One may deduce P2 from the point counts # Xp(𝔽qm) for m ≤b2/2 + 1 = 12.
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 ∈ℤ[x,y,z,w] and X := Z(f) ⊂ℙ3ℤ.
We may consider the surface X𝔽p := Z(f mod p) ⊂ℙ3(𝔽p).
Yesterday, we mostly counted classes, but we also saw the cokernel theorem.
Theorem
If X and X𝔽p are smooth then the specialization map is injective
Pic (Xℚal) ↪Pic (X𝔽pal)
and ρ(Xℚal) = ρ(Xℚpal) ≤ρ(X𝔽pal).
The specialization map has torsion-free cokernel for p ≠2.
Can we use it without computing Pic (X𝔽pal)?
The lifting question
Goal
For a given f and p, improve the inequality ρ(Xℚal) ≤ρ(Xpal).
Which classes in Pic (Xpal) 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.
We will do this by considering the thickenings
Z(f mod pi) ⊂ℙ3ℤ/(p)i i = 1, 2, ...
Not "which prime" but "which classes": lifting, not counting.
1st ingredient: cohomology
Choose a finite extension K'/ℚp, with residue field k'=𝔽pm, so the geometric divisor classes of X are defined over K' and those of Xp over k'.
The last coordinate of the vectors above gives the projection to H2/F1. Therefore, v1 ∉F1 and the corresponding algebraic cycle cannot lift to ℚp.
Thus, we improved rank NS (Aℚal) ≤2 to rank NS (Aℚal) ≤1, and therefore End (Aℚal) = ℤ.
van Luijk's method would have succeeded in this example by using a second prime.
What Frobenius acts on
Via the isomorphism H2crys(X𝔽p/ℤp) ⊗ℚp ≃H2dR(X/ℚ), we have
Frob p : H2dR(X/ℚp) →H2dR(X/ℚp).
The obstruction map [C-Sertöz]
Compute a p-adic approximation of the obstruction map
π: Pic (X𝔽p) ⊂H2crys(X/ℤp) ⟶H2crys(X/ℤp) / F1 H2crys(X/ℤp)
If π(C) ≠0, then C ∉Pic (X). (analogous to Pic (Xℂ) = H1,1(Xℂ) ∩H2(X, ℤ))
compute a p-adic approximation of Frob p
compute an approximation of Pic (X𝔽p)ℚp = ker ( Frob p − p ·id ∣H2dR(X/ℚp))
compute an approximation of πℚp : Pic (X𝔽p)ℚp →H2dR(X/ℚp) / F1 H2dR(X/ℚp)
dim Pic (X) ≤dim ℚp ker πℚp
By picking a basis that respects the Hodge filtration, the map H2dR(X/ℚp) →H2dR(X/ℚp)/F1ℚp is a coordinate projection.
What you actually compute
Frobenius is not known exactly.
Today we use a p-adic toric approach to compute Frobenius modulo pN, for a chosen 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 pN.
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 π(C) ≠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.
K3 surface
X := Z(y4 − x3 z + y z3 + z w3 + w4) ⊂ℙ3ℚ
p=89, N=3, F=89−1Frob 89
+1-eigenspace, polarization: dimension 1. No obstruction; this class lifts.
+1-eigenspace, primitive part: dimension 4. Obstruction found; at most 3 dimensions remain.
−1-eigenspace: dimension 1. Obstructed; no dimension remains.
Primitive eighth-root piece ker (F4+1): dimension 4. Obstructed; no dimension remains.
ρ(X𝔽89al)=5+1+4=10, ρ(Xℚal)≤(1+3)+0+0=4
In fact, ρ(Xℚal) = 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?
Quartic surface
X = Z(y4 − x3 z + y z3 + z w3 + w4) ⊂ℙ3ℂ
p=31, N=5
factor of p−1Frob p
dimension
liftable dimension, at most
t−1
1
1
t−1
1
1
(t+1)2
2
2
ρ(X𝔽31al) = 4
no cycle obstruction found while working 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;
Quintic surface
X := Z(9 x y4 + 3 x4 z + 9 y2 z3 + z5 + 5 w5) ⊂ℙ3
p=23, N=6, ρ(Xℚal) ≤1
factor of p−1Frob p
dimension
liftable dimension, at most
t−1
1
1
t−1
1
0
t+1
1
0
t2+1
2
0
ρ(Xℚal) = 1
p=29, N=20, ρ(Xℚal) ≤3
factor of p−1Frob p
dimension
liftable dimension, at most
t−1
1
1
(t−1)2
2
1
(t+1)2
2
1
The surface has CM by ℚ(ζ5)
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.
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.
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 ℚpL, where L is the largest Frob p-stable subspace of Tev∩F1.
To combine several primes we need at least Pic (Xpal)ℚ, to be able to use
sp (Pic (Xal)ℚ)=Pic (Xpal)ℚ∩F1K'
in its full strength.
At the moment we are only using
sp (Pic (Xal))⊗ℤK'⊆L⊗ℚpK'⊆(Tev⊗ℚpK')∩F1K'
Raising N does not touch the missing rational structure.
We would also love to combine this with methods over ℂ.
Theoretical example
X : w2 = (−y2/8 + yz − z2)(7x2/8 + 5xz + 7z2)(2x2 + 3xy + y2)
The real multiplication example from Lecture 1: X is the minimal resolution of this double cover of ℙ2; the sextic is a product of three conics, singular at 15 points.
Known sublattice: the polarization and the 15 exceptional curves, so ρ(Xal)≥16.
At p = 83 it has χ1 = (t−1)10(t+1)6, and the reduction bound leaves ρ(Xal)=16,17 or 18.
The two extra classes span the piece t2+1, irreducible over ℚ83, so 17 is out: ρ(Xal)=16 or 18. In fact ρ(Xℚal)=16, and X has RM by ℚ(√(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 ℚ, a nonzero obstruction on the t2+1 piece would remove both extra dimensions and prove ρ(Xℚal)=16, without knowing RM.
After base change to ℚ(√(2)), the residue degree at 83 is 2. Using only F2, where F=83−1Frob 83, gives F2=−1 on the extra piece. A kernel line is Frobenius-stable and survives: the bound is at least 17.
"Given a good enough approximation to 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?
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 ℚ, but at least 17 using only the squared Frobenius over ℚ(√(2)).