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

\[ \rho(X) = 16 + \rho(A).\]


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 \(A\):

  1. 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).
  2. 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 (5if 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.

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