The Picard lattice

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

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

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

  • Records the algebraic cycles on $X$: curves modulo linear, algebraic or numerical equivalence.
  • For K3 surfaces the three agree; $\operatorname{Pic}^0 = 0$, and $\operatorname{Pic}$ 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

  • $\operatorname{Pic}(X^{\mathrm{al}})$ 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.
  • $\operatorname{disc}\operatorname{Pic}(X^{\mathrm{al}}) := \det$ of the Gram matrix in any $\Z$-basis; a non-zero integer. A basis change multiplies it by $\det(M)^2 = 1$.
  • $\operatorname{Gal}(k^{\mathrm{al}}/k)$ permutes curves, respects linear equivalence and preserves the pairing: an orthogonal action.

Pic inside $H^2$

  • Over $\Bbb{Q}^{\mathrm{al}}$, viewing $X$ also as a complex manifold,

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

  • 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 $\rho(X^{\mathrm{al}}) \in \{1,2,\dots,20\}$; for a generic K3 surface $\rho(X^{\mathrm{al}}) = 1$.
  • The degree of "difficulty" is negatively correlated with $\rho(X)$.
  • $T(X) := \operatorname{Pic}(X^{\mathrm{al}})^{\perp} \subset H^2(X,\Z)$, the transcendental lattice; equivalently the minimal sub-Hodge structure of $H^2(X,\Bbb{Q})$ whose complexification contains $H^{2,0}(X)$.

    $$H^2(X_{\Bbb{C}},\Bbb{Q}) \simeq \operatorname{Pic}(X^{\mathrm{al}})_{\Bbb{Q}} \oplus T(X)_{\Bbb{Q}}$$

  • The "new and interesting" Galois representations arise from $T(X)$.
  • In characteristic $p$ the bound $20$ fails: $\rho(X_p^{\mathrm{al}})$ 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 $X_p$;
  • $\operatorname{Pic}(\bullet)$, the group of line bundles modulo isomorphism; for K3 surfaces $\operatorname{Pic}^0(\bullet)=0$;
  • $\rho(\bullet) := \operatorname{rk}\operatorname{Pic}(\bullet)$, the arithmetic or geometric Picard number of $\bullet$.
  • $\operatorname{Pic}(X) = \operatorname{Pic}(X^{\mathrm{al}})^{\operatorname{Gal}(k^{\mathrm{al}}/k)}$, so $\rho(X) \leq \rho(X^{\mathrm{al}})$, usually strictly.
  • A class defined only over an extension is invisible over $k$.
  • "The Picard number" in this lecture means the geometric one, $\rho(X^{\mathrm{al}})$.

Goal

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

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

Question

  • How are the geometric Picard numbers $\rho(X^{\mathrm{al}})$ and $\rho(X_p^{\mathrm{al}})$ related?
  • How does the geometric Picard number behave under reduction modulo $p$?

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

What the characteristic polynomial gives you

Theorem (many people)

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

  • $\rho(X_p) = \operatorname{ord}_{t = q} P_2(t)$
  • $\rho(X_p^{\mathrm{al}}) = \sum_{\zeta} \operatorname{ord}_{t = q\zeta} P_2(t)$, where $\zeta$ runs over all roots of unity.
  • For K3 surfaces: $\rho(X_p^{\mathrm{al}})$ is even.
Artin-Tate for K3 surfaces

$$\boxed{ \lim_{t\to q}\frac{P_2(t)}{(t-q)^\rho} =(-1)^{\rho-1}q^{21-\rho}\#\operatorname{Br}(X_p)\,\operatorname{disc}(\operatorname{Pic}(X_p)) }$$

  • $P_2(t) \leadsto \operatorname{disc}(\operatorname{Pic}(X_p)) \bmod \Bbb{Q}^{\times 2}$

$$\begin{gathered} \rho=\operatorname{rk}\operatorname{Pic}(X_p),\quad \#\operatorname{Br}(X_p)\in\Bbb{Q}^{\times 2} \\ \mathrm{Tate}\Rightarrow\mathrm{Artin\!-\!Tate} \end{gathered}$$

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 $8\mid r$; $\rho(X_{89}^{\mathrm{al}}) = 1+1+4+4 = 10$, reached at $r=8$.
  • $\operatorname{Pic}(X_{89}^{\mathrm{al}})_{\Bbb{Q}}$ 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].

Reduction to finite characteristic

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

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

Theorem

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

$$\operatorname{Pic}(X^{\mathrm{al}}) \hookrightarrow \operatorname{Pic}(X_p^{\mathrm{al}}) \quad \text{and} \quad \rho(X^{\mathrm{al}}) \leq \rho(X_p^{\mathrm{al}}).$$

Goal

For a given $f$ and $p$, improve the inequality $\rho(X^{\mathrm{al}}) \leq \rho(X_p^{\mathrm{al}})$.

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 $A$.

$$\operatorname{Pic}(A)/\operatorname{Pic}^0(A) = \operatorname{NS}(A)$$

$$\bigl(\operatorname{Pic}(A)/\operatorname{Pic}^0(A)\bigr)_{\Bbb{Q}} \simeq \{\phi \in \operatorname{End}(A)_{\Bbb{Q}} : \phi^{\dagger} = \phi\}, \qquad \dagger\text{ the Rosati involution}$$

  • "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.

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

  • $\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \hookrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \hookleftarrow \Bbb{Q}(\operatorname{Frob}_p)$.
  • $p\nmid a_p \Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$ is a quadratic field
  • 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, then $\operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \cap \operatorname{End}_{\Bbb{Q}} E_q^{\mathrm{al}} \simeq \Bbb{Q}$ with prob. 1;
    and we expect $\operatorname{Prob}(a_p \equiv 0 \bmod p) \sim 1/\sqrt{p}$

Examples: 11.a2 and 27.a2

$E: y^2 + y = x^3 - x^2 - 10x - 20$ (LMFDB label: 11.a2)

  • $\operatorname{End}_{\Bbb{Q}} E_3^{\mathrm{al}} \simeq \Bbb{Q}(\sqrt{-11})$
  • $\operatorname{End}_{\Bbb{Q}} E_{13}^{\mathrm{al}} \simeq \Bbb{Q}(\sqrt{-1})$
  • $\Rightarrow \operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}$

$E: y^2 + y = x^3 - 7$ (LMFDB label: 27.a2)

  • $p = 2 \bmod 3 \Rightarrow a_p = 0 \Rightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$ is a quaternion algebra
  • $p = 1 \bmod 3 \Rightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \simeq \Bbb{Q}(\sqrt{-3})$
  • $\leadsto \operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}(\sqrt{-3})$

Improving upper bounds: two specializations

$$\operatorname{Pic}(X^{\mathrm{al}}) \hookrightarrow \operatorname{Pic}(X_p^{\mathrm{al}}) \quad \text{and} \quad \rho(X^{\mathrm{al}}) \leq \rho(X_p^{\mathrm{al}})$$

van Luijk

If $p$ and $q$ are two primes of good reduction, and

$$\begin{gathered} \rho(X_p^{\mathrm{al}}) = \rho(X_q^{\mathrm{al}}) = 2r, \\ \operatorname{disc} \operatorname{Pic}(X_p^{\mathrm{al}}) \neq \operatorname{disc} \operatorname{Pic}(X_q^{\mathrm{al}}) \quad \text{in } \Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2. \end{gathered}$$

then

$$\rho(X^{\mathrm{al}}) < 2r.$$

van Luijk, used this technique with $r = 1$, to provide the first known examples of K3 surfaces over $\Bbb{Q}$ such that $\rho(X^{\mathrm{al}}) = 1$

Does this always work?

Let's apply it to a K3 surface with a $\Bbb{Z}/5$ automorphism

$$X : x^3 z + 3x^2 y^2 + 5xw^3 + y^3 w + 3yz^3 - 5z^2 w^2 = 0 \ \subset \ \mathbf{P}^3$$

$p$$\rho(X_p^{\mathrm{al}})$disc
1118$-55$
1318$-85$
Theorem

$$P_2(t) \leadsto \operatorname{disc}\operatorname{Pic}(X_p) \bmod (\Bbb{Q}^{\times})^2.$$

  • disc: $\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})$, the geometric discriminant.
  • Base-field ranks: $1$ at $11$, $5$ at $13$.
  • Artin-Tate over $\Bbb{F}_{11^{30}}$ and $\Bbb{F}_{13^4}$: $\rho(X_p) = \rho(X_p^{\mathrm{al}}) = 18$.
  • Equal ranks; $-55 : -85 = 11/17 \notin (\Bbb{Q}^{\times})^2$.
  • Van Luijk: $\rho(X^{\mathrm{al}}) < 18$, hence $\leq 17$.
  • Symplectic order-5 action: $\rho(X^{\mathrm{al}}) \geq 17$ [Garbagnati-Sarti 2007, Prop. 1.1].

Torsion-free cokernel

Theorem (Elsenhans-Jahnel)

The specialization map

$$\operatorname{Pic}(X^{\mathrm{al}}) \hookrightarrow \operatorname{Pic}(X_p^{\mathrm{al}})$$

has torsion-free cokernel for $p \neq 2$.

Thus, if $\rho(X_p^{\mathrm{al}}) = \rho(X^{\mathrm{al}})$ every invertible sheaf lifts.

For example, if $\rho(X_p^{\mathrm{al}}) = 2$,

  1. compute $\operatorname{Pic}(X_p^{\mathrm{al}})$
  2. estimate the degree of a hypothetical effective divisor of the lift
  3. use Gröbner bases to verify that such a divisor does or does not exist

This approach is only practical if one can compute $\operatorname{Pic}(X_p^{\mathrm{al}})$ and if the obtained estimates are low.

Why the reduction rank is even

  • $X_p/\Bbb{F}_q$ K3; $P_2(t)=\det(t-\operatorname{Frob}\mid H^2)$

$$q^{-22}P_2(qt)=h(t)\prod_i\Phi_{k_i}(t)^{\gamma_i}$$

  • $h\in\Bbb{Q}[t]$: no cyclotomic factor
  • $|z|=1$: $\operatorname{conj}(z)=z^{-1}$
  • Real roots: $+1,-1$, already in $\Phi_1,\Phi_2$
  • Roots of $h$: nonreal pairs; $\deg h$ even
Weil + Tate

$$\rho(X_p^{\mathrm{al}})=\sum_i\gamma_i\deg\Phi_{k_i}=22-\deg h\in2\Z$$

Endomorphisms of the transcendental Hodge structure

  • $X/k$ K3; $k\subset\Bbb{C}$ a number field

$$T:=T(X)_{\Bbb{Q}}=c_1(\operatorname{Pic}(X^{\mathrm{al}}))_{\Bbb{Q}}^{\perp}\subset H^2(X_{\Bbb{C}},\Bbb{Q})$$

$$E:=\operatorname{End}_{\mathrm{Hdg}}(T)=\{a\in\operatorname{End}_{\Bbb{Q}}(T):a_{\Bbb{C}}(T^{i,j})\subset T^{i,j}\}$$

$T$ minimal rational sub-Hodge structure with $H^{2,0}\subset T_{\Bbb{C}}$: $0\neq\alpha\in E\Rightarrow\alpha(H^{2,0})=H^{2,0}\Rightarrow\operatorname{im}\alpha=T\Rightarrow\alpha^{-1}\in E$

Theorem (Zarhin)
  • $E$: a totally real field or a CM field
  • Totally real: every embedding $E\hookrightarrow\Bbb{C}$ lands in $\Bbb{R}$
  • CM: totally imaginary quadratic extension of a totally real field
  • $d:=[E:\Bbb{Q}]$, $m:=\dim_E T$; $dm=22-\rho(X^{\mathrm{al}})$
  • $E$ totally real $\Rightarrow m\geq3$ [van Geemen]
  • $E$ totally real; $V:=T_\ell(1)$; $g=\operatorname{Frob}_p^a$ in connected monodromy

$$V\otimes\Bbb{Q}_\ell^{\mathrm{al}}=\bigoplus_{\sigma:E\hookrightarrow\Bbb{Q}_\ell^{\mathrm{al}}}V_\sigma,\quad\dim V_\sigma=m,\quad g|V_\sigma\in SO(V_\sigma)$$

$$m\text{ odd}\Rightarrow\dim\ker(g-1)\geq d\Rightarrow\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+d$$

Jumping Picard ranks

$$\eta(X^{\mathrm{al}}):=\min_{p\text{ good}}\bigl(\rho(X_p^{\mathrm{al}})-\rho(X^{\mathrm{al}})\bigr)$$

Consider

$$\Pi_{\mathrm{jump}}(X):=\{p\text{ good}:\rho(X_p^{\mathrm{al}})>\rho(X^{\mathrm{al}})+\eta(X^{\mathrm{al}})\}$$

Is this set infinite? What is its density?

What about

$$X/\Bbb{Q}:\quad\gamma(X,B):=\frac{\#\{p\leq B:p\in\Pi_{\mathrm{jump}}(X)\}}{\#\{p\leq B:p\text{ prime}\}}\quad\text{as }B\rightarrow\infty\quad ?$$

K3 surfaces

So far we have been trying to improve the inequality $\rho(X^{\mathrm{al}})\leq\rho(X_p^{\mathrm{al}})$.
Can we use the inequality to our advantage?

Theorem (Li-Liedtke)

If there are infinitely many $p$ primes such that

$$\rho(X^{\mathrm{al}})<\rho(X_p^{\mathrm{al}})\text{ and }\rho(X_p^{\mathrm{al}})\neq22,$$

then $X^{\mathrm{al}}$ contains infinitely many rational curves.

Theorem (Bogomolov-Zarhin)

The set $\{p:\rho(X_p^{\mathrm{al}})\neq22\}$ has positive density (density 1 after finite extension).

Corollary (after Li-Liedtke; C-Elsenhans-Jahnel)
  • $X/k$ K3; $k$ a number field; $e:=[L:k]\in\{1,2\}$

$$J(X):=\{p\text{ good}:\rho(X_p^{\mathrm{al}})>\rho(X^{\mathrm{al}})\}$$

$$S_L:=\begin{cases}\{p\text{ good}\}&L=k,\\\{p\text{ good, inert in }L/k\}&e=2.\end{cases}$$

  • $S_L\subset J(X)$, up to finitely many primes: lower density $\geq1/e$
  • Additionally $L=k$ or $E=\Bbb{Q}$: infinitely many integral rational curves on $X^{\mathrm{al}}$

Product of elliptic curves

  • $X=\operatorname{Km}(A)$; $A/\Bbb{Q}$ an abelian surface
  • $\rho(A^{\mathrm{al}}):=\operatorname{rk}(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))$
  • $\rho(X^{\mathrm{al}})=16+\rho(A^{\mathrm{al}})$
  • $\rho(X_p^{\mathrm{al}})=16+\rho(A_p^{\mathrm{al}})$; $p>2$ good
  • $\eta(X^{\mathrm{al}})=\eta(A^{\mathrm{al}})=\rho(A^{\mathrm{al}})\bmod2$
  • $\Pi_{\mathrm{jump}}(X)=\Pi_{\mathrm{jump}}(A)$
  • Fix a polarization on $A$; $\dagger$ the Rosati involution

$$(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))_{\Bbb{Q}}\simeq\{\phi\in\operatorname{End}(A^{\mathrm{al}})_{\Bbb{Q}}:\phi^\dagger=\phi\}$$

  • $A=E_1\times E_2$; $E_i/\Bbb{Q}$

$$\rho(X^{\mathrm{al}})=18+\operatorname{rk}\operatorname{Hom}(E_1^{\mathrm{al}},E_2^{\mathrm{al}})$$

$X$ $\rho(X^{\mathrm{al}})$ $\gamma(X,B)$, predicted What is known
square of CM 20 $1/2$ $1/2+o(1)$, CM theory
square of non-CM 19 $\sim c_X/\sqrt{B}$ infinitely many [Elkies 1987]
CM times CM 18 $1/4$ $1/4+o(1)$, CM theory
CM times non-CM 18 $\sim c_X/\sqrt{B}$ infinitely many [Charles 2018]
non-CM times non-CM 18 $\sim c_X/\sqrt{B}$ infinitely many [Charles 2018]
  • Product rows: geometrically non-isogenous factors
  • Non-CM rates: Lang-Trotter heuristics; per-prime scale $1/\sqrt{p}$

Remark

$p\in\Pi_{\mathrm{jump}}(X)$ depends uniquely on the pair $(a_{E_1}(p),a_{E_2}(p))$.

Jumping Picard ranks for Kummer surfaces

  • $\rho(A_p^{\mathrm{al}})\geq4\Longleftrightarrow A_p^{\mathrm{al}}\sim E^2$, $E$ an elliptic curve
  • $\rho(A_p^{\mathrm{al}})=6\Longleftrightarrow A_p^{\mathrm{al}}\sim E^2$, $E$ a supersingular elliptic curve
  • If $A^{\mathrm{al}}\sim E^2$, then $p\in\Pi_{\mathrm{jump}}(A)$ iff $p$ is supersingular for $E$.
  • If $A^{\mathrm{al}}\sim E_1\times E_2$ with $E_1^{\mathrm{al}}\not\sim E_2^{\mathrm{al}}$, then $p\in\Pi_{\mathrm{jump}}(A)$ iff $E_{1,p}^{\mathrm{al}}\sim E_{2,p}^{\mathrm{al}}$.
  • If $\operatorname{End}(A^{\mathrm{al}})=\Z$, then $p\in\Pi_{\mathrm{jump}}(A)$ iff $A_p^{\mathrm{al}}\sim E^2$.

O or SO?

  • $V:=T_\ell(1)$; cup-product pairing

$$\tau:\operatorname{Gal}(k^{\mathrm{al}}/k)\longrightarrow O(V)$$

  • $\det\tau=1\Longleftrightarrow\operatorname{im}\tau\subset SO(V)$
  • $\det\tau\neq1$: nontrivial quadratic character
  • An easy way to explain some jumps: $O$ vs $SO$.

What $\det=-1$ costs you

  • $\rho(X^{\mathrm{al}})$ even; $\varphi:=\operatorname{Frob}_p|T_\ell(1)$; $\det\varphi=-1$
  1. Orthogonality: $\lambda$ and $\lambda^{-1}$, with equal multiplicities.
  2. Other pairs: determinant $+1$; multiplicity of $-1$ odd.
  3. $\dim T_\ell(1)$ even: multiplicity of $+1$ odd.
  4. Tate: $+1,-1$ give two new geometric divisor classes.

    $$\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+2$$

Discriminant of a K3 surface

  • $X/\Bbb{Q}$ quartic K3
  • $D_X:=\Delta_{H^2}(X)\in\Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2$: determinant-character square class
  • $D_X\in\Z\setminus\{0\}$ a representative; $p$ good, $p\nmid2D_X$
Theorem (Deligne; C-Elsenhans-Jahnel 2020)

The functional equation of the Frobenius action on $H^2(X)$ has the plus sign if and only if $D_X$ is square mod $p$.

$$\varepsilon_p=\det(-\operatorname{Frob}_p\mid H^2_{\mathrm{et}}(X^{\mathrm{al}},\Bbb{Q}_\ell(1)))=\left(\frac{D_X}{p}\right)$$

  • $\operatorname{Gal}(\Bbb{Q}^{\mathrm{al}}/\Bbb{Q})$ fixes $\operatorname{Pic}(X^{\mathrm{al}})$: $\Delta_{\operatorname{Pic}}(X)=1$
Theorem (C-Elsenhans-Jahnel)

$$\rho(X^{\mathrm{al}})=2r,\quad\left(\frac{D_X}{p}\right)=-1\quad\Rightarrow\quad\rho(X_p^{\mathrm{al}})\geq2r+2$$

We can explain the $1/2$

  • $X/\Bbb{Q}$ K3; $r:=\rho(X^{\mathrm{al}})$ even; $\eta(X^{\mathrm{al}})=0$
  • $d_X:=\Delta_{H^2}(X)\Delta_{\operatorname{Pic}}(X)$ modulo squares; $d_X\in\Z\setminus\{0\}$
Theorem (C-Elsenhans-Jahnel)

$$p\text{ good},\ p\nmid2d_X:\quad\det(\operatorname{Frob}_p\mid T_\ell(1))=\left(\frac{d_X}{p}\right)=-1\Rightarrow\rho(X_p^{\mathrm{al}})\geq r+2$$

Corollary
  • $d_X$ nonsquare: $L=\Bbb{Q}(\sqrt{d_X})$, $[L:\Bbb{Q}]=2$
  • $S_L\subset J(X)=\Pi_{\mathrm{jump}}(X)$, up to finitely many primes
  • $\displaystyle\liminf_{B\rightarrow\infty}\gamma(X,B)\geq1/2$
  • $E=\Bbb{Q}$: infinitely many integral rational curves on $X^{\mathrm{al}}$
  • Example: Costa-Tschinkel 2014, Ex. 3.3

$$\begin{aligned}d_X={}&-1\cdot5\cdot151\cdot22490817357414371041\\&\cdot387308497430149337233666358807996260780875056740850984213276970343278935342068889706146733313789\end{aligned}$$

Computing $\rho(X^{\mathrm{al}})$

  • $T=T(X)_{\Bbb{Q}}$; $E=\operatorname{End}_{\mathrm{Hdg}}(T)$; $d=[E:\Bbb{Q}]$; $m=\dim_E T$
Theorem (Charles 2014)

$$\rho(X_p^{\mathrm{al}})\geq\begin{cases}\rho(X^{\mathrm{al}})&\text{if }E\text{ is CM or }m\text{ is even,}\\\rho(X^{\mathrm{al}})+d&\text{if }E\text{ is totally real and }m\text{ is odd.}\end{cases}$$

  • Equality occurs infinitely often (density $1$ after some finite extension).

Further, assume that we are in the second case, then exist infinitely many pairs $(p,q)$ such that the equality holds and

$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$

When every prime overshoots

  • $r:=\rho(X^{\mathrm{al}})$; $d:=[E:\Bbb{Q}]$; $m:=\dim_E T$
  • $E=\Bbb{Q}$, $m$ odd: $\eta=1$; van Luijk succeeds [Charles 2014]

    $$\text{Order-5 example:}\quad17\leq\rho(X^{\mathrm{al}})<18$$

  • $E$ totally real, $E\neq\Bbb{Q}$, $m$ odd: $\eta=d\geq2$

    $$\min_p\rho(X_p^{\mathrm{al}})=r+d;\qquad\text{two-prime upper bound: }r+d-1$$

Certified quadratic RM

$$F\hookrightarrow E,\quad[F:\Bbb{Q}]=2;\qquad\rho(X_p^{\mathrm{al}})=\rho(X_q^{\mathrm{al}})=18$$

$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$

$$\rho(X^{\mathrm{al}})\leq17,\quad\rho(X^{\mathrm{al}})\text{ even}\quad\Rightarrow\quad\rho(X^{\mathrm{al}})\leq16$$

A real multiplication example

  • Elsenhans-Jahnel [2014, Thms. 5.12 and 6.6]
  • $X$: minimal resolution of

$$w^2=(-y^2/8+yz-z^2)(7x^2/8+5xz+7z^2)(2x^2+3xy+y^2)$$

  • $6$ lines; $15=\binom{6}{2}$ nodes; $15$ exceptional $(-2)$-curves
  • $H,E_{ij}$: $16$ independent classes
  • $\rho(X^{\mathrm{al}})=16$
  • RM: $E=\Bbb{Q}(\sqrt{2})$; $\dim_E T=(22-16)/2=3$
  • $\eta=2$; $\rho(X_p^{\mathrm{al}})\geq18$ at every good prime
  • Rank-$18$ pair, unequal square classes, certified RM: $\rho(X^{\mathrm{al}})\leq16$