Computing Picard Lattices of K3 Surfaces

Where we got to yesterday

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

Jumping is frequent; extra endomorphisms complicate the Kloosterman-van Luijk approach.

Picard lattice, over finite fields

$$\det(1 - t\operatorname{Frob}_q \mid H^2_{\mathrm{et}}(X_p^{\mathrm{al}}, \mathbf{Q}_\ell)) = t^{22}P_2(1/t).$$

Reduction to finite characteristic

In this lecture we will focus on projective hypersurfaces, with quartic K3s in mind, but the methods are more generic, as at some point we will need to do explicit computations.

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

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

Yesterday, we mostly counted classes, but we also saw the cokernel theorem.

Can we use it without computing $\operatorname{Pic}(X_p^{\mathrm{al}})$?

The lifting question

Which classes in $\operatorname{Pic}(X_p^{\mathrm{al}})$ actually come from $X$?

• Every excess counted yesterday is a class that exists in the special fibre and doesn't lift.
• Reduction ranks cannot see the difference: they count the special fibre and nothing else.

Not "which prime" but "which classes": lifting, not counting.

1st ingredient: cohomology

Choose a finite extension $K/\mathbf{Q}_p$, with residue field $k=\mathbf{F}_{p^m}$, so $\rho(X_{K})=\rho(X^{\mathrm{al}})$ and $\rho(X_{p,k})=\rho(X_p^{\mathrm{al}})$.

Over characteristic zero we have:

• $H:=H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) = F^0 \supset F^1 \supset F^2$, the Hodge filtration
• $\operatorname{Pic}(X^{\mathrm{al}})_{\mathbf{Q}} \hookrightarrow F^1_{K}:=F^1\otimes_{\mathbf{Q}_p}K$
• For $d = 4$, $\dim F^i(X) = 22, 21, 1$.

Over characteristic $p$ we have:

• $\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}} \hookrightarrow H^2_{\mathrm{crys}}(X_p\times_{\mathbf{F}_p}k/W(k))\otimes_{W(k)}K \simeq H\otimes_{\mathbf{Q}_p}K=:H_{K}=F^0_{K}\supset F^1_{K}\supset F^2_{K}$

Berthelot-Ogus-Raynaud

Let $C \in \operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}$.

Classes on $X_p^{\mathrm{al}}$ need not lie in $F^1_{K}$.

Classes from characteristic zero must lie in $F^1_{K}$.

Abelian surface

$A = \operatorname{Jac}(y^2 = 4x^5 - 36x^4 + 56x^3 - 76x^2 + 44x - 23)$

Abelian surface, continued

SageMath package: crystalline_obstruction.

$A = \operatorname{Jac}(y^2 = 4x^5 - 36x^4 + 56x^3 - 76x^2 + 44x - 23)$

$\det(t - 31^{-1}\operatorname{Frob} \mid H^2_{\mathrm{dR}}(A/\mathbf{Q}_p)) = (t-1)^2(31t^4 - 6t^3 + 26t^2 - 6t + 31)/31$

Thus, $\rho\bigl(A_{\mathbf{F}_p^{\mathrm{al}}}\bigr) = 2$.

Compute 2 eigenvectors

$\begin{aligned} v_1 \equiv{}& \left(0,\,839,\,568,\,649,\,1,\,372\right) \pmod{31^2} \\ v_2 \equiv{}& \left(1,\,136,\,618,\,206,\,0,\,0\right) \pmod{31^2}. \end{aligned}$

The last coordinate of the vectors above gives the projection to $H^2/F^1$.

Therefore, $v_1 \notin F^1$ and the corresponding algebraic cycle cannot lift to $\mathbf{Q}_p$.

Thus, we improved $\operatorname{rank} \operatorname{NS}(A_{\mathbf{Q}^{\mathrm{al}}}) \leq 2$ to $\operatorname{rank} \operatorname{NS}(A_{\mathbf{Q}^{\mathrm{al}}}) \leq 1$, and therefore $\operatorname{End}(A_{\mathbf{Q}^{\mathrm{al}}}) = \mathbf{Z}$.

van Luijk's method would have succeeded in this example by using a second prime.

What Frobenius acts on

Via the isomorphism $H^2_{\mathrm{crys}}(X_p/\mathbf{Z}_p) \otimes \mathbf{Q}_p \simeq H^2_{\mathrm{dR}}(X/\mathbf{Q})$, we have

$$\operatorname{Frob}_p : H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) \rightarrow H^2_{\mathrm{dR}}(X/\mathbf{Q}_p).$$

The obstruction map [C-Sertöz]

Compute a $p$-adic approximation of the obstruction map

$\pi : \ker(\operatorname{Frob}_p^m-p^m\operatorname{id}\mid H) \subset H \longrightarrow H/F^1$

If $\pi(C) \neq 0$, then $C \notin \operatorname{sp}\bigl(\operatorname{Pic}(X^{\mathrm{al}})_{\mathbf{Q}}\bigr)$.   (cf. $\operatorname{Pic}(X_{\mathbf{C}}) = H^{1,1}(X_{\mathbf{C}}) \cap H^2(X, \mathbf{Z})$)

1. compute a $p$-adic approximation of $\operatorname{Frob}_p$
2. compute an approximation of $\ker(\operatorname{Frob}_p^m-p^m\operatorname{id}\mid H) = \bigl(\bigoplus_i\ker\Phi_i(p^{-1}\operatorname{Frob}_p)\bigr)$, where $\Phi_i$ runs over the distinct cyclotomic factors of $p^{-22}P_2(pt)$.
3. compute an approximation of $\pi : \ker(\operatorname{Frob}_p^m-p^m\operatorname{id}\mid H) \rightarrow H/F^1$
4. $\rho(X^{\mathrm{al}}) \leq \dim_{\mathbf{Q}_p} \ker \pi_{\mathbf{Q}_p}$

By picking a basis that respects the Hodge filtration, the map $H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) \rightarrow H^2_{\mathrm{dR}}(X/\mathbf{Q}_p)/F^1_{\mathbf{Q}_p}$ is a coordinate projection;
$\pi_{\mathbf{Q}_p}=\pi$ is its restriction to $\ker(\operatorname{Frob}_p^m-p^m\operatorname{id}\mid H)$.

What you actually compute

• Frobenius is not known exactly.
• Today we use a $p$-adic toric approach to compute Frobenius modulo $p^N$, for a precision $N$.
• For a hypersurface, use a basis of differential forms on its affine complement.
• Apply Frobenius and reduce its images back to the basis to obtain the matrix modulo $p^N$.
• What matters here is the shape of the output: an approximation whose error you control.

Why finite precision still proves something

• An approximate Frobenius gives an approximate eigenspace.
• But $\pi(C) \neq 0$ is an open condition.
• Establishing it to finite precision establishes it.
• Every dimension the method removes is removed rigorously: a genuine upper bound at any $N$.

Raising $N$ can only remove more. It never puts a dimension back.

Making the bound sharp

$X := Z(y^4 - x^3 z + y z^3 + z w^3 + w^4) \subset \mathbf{P}^3_{\mathbf{Q}}$

$\rho(X_{89}^{\mathrm{al}})=5+1+4=10,\qquad \rho(X_{\mathbf{Q}^{\mathrm{al}}})\leq (1+3)+0+0=4$

In fact, $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 4$ as there are four lines in $z = 0$.

previous approaches would not have used $p = 89$

at $p = 89$ the obstruction improves 10 to 4, and 4 is sharp; Does this always work?

The bound is already sharp

$X = Z(y^4 - x^3 z + y z^3 + z w^3 + w^4) \subset \mathbf{P}^3_{\mathbf{C}}$

$\rho(X_{31}^{\mathrm{al}}) = 4$

no cycle obstruction found while working $\mathbf{Z}/(p)^5$

by searching for lines Elsenhans-Jahnel's method would have succeeded in this example

at $p = 31$ the reduction bound is already the sharp value 4, so there is nothing left to remove;

The bound stays too high

$X := Z(9 x y^{4} + 3 x^{4} z + 9 y^{2} z^{3} + z^{5} + 5 w^{5}) \subset \mathbf{P}^3$

At $p = 29$: $5 \to 3$, but no further at any precision.

The involution permits only one condition per $2$-dimensional primitive Frobenius eigenspace.

What the three examples say

Three outcomes, and they are different in kind:

1. at $p = 89$ the obstruction improves 10 to 4, and 4 is sharp;
2. at $p = 31$ the reduction bound is already the sharp value 4, so there is nothing left to remove;
3. at $p = 29$ on the quintic the obstruction fires, improving 5 to 3, but 3 is not sharp and no precision brings it to 1.

Only the third is a statement about the method rather than about an attempt.

The difficulty is proving that some prime works, not finding one.

Does this always work?

No.

Is there a prime for which the bound will be tight?

What is being computed, exactly

Can we combine both approaches?

• At the moment we only certify an upper bound for $\dim_{\mathbf{Q}_p}L$, where $L$ is the largest $\operatorname{Frob}_p$-stable subspace of $H\cap\bigl(\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}\otimes_{\mathbf{Q}}K\bigr)\cap F^1$.
• To combine several primes we need at least $\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}$, to be able to use $\operatorname{sp}\bigl(\operatorname{Pic}(X^{\mathrm{al}})_{\mathbf{Q}}\bigr)=\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}\cap F^1_{K}$ in its full strength.
• At the moment we are only using $\operatorname{sp}\bigl(\operatorname{Pic}(X^{\mathrm{al}})\bigr)\otimes_{\mathbf{Z}}K\subseteq L\otimes_{\mathbf{Q}_p}K\subseteq\bigl(\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}\otimes_{\mathbf{Q}}K\bigr)\cap F^1_{K}$

Raising $N$ does not touch the missing rational structure.

We would also love to combine this with methods over $\mathbf{C}$.

Theoretical example

• The real multiplication example from Lecture 1.
• Known sublattice: the polarization and the 15 exceptional curves, so $\rho(X^{\mathrm{al}})\geq16$.
• The two extra classes span the piece $t^2+1$, irreducible over $\mathbf{Q}_{83}$, so 17 is out: $\rho(X^{\mathrm{al}})=16$ or $18$. In fact $\rho(X_{\mathbf{Q}^{\mathrm{al}}})=16$, and $X$ has RM by $\mathbf{Q}(\sqrt{2})$.
• An obstruction on the $t^2+1$ block would prove $\rho(X_{\mathbf{Q}^{\mathrm{al}}})=16$, without knowing RM.
• After base change to $\mathbf{Q}(\sqrt{2})$, the residue degree at $83$ is $2$. Using only $F^2$, where $F=83^{-1}\operatorname{Frob}_{83}$, gives $F^2=-1$ on the extra piece. A kernel line is Frobenius-stable and survives: the bound is at least $17$.

Questions the examples raise

Is there a prime for which the bound will be tight?

• Does it give sharp bounds? Yes: the quartic at $p=89$, from $10$ to $4$.
• Does it always improve the reduction bound? No: at $p=31$ the quartic already has the sharp bound $4$.
• Does more precision always make the bound sharp? No: the quintic at $p=29$ stays at $3$, although its Picard number is $1$.
• Can base change lose sharpness? Yes: the real multiplication example at $83$ would give $16$ over $\mathbf{Q}$, but at least $17$ using only the squared Frobenius over $\mathbf{Q}(\sqrt{2})$.

Is there always a sharp prime over $\mathbf{Q}$?

What conditions on RM would we need for this not to fail?