) \to \pi\_1(X^o,x\_1), \ \ \pi\_1(e^o\_1): \pi\_1(M^o,m\_1) \to \pi\_1(X^o,x\_1).$$ The index of the second map equals $m$, since $e\_1^o$ is finite, étale of degree $m$ with normal, connected domain. The kernel of the corresponding "monodromy representation" from $\pi\_1(X^o,x\_1)$ to $\text{Aut}((e\_1^o)^{-1}(x\_1))$ is a normal subgroup $N$ of finite index $\mu$, where $\mu \leq m!$, and I write these words to avoid ending the sentence with an exclamation point.


Associated to $f^o$, there is also a monodromy representation $\rho$ of $\pi\_1(X^o,x\_1)$ on $(f^o)^{-1}(x\_1)$, and the action is again transitive since $Y$ is normal and connected. Consider the induced action of $N$ on $(f^o)^{-1}(x\_1)$. Since $\rho$ is transitive, and since $N$ has index $\mu$, there are at most $\mu$ orbits for $\rho|\_N$.  

For every pair $y'$, $y''$ of points of $(f^o)^{-1}(x\_1)$ that are in the same orbit of $N$, there is a real continuous path $\gamma$ in $Y^o$ from $y'$ to $y''$ whose image under $f^o$ is a loop $f(\gamma)$ in $X^o$ based at $x\_1$ and whose homotopy class is in $N$. In particular, the loop lifts to a loop $\gamma\_M$ in $M^o$ based at $m\_1$.


Denote by, $$u\_{m\_1}:(\mathbb{P}^1,0,1) \to (X,x\_0,x\_1),$$ the morphism parameterized by $m\_1$. Denote by $\widetilde{C}$ the inverse image of this curve in $Y$. This decomposes into irreducible components $\widetilde{C} = \widetilde{C}\_1 \cup \dots \cup \widetilde{C}\_n$. By the hypotheses on $x\_0$ and $x\_1$, the fibers $f^{-1}(x\_0)$ and $f^{-1}(x\_1)$ are each reduced with $d$ elements. The decomposition of $\widetilde{C}$ into irreducible components induces partitions of $(f^o)^{-1}(x\_0)$ and $(f^o)^{-1}(x\_1)$ into $n$ distinct sets. Moreover, there is a natural bijection between these sets of size $n$: the partition set of an element $y$ in $(f^o)^{-1}(x\_0)$ corresponds to the partition set of an element $y'$ of $(f^o)^{-1}(x\_1)$ if and only if $y$ and $y'$ are in a common irreducible component of $\widetilde{C}$. 


As we vary the point $m\_1$ in $M^o$, the point $x\_0$ and the corresponding partition of $(f^o)^{-1}(x\_0)$ do not vary. Thus, if we take the loop $\gamma\_M$ in $M^o$ based at $m\_1$, and if we analytically continue $y'$ along this map according to the path $\gamma$, if $y'$ was in the same irreducible component of $\widetilde{C}$ as $y$ at the beginning of the analytic continuation, then the conjugate point $y''$ of $y'$ must be in the same irreducible component of $\widetilde{C}$ as $y$ at the end of the analytic continuation. Therefore, every pair of $N$-conjugate points of $(f^o)^{-1}(x\_1)$ are in the same irreducible component of $\widetilde{C}$.


Finally, since there were at most $\mu$ distinct orbits of $N$ acting on $(f^o)^{-1}(x\_1)$, it follows that $\widetilde{C}$ has at most $\mu$ distinct irreducible components. For each irreducible component $\widetilde{C}\_i$, denoting by $d\_i$ the degree of that component over $\text{Image}(u\_{m\_1})$, denoting by $g\_i$ the arithmetic genus of (the normalization of) that component, and denoting by $b\_i$ the branch number of that component, then Riemann-Hurwitz gives
$$
2g\_i-2 = d\_i(-2) + b\_i, \ \ b\_i = 2d\_i - 2 + 2g\_i \geq 2d\_i - 2.
$$
Summing up over all irreducible components, we have,
$$
\langle [D\_f], \beta \rangle \geq 2d - 2\mu \geq 2d - 2(m!).
$$


**Edit.** I realize now that there is no need to pass from $\pi\_1(M^o,m\_1)$ to the normal subgroup $N$. So the correct bound is not $m!$, but $m$. Also, I actually can think of no example [added: in characteristic zero] where $\widetilde{C}$ is disconnected. So it might be that the correct inequality is $$\langle [D\_f], \beta \rangle \geq 2d-2.$$ [Added later:] The argument above seems to be valid even in positive characteristic (after excising all mentions of "loops"), even when $X$ is rationally connected but not separably rationally connected. Such $X$ may have finite, nonzero $\pi\_1(X,x\_1)$. Thus, it is probably impossible to improve much on the inequality $\geq 2d - 2m$ in positive characteristic. As far as I know, the true inequality in characteristic zero may be $\geq 2d-2$.


