The Picard number of a Kummer K3 surface

Let $k$ be a separably closed field of characteristic not $2$ , and $A/k$ an abelian surface. Then it is a basic fact (e.g. see Example 1.3 (iii) of Huybrechts’ “K3 Surfaces”) that one can make a K3 surface out of $A$ . The construction is as follows. Consider the involution $\iota : A \to A$ given by $x \mapsto -x.$ The fixed locus of this involution is exactly $A[2]$ , a finite constant closed $k$ -subgroup scheme of $A$ with $16$ $k$ -points. Let $Z := A[2]$ , and consider the blow-up $p : \operatorname{Bl}_Z A \to A$ . By the universal property of the blow-up, the involution $\iota$ lifts to a map $\tilde{\iota} : \operatorname{Bl}_Z A \to \operatorname{Bl}_Z A$ such that the diagram

$\begin{tikzcd} \text{Bl}_Z A \ar{r}{\tilde{\iota}}\ar{d}{p} & \text{Bl}_Z A\ar{d}{p} \\ A \ar{r}{\iota} & A \end{tikzcd}$

commutes. Furthermore, by the uniqueness part of the statement of the universal property of $\operatorname{Bl}_Z A$ , we deduce easily that $\tilde{\iota}$ is also an involution.

Now the blow-up $\operatorname{Bl}_Z A$ is a projective variety over $k$ with an action of $\mathbf{Z}/2\mathbf{Z}$ via the involution $\tilde{\iota}$ . Therefore, the categorical quotient $X := \operatorname{Bl}_Z A/(\mathbf{Z}/2\mathbf{Z})$ exists in the category of schemes. The scheme $X$ constructed in this way is called the Kummer surface associated to $A$ , and turns out to be a K3 surface. In particular, $H^1(X, \mathcal{O}_X) = 0$ , so the Picard scheme ${\rm{Pic}}_{X/k}$ of $X$ is étale, and ${\rm{Pic}}(X)$ is therefore a finitely-generated abelian group.

For an arbitrary proper scheme $Y/k$ , recall that the Néron–Severi group of $Y$ is a finitely generated abelian group (SGA 6, Exposé XIII, Théorème 5.1). Therefore, we may consider the Picard number $\rho(Y)$ , defined as the rank of the Néron–Severi group of $Y$ (which for $X$ is just the rank of the Picard group). It is proven in Shioda’s paper “Supersingular K3 surfaces” that the Picard number of $X$ is given by the formula $\rho(X) = 16 + \rho(A)$ . However, we find his calculation unclear and difficult to follow. In this note, we give an explicit proof of this fact that is entirely self-contained. We do not use any Hodge theory, e.g. we do not study the complement of $\operatorname{Pic}(X)$ in $H^2(X, \mathbf{Z})$ , i.e. the transcendental lattice $T(X)$ .

Main Theorem: Let $k$ be a separably closed field of characteristic not $2$ , and $A/k$ an abelian surface. Let $X$ denote the Kummer surface associated to $A$ . Then the Picard number of $X$ is given by

$\rho(X) = 16 + \rho(A).$

Preliminaries

In this section, we record a crucial result about abelian varieties that we will need. Let $k$ be a separably closed field of characteristic not $2$ , and let $A/k$ be an abelian variety (of arbitrary dimension). The group $\mathbf{Z}/2\mathbf{Z}$ acts on $\operatorname{Pic}(A)$ by $\mathcal{L} \mapsto [-1]^\ast \mathcal{L}$ , and this action descends to one on the subgroup of numerically trivial line bundles $\operatorname{Pic}^0(A)$ . In particular, the sequence

(4) $\begin{equation*} 0 \to \operatorname{Pic}^0(A) \to \operatorname{Pic}(A) \to \operatorname{NS}(A) \to 0 \end{equation*}$

is an exact sequence of $\mathbf{Z}/2\mathbf{Z}$ -modules.

Proposition 1: Taking $\mathbf{Z}/2\mathbf{Z}$ -invariants in (4), we obtain an exact sequence

$0 \to \widehat{A}[2](k) \to \operatorname{Pic}(A)^{\mathbf{Z}/2\mathbf{Z}} \to \operatorname{NS}(A) \to 0,$

where $\widehat{A}$ is the dual abelian variety of $A$ . In particular, $\operatorname{Pic}(A)^{\mathbf{Z}/2\mathbf{Z}}$ is a finitely generated abelian group of rank equal to the Picard number $\rho(A)$ of $A$ .

Let us first recall several facts about multiplication by $n$ on $A$ :

For $\mathcal{L} \in \operatorname{Pic}^0(A)$ , $[n]^\ast \mathcal{L} \simeq \mathcal{L}^{\otimes n}$ (See the proof of Lemma 5.2.5 in Brian Conrad’s abelian varieties notes).
For $\mathcal{L} \in \operatorname{Pic}(A)$ , we have
$[n]^\ast \mathcal{L} \simeq \mathcal{L}^{\otimes n^2} \mod{\operatorname{Pic}^0(A)}.$
In other words, $[n]^\ast$ has the effect of multiplication by $n^2$ on $\operatorname{NS}(A)$ (Lemma 7.5.2 of loc. cit.).

Granting these facts, let us first compute $\operatorname{Pic}^0(A)^{\mathbf{Z}/2\mathbf{Z}}$ . By (I), a line bundle $\mathcal{L}\in \operatorname{Pic}^0(A)$ satisfies $[-1]^\ast \mathcal{L} = \mathcal{L}$ precisely when $\mathcal{L}^{\otimes 2} \simeq 0$ . In other words,

$\operatorname{Pic}^0(A)^{\mathbf{Z}/2\mathbf{Z}} = \widehat{A}[2](k).$

Next, by (II) above, the $\mathbf{Z}/2\mathbf{Z}$ -action on $\operatorname{NS}(A)$ is trivial, so $\operatorname{NS}(A)^{\mathbf{Z}/2\mathbf{Z}} = \operatorname{NS}(A)$ . Finally, we show that $H^1( \mathbf{Z}/2\mathbf{Z}, \operatorname{Pic}^0(A)) = 0$ , which will complete the proof of the proposition. Write $\sigma$ for the generator of $\mathbf{Z}/2\mathbf{Z}$ , and let $N$ denote the “norm” map

$\begin{eqnarray*} N : &\operatorname{Pic}^0(A)& \to \operatorname{Pic}^0(A) \\ &\mathcal{L} & \mapsto \mathcal{L} \otimes [-1]^\ast\mathcal{L}. \end{eqnarray*}$

By the calculation of the cohomology of finite cyclic groups,

$H^1( \mathbf{Z}/2\mathbf{Z}, \operatorname{Pic}^0(A)) \simeq \ker N/(\sigma - 1)\operatorname{Pic}^0(A).$

By (I) above, we have $\ker N = \operatorname{Pic}^0(A)$ . On the other hand, for $\mathcal{L} \in \operatorname{Pic}^0(A)$ ,

$(\sigma - 1)(\mathcal{L}) = [-1]^\ast \mathcal{L} \otimes \mathcal{L}^{\vee} \simeq (\mathcal{L}^{\vee})^{\otimes 2}.$

In other words, $(\sigma -1)$ has the effect of multiplication by $-2$ on $\operatorname{Pic}^0(A)$ . But now recall that $\operatorname{Pic}^0(A) = \widehat{A}(k)$ , and $[-2] : \widehat{A} \to \widehat{A}$ is surjective étale. Since $k$ is separably closed, the map on $k$ -points is surjective, and therefore $(\sigma - 1)\operatorname{Pic}^0(A) = \operatorname{Pic}^0(A)$ , from which the vanishing of the cohomology group in question follows.

Proof of Main Theorem

Let $p_1, \dots, p_{16}$ be the points in $A[2](k)$ , let $E_i$ be the preimage of $p_i$ in $\operatorname{Bl}_Z A$ (the exceptional divisors), and let $\widetilde{E}_i$ denote the image of $E_i$ in the quotient $X$ . Define

$\begin{eqnarray*} \widetilde{E} &:=& \bigcup_{i=1}^{16} \widetilde{E}_i.\\ E &:=& \bigcup_{i=1}^{16} E_i.\end{eqnarray*}$

It is proven in Theorem 10.6 of Badescu’s “Algebraic Surfaces” that the $\widetilde{E}_i$ ’s are irreducible divisors in $X$ , and are furthermore $\mathbf{Z}$ -linearly independent in $\operatorname{Pic}(X)$ . Therefore, identifying the Weil class group of $X$ with its Picard group (by smoothness), we obtain the exact sequence

(5) $\begin{equation*} 0 \to \bigoplus_{i=1}^{16} \mathbf{Z}[\widetilde{E}_i] \to \operatorname{Pic}(X) \to \operatorname{Pic}(X - \widetilde{E}) \to 0\end{equation*}$

with the group on the left isomorphic in the obvious way to $\mathbf{Z}^{16}$ .

Now the formation of the quotient $\pi : \operatorname{Bl}_Z A \to X$ commutes with open immersions. More precisely, for any open subscheme $\widetilde{V} \subset X$ , if we let $V := \pi^{-1}(\widetilde{V})$ , then the map $\pi\colon V \rightarrow \widetilde{V}$ is the categorical quotient of $V$ by $\mathbf{Z}/2\mathbf{Z}$ . Therefore,

$X - \widetilde{E} \simeq \pi^{-1}(X-\widetilde{E})/(\mathbf{Z}/2\mathbf{Z}).$

But now observe that

$\pi^{-1}(X- \widetilde{E}) \simeq \operatorname{Bl}_Z A -E \simeq A - A[2],$

$(A - A[2])/(\mathbf{Z}/2\mathbf{Z}) \simeq X - \widetilde{E}.$

Therefore, the result $\rho(X) = 16 + \rho(A)$ will follow from (5) if we can show that the abelian group $\operatorname{Pic}((A - A[2])/(\mathbf{Z}/2\mathbf{Z}) )$ is finitely generated of rank equal to $\rho(A)$ . To this end, define $U := A - A[2]$ and $G := \mathbf{Z}/2\mathbf{Z}$ . Since the action of $G$ on $U$ is free, the quotient map $U \to U/G$ is a Galois cover (SGA3, Exposé V, Théorème 4.1(iii) and (iv)), and we have an associated Hochschild-Serre spectral sequence

$H^i(G, H^j(U,\mathbf{G}_m)) \implies H^{i+j}(U/G, \mathbf{G}_m).$

The low-degree terms of this spectral sequence are

$0 \to H^1(G, H^0 (U, \mathbf{G}_m)) \to \operatorname{Pic}(U/G) \to \operatorname{Pic}(U)^G \to H^2(G, H^0 (U, \mathbf{G}_m)).$

Now before we calculate any cohomology, we make the observation that

$H^0(U, \mathbf{G}_m) = H^0(A,\mathbf{G}_m) = k^\times.$

Indeed, this is true by Hartogs’ Lemma since $A$ is smooth (a fortiori normal!) and $A \setminus U$ is codimension $2$ in $A$ . Also, observe that the Galois action of $G$ on $k^\times$ is trivial.

We now compute $H^1(G, H^0(U,\mathbf{G}_m))$ . By the discussion above, this is isomorphic to $\operatorname{Hom}(G, k^\times) = \mu_2(k).$ On the other hand, by the calculation of the cohomology of finite cyclic groups, $H^2(G, k^\times) \simeq k^\times/(k^{\times})^2$ . Since $k$ is separably closed, this is zero and so $\operatorname{Pic}(U/G)$ sits in an exact sequence

(6) $\begin{equation*} 0 \to \mu_2(k) \to \operatorname{Pic}(U/G) \to \operatorname{Pic}(U)^G \to 0.\end{equation*}$

By the equivalence of the Picard group with the Weil divisor class group for regular schemes, and because $A \backslash U$ has codimension $2$ in $A$ , $\operatorname{Pic}(U) = \operatorname{Pic}(A)$ . By Proposition 1, $\operatorname{rk} \operatorname{Pic}(A)^G = \rho(A).$ Combining this with (6) yields the equality

$\operatorname{rk} \operatorname{Pic}((A- A[2])/(\mathbf{Z}/2\mathbf{Z})) = \rho(A),$

as desired.