Computing Picard Lattices of K3 Surfaces
Lecture 1: Reduction methods
Edgar Costa (MIT)
ICERM: Arithmetic, Geometry and Computations on K3 surfaces
September 14, 2026
Supported by the Simons Foundation
Slides available at edgarcosta.org
The intersection pairing
- $\operatorname{Pic}(X^{\mathrm{al}})$ with the intersection pairing $(D,D') \mapsto D \cdot D'$.
- Even symmetric bilinear form: $D \cdot D = 2g(D) - 2$ for every curve $D$.
- $\operatorname{disc}\operatorname{Pic}(X^{\mathrm{al}}):=\det(D_i\cdot D_j)$ in a $\Z$-basis.
- Signature $(1,\rho(X^{\mathrm{al}})-1)$: Hodge index theorem.
- $\operatorname{Gal}(k^{\mathrm{al}}/k)$ permutes curves, respects linear equivalence and preserves the pairing.
What the characteristic polynomial gives you
- $X_p/\Bbb{F}_q$ an abelian surface or a K3 surface; $q=p^n$, $\ell\neq p$.
- For abelian surfaces: $\rho:=\operatorname{rk}(\operatorname{Pic}/\operatorname{Pic}^0)$.
- $\rho(X_p) = \operatorname{ord}_{t = q} P_2(t)$
- $\rho(X_p^{\mathrm{al}}) = \sum_{\zeta} \operatorname{ord}_{t = q\zeta} P_2(t)$, where $\zeta$ runs over all roots of unity.
- For K3 surfaces $\Rightarrow$ $\rho(X_p^{\mathrm{al}})$ is even.
$$\boxed{ \lim_{t\to q}\frac{P_2(t)}{(t-q)^\rho} =(-1)^{\rho-1}q^{21-\rho}\#\operatorname{Br}(X_p)\,\operatorname{disc}(\operatorname{Pic}(X_p)) }$$
- $\#\operatorname{Br}(X_p)\in\Bbb{Q}^{\times 2}$ and $P_2(t)$ give $\operatorname{disc}(\operatorname{Pic}(X_p)) \bmod \Bbb{Q}^{\times 2}$
Why the reduction rank is even
- $X_p/\Bbb{F}_q$ K3; $P_2(t)=\det(t-\operatorname{Frob}\mid H^2_{\mathrm{et}}(X_p^{\mathrm{al}},\Bbb{Q}_\ell))$
$$q^{-22}P_2(qt)=h(t)\prod_i\Phi_{k_i}(t)^{\gamma_i}$$
- $h\in\Bbb{Q}[t]$: no cyclotomic factor
- $|z|=1$ $\Rightarrow$ $\operatorname{conj}(z)=z^{-1}$
- Real roots: $+1,-1$, already in $\Phi_1,\Phi_2$
- Roots of $h$: nonreal pairs; $\deg h$ even
$$\rho(X_p^{\mathrm{al}})=\sum_i\gamma_i\deg\Phi_{k_i}=22-\deg h\in2\Z$$
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 $X_p := Z(f \bmod{p}) \subset \mathbf{P}^3_{\Bbb{F}_p}$.
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 $A$.
$$\operatorname{Pic}(A)/\operatorname{Pic}^0(A) = \operatorname{NS}(A)$$
- $\rho(A):=\operatorname{rk}(\operatorname{Pic}(A)/\operatorname{Pic}^0(A))$; also after base change.
$$\bigl(\operatorname{Pic}(A)/\operatorname{Pic}^0(A)\bigr)_{\Bbb{Q}} \simeq \{\phi \in \operatorname{End}(A)_{\Bbb{Q}} : \phi^{\dagger} = \phi\}, \qquad \dagger\text{ the Rosati involution}$$
- "Compute the Picard lattice of a K3 surface" ↭ "compute $\operatorname{End}(A)$".
- We also stratify moduli of abelian varieties via $\operatorname{End}(A)$.
- $A$ an abelian surface, $\operatorname{char}k\neq2$ $\Rightarrow$ $\rho(\operatorname{Kum}(A)^{\mathrm{al}}) = \rho(A^{\mathrm{al}})+16$.
- $\operatorname{Kum}(A)$ is the K3 surface obtained by resolving $A/\{\pm1\}$.
Proving that an elliptic curve does not have CM
$$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q} \quad\text{or}\quad \Bbb{Q}(\sqrt{-d})\;(\mathrm{CM})$$
- $\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \hookrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \hookleftarrow \Bbb{Q}(\operatorname{Frob}_p)$.
- $a_p := \operatorname{Tr}(\operatorname{Frob})$
- $p\nmid a_p \Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$ 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 $\operatorname{Prob}(a_p \equiv 0 \bmod p) \sim 1/\sqrt{p}$
Two universal examples: 11.a2 and 27.a3
$E: y^2 + y = x^3 - x^2 - 10x - 20$ (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.a3)
- $p = 2 \bmod 3 \Rightarrow a_p = 0 \Rightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$ is a quaternion algebra
- $p = 1 \bmod 3 \Rightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \simeq \Bbb{Q}(\sqrt{-3})$
- $\Rightarrow \operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q}(\sqrt{-3})$
Improving upper bounds: two specializations
$$\operatorname{Pic}(X^{\mathrm{al}}) \hookrightarrow \operatorname{Pic}(X_p^{\mathrm{al}}) \quad \text{and} \quad \rho(X^{\mathrm{al}}) \leq \rho(X_p^{\mathrm{al}})$$
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 (2005): first explicit K3 surfaces $X/\Bbb{Q}$ with $\rho(X^{\mathrm{al}})=1$.
Does this always work?
Torsion-free cokernel
$X/\Bbb{Q}$ K3; $p>2$ a prime of good reduction.
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 $\rho(X_p^{\mathrm{al}}) = 2$, Elsenhans—Jahnel approach is
- compute $\operatorname{Pic}(X_p^{\mathrm{al}})$
- estimate the degree of a hypothetical effective divisor of the lift
- 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.
- Elsenhans-Jahnel: this becomes the proof of a generic example using a single prime.
Computing $\rho(X^{\mathrm{al}})$
- $T=T(X)_{\Bbb{Q}}$; $E=\operatorname{End}_{\mathrm{Hdg}}(T)$; $d=[E:\Bbb{Q}]$; $m=\dim_E T$
$$\rho(X_p^{\mathrm{al}})\geq\begin{cases}\rho(X^{\mathrm{al}})&\text{if }E\text{ is CM or }m\text{ is even,}\\\rho(X^{\mathrm{al}})+d&\text{if }E\text{ is totally real and }m\text{ is odd,}\end{cases}$$
Equality occurs infinitely often (density $1$ after some finite extension).
If $E$ is totally real and $m$ is odd, infinitely many good ordinary prime pairs $(p,q)$ satisfy $\rho(X_p^{\mathrm{al}})=\rho(X_q^{\mathrm{al}})=\rho(X^{\mathrm{al}})+d$ and
$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$
The Kloosterman—van Luijk method works, if it is aware of $E$.
A real multiplication example
- Elsenhans-Jahnel
- $X$: 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 at $p=17$, $q=23$, unequal square classes, certified RM $\Rightarrow$ $\rho(X^{\mathrm{al}})\leq16$
Without certified real multiplication, one could only prove $\rho(X^{\mathrm{al}})\leq17$.
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
$$X/\Bbb{Q}:\quad\gamma(X,B):=\frac{\#\{p\leq B:p\in\Pi_{\mathrm{jump}}(X)\}}{\#\{p\leq B:p\text{ prime}\}}\quad\text{as }B\rightarrow\infty\quad ?$$
Numerical experiments for $\rho(X^{\mathrm{al}})=2$
$\rho(X)=\rho(X^{\mathrm{al}})=2$ and $E=\Bbb{Q}$ or CM
No obvious trend …