A compact complex surface without divisors

Why I decided to write this post
Complex tori and abelian surfaces
An explicit example
Why this example is generic
References

1 Why I decided to write this post

This post (as most others in this blog will be) is an example that I have to recall quite often, but can never remember the details. On one hand, it is a very simple example, on the other hand, it shows a huge difference between the algebraic and complex geometry. For instance, it shows that some naïve analogies between algebraic and holomorphic “toric” geometries can not literally be true. In particular, it shows how the relationship between Cartier divisors and line bundles breaks down in the non-algebraic case.

This example is nice in several ways: it provides a simple example of a non-algebraic Kähler complex manifold that does not contain any compact complex submanifolds of positive dimension.

2 Complex tori and abelian surfaces

Definition 2.1. A complex torus of dimension two is a quotient

X = ℂ^{2} ∕ Λ,

where $Λ \subset ℂ^{2}$ is a lattice of real rank four. An abelian surface is a two-dimensional complex torus which is projective. A divisor on a smooth complex surface is a finite integral linear combination of irreducible codimension-one analytic subsets, hence of compact analytic curves.

Proposition 2.2. Every abelian surface has a nonzero effective divisor.

Proof

Proof. Embed the surface in projective space and intersect it with a hyperplane which does not contain it. The resulting hyperplane section is a nonempty effective divisor. Equivalently, an ample line bundle has a very ample positive power, and a hyperplane section of that power supplies the divisor. □

Remark 2.3. If a complex torus is algebraic then it is projective, and hence an abelian surface. Therefore a complex torus with no nonzero divisors is not algebraic.

Thus, our example must be a nonprojective complex torus. It is Shafarevich’s Example 8.4 [Sha13, Chapter 8, §1.4, Example 8.4].

3 An explicit example

It is not that hard to write down a nonprojective complex torus (via the Kodaira embedding theorem we know that the class of the Kähler form has to be non-integral), but it is a bit tricky to make sure that it has no divisors. The following example is due to Shafarevich (as far as I know).

Fix a transcendental real number $t$ , for example $t = π$ , and put

v_{1} = (\binom{1}{0}), v_{2} = (\binom{0}{1}), v_{3} = (\binom{i}{t^{4}}), v_{4} = (\binom{t^{2}}{𝑖𝑡}) .

The four vectors are linearly independent over $ℝ$ : in period-matrix form they are the columns of

Π = (I_{2} Z), Z = (\begin{matrix} i & t^{2} \\ t^{4} & 𝑖𝑡 \end{matrix}), \det (Im Z) = t \neq 0 .

Consequently $Λ_{t} = \sum_{j = 1}^{4} ℤ v_{j}$ is a lattice and $X_{t} = ℂ^{2} ∕ Λ_{t}$ is a compact complex surface.

Proposition 3.1. The surface $X_{t}$ contains no compact analytic curve. In particular, $Div (X_{t}) = 0$ .

Proof

Proof. Let $S_{𝑗𝑘}$ be the image in $X_{t}$ of the oriented parallelogram spanned by $v_{j}, v_{k}$ . The six classes $[S_{𝑗𝑘}]$ , for $j < k$ , form a basis of

H_{2} (X_{t}, ℤ) ≃ \overset{2}{\land} Λ_{t} .

Suppose that $C \subset X_{t}$ is an irreducible compact curve and let $ν : \tilde{C} \to C$ be its normalization. The flat Kähler form

ω = \frac{i}{2} (d z_{1} \land d {\bar{z}}_{1} + d z_{2} \land d {\bar{z}}_{2})

descends to $X_{t}$ , and $\int_{\tilde{C}} ν^{*} ω > 0$ . Hence $[C] \neq 0$ , so

[C] = \sum_{j < k} a_{𝑗𝑘} [S_{𝑗𝑘}], a_{𝑗𝑘} \in ℤ,

with not all $a_{𝑗𝑘}$ zero.

Now consider the holomorphic two-form $σ = d z_{1} \land d z_{2}$ . Its pullback to the curve $\tilde{C}$ vanishes, while

\int_{S_{𝑗𝑘}} σ = \det (v_{j}, v_{k}) .

The six determinants are

1, t^{4}, 𝑖𝑡, - i, - t^{2}, - t - t^{6} .

It follows that

0 = \int_{C} σ = a_{12} + a_{13} t^{4} + i a_{14} t - i a_{23} - a_{24} t^{2} - a_{34} (t + t^{6}) . (1)

The imaginary part of (1) gives $a_{14} t - a_{23} = 0$ , hence $a_{14} = a_{23} = 0$ . Its real part is a polynomial relation over $ℤ$ for the transcendental number $t$ , so all the remaining coefficients vanish as well. This contradicts $[C] \neq 0$ . Therefore there is no curve, and on a surface there can consequently be no prime divisor and no nonzero divisor. □

Corollary 3.2. Every meromorphic function on $X_{t}$ is constant.

Proof

Proof. The divisor of a meromorphic function is zero by the proposition above. On a compact complex manifold a meromorphic function is determined up to a nonzero scalar by its divisor, so it is constant. See again Shafarevich’s [Sha13, §2.2, p. 170] book. □

4 Why this example is generic

There is a useful slightly more modern translation of the same calculation. For a compact Kähler manifold $X$ , its Néron–Severi group is

NS (X) = H^{1, 1} (X) \cap H^{2} (X, ℤ) = c_{1} (Pic (X)) .

If $D$ is a nonzero effective divisor on a Kähler surface, then

\int_{X} c_{1} (𝒪_{X} (D)) \land ω = \int_{D} ω > 0 . (2)

Thus $NS (X) = 0$ implies that $X$ has no divisors. In the proof above, the absence of an integral relation among the six periods of $σ$ says exactly that $NS (X_{t}) = 0$ .

Proposition 4.1. A very general¹ two-dimensional complex torus has Néron–Severi group zero, and hence has no nonzero divisors.

Proof

Proof. Let $Π = (v_{1}, \dots, v_{4})$ vary through the open set of $2 \times 4$ complex matrices whose columns are independent over $ℝ$ . For each fixed nonzero tuple $(a_{𝑗𝑘}) \in ℤ^{6}$ , the equation

\sum_{j < k} a_{𝑗𝑘} \det (v_{j}, v_{k}) = 0

cuts out a proper analytic locus in period space. There are only countably many such tuples. Outside their union, no nonzero integral two-cycle annihilates the holomorphic two-form. By Poincaré duality, no nonzero integral cohomology class is then of type $(1, 1)$ . Therefore $NS (X) = 0$ . This is the usual meaning of “very general” here; compare [Dem07, §2.4]. □

Remark 4.2. Merely being nonprojective is not enough. Shafarevich’s also gives an example ([Sha13, Example 8.3]) of a nonprojective two-torus mapping onto an elliptic curve; its fibres are curves, and hence divisors. Simplicity is also not enough: a simple abelian surface still has hyperplane sections. What implies absence of divisors is the stronger condition $NS (X) = 0$ .

Remark 4.3 (Divisors are not the same as line bundles). The conclusion is $Div (X_{t}) = 0$ , not $Pic (X_{t}) = 0$ . The exponential sequence gives an exact segment

H^{1} (X_{t}, ℤ) \to H^{1} (X_{t}, 𝒪_{X_{t}}) \to Pic (X_{t}) \to_{}^{c_{1}} H^{2} (X_{t}, ℤ) .

Since $NS (X_{t}) = 0$ , every holomorphic line bundle is topologically trivial:

Pic (X_{t}) = {Pic}^{0} (X_{t}) ≃ H^{1} (X_{t}, 𝒪_{X_{t}}) ∕ H^{1} (X_{t}, ℤ) .

This is the dual two-dimensional complex torus, so it is far from trivial; see [BL99, Chapter 1, §4, Proposition 4.2]. If a nontrivial $L \in {Pic}^{0} (X_{t})$ had a nonzero holomorphic section, its zero locus would be a divisor. Since there are none, the section would be nowhere vanishing and would trivialize $L$ , a contradiction. Thus $H^{0} (X_{t}, L) = 0$ for every nontrivial $L \in {Pic}^{0} (X_{t})$ .

Corollary 4.4. The algebraic dimension² of $X_{t}$ is zero.

Proof

Proof. By the first corollary, its field of meromorphic functions is $ℂ$ , whose transcendence degree is zero. □

References

[BL99]: Christina Birkenhake and Herbert Lange. Complex Tori. Vol. 177. Progress in Mathematics. Boston: Birkhäuser Boston, 1999. doi: 10.1007/978-1-4612-1566-0.
[Dem07]: Jean-Pierre Demailly. “Kähler manifolds and transcendental techniques in algebraic geometry”. In: International Congress of Mathematicians, Vol. I. Zürich: European Mathematical Society, 2007, pp. 153–186. doi: 10.4171/022-1/8.
[Sha13]: Igor R. Shafarevich. Basic Algebraic Geometry 2. Schemes and Complex Manifolds. Trans. by Miles Reid. 3rd ed. Heidelberg: Springer, 2013. doi: 10.1007/978-3-642-38010-5.

A compact complex surface without divisors

Contents

1 Why I decided to write this post

2 Complex tori and abelian surfaces

3 An explicit example

4 Why this example is generic

References

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