K3 surfaces over $\mathbb{Q}$ of degree $10$that have Picard rank$1$

This paper presents explicit examples of K3 surfaces over $\mathbb{Q}$ with geometric Picard rank 1, specifically constructing degree 10 surfaces as intersections of hyperplanes, a quadric, and the Grassmannian $\mathrm{Gr}(2,5)$ in $\mathbb{P}^9$, as well as a degree 6 example.

The Gist

Imagine you are an architect trying to build a very special, perfectly symmetrical house. In the world of mathematics, this "house" is called a K3 surface. It's a complex, multi-dimensional shape that mathematicians love because it sits right at the edge of what we can understand and what is completely mysterious.

The author of this paper, Victor De Vries, is trying to build one of these houses using only rational numbers (fractions like 1/2, 3/4, etc., which are the "building blocks" of our everyday arithmetic).

Here is the simple story of what he did, using some analogies:

1. The Goal: The "Lonely" House

Every K3 surface has a hidden feature called its Picard Rank. Think of this as the number of "structural beams" or "blueprints" that define the house's geometry.

• High Rank (Many beams): The house is very structured. We know a lot about it, and it's easy to find "furniture" (rational points) inside it.
• Low Rank (Few beams): The house is wild and chaotic. It's very hard to predict where furniture will be.

Mathematicians have been looking for a K3 surface with a Picard Rank of 1. This is the "ultimate lonely house"—a shape so simple in its structure that it's incredibly difficult to understand its arithmetic. It's like trying to find a needle in a haystack, but the needle is made of pure math.

2. The Problem: The "Too Big" Blueprint

For small houses (low degrees), mathematicians already knew how to build these "lonely" houses. But for a specific size called Degree 10, the blueprint is too big and complex.

• Imagine trying to build a house in a 10-dimensional room. You can't just draw it on paper.
• The shape is formed by the intersection of a giant, complex grid (called a Grassmannian) and some other curves. It's like trying to find the exact spot where three invisible walls and one curved glass dome meet in a 10D room.

3. The Strategy: The "Modular" Trick

Since building the house directly in the complex "real world" (over the rational numbers) is too hard, De Vries used a clever trick: The Modular Test.

Imagine you want to prove a house is stable. Instead of building the whole thing in the rain, you build two tiny, cheap models:

1. Model A: Built using "Mod 2" bricks (only even and odd numbers).
2. Model B: Built using "Mod 3" bricks (numbers that leave remainders of 0, 1, or 2).

The Logic:

• If you build a house over the real numbers, it must look like Model A when you view it through a "Mod 2" lens, and like Model B through a "Mod 3" lens.
• De Vries built a Model A and a Model B.
• He checked Model A and found it was "geometrically reduced" (it didn't fall apart or look weird).
• He checked Model B and found it had exactly 2 structural beams (Picard Rank 2).

4. The "Double-Check" Logic

Here is the brilliant part of the proof:

• If the final "Real House" (over the rational numbers) had 2 beams, it would have to match the structure of Model B perfectly.
• However, Model A has a very specific, rigid property: it forbids certain types of beams from existing together in a specific way.
• De Vries proved that if the Real House had 2 beams, it would create a contradiction with Model A. It would be like trying to fit a square peg into a round hole that Model A is guarding.
• Therefore, the Real House cannot have 2 beams.
• Since we know it must have at least 1 beam, and it can't have 2, it must have exactly 1.

5. The Result

De Vries successfully constructed:

1. A Degree 10 K3 surface over the rational numbers with Picard Rank 1. (This was the main goal).
2. A Degree 6 K3 surface over the rational numbers with Picard Rank 1. (This was a bonus to "fill in a gap" in previous research).

The Big Picture Analogy

Think of the universe of K3 surfaces as a vast forest.

• Most trees (surfaces) are easy to study; they have many branches (high rank).
• Some trees are rare and have very few branches (low rank).
• De Vries found a specific, very rare tree (Degree 10) that has only one branch.
• He didn't find it by looking at the tree directly (which is too far away). Instead, he looked at its shadow on the ground under two different colored lights (Mod 2 and Mod 3). By analyzing how the shadows behaved, he proved with certainty that the real tree must have exactly one branch.

Why does this matter?
These "lonely" houses (Rank 1) are the hardest to understand. By proving they exist, De Vries gives mathematicians a new playground to test their theories about how numbers behave in complex geometric shapes. It's like finding a new, uncharted island on a map of the mathematical world.

Technical Summary

Here is a detailed technical summary of the paper "K3 Surfaces over $\mathbb{Q}$ of Degree 10 That Have Picard Rank 1" by Victor De Vries.

1. Problem Statement

The central problem addressed is the explicit construction of K3 surfaces defined over the rational numbers ($\mathbb{Q}$) with Picard rank 1.

• Context: For a K3 surface $S$ over a number field, the Picard rank $\rho(S)$ (the rank of the geometric Picard group) satisfies $1 \leq \rho \leq 20$.
• Significance: Surfaces with high Picard rank ($\rho \geq 5$) often have potentially dense rational points (Bogomolov-Tschinkel). Conversely, surfaces with low Picard rank (specifically $\rho=1$) are arithmetically complex and difficult to analyze.
• Gap: While examples of Picard rank 1 K3 surfaces existed for degrees $2d \in {2, 4, 8}$, and general existence was known for$2d \leq 18$ via non-constructive arguments (Terasoma), explicit examples were missing for degree 10 and degree 6.
• Goal: To provide explicit, constructive examples of K3 surfaces over $\mathbb{Q}$ of degree 10 and degree 6 with Picard rank 1.

2. Methodology

The author employs a constructive approach inspired by van Luijk's method, utilizing reduction modulo primes and lattice lifting. The strategy involves:

1. Construction of Modulo $p$ Examples:

   • Construct specific K3 surfaces $S_2$ over $\mathbb{F}_2$ and $S_3$ over $\mathbb{F}_3$.
   • These surfaces are defined as complete intersections within the Grassmannian $\text{Gr}(2, 5)$ (embedded in $\mathbb{P}^9$ via the Plücker embedding).
   • Specifically, they are intersections of $\text{Gr}(2, 5)$ with three hyperplanes and one quadric hypersurface (type $(1, 1, 1, 2)$).

2. Lifting to $\mathbb{Q}$:

   • Lift the defining equations of $S_2$ and $S_3$ from $\mathbb{F}_2$ and $\mathbb{F}_3$ to $\mathbb{Z}$ simultaneously.
   • This creates a scheme $S$ over $\mathbb{Q}$ that reduces to $S_2$ modulo 2 and $S_3$ modulo 3.

3. Picard Rank Determination via Specialization:

   • Upper Bound: Use the specialization map (Lemma 1.1) which injects $\text{Pic}(S)$ into $\text{Pic}(S_p)$ for primes of good reduction. Thus, $\rho(S) \leq \min(\rho(S_2), \rho(S_3))$.
   • Lower Bound/Contradiction: Assume $\rho(S) = 2$. This implies $\text{Pic}(S)$ is a primitive sublattice of $\text{Pic}(S_3)$ (by Elsenhans-Jahnel).
   • Geometric Analysis: Analyze the specific lattice structures of $S_2$ and $S_3$. If $\rho(S)=2$, the lattice generators must satisfy specific intersection properties that lead to a contradiction when examining the geometry of hyperplane sections over $\mathbb{F}_2$ (specifically regarding geometric reducedness and component degrees).

3. Key Contributions and Technical Details

A. The Degree 10 Case (Main Result)

• Geometry: The surfaces are constructed in $\text{Gr}(2, 5) \subset \mathbb{P}^9$. The canonical sheaf of $\text{Gr}(2, 5)$ is $\mathcal{O}_G(-5)$, ensuring that a $(1,1,1,2)$ complete intersection has a trivial canonical bundle, making it a K3 surface (Lemma 1.10).
• Surface $S_2$ (over $\mathbb{F}_2$):
  • Defined by specific linear forms $l_i$ and a quadratic form $q$.
  • Key Property (Lemma 2.2): Every hyperplane section is geometrically reduced. Furthermore, for any hyperplane section $H$, if $H_{\text{sing}}$ (singular locus) has an irreducible component of degree $\geq 6$, it leads to a contradiction with the specific equations chosen.
• Surface $S_3$ (over $\mathbb{F}_3$):
  • Defined by specific forms $l'_i$ and $q'$.
  • Picard Rank: Using point counting over $\mathbb{F}_{3^n}$ and the Lefschetz fixed-point formula, the author proves $\rho(S_3) = 2$ (Lemma 2.7).
  • Lattice Structure: The Picard lattice is spanned by the hyperplane class $H$ and an elliptic fiber class $M$, with intersection matrix $\begin{pmatrix} 10 & 5 \\ 5 & 0 \end{pmatrix}$.
• The Contradiction (Theorem 3.3):
  • Assume $\rho(S) = 2$. Then $\text{Pic}(S) \cong \text{Pic}(S_3)$.
  • This implies the existence of a divisor class $M$ in $\text{Pic}(S)$ with $M^2=0$ and $M \cdot H = 5$.
  • Reducing modulo 2, this class must correspond to a divisor on $S_2$. The Galois orbit of this class forces the existence of an effective divisor $D$ on $S_2$ (over $\mathbb{F}_4$) such that $D$ lies in a hyperplane section defined over $\mathbb{F}_2$.
  • Geometric analysis shows that the intersection of such divisors would imply a singular locus of degree $\geq 5$, which contradicts the specific geometric properties of $S_2$ (Lemma 2.2).
  • Conclusion: $\rho(S) \neq 2$, so $\rho(S) = 1$.

B. The Degree 6 Case (Sidenote)

• Geometry: Constructed as a complete intersection of a quadric and a cubic in $\mathbb{P}^4$.
• Method: Similar to the degree 10 case.
  • $X_2$ over $\mathbb{F}_2$: Picard rank 2, lattice $\begin{pmatrix} 6 & 4 \\ 4 & 0 \end{pmatrix}$.
  • $X_3$ over $\mathbb{F}_3$: Picard rank 2, lattice $\begin{pmatrix} 6 & 3 \\ 3 & 0 \end{pmatrix}$.
• Proof of Rank 1:
  • If $\rho(X) = 2$, then $\text{Pic}(X)$ must embed into both $\text{Pic}(X_2)$ and $\text{Pic}(X_3)$.
  • The discriminant of $\text{Pic}(X_2)$ is $-16$, and for $\text{Pic}(X_3)$ is $-9$.
  • A lattice of rank 2 cannot have discriminants $-16$ and $-9$ simultaneously (as the discriminant of the lifted lattice must divide both, or the lattices must be isometric, which they are not).
  • Conclusion: $\rho(X) = 1$.

4. Results

1. Theorem 0.1 / Theorem 3.3: Explicitly constructed a K3 surface $S$ over $\mathbb{Q}$ of degree 10 (intersection of $\text{Gr}(2, 5)$, 3 hyperplanes, and 1 quadric) with Picard rank 1.
2. Theorem 4.5: Explicitly constructed a K3 surface $X$ over $\mathbb{Q}$ of degree 6 (intersection of a quadric and cubic in $\mathbb{P}^4$) with Picard rank 1.
3. Computational Verification: Used Magma to verify smoothness, dimension, point counts, and specific geometric properties (reducedness of sections) for the finite field models.

5. Significance

• Filling the Gap: This work provides the first explicit examples for degree 10 and degree 6, completing the set of known explicit examples for small degrees where the general K3 surface is a complete intersection in a Grassmannian or projective space.
• Arithmetic Complexity: These examples serve as test cases for studying the arithmetic of K3 surfaces with minimal Picard rank, where rational points are expected to be sparse and the Brauer-Manin obstruction is more subtle.
• Methodological Extension: The paper demonstrates that the "mod $p$ specialization + lattice discriminant/geometry" method is robust enough to handle higher-degree K3 surfaces (degree 10) that are not complete intersections in $\mathbb{P}^n$ but rather in Grassmannians.
• Future Directions: The author suggests this method could potentially be extended to degrees 12, 14, 16, and 18, provided the computational complexity of the mod $p$ calculations remains feasible.

In summary, De Vries successfully bridges the gap between theoretical existence and explicit construction for low-rank K3 surfaces of degree 10 and 6, utilizing a sophisticated interplay between algebraic geometry over finite fields, lattice theory, and computational verification.