Examples: 11.a2 and 27.a2

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

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

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

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

Improving upper bounds: two specializations

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

van Luijk

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

$$\begin{gathered} \rho(\overline{X}_p) = \rho(\overline{X}_q) = 2r, \\ \operatorname{disc} \operatorname{Pic}(\overline{X}_p) \neq \operatorname{disc} \operatorname{Pic}(\overline{X}_q) \quad \text{in } \mathbf{Q}^{\times}/(\mathbf{Q}^{\times})^2. \end{gathered}$$

then

$$\rho(X_{\mathbf{Q}^{\mathrm{al}}}) < 2r.$$

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

Under the right conditions we know that this method will not succeed to give a tight upper bound (parity + endomorphisms of the transcendental lattice).

The quartic, worked

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

A quartic with an automorphism of order 5. Two primes, read off the recorded data:

$p$$\rho(\overline{X}_p)$disc
1118$-55$
1318$-85$
Theorem (Artin-Tate)

$$P_2(T) \leadsto \operatorname{disc}\operatorname{NS}(X_p) \bmod (\mathbf{Q}^{\times})^2.$$

The disc column is the geometric discriminant $\operatorname{disc}\operatorname{Pic}(\overline{X}_p)$. The base-field ranks are 1 at 11 and 5 at 13, so Artin-Tate is applied over $\mathbf{F}_{11^{30}}$ and $\mathbf{F}_{13^4}$, where $\rho(X_p) = \rho(\overline{X}_p) = 18$.

Equal ranks, and the ratio $-55 : -85$ is $11/17$, not a square. Van Luijk gives $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) < 18$, hence $\leq 17$.

Torsion-free cokernel

Theorem (Elsenhans-Jahnel)

Elsenhans-Jahnel showed that the specialization map

$$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p)$$

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

Thus, if $\rho(\overline{X}_p) = \rho(X_{\mathbf{Q}^{\mathrm{al}}})$ every invertible sheaf lifts.

For example, if $\rho(\overline{X}_p) = 2$, Elsenhans-Jahnel approach is

  1. compute $\operatorname{Pic}(\overline{X}_p)$
  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 $\operatorname{Pic}(\overline{X}_p)$ and if the obtained estimates are low.

Jumping Picard ranks

Theorem (Charles)

We have

$$\rho(\overline{X}) + \eta(\overline{X}) \leq \rho(\overline{X}_p)$$

for some $\eta(\overline{X}) \geq 0$. Equality occurs infinitely often (density 1 after some finite extension).

Consider

$$\Pi_{\mathrm{jump}}(X) := \lbrace\, p : \rho(\overline{X}_p) > \rho(\overline{X}) + \eta(\overline{X}) \,\rbrace$$

Is this set infinite? What is its density?

What about

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

K3 surfaces

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

Theorem (Li-Liedtke)

If there are infinitely many $p$ primes such that

$$\rho(\overline{X}) < \rho(\overline{X}_p) \text{ and } \rho(\overline{X}_p) \neq 22,$$

then $\overline{X}$ contains infinitely many rational curves.

Theorem (Bogomolov-Zarhin)

The set $\lbrace p : \rho(\overline{X}_p) \neq 22 \rbrace$ has positive density (density 1 after finite extension).

Corollary (Li-Liedtke)

If $\rho(\overline{X})$ is odd, then $\overline{X}$ contains infinitely many rational curves.

Product of elliptic curves

$X \simeq \operatorname{Km}(A)$, $A$ an abelian surface.

$$\begin{gathered} \rho(\overline{X}) = \rho(\overline{A}) + 16, \quad \rho(\overline{X}_p) = \rho(\overline{A}_p) + 16, \\ \eta(\overline{X}) = \eta(\overline{A}) = \rho(\overline{A}) \bmod 2, \\ \Pi_{\mathrm{jump}}(X) = \Pi_{\mathrm{jump}}(A). \end{gathered}$$

$\operatorname{NS}(A)_{\mathbf{Q}} \simeq \lbrace \phi \in \operatorname{End}(A)_{\mathbf{Q}} : \phi^{\dagger} = \phi \rbrace$, $\dagger$ the Rosati involution.

$A = E_1 \times E_2$, $E_1, E_2$ elliptic curves over $\mathbf{Q}$.

$$\begin{aligned} \rho(\operatorname{Km}(E_1 \times E_2)) &= 18 + \operatorname{rk}\operatorname{Hom}(E_1,E_2) \\ &= 18 + \begin{cases} 0 & E_1 \not\sim E_2; \\ \operatorname{rk}\operatorname{End}(E_1) & E_1 \sim E_2. \end{cases} \end{aligned}$$

$X$$\rho(\overline{X})$$P(p \in \Pi_{\mathrm{jump}}(X))$$\gamma(X,B)$,
predicted		$\gamma(X,B)$,
what is known
square of CM20$1/2$$1/2$$1/2 + o(1)$, CM theory
square of non-CM19$\sim 1/\sqrt{p}$ $\dagger$$c/\sqrt{B}$ $\dagger$$\tfrac{c(\log\log B)(\log B)}{B} < \cdot < \tfrac{C\log B}{B^{1/4}}$ $\ddagger$ [Elkies]
CM times CM18$1/4$$1/4$$1/4 + o(1)$, CM theory
CM times non-CM18$\sim 1/\sqrt{p}$ $\dagger$$c/\sqrt{B}$ $\dagger$?
non-CM times non-CM18$\sim 1/\sqrt{p}$ $\dagger$$c/\sqrt{B}$ $\dagger$infinitely many jump primes [Charles]

$\dagger$ Conjectural, the Lang-Trotter heuristic. The CM rows are unconditional. No row is 1; every bound is attainable.

$\ddagger$ Lower bound conditional on ERH, with an absolute implied constant $c > 0$; upper bound unconditional.

For a geometrically non-isogenous non-CM pair over $\mathbf{Q}$, $\gamma(X,B) \to 0$; Charles gives infinitely many jump primes.

Remark

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

Jumping Picard ranks for Kummer surfaces

  • $\rho(\overline{A}_p) \geq 4 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ an elliptic curve
  • $\rho(\overline{A}_p) = 6 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ a supersingular elliptic curve
  • If $A \sim E^2$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $p$ is supersingular for $E$. This is related to the Lang-Trotter conjecture. For non-CM $E/\mathbf{Q}$, it predicts that $p$ is supersingular with probability proportional to $1/\sqrt{p}$. Elkies has shown that there are infinitely many supersingular primes for $E/\mathbf{Q}$. For CM $E$, the supersingular primes are the inert primes, of density $1/2$.
  • If $A \sim E_1 \times E_2$ with $E_1 \not\sim E_2$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $E_1 \sim E_2$ over $\overline{\mathbf{F}_p}$. Charles has shown that there are also infinitely many such primes.
  • If $\operatorname{End}(\overline{A}) = \mathbf{Z}$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $\overline{A}_p \sim E^2$. What do you think it should happen in this case?

O or SO?

  • Galois acts on the Tate-twisted $\ell$-adic realisation $T_{\ell}(1)$, where the cup product is an orthogonal pairing.

$$\tau : \operatorname{Gal}(\overline{K}/K) \longrightarrow O(T_{\ell}(1)).$$

  • There is no reason for the image to sit inside $SO(T_{\ell}(1))$.
  • When it does not, $\det \tau$ is a non-trivial quadratic character and the image meets both determinant components of $O(T_{\ell}(1))$.

The determinant records only a quotient of order two; the monodromy group itself can have more components than two.

The jump criterion on the slides that immediately follow is a consequence of that one fact.

What $\det = -1$ costs you

Let $\varphi = \operatorname{Frob}_p$ on $T(1)$, and suppose $\det \varphi = -1$.

  1. Every eigenvalue has absolute value 1, so those other than $\pm 1$ come in conjugate pairs $\lbrace z, \bar z \rbrace$ with $z \bar z = 1$.
  2. Those pairs contribute $+1$ to the determinant, so some eigenvalue must be $-1$.
  3. If $\operatorname{rk}\operatorname{Pic} X_{\overline{K}}$ is even, so is $\dim T = 22 - \operatorname{rk}\operatorname{Pic} X_{\overline{K}}$; then one more eigenvalue is forced, and it is $+1$.
  4. Both are roots of unity, so both are new Tate classes. The rank jumps by at least 2.

Discriminant of a K3 surface

Let $X$ be a quartic K3 surface over $\mathbf{Q}$ and $D_X$ be its discriminant.

Theorem (Costa-Elsenhans-Jahnel)

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

Assume $\operatorname{NS}(X/k) = \operatorname{NS}(X/\bar k)$, so the Picard representation is trivial, $\Delta_{\operatorname{Pic}} = 1$, and a single discriminant $D_X$ does both jobs.

Theorem

If $\rho(\overline{X}) = 2r$, then $\rho(\overline{X}_p) \geq 2r + 2$ at every prime $p$ of good reduction with $p \nmid 2 D_X$ at which $D_X$ is not a square mod $p$.

We can explain the $1/2$

Theorem (C, C-Elsenhans-Jahnel)

If $\rho(\overline{X}) = \min_p \rho(\overline{X}_p)$, then there is $d_X \in \mathbf{Z}$ such that:

$$\bigl\lbrace p > 2 : p \text{ inert in } \mathbf{Q}(\sqrt{d_X}) \bigr\rbrace \subset \Pi_{\mathrm{jump}}(X).$$

$d_X$ represents the quadratic character $p \mapsto \det(\operatorname{Frob}_p | T(X)(1)) \in \lbrace \pm 1 \rbrace$.

Corollary

If $d_X$ is not a square:

  • $\liminf_{B \rightarrow \infty} \gamma(X,B) \geq 1/2$
  • $\overline{X}$ has infinitely many rational curves.

$d_{X_3} = -1 \cdot 5 \cdot 151 \cdot 22490817357414371041 \cdot 387308497430149337233666358807996260780875056740850984213276970343278935342068889706146733313789$

Computing $\rho(\overline{X})$

Let $T_X$ be the orthogonal complement of $\operatorname{NS}(X_{\mathbf{C}}^{\mathrm{top}})$ in $H^2(X_{\mathbf{C}}^{\mathrm{top}}, \mathbf{Q})$.
Let $E_X$ be the endomorphism algebra of $T_X$ that respects the Hodge structure.
$E_X$ is a totally real field or a CM-field.

Theorem (Charles)

$$\rho(\overline{X}_p) \geq \begin{cases} \rho(\overline{X}) & \text{if } E_X \text{ is CM or } \dim_{E_X}(T_X) \text{ is even,} \\ \rho(\overline{X}) + [E_X:\mathbf{Q}] & \text{if } E_X \text{ is totally real and } \dim_{E_X}(T_X) \text{ is odd.} \end{cases}$$

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

$$\operatorname{disc}(\operatorname{NS}(X_p)) \not\equiv \operatorname{disc}(\operatorname{NS}(X_q)) \bmod \mathbf{Q}^{\times 2}$$

When every prime overshoots

Forced excess is a different question, and there are exactly two cases with $\eta(\overline{X}) > 0$, so that every good reduction has rank strictly above $\rho(\overline{X})$:

  1. $E_X = \mathbf{Q}$ with $\dim T_X$ odd: $\eta = 1$, van Luijk survives;
  2. $E_X$ totally real, $E_X \neq \mathbf{Q}$, $\dim_{E_X} T_X$ odd: excess $[E_X:\mathbf{Q}] \geq 2$, a genuine wall.

And these are the only two, by a theorem of Charles.

Only the second is a wall. In case 1 parity hands the excess straight back: the quartic on slide 12 has $\rho(\overline{X}) = 17$, every good prime overshoots to 18 or more, and cutting 18 to below 18 gives the upper bound 17.

Case 2 is different. The excess is $[E_X:\mathbf{Q}] \geq 2$, nothing tells you it is exactly that, and no pair of primes cuts far enough on reduction data alone.

A surface where that happens

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

  • $X$ is the minimal desingularization of this double cover: the branch locus is a union of six lines, so the cover is singular at their 15 intersection points.
  • A double cover of $\mathbf{P}^2$, $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 16$, RM by $\mathbf{Q}(\sqrt{2})$. The known classes are the polarization and the 15 exceptional curves.
  • Case 2 exactly: $(22 - 16)/2 = 3$, odd. Reduction will overshoot at every good prime, and unaugmented reduction, van Luijk's two-prime argument included, stalls here; only the certified RM of slide 23 cuts further.