Surfaces with Kodaira dimension zero

• Some numerical invariants
• Proof strategy for the Enriques surface

Recall how we did the classification of surfaces. We started with $X/\bar{k}$ a smooth projective surface, and then we found a minimal model $$ X \to \dotsb \to X^\mathrm{min} $$ in the sense that it has no $(-1)$-curves, and moreover it is unique if $K(X) \ge 0$. (If not, we knew that $X$ is ruled.)

Theorem 1 (Castelnuovo). The surface $X$ is rational if and only if $g = \dim \operatorname{Pic}_ X^0 = 0$ and $p_2(X) = \dim H^0(X, \omega_X^{\otimes 2}) = 0$.

Now $K_{X^\mathrm{min}}$ is nef if and only if $X$ is not ruled, and in this case it is semi-ample in the sense that $\lvert n K_{X^\mathrm{min}} \rvert$ has no base-points when $n \gg 0$. For $\kappa(X) = 2$ this is of general type, and in the other case $\kappa(X) = 1$ there is an elliptic fibration $X^\mathrm{min} \to X^\mathrm{can}$. We will focus on the case when $\kappa(X) = 0$ where $\omega_X$ is torsion.

Some numerical invariants

One obvious numerical invariant is $$ h^{p,q} = \dim H^q(X, \Omega_X^p). $$ But note that in positive characteristic the Hodge symmetry $h^{p,q} = h^{q,p}$ fails. We will see examples of $h^{0,1} \neq h^{1,0}$. But at least by Serre duality we have $$ h^{0,2} = h^{2,0}. $$ Another thing to be careful about is that we might have $\operatorname{Pic}_ X^0$ that is singular. So we have $$ q = \dim \operatorname{Pic}_ X^0 \le h^{0,1} $$ but this is not an equality in general. But the obstructions live in $H^2(X, \mathscr{O})$, so $p_g = h^{0,2} = h^{2,0} = 0$ then this is an equality.

The next thing we have are Chern classes $c_i = c_i(T_X) \in \operatorname{CH}^i(X)$. If we fix a nice Weil cohomology theory, we can check that $$ c_2 = \sum_{i=0}^4 (-1)^i \dim_K H^i(X, K). $$ Another thing that is a bit harder to check is Noether’s formula $$ 12 \chi(\mathscr{O}_ X) = c_1^2 + c_2 $$ which is something that follows from Grothendieck–Riemann–Roch if you’d like. It also turns out that $$ b_1 = \dim_K H^1(X, K) = 2q. $$ This is because basically $H^1(X, \mathbb{Z}/\ell^n \mathbb{Z})$ is like $(\mathbb{Z}/\ell^n \mathbb{Z})^{b_1}$ and this is like the $\ell^n$-torsion of $\operatorname{Pic}_ X^0$.

We then have $$ 12 (1 - h^{0,1} + p_g) = c_1^2 + c_2, \quad c_2 = 1 - 2q + b_2 - 2q + 1 = 2 (1-2q) + b_2. $$ If $X$ is minimal and $\kappa(X) = 0$, then $n K_X = 0$ and so $c_1^2 = 0$. Moreover, $p_g = \dim H^0(X, \omega_X)$ is either $0, 1$ because otherwise we will have more and more sections. Then we have $$ 10 + 12 p_g = 8 h^{0,1} + 2 (2 h^{0,1} - b_1) + b_2 $$ where this quantity $$ \Delta = 2h^{0,1} - b_1 = 2 (h^{0,1} - q) $$ is at least zero. We also have $$ b_2 = 10 + 12 p_g - 8 h^{0,1} - 2 \Delta \ge 0. $$ At this point, there are only finitely many possibilities.

Proposition 2. If $X$ is minimal with $n K_X = 0$ then $c_2 \neq 0$ then $\pi_1^\mathrm{et}(X)$ is finite.

If $\pi \colon X^\prime \to X$ is a finite étale cover, then $X^\prime$ is also minimal with torsion canonical bundle, so it appears on the list. On the other hand, $c_2(X^\prime) = (\deg \pi) c_2(X)$.

It also follows that K3 surfaces are simply connected and the Enriques surfaces can only have $\pi_1 = \mathbb{Z}/2\mathbb{Z}$ covered by a K3 surface.

Corollary 3. A DNE surface does not exist.

This is because it will have finite $\pi_1$, but it has nontrivial $\operatorname{Pic}_ X^0$ and hence it has to have lots of finite étale covers. For example, we can take a torsion line bundle $\mathscr{L}$ and take the relative spectrum of $\mathscr{O} \oplus \dotsb \oplus \mathscr{L}^{n-1}$.

Remark 4. Fontaine told us that if either $h_{0,1}$ or $h^{2,0} = p_g$ is nonzero on the generic fiber, then this cannot exist over $\mathbb{Z}$. In the K3 and abelian surface cases, $p_g = 1$ and so this is ruled out. In the hyperelliptic case, $h^{0,1} \ge 1$ and this is also ruled out. So the only remaining case is the Enriques surface case.

Proof strategy for the Enriques surface

Theorem 5. There is no Enriques surface over $\mathbb{Z}$.

Definition 6. A smooth projective surface $X/k$ is Enriques if $K_X$ is numerically trivial and $b_2 = 10$. A family of Enriques surface is a smooth projective $f \colon \mathfrak{X} \to S$ with fibers being Enriques surfaces. (We can even allow $\mathfrak{X}$ an algebraic space and $f$ proper.)

In this case, it will turn out that the numerical curve class in $X_{\bar{k}}$ will be isomorphic to $\mathbb{Z}^{10}$ with some Galois action, and moreover if $\operatorname{char} k \neq 2$ then we have $$ \operatorname{Pic}_ {X_{\bar{k}}}(\bar{k}) = \mathbb{Z}^{10} \oplus (\mathbb{Z}/2\mathbb{Z}), $$ with the torsion generated by $\omega_X$.

The idea is as follows. Consider the geometric fiber $X = \mathfrak{X}_ {\bar{\mathbb{Q}}}$. Then $$ \operatorname{Pic}_ X = (\mathbb{Z}/2\mathbb{Z}) \oplus \mathbb{Z}^{10} $$ has a trivial Galois action, because $\mathbb{Z}$ doesn’t have finite étale covers. Moreover, we will have that $$ \operatorname{Pic}_ {\mathfrak{X}/\mathbb{Z}} = (\mathbb{Z}/2) \oplus \mathbb{Z}^{10} $$ as a constant group scheme over $\mathbb{Z}$.

Next, we will have from the fixed point formula that $$ \lvert \mathfrak{X}(\mathbb{F}_ p) \rvert = 1 + 10p + p^2, $$ and in particular, this gives $\lvert \mathfrak{X}(\mathbb{F}_ 2) \rvert = 25$. On the other hand, there will exists a map $$ X_2 \to \mathbb{P}_ {\mathbb{F}_ 2}^1 $$ defined over $\mathbb{F}_ 2$, and then these 25 points should arrange themselves over $\mathbb{P}^1(\mathbb{F}_ 2) = \lbrace 0, 1, \infty \rbrace$. It also turns out that the only reducible fibers can occur over these three points, because the Galois action on the numerical classes is trivial. We are left with a fibration with at most three reducible fibers with 25 points in these fibers. This leaves us with only the “exceptional” case, and this will not lift to $\mathbb{Z}/4\mathbb{Z}$.