Where we got to yesterday

Reduction gives

$$\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p) \quad \text{for every good } p.$$

  • Jumping is frequent: a non-trivial jump character makes a density-one-half set of primes overshoot.
  • Forced excess is different: every good prime overshoots. Exactly two cases:
    • $E_X = \mathbf{Q}$, $\dim T_X$ odd: forced excess $\eta = 1$, van Luijk survives;
    • $E_X$ totally real, $E_X \neq \mathbf{Q}$, $\dim_{E_X} T_X$ odd: excess $[E_X:\mathbf{Q}] \geq 2$, a wall.

Only the second defeats the unaugmented two-prime argument.

Picard lattice, over finite fields

Tate conjecture

$$\operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}_\ell} = \ker\left( \operatorname{Frob}_p - p \cdot \operatorname{id} \mid H^2_{\mathrm{et}}(\overline{X}_p, \mathbf{Q}_\ell)\right)$$

Tate conjecture is known for $d \leq 4$ over finite fields.

The Hasse-Weil zeta function $Z_X(t)$ for a surface $X$ can be written as

$$Z_X(t) := \exp\left( \sum_{m=1}^{\infty} \frac{\# X(\mathbf{F}_{p^m})}{m} t^m \right) = \frac{1}{(1-t)\, \chi(t)\, (1-p^2 t)},$$

where $\chi(t) := \det(1 - t \operatorname{Frob} \mid H^2_{\mathrm{et}}(\overline{X}_p, \mathbf{Q}_\ell)) \in \mathbf{Z}[t]$. One may deduce $\chi$ by naively computing $\# X(\mathbf{F}_{p^m})$ for $m \leq b_2/2 + 1$.

Since $\operatorname{Frob}_p$ acts semisimply, we have:

$$\rho\bigl(X_{\mathbf{F}_{p^n}}\bigr) = \#\{ z : \chi(1/z) = 0 \text{ and } z^n = p^n \}.$$

Note: $\rho(\overline{X}_p) \equiv b_2 \bmod 2$

For $p > 7$ computing $\chi(t)$ by naive point counting is not practical. Instead, one relies in a infrastructure of methods in crystalline cohomology [Abbott-Kedlaya-Roe, C, C-Harvey-Kedlaya, Tuitman-Pancratz]

Reduction to finite characteristic

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

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

Theorem

If $X$ and $X_{\mathbf{F}_p}$ are smooth then $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = \rho(X_{\mathbf{Q}_p^{\mathrm{al}}}) \leq \rho(\overline{X}_p)$.

Goal

For a given $f$ and $p$, improve the inequality above.

Idea, try to lift algebraic cycles (curves) from $\mathbf{F}_p^{\mathrm{al}}$ to $\mathbf{Q}_p^{\mathrm{al}}$.

We will do this by considering the thickenings

$$Z(f \bmod p^i) \subset \mathbf{P}^3_{\mathbf{Z}/(p)^i} \quad i = 1, 2, \ldots$$

The lifting question

Which classes in $\operatorname{Pic}(\overline{X}_p)$ actually come from $X$?

  • Every excess counted yesterday is a class that exists in the special fibre and has no ancestor upstairs.
  • 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

For simplicity, assume that all curve classes are defined over the base field, i.e.,

$$\rho(X) = \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \quad \text{and} \quad \rho(X_{\mathbf{F}_p}) = \rho(\overline{X}_p)$$

Over characteristic zero we have:

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

Over characteristic $p$ we have:

  • $\operatorname{Pic}(X_{\mathbf{F}_p}) \hookrightarrow H^2_{\mathrm{crys}}(X_{\mathbf{F}_p}/\mathbf{Z}_p) \otimes \mathbf{Q}_p \simeq H^2_{\mathrm{dR}}(X/\mathbf{Q}) \otimes_{\mathbf{Q}} \mathbf{Q}_p = F^0_{\mathbf{Q}_p} \supset F^1_{\mathbf{Q}_p} \supset F^2_{\mathbf{Q}_p}$

Berthelot-Ogus-Raynaud

Theorem (Berthelot-Ogus, F-isocrystals and de Rham cohomology I, Invent. Math. 1983, §3; Raynaud 1979)

$$\operatorname{Pic}(X)_{\mathbf{Q}} = \operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}} \cap F^1_{\mathbf{Q}_p}$$

Geometric version, the one used below: replace the $p$-eigenspace by the span of the eigenspaces for all eigenvalues $\zeta p$, $\zeta$ a root of unity, i.e. all cyclotomic factors of $P_2(pT)$, and read the equality after the finite extension over which the classes are defined.

What Frobenius acts on

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

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

Tate over a finite field

Tate conjecture

$\operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}_p} = \ker\!\left( \operatorname{Frob}_p - p \cdot \operatorname{id} \ \big| \ H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) \right)$

The obstruction map [C-Sertöz]

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

$\pi : \operatorname{Pic}(X_{\mathbf{F}_p}) \subset H^2_{\mathrm{crys}}(X/\mathbf{Z}_p) \longrightarrow H^2_{\mathrm{crys}}(X/\mathbf{Z}_p) / F^1 H^2_{\mathrm{crys}}(X/\mathbf{Z}_p)$

If $\pi(C) \neq 0$, then $C \notin \operatorname{Pic}(X)$.   (analogous to $\operatorname{Pic}(X_{\mathbf{C}}) = H^{1,1}(X_{\mathbf{C}}) \cap H^2(X, \mathbf{Z})$)

  1. compute a $p$-adic approximation of $\operatorname{Frob}_p$
  2. compute an approximation of $\operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}_p} = \ker( \operatorname{Frob}_p - p \cdot \operatorname{id} \mid H^2_{\mathrm{dR}}(X/\mathbf{Q}_p))$
  3. compute an approximation of $\pi_{\mathbf{Q}_p} : \operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}_p} \rightarrow H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) / F^1 H^2_{\mathrm{dR}}(X/\mathbf{Q}_p)$
  4. $\dim \operatorname{Pic}(X) \leq \dim_{\mathbf{Q}_p} \ker \pi_{\mathbf{Q}_p}$

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.

What you actually compute

  • Frobenius is not known exactly.
  • It is computed modulo $p^N$, for a chosen precision $N$, by point counting and $p$-adic cohomology.
  • The machinery is Kedlaya-style, and not this course's subject.
  • 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.

Abelian surface

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

$\begin{aligned} \operatorname{Frob}|_{H^1_{\mathrm{dR}}(A/\mathbf{Q}_p)} \equiv{}& \begin{pmatrix} 31 \cdot 482 & 31 \cdot 284 & 16241 & 3075 \\ 31 \cdot 386 & 31 \cdot 886 & 2644 & 12126 \\ 31 \cdot 284 & 31 \cdot 659 & 6336 & 9750 \\ 31 \cdot 194 & 31 \cdot 876 & 27408 & 10841 \end{pmatrix} \pmod{31^3}, \\ L(t) ={}& \det(1 - t\operatorname{Frob} \mid H^1) = 1 - 3t + 14t^2 - 93t^3 + 961t^4. \end{aligned}$

From this we deduce $\operatorname{Frob}|_{H^2_{\mathrm{dR}}(A/\mathbf{Q}_p)}$ and

$\det(1 - t\,31^{-1}\operatorname{Frob} \mid H^2_{\mathrm{dR}}(A/\mathbf{Q}_p)) = (t-1)^2(31t^4 + 48t^3 + 43t^2 + 48t + 31)/31$

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

Since the basis of $H^1$ respects the Hodge filtration, the induced basis in $H^2$ will also respect it.

Abelian surface, continued

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

$\det(1 - t\,31^{-1}\operatorname{Frob} \mid H^2_{\mathrm{dR}}(A/\mathbf{Q}_p)) = (t-1)^2(31t^4 + 48t^3 + 43t^2 + 48t + 31)/31$

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

Compute 2 eigenvectors

$\begin{aligned} v_1 \equiv{}& \left(356,\,37,\,831,\,0,\,295,\,31\right) \pmod{31^2} \\ v_2 \equiv{}& \left(4,\,957,\,3,\,1,\,0,\,0\right) \pmod{31^2}. \end{aligned}$

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.

K3 surface

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

sage: crystalline_obstruction(f, p=89, precision=3)
(4,
 {'rank T(X_Fpbar)': 10,
  'factors': [(t - 1, 1), (t + 1, 1), (t - 1, 4), (t^4 + 1, 1)],
  'dim Ti': [1, 1, 4, 4],
  'dim Li': [1, 0, 3, 0]},
  'precision': 3, 'p': 89})
  • $\rho(X_{\mathbf{F}_{89}^{\mathrm{al}}}) = 10$
  • $\operatorname{Pic}(X_{\mathbf{F}_{89}^{\mathrm{al}}})$ decomposes as $P_{\zeta_1} \oplus P_{\zeta_2} \oplus P_{\zeta_8}$
  • By studying each factor independently, we show $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq 4$
  • In fact, $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 4$ as there are four lines in $z = 0$.
  • previous approaches would not have used $p = 89$

Quartic surface

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

sage: crystalline_obstruction(f, p=31, precision=5)
(4,
 {'rank T(X_Fpbar)': 4,
  'factors': [(t - 1, 1), (t - 1, 1), (t + 1, 2)],
  'dim Ti': [1, 1, 2],
  'dim Li': [1, 1, 2]},
  'precision': 5, 'p': 31})
  • $\rho(X_{\mathbf{F}_{31}^{\mathrm{al}}}) = 4$
  • no cycle obstruction found while working $\mathbf{Z}/(p)^5$
  • $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq 4$, with some extra confidence that the equality might hold.
  • by searching for lines Elsenhans-Jahnel's method would have succeeded in this example

Quintic surface

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

sage: crystalline_obstruction(f, p=23, precision=6)
(1, {'rank T(X_Fpbar)': 5,
     'factors': [(t - 1, 1), (t - 1, 1), (t + 1, 1), (t^2 + 1, 1)],
     'dim Ti': [1, 1, 1, 2],
     'dim Li': [1, 0, 0, 0],
    'precision': 6, 'p': 23})
sage: crystalline_obstruction(f, p=29, precision=20)
(3, {'rank T(X_Fpbar)': 5,
     'factors': [(t - 1, 1), (t - 1, 2), (t + 1, 2)],
     'dim Ti': [1, 2, 2],
     'dim Li': [1, 1, 1]})
     'precision': 20, 'p': 29})
  • $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 1$
  • However, we cannot deduce this from $p = 29$, not even with infinite precision.
  • The surface has $CM$ by $\mathbf{Q}(\zeta_5)$

What the three examples say

Three outcomes, and they are different in kind:

  • at $p = 89$ the obstruction improves 10 to 4, and 4 is sharp;
  • at $p = 31$ the reduction bound is already the sharp value 4, so there is nothing left to remove;
  • 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.

So the open question is not whether some prime works, but whether one can prove that some prime works.

What is being computed, exactly

Can we combine both approaches?

  • At the moment we are only computing an approximation of $\operatorname{Pic}(X)_{\mathbf{Q}_p}$.
  • To combine several primes we need at least $\operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}}$, to be able to use $\operatorname{Pic}(X)_{\mathbf{Q}} = \operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}} \cap F^1_{\mathbf{Q}_p}$ in its full strength.
  • At the moment we are only using $\operatorname{Pic}(X)_{\mathbf{Q}_p} \subset \operatorname{Pic}(X_{\mathbf{F}_p})_{\mathbf{Q}_p} \cap F^1_{\mathbf{Q}_p}$

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

Theoretical example

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

  • $X$ is the minimal resolution of this double cover of $\mathbf{P}^2$; the sextic is a product of three conics, singular at 15 points.
  • Known sublattice: the polarization and the 15 exceptional curves, so $\rho \geq 16$.
  • At $p = 83$ it has $\chi_1 = (t-1)^{10}(t+1)^6$, and the reduction bound leaves $\rho = 16, 17$ or $18$.
  • The two extra classes span the single $\mathbf{Q}$-irreducible piece $t^2 + 1$, so 17 is out: $\rho = 16$ or $18$. In fact $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 16$, and $X$ has RM by $\mathbf{Q}(\sqrt{2})$.
  • This is the wall from Lecture 1: RM by a field of degree 2 with $(22 - 16)/2 = 3$ odd, so every good prime overshoots.
  • Over $\mathbf{Q}$ the two extra classes at $83$ are Galois conjugate: obstruct one and both go, so the bound would be $16$. Over $\mathbf{Q}(\sqrt{2})$, where the RM is defined, they are not, and one always survives: $17$ at best.

"Given a good enough approximation to $\operatorname{Frob}_p$ we will obstruct 2 extra cycles. However, we don't yet know how to compute such approximation!"

The question

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

  • In general, no.
  • For example, take a K3 surface $X$ with real multiplication, defined over a number field where all the algebraic cycles in $X$ and $X \times X$ are defined.
  • However, we are hopeful for K3 surfaces and abelian 3folds defined over $\mathbf{Q}$.

Not a claim that the method always works, and not a claim that it cannot be made to.