# Are rationally connected varieties robustly simply connected?

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.

• It sounds to me like your argument should work in general. You have an a priori bound on the "degree" of the map $\mathcal{M}_{0,2}(X,[C])\to X\times X$, i.e., the degree when restricted to a projective subvariety of the base that dominates $X\times X$. For lines, that degree is $1$, but since you have $o(d)$ in your formula, I think the argument should work in any case . . . – Jason Starr Sep 6 '15 at 19:03
• @JasonStarr What if $\mathcal M_{0,2}(X,[C])$ isn't simply connected? – Will Sawin Sep 6 '15 at 19:04
• I don't think simple connectedness of $\overline{M}_{0,2}(X,[C])$ is the issue. I think the issue is the index of the image of $\pi_1(\overline{M}_{0,2}(X))$ in $\pi_1(X\setminus D)$. – Jason Starr Sep 7 '15 at 1:13
• @JasonStarr What's the replacement for "thus a general line has a connected inverse image" without showing that a finite etale cover of $\overline{M}_{0,2}(X)$ is trivial? – Will Sawin Sep 7 '15 at 2:34
• I only have to bound the number of inverse images independent of the degree $d$. The bound should be something like $m!$ where $m$ is the "degree" of the evaluation morphism. I will try to write it carefully. – Jason Starr Sep 7 '15 at 9:57

## 2 Answers

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 $\geq 2d-2$.

For my own edification I'm rewriting Jason's answer so as to excise all mention of loops (as he suggests), and trying to get the best possible bound from this technique.

Let $M$ be a variety with a $\mathbb P^1$-bundles $U \to M$ and two sections $s_1, s_2$ with a map $f: U \to X$. Let $Y \to X$ be a finite covering and let $x$ be a general unramified point. We want to bound the degree of the finite part of the Stein factorization of that map $U \times_X Y \to M$ by a constant $n$.

Once we do that, we get that a general fiber of that map has at most $n$ connected components and hence has branch divisor of degree at least $2d-2n$.

It is sufficient to bound it restricted to the fiber of $f \circ s_1$ over a general point $x_0$. But then because $s_1$ has a constant map to $X$, the covering $U_{x_0} \times_X Y$ restricted to $M_{x_0} \times_X Y$ by $s_1: M_{x_0} \to U_{x_0}$ is a finite union of copies of $M_{x_0}$. So we are taking the Stein factorization of a map with many sections - in fact one section in each connected component. Hence the finite part will have a section in each connected component, so because it is finite each connected component will actually be a section (maybe after pasing to an open set). But now the fiber $U_{x_0} \times_X Y$ restricted to $M_{x_0} \times_X Y$ by $s_2$ has one connected component meeting each connected component of $U_{x_0} \times_X Y$. So we may take $n$ to be an upper bound for the connected components, which we can take to be the degree of the Stein factorization of the map $f \circ s_2 : M_{x_0} \to X$.

So if $M$ is the space of all rational curves of class $\beta$ with two marked points, we may take $n$ to be the number of connected components of the space of curves of class $\beta$ between two general points. In fact we can do a little better - it's the index of $\operatorname{Gal}(M_{0,2}(X, \beta))$ in the image of $\operatorname{Gal}(X)\times \operatorname{Gal}(X)$, where by the Galois group of a variety I mean the Galois group of its function field.