Degree of irrationality of very general abelian surfaces

TL;DR

This paper proves that the degree of irrationality for very general polarized abelian surfaces is uniformly bounded by 4, regardless of polarization degree, challenging previous conjectures and extending understanding of their geometric complexity.

Contribution

It establishes a uniform upper bound of 4 for the degree of irrationality of very general polarized abelian surfaces, regardless of polarization degree.

Findings

Degree of irrationality is bounded above by 4 for very general abelian surfaces.

Disproves part of a previous conjecture on irrationality bounds.

Shows the bound is independent of polarization degree.

Abstract

The degree of irrationality of a projective variety $X$ is defined to be the smallest degree rational dominant map to a projective space of the same dimension. For abelian surfaces, Yoshihara computed this invariant in specific cases, while Stapleton gave a sublinear upper bound for very general polarized abelian surfaces $(A, L)$ of degree $d$. Somewhat surprisingly, we show that the degree of irrationality of a very general polarized abelian surface is uniformly bounded above by $4$, independently of the degree of the polarization. This result disproves part of a conjecture of Bastianelli, De Poi, Ein, Lazarsfeld and Ullery.

Equations

$$\mathrm{irr}(X)=\min \{\delta >0\mid \exists \text{ a degree δ rational dominant map }X⇢{\mathbb{P}}^n\}.$$

$$\mathrm{irr}(A)\le C\cdot \sqrt{d}$$

$$\mathrm{irr}(A)\le 4.$$

$$\iota :A\to A,x\mapsto -x$$

$$h^0(A,2L{)}^{+}=2d+2$$

$$1+3+\cdots +(2m-1)=m^2$$

$$\sum _{i=1}^{16}{a}_i^2=2d-2,$$

$$\dim V\ge 2d+2-\sum _{i=1}^{16}{a}_i^2\ge 4.$$

$$d_i:={\mathrm{mult}}_{p_i}D$$

$$\mathrm{deg}\phi \cdot \mathrm{deg}S\le D^2.$$

$$\mathrm{deg}\phi \cdot \mathrm{deg}S\le 8.$$

$$\hat{D}{\sim }_{lin}{\pi }^{\ast }D-\sum _{i=1}^{16}{d}_i{E}_i,$$

$$\deg\varphi\cdot\deg S=\deg\psi\cdot\deg S\leq\widehat{D}^{2}=4L^{2}-\sum_{i=1}^{16}d_{i}^{2}\leq 4\Bigg{(}2d-\sum_{i=1}^{16}a_{i}^{2}\Bigg{)}=8.\qed$$

$$D=F+M\in \mathfrak{d}\text{and}\mathfrak{d}=F+\mathfrak{b},$$

$$m_i:={\mathrm{mult}}_{p_i}M\text{and}{f}_i:={\mathrm{mult}}_{p_i}F,$$

$$\iota (M)+\iota (F)=\iota (D)=D=M+F\text{for all}D\in \mathfrak{d}.$$

$$\sum _{i=1}^{16}{m}_i^2\ge 2d-8.$$

$$\hat{F}={\pi }^{\ast }F-\sum _{i=1}^{16}{f}_i{E}_i,$$

$$h^0(K,{\mathcal{O}}_K(\bar{F}))=1.$$

$$1=h^0(K,{\mathcal{O}}_K(\bar{F}))=\chi ({\mathcal{O}}_K,{\mathcal{O}}_K(\bar{F}))=\frac{1}{2}(\bar{F}{)}^2+2$$

$$-4=2(\bar{F}{)}^2=({\pi }^{\ast }\bar{F}{)}^2=(\hat{F}{)}^2=F^2-\sum _{i=1}^{16}{f}_i^2=2d-\sum _{i=1}^{16}{f}_i^2$$

$$\sum _{i=1}^{16}{d}_i^2=\sum _{i=1}^{16}(f_i^2+m_i^2+2f_i{m}_i)\le 2d+4+\sum _{i=1}^{16}{m}_i^2+\frac{1}{2}\sum _{i=1}^{16}{d}_i^2.$$

$$\sum _{i=1}^{16}{m}_i^2\ge -2d-4+\frac{1}{2}\sum _{i=1}^{16}{d}_i^2\ge -2d-4+2\sum _{i=1}^{16}{a}_i^2\ge 2d-8$$

$$\mathrm{deg}\phi \cdot \mathrm{deg}S\le 8.$$

$$\mathrm{deg}\phi \cdot \mathrm{deg}S\le M^2-\sum _{i=1}^{16}{m}_i^2\le 2d-(2d-8)=8.\text{qed}$$

$$\{\mathrm{deg}\phi =2,\mathrm{deg}S=4\}\text{and}\{\mathrm{deg}\phi =4,\mathrm{deg}S=2\}.$$

Full text

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chapsec \sethead[0][NATHAN CHEN][]DEGREE OF IRRATIONALITY OF VERY GENERAL ABELIAN SURFACES0

DEGREE OF IRRATIONALITY OF VERY GENERAL ABELIAN SURFACES

Nathan Chen

Introduction

Given a projective variety $X$ of dimension $n$ which is not rational, one can try to quantify how far it is from being rational. When $n=1$, the natural invariant is the gonality of a curve $C$, defined to be the smallest degree of a branched covering $C^{\prime}\rightarrow\mathbb{P}^{1}$ (where $C^{\prime}$ is the normalization of $C$). One generalization of gonality to higher dimensions is the degree of irrationality, defined as:

[TABLE]

Recently, there has been significant progress in understanding the case of hypersurfaces of large degree (cf. [2], [3], [4]). The history behind the development of these ideas is described in [4]. The results of [2], [3], [4] depend on the positivity of the canonical bundles of the varieties in question, so it is interesting to consider what happens in the $K_{X}$-trivial case. Our purpose here is to prove the somewhat surprising fact that the degree of irrationality of a very general polarized abelian surface is uniformly bounded above, independently of the degree of the polarization.

To be precise, let $A=A_{d}$ be an abelian surface carrying a polarization $L=L_{d}$ of type $(1,d)$ and assume that $\operatorname{NS}(A)=\mathbb{Z}[L]$. An argument of Stapleton [9] showed that there is a constant $C$ such that

[TABLE]

for $d\gg 0$, and it was conjectured in [4] that equality holds asymptotically. Our main result shows that this is maximally false:

Theorem 1.1**.**

For an abelian surface $A=A_{d}$ with Picard number $\rho=1$, one has

[TABLE]

We conjecture that in general equality holds. However, as far as we can see, the conjecture of [4] for polarized K3 surfaces $(S_{d},B_{d})$ of genus $d$ - namely, that there exist constants $C_{1},C_{2}$ such that $C_{1}\cdot\sqrt{d}\leq\operatorname{irr}(S_{d})\leq C_{2}\cdot\sqrt{d}$ for $d\gg 0$ - remains plausable.111In other words, $B_{d}$ is an ample line bundle on $S_{d}$ with $B_{d}^{2}=2d-2$.

For an abelian variety $A$ of dimension $n$, it has been shown in [1] that $\operatorname{irr}(A)\geq n+1$ (for $n=2$, one can also see this via Lemma 3.5). When $n=2$, Yoshihara proved that $\operatorname{irr}(A)=3$ for abelian surfaces $A$ containing a smooth curve of genus 3 (cf. [11]). On a related note, Voisin [10] showed that the covering gonality of a very general abelian variety $A$ of dimension $n$ is bounded from below by $f(n)$, where $f(n)$ grows like $\log(n)$, and this lower bound was subsequently improved to $\lceil\frac{1}{2}n+1\rceil$ by Martin [8].222Covering gonality is defined as the minimum integer $c>0$ such that given a general point $x\in A$, there exists a curve $C$ passing through $x$ with gonality $c$.

In the proof of our theorem, assuming as we may that $L$ is symmetric, we consider the space $H^{0}(A,\mathcal{O}_{A}(2L))^{+}$ of even sections of $\mathcal{O}_{A}(2L)$. By imposing suitable multiplicities at the two-torsion points of $A$, we construct a subspace $V\subset H^{0}(A,\mathcal{O}_{A}(2L))^{+}$ which numerically should define a rational map from $A$ to a surface $S\subset\mathbb{P}^{N}$. Using bounds on the degree of the map and the degree of $S$, as well as projection from linear subspaces, we construct a degree 4 rational covering $A\dashrightarrow\mathbb{P}^{2}$. The main difficulty is to deal with the possibility that $\mathbb{P}_{\text{sub}}(V)$ has a fixed component; this approach was inspired in part by the work of Bauer in [5], [6].

Acknowledgments. I would like to thank my advisor Robert Lazarsfeld for suggesting the conjecture and for his encouragement and guidance throughout the formulation of the results in this paper. I would also like to thank Frederik Benirschke, Mohamed El Alami, François Greer, Samuel Grushevsky, Ljudmila Kamenova, Yoon-Joo Kim, Radu Laza, John Sheridan, and Ruijie Yang for engaging in valuable discussions.

Set-up

Let $A=A_{d}$ be an abelian surface with $\rho(A)=1$. Assume $\operatorname{NS}(A)\cong\mathbb{Z}[L]$ where $L$ is a polarization of type $(1,d)$ for some fixed $d\geq 1$, so that $L^{2}=2d$ and $h^{0}(L)=d$. Let

[TABLE]

be the inverse morphism, and let $Z=\{p_{1},\ldots,p_{16}\}$ be the set of two-torsion points of $A$ (fixed points of $\iota$). We may assume that $L$ is symmetric – that is, $\iota^{\ast}\mathcal{O}_{A}(L)\cong\mathcal{O}_{A}(L)$ – by replacing $L$ with a suitable translate. In particular, the cyclic group of order two acts on $H^{0}(A,\mathcal{O}_{A}(2L))$. The space of even sections $H^{0}(A,\mathcal{O}_{A}(2L))^{+}$ of the line bundle $\mathcal{O}_{A}(2L)$ (sections $s$ with the property that $\iota^{\ast}s=s$) has dimension

[TABLE]

(see [7, Corollary 4.6.6]). An even section of $\mathcal{O}_{A}(2L)$ vanishes to even order at any two-torsion point, so we need to impose at most

[TABLE]

conditions for every even section to vanish to order $2m$ at any fixed point $p\in Z$ (see [5] for more details).

Fix any integer solutions $a_{1},\ldots,a_{16}\geq 0$ to the equation

[TABLE]

with $a_{15}=0=a_{16}$.333This assumption will be useful in Corollary 3.4. For larger values of $d$, note that there are many solutions. This is possible by Lagrange’s four-squares theorem. Let $V\subset H^{0}(A,\mathcal{O}_{A}(2L))^{+}$ be the space of even sections vanishing to order at least $2a_{i}$ at each point $p_{i}$, such that

[TABLE]

Let $\mathfrak{d}=\mathbb{P}_{\text{sub}}(V)\subseteq\absolutevalue{2L}^{+}$ be the corresponding linear system of divisors, whose dimension is $N\coloneqq\dim\mathfrak{d}\geq 3$. Write

[TABLE]

for a general divisor $D\in\mathfrak{d}$, so that $d_{i}\geq 2a_{i}$.

Remark 2.1**.**

From [7, Section 4.8], it follows that sections of $V$ are pulled back from the singular Kummer surface $A/\iota$, so any divisor $D\in\mathfrak{d}$ is symmetric, i.e. $\iota(D)=D$.

Let $\varphi:A\dashrightarrow\mathbb{P}^{N}$ be the rational map given by the linear system $\mathfrak{d}$ above, and write $S\coloneqq\overline{\imaginary(\varphi)}$ for the image of $\varphi$. Regardless of whether or not $\mathfrak{d}$ has a fixed component, we find that:

Proposition 2.2**.**

$S\subset\mathbb{P}^{N}$* is an irreducible and nondegenerate surface.*

Proof.

Suppose for the sake of contradiction that $\overline{\imaginary(\varphi)}$ is a nondegenerate curve $C$. Then $\deg C\geq 3$ since $N\geq 3$, and a hyperplane section of $C\subset\mathbb{P}^{N}$ pulls back to a divisor with at least three irreducible components. This contradicts the fact that any divisor $D(\sim_{lin}2L)\in\mathfrak{d}$ has at most two irreducible components since $\operatorname{NS}(A)=\mathbb{Z}[L]$. So the image of $\varphi$ is a surface. ∎

Lemma 2.3**.**

Let $\varphi:X\dashrightarrow\mathbb{P}^{n}$ be a rational map from a surface $X$ to a projective space of dimension $n\geq 2$, and suppose that its image $S\coloneqq\overline{\imaginary(\varphi)}\subset\mathbb{P}^{n}$ has dimension 2. Let $\mathfrak{d}$ be the linear system corresponding to $\varphi$ (assuming $\mathfrak{d}$ has no base components). Then for any $D\in\mathfrak{d}$,

[TABLE]

Proof.

The indeterminacy locus of $\varphi$ is a finite set. ∎

Degree bounds

We now study the numerical properties of the linear series $\mathfrak{d}$ constructed above. Keeping the notation as in $\S$2:

Lemma 3.1**.**

If $\mathfrak{d}$ has no fixed components, then

[TABLE]

Proof.

By applying Proposition 2.2 and blowing-up $A$ along the collection of two-torsion points $Z$ to resolve some of the base points of $\mathfrak{d}$, we arrive at the diagram

{\widehat{A}}$${\operatorname{Bl}_{Z}A}$${A}$${S}$${\mathbb{P}^{N}}$${\coloneqq}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$$\scriptstyle{\varphi}$${\subset}

The linear system corresponding to $\psi$ has no fixed components, so its divisors are of the form

[TABLE]

where $\widehat{D}$ denotes the strict transform of $D$. By Lemma 2.3 applied to $\psi$,

[TABLE]

The main work is to treat the case when $\mathfrak{d}$ has a fixed divisor $F\not=0$. In this situation, we may write:

[TABLE]

where $F$ and $M$ are the fixed and movable components of $\mathfrak{d}$, respectively. By definition, $\dim\mathfrak{d}=\dim\mathfrak{b}$. Note that $D\sim_{lin}2L$ implies $F,M\sim_{alg}L$ for all $M\in\mathfrak{b}$. Choose a general divisor $M\in\mathfrak{b}$ and write

[TABLE]

so that $d_{i}=m_{i}+f_{i}\geq 2a_{i}$ for all $i$. We claim that $F$ must be symmetric as a divisor. If not, then

[TABLE]

This implies that $M=\iota(F)$ and $F=\iota(M)$ for all $M\in\mathfrak{b}$, which would mean that $M$ must also be fixed, leading to a contradiction. Hence, $F$ must be symmetric, and likewise for all $M\in\mathfrak{b}$.

We first need an intermediate estimate:

Proposition 3.2**.**

Assume $\mathfrak{d}$ has a fixed component $F\not=0$. Keeping the notation as above,

[TABLE]

Proof.

The idea here is to use the Kummer construction to push our fixed curve $F$ onto a K3 surface and apply Riemann-Roch. This is analagous to a proof of Bauer’s in [6, Theorem 6.1]. Consider the smooth Kummer K3 surface $K$ associated to $A$:

{E}$${\widehat{A}}$${\widehat{A}/\{1,\sigma\}}$${K}$${Z}$${A}$${\subset}$$\scriptstyle{\pi}$$\scriptstyle{\gamma}$${\eqqcolon}$${\subset}

where $\pi$ is the blow-up of $A$ along the collection of two-torsion points $Z$. Since the points in $Z$ are $\iota$-invariant, $\iota$ lifts to an involution $\sigma$ on $\widehat{A}$ and the quotient $K$ is a smooth K3 surface. Let $E_{i}$ denote the exceptional curve over $p_{i}\in Z$, so that $E=\sum_{i=1}^{16}E_{i}$ is the exceptional divisor of $\pi$. Since $F$ is symmetric, its strict transform

[TABLE]

descends to an irreducible curve $\bar{F}\subset K$. We claim that

[TABLE]

In fact, if the linear system $\absolutevalue{\mathcal{O}_{K}(\bar{F})}$ were to contain a pencil, then this would give us a pencil of symmetric curves in $\absolutevalue{\mathcal{O}_{A}(F)}$ with the same multiplicities at the two-torsion points, which contradicts $F$ being a fixed component of $\mathfrak{d}$.

From the exact sequence $0\rightarrow\mathcal{O}_{K}(-\bar{F})\rightarrow\mathcal{O}_{K}\rightarrow\mathcal{O}_{\bar{F}}\rightarrow 0$, it follows that $H^{i}(K,\mathcal{O}_{K}(\bar{F}))=0$ for $i>0$, so by Riemann-Roch

[TABLE]

and therefore $(\bar{F})^{2}=-2$. On the other hand, the equality

[TABLE]

combined with $\sum_{i=1}^{16}f_{i}m_{i}\leq\sum_{i=1}^{16}(\frac{d_{i}}{2})^{2}$ yields

[TABLE]

After rearranging the terms, we find that

[TABLE]

for a general divisor $D=F+M\in\mathfrak{d}$, which is the desired inequality. ∎

As an immediate consequence:

Theorem 3.3**.**

Assume $\mathfrak{d}$ has a fixed component $F\not=0$, and let $\mathfrak{b}=\mathfrak{d}-F$ be the linear system defining $\varphi:A\dashrightarrow S\subseteq\mathbb{P}^{N}$. Then

[TABLE]

Proof.

As we saw in the proof of Lemma 3.1,

[TABLE]

Corollary 3.4**.**

There exists a 4-to-1 rational map $\varphi:A\dashrightarrow\mathbb{P}^{2}$.

Proof.

Recall that we chose the $a_{i}$ so that $a_{15}=0=a_{16}$. From Remark 2.1, it follows that $\varphi:A\dashrightarrow S\subset\mathbb{P}^{N}$ factors through the quotient $A\rightarrow A/\iota$, so $\deg\varphi$ must be even. The surface $S$ is nondegenerate, so $\deg S\geq 2$. By Lemma 3.5 below, it is impossible for $S$ to be rational together with $\deg\varphi=2$, so $\{\deg\varphi=2,\deg S=2,3\}$ is ruled out by the classification of quadric and cubic surfaces (using the fact that $\rho(A)=1$).

Together with the upper bound $\deg\varphi\cdot\deg S\leq 8$ given by Lemma 3.1 and Theorem 3.3, there are two possibilities:

[TABLE]

Either of these imply equality throughout (1) or (2), so that there is a morphism $\operatorname{Bl}_{Z}A\rightarrow S$ which fits into the diagram:

{E_{i}}$${\operatorname{Bl}_{Z}A}$${K}$${G_{i}}$${A}$${S}$${\mathbb{P}^{3}}$${\subset}$$\scriptstyle{\pi}$$\scriptstyle{\gamma}$$\scriptstyle{\alpha}$${\supset}$$\scriptstyle{\varphi}$${\subset}

where $K$ is the smooth Kummer K3 surface, $\gamma$ is a branched cover of degree 2, and $G_{i}\coloneqq\gamma(E_{i})$.

In the first case where $\deg\varphi=2$ and $\deg S=4$, from (1) and (2) it follows that $d_{15}=0=d_{16}$ or $m_{15}=0=m_{16}$. This implies that the curves $G_{15},G_{16}$ are contracted and their images $q_{15},q_{16}$ under $\alpha$ are double points on $S$ since $\alpha$ is a birational morphism. Projection from a general $(N-3)$-plane containing one but not both of the $q_{i}$ defines a rational map $A\dashrightarrow\mathbb{P}^{2}$ of degree 2 (if $q_{15}$ is a cone point of $S$, pick a general plane passing through $q_{16}$, and vice versa). In the second case where $\deg\varphi=4$ and $\deg S=2$, note that $S$ is rational. ∎

This immediately leads to Theorem 1.1. It is natural to ask what $\operatorname{irr}(A_{d})$ is equal to for a very general polarized abelian surface. At least one can see geometrically:

Lemma 3.5**.**

There are no rational dominant maps $A\dashrightarrow\mathbb{P}^{2}$ of degree 2.

Proof.

Suppose there exists such a map. We have the following diagram

{A^{[2]}}$${A}$${A}$${\mathbb{P}^{2}}$${K^{[2]}(A)}$${\Sigma^{-1}(0)}$$\scriptstyle{\Sigma}$$\scriptstyle{f}$$\scriptstyle{h}$$\scriptstyle{g}$${\eqqcolon}

where $g$ is the pullback map on 0-cycles and $A^{[2]}$ is the Hilbert scheme of 2 points on $A$. Since the rational map $\Sigma\circ g$ can be extended to a morphism, it must be constant. So $\overline{\imaginary(g)}$ is contained in a fiber $\Sigma^{-1}(0)$, which is a smooth Kummer K3 surface $K^{[2]}(A)$. Since $g$ is injective, it descends to an injective (and hence birational) map $h:\mathbb{P}^{2}\dashrightarrow K^{[2]}(A)$, yielding a contradiction. ∎