The Picard lattice

$X$ an algebraic K3 surface; its Picard lattice:

$\operatorname{Pic}(X) = \operatorname{NS}(X) \simeq \Z^{\rho}, \qquad \rho(X) := \operatorname{rk} \operatorname{Pic}(X)$

$\operatorname{NS}(\overline{X}) \simeq \operatorname{Pic}(\overline{X}) \simeq \Z\langle \text{algebraic curves in } X \rangle / \langle \text{linear equivalences} \rangle \subset H_2(X, \Z)$

Records the algebraic cycles on $X$ : curves modulo linear, algebraic or numerical equivalence.
For K3 surfaces the three agree; $\operatorname{Pic}^0 = 0$ , and $\operatorname{Pic} = \operatorname{NS}$ is finite free.
$\operatorname{Pic}(\mathbf{P}^2) = \Z$ , $\operatorname{Pic}(\mathbf{P}^1 \times \mathbf{P}^1) = \Z^2$ , $\operatorname{Pic}(\text{cubic surface}) = \Z^7$ .
$\rho$ , and more precisely the Picard lattice, is a coarse invariant.
K3 theorems are stated by lattice or rank, not equation; as for abelian varieties.

The lattice structure

$\operatorname{Pic}(\overline{X})$ with the intersection pairing $(D,D') \mapsto D \cdot D'$ .

Even symmetric bilinear form: $D \cdot D = 2p_a(D) - 2$ for every curve $D$ , by adjunction ( $K_X = 0$ ).
Signature $(1,\rho - 1)$ (Hodge index theorem): one positive direction, negative definite complement; hence non-degenerate.
$\operatorname{disc}\operatorname{Pic}(\overline{X}) := \det$ of the Gram matrix in any $\Z$ -basis; a non-zero integer. A basis change multiplies it by $\det(M)^2 = 1$ .
$\operatorname{Gal}(\overline{k}/k)$ permutes curves, respects linear equivalence and preserves the pairing: an orthogonal action.

Pic inside $H^2$

Over $\overline{\Bbb{Q}}$ , viewing $X$ also as a complex manifold,

$\begin{aligned} \operatorname{Pic}(X_{\overline{\Bbb{Q}}}) &\simeq H^{1,1}(X_{\Bbb{C}}) \cap H^2(X_{\Bbb{C}}, \Z) \\ &\subset H^2(X_{\Bbb{C}}, \Z) \simeq (-E_8)^2 \oplus U^3 \simeq \Z^{22} \end{aligned}$

Lefschetz $(1,1)$ : an integral class is algebraic exactly when it has type $(1,1)$ .
$H^2(X_{\Bbb{C}},\Z)$ is the unique even unimodular lattice of signature $(3,19)$ ; the embedding is primitive.
$\dim H^{1,1}(X) = 20$ , so $\rho(X_{\overline{\Bbb{Q}}}) \in \{1,2,\dots,20\}$ ; for a generic K3 surface $\rho(X_{\overline{\Bbb{Q}}}) = 1$ .
The degree of "difficulty" is negatively correlated with $\rho(X)$ .
$T(X) := \operatorname{Pic}(\overline{X})^{\perp} \subset H^2(X,\Z)$ , the transcendental lattice; equivalently the minimal sub-Hodge structure of $H^2(X,\Bbb{Q})$ whose complexification contains $H^{2,0}(X)$ .
$H^2(X_{\Bbb{C}},\Bbb{Q}) \simeq \operatorname{Pic}(X_{\overline{\Bbb{Q}}})_{\Bbb{Q}} \oplus T(X)_{\Bbb{Q}}$
The "new and interesting" Galois representations arise from $T(X)$ .
In characteristic $p$ the bound $20$ fails: $\rho(\overline{X}_p)$ can be as large as $22$ .

Geometric versus ground-field Picard group

$k$ a number field, $X$ a K3 surface over $k$ ;
$p$ a prime of $k$ where $X$ has good reduction $X_p$ ;
$\operatorname{NS}(\bullet) := \operatorname{Pic}(\bullet)/\operatorname{Pic}^0(\bullet)$ , the Néron-Severi group, a $\Z$ -lattice geometrically attached to $\bullet$ ;
$\rho(\bullet) := \operatorname{rk}\operatorname{NS}(\bullet)$ , the arithmetic or geometric Picard number of $\bullet$ .

$\operatorname{Pic}(X) = \operatorname{Pic}(\overline{X})^{\operatorname{Gal}(\overline{k}/k)}$ , so $\rho(X) \leq \rho(\overline{X})$ , usually strictly.
A class defined only over an extension is invisible over $k$ .
"The Picard number" in this lecture means the geometric one, $\rho(\overline{X})$ .

Goal

From the equations of $X$ , compute $\operatorname{Pic}(\overline{X}) \subset H_2(X,\Z)$ as a $\operatorname{Gal}(\overline{k}/k)$ -module.

"The evaluation of $\rho$ for a given surface presents in general grave difficulties." (Zariski)

How are the geometric Picard numbers $\rho(\overline{X})$ and $\rho(\overline{X}_p)$ related?
How does the geometric Picard number behave under reduction modulo $p$ ?

$\begin{aligned} H^1(\operatorname{Gal}(\overline{k}/k), \operatorname{Pic}\overline{X}) &\simeq \operatorname{Br}_1(X)/\operatorname{Br}_0(X) \\ X(k) &\subset X(\mathbf{A}_k)^{\operatorname{Br}} \subset X(\mathbf{A}_k) \end{aligned}$

Picard lattice, over finite fields

$P_2(T) = \det(T - \operatorname{Frob} \mid H^2)$ ; roots $\alpha_i$ , $|\alpha_i| = q$ .
$q^{-22}P_2(qT)$ monic; roots $\zeta_i := \alpha_i/q$ , $|\zeta_i| = 1$ .

Tate classes correspond to roots of unity (Tate, a theorem for K3 surfaces over finite fields).

$\begin{gathered} q^{-22}P_2(qT) = h(T)\prod_i\Phi_{k_i}(T)^{\gamma_i} \\ \Phi_k\text{ the }k\text{-th cyclotomic polynomial};\quad h\text{ has no cyclotomic factor} \\ \rho(X_{\Bbb{F}_{q^r}}) = \sum_{k_i\mid r}\gamma_i\deg\Phi_{k_i} \end{gathered}$

Example: $X := Z(y^4 - x^3z + yz^3 + zw^3 + w^4) \subset \mathbf{P}^3$ , $p = 89$ .

$p^{-22}P_2(pT) = (T-1)(T+1)(T-1)^4(T^4+1)h(T), \qquad \deg h = 12$

$(T-1) = \Phi_1$ , degree $1$ ;
$(T+1) = \Phi_2$ , degree $1$ ;
$(T-1)^4 = \Phi_1^4$ , degree $4$ ;
$(T^4+1) = \Phi_8$ , degree $4$ .

Over $\Bbb{F}_{89}$ : only $k=1$ divides $r=1$ , so $\rho(X_{\Bbb{F}_{89}}) = 1+4 = 5$ .
Over $\Bbb{F}_{89^r}$ : $\Phi_2$ joins when $2\mid r$ , $\Phi_8$ when $8\mid r$ ; $\rho(\overline{X}_{89}) = 1+1+4+4 = 10$ , reached at $r=8$ .
$\operatorname{Pic}(\overline{X}_{89})$ decomposes as $P_{\zeta_1}\oplus P_{\zeta_2}\oplus P_{\zeta_8}$ .

For $p > 7$ , naive point counting is impractical; crystalline methods [Abbott--Kedlaya--Roe, C, C--Harvey--Kedlaya, Tuitman--Pancratz].

What the characteristic polynomial gives you

Theorem (Tate) [Tate], [Charles], [Madapusi Pera], [Maulik], [Kim-Madapusi Pera]

Let $X/\mathbf{F}_q$ , where $q = p^n$ , be an abelian surface or a K3 surface. Then:

$\rho(X_p) = \operatorname{ord}_{T = 1/q} P_2(T)$
$\rho(\overline{X}_p) = \sum_{\zeta} \operatorname{ord}_{T = \zeta/q} P_2(T)$ , where $\zeta$ runs over all roots of unity.
(Artin-Tate, a theorem here) $P_2(T) \leadsto \operatorname{disc}(\operatorname{NS}(X_p)) \bmod \mathbf{Q}^{\times 2}$

Reduction to finite characteristic

Take $f \in \mathbf{Z}[x,y,z,w]$ and $X := Z(f) \subset \mathbf{P}^3_{\mathbf{Q}}$ .

We may consider the surface $X_p := Z(f \bmod{p}) \subset \mathbf{P}^3(\mathbf{F}_p)$ .

Theorem

If $X$ and $X_p$ are smooth then the specialization map is injective

$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p) \quad \text{and} \quad \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p).$

Goal

For a given $f$ and $p$ , improve the inequality $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p)$ .

Parity reasons might already force the inequality to not be sharp.

Endomorphisms of the transcendental lattice can complicate things even further.

Pic plays the role of $\operatorname{End}(A)$

$\operatorname{Pic}$ for a K3 surface plays a similar role as $\operatorname{End}(A)$ for an abelian variety $A$ .

$\operatorname{NS}(A)_{\Bbb{Q}} \simeq \{\phi \in \operatorname{End}(A)_{\Bbb{Q}} : \phi^{\dagger} = \phi\}, \qquad \dagger\text{ the Rosati involution}$

"Compute the Picard lattice of a K3 surface" is the same kind of question as "compute the endomorphism algebra of an abelian variety".
Slides 9 and 10 answer the second one, for elliptic curves, by reduction mod $p$ .
For Kummer surfaces: $\rho(\operatorname{Km}(A)) = \rho(A)+16$ , the identity used in section 1.4.

How to distinguish an elliptic curve with CM from one without?

non-CM	CM
$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}$	$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}(\sqrt{-d})$

$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \hookrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \hookleftarrow \Bbb{Q}(\operatorname{Frob}_p)$ .
$p\nmid a_p \Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \simeq \Bbb{Q}[T]/(c_p(T))$ is a quadratic field; otherwise a quaternion algebra.
If $E$ has CM by $\Bbb{Q}(\sqrt{-d})$ , then
$\begin{aligned} a_p \equiv 0 \bmod p &\Longleftrightarrow p\text{ inert or ramified in }\Bbb{Q}(\sqrt{-d}) \\ &\Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \not\simeq \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \end{aligned}$
If $E$ is non-CM, $\operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \cap \operatorname{End}_{\Bbb{Q}} E_q^{\mathrm{al}} \simeq \Bbb{Q}$ with probability $1$ ;
heuristically, $\operatorname{Prob}(a_p \equiv 0 \bmod p) \sim 1/\sqrt{p}$ .

$p$	11.a2: $\operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$	27.a2: $\operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$
$2$	$B_{2,\infty}$ , $a_2=-2$ , supersingular	$B_{2,\infty}$ , $2\equiv 2\bmod 3$
$3$	$\Bbb{Q}(\sqrt{-11})$	bad reduction
$13$	$\Bbb{Q}(\sqrt{-1})$	$\Bbb{Q}(\sqrt{-3})$ , $13\equiv 1\bmod 3$
$\Rightarrow$	$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}$	$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}(\sqrt{-3})$

Left column wanders: two non-isomorphic quadratic fields force $\Bbb{Q}$ .
Right column is constant where it is a field: CM by $\Bbb{Q}(\sqrt{-3})$ .