Picard Lecture 2: written out

1 Where we got to yesterday

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Reduction gives ρ(X_Qal) ≤ ρ(X̄_p)  for every good p, and Lecture 1 closed by separating two things. Jumping is frequent: a non-trivial jump character makes a density-one-half set of primes overshoot, so half your primes are wasted, but the bound is still attainable at the others. Forced excess is different: there are exactly two cases in which every good prime overshoots. Either E_X = Q with dim T_X odd, where the forced excess is η = 1 and van Luijk survives; or E_X totally real with E_X ≠ Q and dim_{E_X} T_X odd, where the excess is [E_X:Q] ≥ 2 and no better prime helps. Odd geometric Picard rank alone is not the first case: totally real E_X of odd degree ≥ 3 also gives odd rank, and is a wall. Every good prime overshoots in both cases; what separates them is that only the second defeats the unaugmented two-prime argument.

>>[minima-uv3.209] AUTHOR (2026-09-15): Slide 1 we should put the theorem of Charles, as we had in Slide 17. And then add the line saying, that is very hard to prove RM! We are now not only asking to find algebraic cycles in X but also in X x X.

[C] Concrete form. Lecture 1 deck position 17 is artifacts/draft/s16.html, "Computing rho(X^{al})": the notation line T = T(X)_Q, E = End_Hdg(T), d = [E:Q], m = dim_E T, then

with "Equality occurs infinitely often (density 1 after some finite extension)." The closing line to add: proving RM is very hard, because E acts on T through Hodge correspondences, so certifying it asks for algebraic cycles on X x X and not only on X. That is what makes the second case a wall rather than a nuisance, and it is why slide 19's example needs Elsenhans-Jahnel's certified RM rather than a numerical guess.

Deck (governs). The visible slide reduces this to the specialization inequality, two nested bullets for the two kinds of excess, and the closing claim that only the totally real E_X ≠ Q case defeats the unaugmented two-prime argument.

[C] DECIDED (jury, 2026-09-15). On slide 1 show Charles's notation line, the two-case minimum, and the equality/density line, then add the author's sentence that proving RM is very hard because we seek algebraic cycles on X x X as well as on X. Leave the prime-pair discriminants and the Kloosterman-van Luijk corollary in Lecture 1.

What it does. Re-enters after the break, and re-draws the one distinction the whole lecture rests on: a prime that overshoots is a nuisance, a surface where every prime overshoots is a wall.

2 PROPOSED DROP: Picard lattice, over finite fields

O:L1108–1142

Reminder only. The Tate conjecture is a theorem for K3 surfaces over finite fields, so the geometric Picard rank is read off P₂(T) = det(T − Frob | H²), exactly as in Lecture 1 (a slide that calls it χ(t) means the same object): ρ(X̄_p) = #{ roots of P₂(pT) that are roots of unity, with multiplicity } = ∑i deg Φi,

the sum of the degrees of the cyclotomic factors Φi of the normalised polynomial, not the number of those factors. Slide 14 is the example to keep in mind: four factors there, and rank 10. Compressed from Lecture 1, where it was stated properly.

>>[minima-uv3.210] AUTHOR (2026-09-15): Slide 2, the notation should try to match of lecture 1, about Tate conjecture. I am okay to use chi instead, but we should not repeat the comment about harding to compute it, it should just be a recall. But in lecture 1, we never defined P_2 in terms of point counts, so it would be good to do it here, even if takes 2 slides now.

[C] Concrete form. Lecture 1 states it non-reciprocally on s05.html and s06.html: P_2(t) = det(t - Frob_q | H^2_et(X_p^{al}, Q_ell)) with q = p^n, the normalisation q^{-22} P_2(qt) monic with roots on the unit circle, and Tate as rho(X_p) = ord_{t=q} P_2(t), rho(X_p^{al}) = sum_zeta ord_{t=q.zeta} P_2(t). Both names are approved and mean different things (see minima-uv3.49, quoted on slide 6 below): P_2 is det(t - Frob), chi is its reciprocal t^22 P_2(1/t). The zeta function takes chi; the root-counting and the cyclotomic factorisations on slides 14 and 19 take P_2. Slide 2 needs both, correctly labelled, rather than a choice between them.

[C] The point-count definition Lecture 1 never gave. For a K3 surface b_1 = b_3 = 0, so #X_p(F_{q^n}) = 1 + q^{2n} + tr(Frob^n | H^2), and

[C] The traces determine chi by Newton's identities, and the functional equation cuts the range to m <= b_2/2 + 1 = 12, which is the bound the current deck slide already states. Expanding this is what earns the second slide: the deck compresses it to one sentence, "One may deduce chi by naively computing #X(F_{p^m}) for m <= b_2/2 + 1", with the zeta function displayed above it and nothing connecting the two.

[C] The line to cut. The deck slide's closing paragraph is "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]". Lecture 1 already says this on s05.html ("Away from small p, naive point counting is impractical, more on Kedlaya's talk"), so it comes off slide 2. Slide 10 of this lecture is where the p-adic machinery is motivated, and it is the home for the citation list if it is kept at all.

[C] Consequences. This expands slide 2 rather than dropping it, so the PROPOSED DROP below is rejected. The count goes the other way from the target: 20 spoken slides plus the title, and 21 if slide 2 takes two. The author's budget in artifacts/plan/plan.md is 45 minutes and roughly 19 slides, so the slides this comment adds have to be paid for by merges elsewhere. The first item under "Open on this lecture" is settled by this comment.

Deck (governs). The current slide uses the reciprocal polynomial χ(t) = det(1 − t Frob | H²), gives the zeta-function formula and finite-extension rank count, and states ρ(X̄_p) ≡ b₂ mod 2.

[C] DECIDED (jury, 2026-09-15). Use two objects, not two names: P_2(t) = det(t - Frob); chi(t) = t^22 P_2(1/t). chi belongs in the zeta function, P_2 in root counting and cyclotomic factors, and L(t) only in the abelian surface's H^1. Repair slides 2, 6, 8, 12, 14 and 19 consistently: remove slide 2's false parenthetical that chi and P_2 are the same object, rename slide 19's chi_1 when it is taken from a P_2 expression, and name slide 12's det(1 - t*31^{-1} Frob | H^2_dR). The objects have different jobs, so keep both; this is a cleanup, not a choice of one notation.

What it does. Recalls Tate and the zeta function before the obstruction.

PROPOSED DROP: Lecture 1 now establishes the finite-field Picard group, Tate, both polynomial normalisations and the p = 89 factorisation; slides 1, 3 and 8 retain the specialization, p-adic comparison and Tate-kernel facts this lecture needs. Removing this repetition gives 19 spoken slides.

3 Reduction to finite characteristic

O:L1205–1222

If X and XFp are smooth, then ρ(X_Qal) = ρ(X_Qpal) ≤ ρ(X̄_p).

The middle equality is the one to notice. It says nothing is lost by working p-adically rather than over Q, which is what makes a p-adic obstruction a legitimate tool rather than a weaker substitute.

Then the geometric device: a divisor class on the special fibre lifts, or fails to, through the thickenings Z(f mod p^i).

>>[minima-uv3.211] AUTHOR (2026-09-15): Slide 3, it was much better framed in Lecture 1, we have the map and say that the cokernel is free. And there we should try to empahsize. Yesterday, we mostly counted classes, but we also saw the cokernel theorem (do we really need p=!2), can we use it in a way that that doesnt require to compute the Pic(Xp^al)?

[C] Concrete form. Lecture 1 splits this over two slides. s07.html displays the map itself, Pic(X^{al}) ↪ Pic(X_p^{al}) and rho(X^{al}) <= rho(X_p^{al}), under smoothness of X and X_p, then a Goal box and two warnings (parity, endomorphisms of the transcendental lattice). s13.html, "Torsion-free cokernel", carries Theorem (Elsenhans-Jahnel): for X/Q a K3 and p > 2 of good reduction the specialization map has torsion-free cokernel, hence "if rho(X_p^{al}) = rho(X^{al}) every invertible sheaf lifts". This lecture's slide 3 currently prints only the rank inequality, with no map and no cokernel.

[C] On p != 2. Answered in Lecture 1's own record, not an open question: the published theorem is for a proper smooth scheme over a mixed-characteristic DVR with perfect residue field and ramification degree e < p - 1. Over Q, e = 1, so e < p - 1 is exactly p > 2. Source: Elsenhans-Jahnel, "The Picard group of a K3 surface and its reduction modulo p", Algebra & Number Theory 5 (2011), 1027-1040, Theorem 1.4 and Remarks 1.5(a), p. 1028. Over a number field the condition is e < p - 1 at the place, so it is a ramification hypothesis rather than a special role for 2.

[C] Open, and the author's. Whether the cokernel theorem can be used without computing Pic(X_p^{al}). Elsenhans-Jahnel's recipe on s13.html opens with "compute Pic(X_p^{al})", and that slide's own closing line concedes "This approach is only practical if one can compute Pic(X_p^{al}) and if the obtained estimates are low". The question is whether the torsion-free conclusion can be run off the rank alone, or off the obstruction of this lecture, rather than off the lattice.

Deck (governs). The slide begins with f ∈ Z[x,y,z,w] and X := Z(f), states the theorem under smoothness, asks to improve the inequality for the chosen f and p, and displays the thickenings Z(f mod p^i).

What it does. Restates specialization with its Q_p refinement and introduces the thickenings through which a divisor class must lift.

4 The lifting question

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Which classes in Pic(X̄p) 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, because they count the special fibre and nothing else.

So the question is not "which prime" but "which classes", and it is a question about lifting rather than about counting.

>>[minima-uv3.212] AUTHOR (2026-09-15): I see that some of my comments are already addressed in Slide 4. We should do this more smotthly, maybe a recall, and then what is the question of today.

[C] Concrete form. The overlap is real: deck slide 3 already closes with "Idea, try to lift algebraic cycles (curves) from F_p^{al} to Q_p^{al}" and the thickenings Z(f mod p^i), and slide 4 then asks the lifting question again and closes on "lifting, not counting". Section 2.1 currently runs four beats, recap (1), Tate recall (2), specialization plus thickenings plus the lifting idea (3), lifting question (4), and the lifting idea is stated twice across 3 and 4.

[C] The requested shape is two beats: one recall, then today's question. The recall carries what Lecture 1 established, the specialization map with its torsion-free cokernel, and Tate with P_2 by point counts; today's question is "which classes lift", stated once, with the thickenings as its picture. This comment therefore governs the arrangement of slides 1, 3 and 4 together, not slide 4 alone, and the two-slide expansion of slide 2 has to sit inside that same recall.

Deck (governs). The slide carries the question, two bullets separating special-fibre classes from ancestors upstairs, and the closing line "lifting, not counting".

What it does. Turns yesterday's dead end into today's question. Without it the cohomology in slide 5 arrives unmotivated.

5 1st ingredient: cohomology

O:L1224–1243

The Hodge filtration, introduced here rather than in Lecture 1, because this is the first place it does any work: F⁰ H²_dR ⊃ F¹ H²_dR ⊃ F² H²_dR.

For a K3 surface the graded pieces have dimensions 1, 20, 1, and F¹ is the part of codimension one, cut out by the holomorphic two-form.

>>[minima-uv3.225] AUTHOR (2026-09-15): let's drop teh simplication. Ask 5.6 sol and gpt-6 astra a way to fix it

[C] DECIDED. The line "For simplicity, assume that all curve classes are defined over the base field" comes off deck slide 5, and the lecture runs the geometric version throughout. This settles the framing of slides 5, 6, 8, 9, 14, 18 and 19 together; the per-eigenspace account on 14 and the Q versus Q(sqrt 2) contrast on 19 both depend on it and are now statable as written.

[C] The mathematics is already in this plan, it is the deck that hides it: slide 8 has the eventual Tate space T_ev with K' = Frac W(F_{p^m}), and slide 9 has the scalar extension to W(F_{p^k}) tensor Q_p. Both currently sit in speaker notes under their Deck lines. Making them visible is the fix, not new content. Concrete wording dispatched to gpt-5.6-sol and gpt-6-astra on the author's instruction.

Deck (governs). The slide assumes ρ(X) = ρ(X_Qal) and ρ(X_Fp) = ρ(X̄_p), gives the Picard embeddings in de Rham and crystalline cohomology, and writes dim F^i = 22, 21, 1.

What it does. Introduces the Hodge filtration and the crystalline comparison in the first place where they do work.

6 Berthelot–Ogus–Raynaud

O:L1240–1242

Over the base fields, for classes defined over Q_p and over F_p:

Pic(X)_Q = Pic(X_Fp)_Q ∩ F¹_Qp.

Geometric version, the one the K3 examples actually use: replace the p-eigenspace by the span of the eigenspaces for all eigenvalues ζp with ζ a root of unity, i.e. the eigenspaces of all cyclotomic factors of P₂(pT), and read the equality over the finite extension of the base fields over which those classes are defined.

Quoted, not unpacked. It is the whole method in one line: a class in the special fibre comes from characteristic zero exactly when it lies in the filtration step. Two things to keep honest. The eigenvalues in the examples below are −p and ζ₈p, so what is being bounded is Pic(X̄_p) and not Pic(X_Fp), which is why the geometric version is stated. And the equality carries a descent hypothesis: local liftability on its own does not put a class in Pic(X)_Q, so either assume the classes are defined over the base fields, or state the lifting locally and pass to a finite extension before speaking of geometric classes. Everything that follows is the work of making both sides of that intersection computable. PANEL FLAG. The panel could not establish what the "Raynaud 1979" half of this attribution supplies; the line-bundle lifting criterion itself is Berthelot–Ogus 1983, §3 (the 1978 reference is their book, not the criterion), and that has been corrected. The Raynaud citation is carried over from the source deck unchanged, pending the author saying what it is for.

>>[minima-uv3.214] AUTHOR (2026-09-15): we need to improve "Berthelot-Ogus-Raynaud", it is a bit hard to parse, we should perhaps use bullet points, and once again, we are mixing chi and P_2, we should stick to one

[C] Bullet skeleton. The slide is currently one theorem line followed by two dense paragraphs. The six things it says, in order:

[C] The last four are speaker-note material under the current Deck line; the author's request is that whatever stays visible be bulleted rather than prose.

[C] ALREADY SETTLED, and this supersedes the "stick to one" reading. Bead minima-uv3.49 records the author's approved convention across all three lectures: P_2(t) = det(t - Frob) and chi(t) = t^22 P_2(1/t) for K3, alongside Pic, explicit Pic/Pic^0 for abelian varieties, superscript al for varieties and fields, and capital T for the transcendental lattice. The two symbols are not alternatives to choose between; they name two different objects, the non-reciprocal characteristic polynomial and its reciprocal. So the fix is not to pick one, it is to use each for its own object and never for the other's.

[C] Notation inventory, measured against that convention. P_2 is used on slides 2, 6, 8, 12, 14 and 19. chi is used on the deck's slide 2 and on slide 19. Two places are actually wrong under the convention, not merely mixed. Slide 2's plan text says "P_2(T) = det(T - Frob | H^2) ... (a slide that calls it chi(t) means the same object)", which is false: they are reciprocals of each other, not the same object, and that sentence has to go. Slide 19 writes "its characteristic polynomial is chi_1 = (t-1)^10 (t+1)^6, the factor of p^{-22} P_2(pT)", which names a chi but takes it from a P_2 expression; one of the two has to change so the object and its name agree. A third convention sits on slide 12, where L(t) = det(1 - t Frob | H^1) is reciprocal and the H^2 line prints det(1 - t*31^{-1} Frob | H^2_dR) with no name at all. Applying the approved convention means fixing all three spots, not just this slide. The author's comment above asks for the slide's prose to be bulleted; the notation half of it is an already-decided cleanup, tracked as minima-uv3.49, not a fresh choice.

Deck (governs). The visible theorem includes the geometric-version paragraph but leaves the descent and eigenvalue qualifications to the speaker notes.

What it does. The room should leave knowing this statement and not its proof. It is the only thing black-boxed in the lecture.

>>[minima-uv3.215] AUTHOR (2026-09-15): Unclear to me what is the goal of Slude 7: "What Frobenius acts on". And why are we repeating Slude 8 again Tate, furthermore, Slide 9 is a monster, so we should either build some blocks, or maybe do some examples first, before putting this monster, certainly we have plenty of space. You should be marking your edits as inteprretation of my requests! I am afraid that you are misinterpreting my comments

[C] Slide 7's stated goal. The plan says it "answers the obvious objection before it is raised": Frobenius lives in characteristic p, F^1 lives in characteristic zero, and the crystalline comparison is what puts them on the same space so that "project Pic into H^2/F^1" is even a well-formed instruction. It has no Deck line, so it may not be a visible slide at all.

[C] Slide 8 against slide 2. The two Tate statements are different forms, not the same one twice: slide 2 is the counting form, rank read off the roots of unity of the characteristic polynomial, and slide 8 is the kernel form, Pic(X_Fp)_Qp = ker(Frob_p - p) on H^2_dR, which is the left factor of the Berthelot-Ogus intersection. Slide 8 is where Tate becomes a subspace you can intersect. Expanding slide 2 to two slides, as this session's earlier comment asks, makes the two statements sit further apart and the repetition more audible.

[C] Slide 9's load. It carries the projection formula, the scalar extension to W(F_{p^k}) tensor Q_p, the rank-at-most-one basic bound, the stronger iterated bound over rational Frobenius-invariant subspaces, and the per-factor application that slide 14 later uses. That is five things. The natural split is the map and the basic bound first, then the per-factor refinement after an example.

[C] On "we have plenty of space", corrected. There is a budget, and it is the author's own: artifacts/plan/plan.md line 3 records "50 + 45 + 45 minutes (author, 2026-09-13); at about 25 slides per hour that budgets roughly 21 + 19 + 19 slides". So 19 spoken slides is his arithmetic against his 45 minutes, not an invented target. Rejecting the slide 2 drop removes one way of reaching 19, not the 45 minutes. Read in context, "plenty of space" was said about slide 9 being overloaded and rejects the slide-count target, not the clock; splitting 9 is affordable, but every split has to be paid for somewhere. Separately, the deck's own header comment at talks/picard-lattices-2.html line 11 says "this one runs about 50 minutes", which contradicts the author's 45 and should be corrected to match him.

[C] DECIDED (jury, 2026-09-15). Runtime is 50 + 45 + 45 minutes (author, 2026-09-13), not a slide-count target: Lecture 1 took 50 minutes as written across 24 frames, about 2.1 minutes per frame, so "plenty of space" rejects the slide-count target, not the clock. Correct the contradictory talks/picard-lattices-2.html line 11, "about 50 minutes," to the author's 45-minute Lecture 2 budget.

7 What Frobenius acts on

O:L1245–1247, split

Frobenius lives in characteristic p and the filtration lives in characteristic zero, so they need a common home. The comparison supplies it: H²_crys(X_Fp / Z_p) ⊗ Q_p  ≃  H²_dR(X / Q_p),

and transporting Frobenius across gives Frob_p acting on H²_dR(X/Q_p), where F¹ also lives.

What it does. Answers the obvious objection before it is raised. First of the four slides split out of the old dense one.

[C] DECIDED (jury, 2026-09-15). Slide 7 IS a visible section in talks/picard-lattices-2.html, and deck numbers run 1-20 matching the plan, with no title slide yet. Reverse the prior claim that it may not be a visible slide: two jurors would remove it as a separate slide, one merging it into 8 because the deck header records 7 and 8 as a split of the source's single "2nd ingredient" frame, and one folding it into 5 because 5 already shows the comparison. Recommend re-merging it into 8 to restore the source's own frame.

8 Tate over a finite field

O:L1249–1251, split

Over the base field, Pic(X_Fp)_Qp = ker( Frob_p − p·id  |  H²_dR(X/Q_p) ), and geometrically, over F̄_p, write T_ev = ker( Frob_p^m − p^m·id ) = ∑i ker Φi(Frob_p / p) for the eventual Tate space, where m is any exponent with all the relevant roots of unity of order dividing it and the Φi are the cyclotomic factors of P₂(pT). Then, with K' = Frac W(F_{p^m}) a field over which all the classes are defined, Pic(X̄_p) ⊗_Z K' = T_ev ⊗_Qp K',   and in particular   ρ(X̄_p) = dim_Qp T_ev.

Writing the eigenvalues as ζp is convenient but the ζ need not lie in Q_p, which is why the kernels are taken of Q-irreducible cyclotomic polynomials in Frob_p/p; and the extension of scalars is not cosmetic: geometric divisor classes need not be defined over Q_p, so the identification above is an equality of K'-spaces, and what survives over Q_p is the dimension count.

The left-hand factor of the Berthelot–Ogus–Raynaud intersection is now a kernel, or a sum of kernels, of a matrix you can write down. The unbarred group is the one over F_p; the examples below need the barred one.

Deck (governs). The visible slide states only the base-field kernel; the eventual Tate space, cyclotomic kernels and scalar extension are in the speaker notes.

What it does. Puts the Tate kernel where the filtration lives.

[C] DECIDED (jury, 2026-09-15). Keep slides 2 and 8, with different jobs: slide 2 reads a rank off roots of unity, while slide 8 realises those classes as a Frobenius kernel that can be intersected with F^1. Retitle slide 8 "Tate as a Frobenius kernel" and say on it that it supplies the left-hand input to Berthelot-Ogus. Dissent: one juror finds the duplication worse because deck slide 2 already shows an etale Tate-kernel formula and would absorb 8 into 9.

9 The obstruction map [C-Sertöz]

V:L500–516; O:L1253–1264

Both sides are in hand, so intersect them by projecting: π : Pic(X̄_p) ⊂ H²_crys(X/W(F_{p^k})) ⊗ Q_p ⟶ (H²_crys ⊗ Q_p) / F¹,

the scalars being extended to W(F_{p^k}) ⊗ Q_p, the field over which the geometric classes of slide 8 are defined; over Z_p alone the inclusion on the left is not available.

If π(C) ≠ 0 then C ∉ Pic(X): the class does not lift. Two bounds, and they are not the same size. The basic one is the rank of π on a single eigenspace, and for a K3 that rank is at most 1, because H²/F¹ is one-dimensional. The stronger one, and the one that produces the multi-dimensional drops below, works with the rational Frobenius-invariant subspaces and iterates the obstruction map (Costa–Sertoz, §2.4 and §4.1.9).

In particular the obstruction is applied to each Frobenius-stable factor separately, one invariant subspace at a time, rather than to Pic(X̄_p)_Qp all at once. Slide 14 is that refinement in action.

Deck (governs). The visible slide uses Pic(X_Fp), displays the basic projection and gives four computational steps; the geometric scalar extension and iterated per-factor refinement are in the speaker notes.

What it does. This is the sentence the lecture is built around. Everything before it is setup and everything after is evidence.

>>[minima-uv3.217] AUTHOR (2026-09-15): I am confused about the transition between slide 9 and 10. I think we should move the abelian surface example (the 2 slides) shoudld come after the theorem. We should first present the characteristic polynomial and then the Frob matrix approxmation!

[C] Resulting order: 9 (the obstruction map) -> 12, 13 (abelian surface) -> 10 (what you actually compute) -> 11 (why finite precision still proves something) -> 14 (K3 at p = 89) and on. The theorem is followed immediately by the one example computed by hand, and the precision discussion then arrives with something concrete to be about. This also answers the earlier comment on slide 9 being a monster: the example is the block that breaks it up.

[C] Within slide 12, the order flips. It currently prints the 4x4 Frobenius matrix mod 31^3 first, then L(t) = det(1 - t Frob | H^1), then det(1 - t.31^{-1} Frob | H^2_dR) and rho(A_F31^{al}) = 2. The requested order is the characteristic polynomial first, giving the reduction bound 2, and the Frobenius matrix approximation after, as the extra input the obstruction needs. That matches the logic of the lecture even though it inverts the order of computation, since L(t) is in fact deduced from the matrix.

[C] The mismatch is narrower than the panel flag says, and a candidate fix is already on record. Bead minima-mo4: the displayed 4x4 matrix has characteristic polynomial T^4 - 3T^3 + 14T^2 - 93T + 961 mod 31^3, which matches the displayed L(t). The matrix and the polynomial agree with each other; it is the curve that does not fit, since exhaustive counting gives #C(F_31) = 42 and hence a linear coefficient of +10. So nothing printed is internally inconsistent: the genus 2 equation on the slide is simply not the Jacobian this computation was run on. artifacts/decisions/jury-fable.md proposes the replacement A = Jac(y^2 = x^5 - 2x^4 + 4x^3 - 4x^2 + 3x - 1), LMFDB 1104.a.17664.1, and reports that crystalline_obstruction at p = 31, N = 3 on that model reproduces the displayed matrix in under a second, keeping the matrix, L(t) and the H^2 line as they stand.

[C] Consequence for the open items. Putting L(t) first makes it the headline of the slide, so the unresolved PANEL FLAG on slide 12, the curve and L-polynomial that the panel could not reproduce, has to be settled before this reordering is drafted rather than after. That is item 3 of "Open on this lecture".

[C] Consequence for slide 10. With the example ahead of it, slide 10 stops being an abstract preamble and becomes the answer to a question the room has just seen: the matrix on 12 was printed mod 31^3, and 10 says where such a matrix comes from and what N means. The source frames named in the previous comment, O:L335-468, fit that slot directly.

10 What you actually compute

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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 is not this course's subject. What matters here is the shape of the output: an approximation whose error you control.

>>[minima-uv3.216] AUTHOR (2026-09-15): certainly, I have other talks about computing Frob on H^2, so we shoould say something about in this slides!

[C] Reading: slide 10 is the place. It currently disclaims the subject outright, "the machinery is Kedlaya-style and is not this course's subject", which is what this comment softens: say what the computation is and point at where it is done properly, rather than waving it away. This is a pointer, not an unpacking, unless the author says otherwise; the stated ceiling for this lecture is unchanged.

[C] Correction to an earlier claim in this session: there is a source archive, artifacts/picard_minicourse/_sources/, and it holds the material. The relevant talks are frobenious-dist/nyc-jnts.tex ("Effective obstruction to lifting algebraic classes from positive characteristic", the O prefix this lecture is already cut from), frobenious-dist/frobenius-dist-around.tex ("From Frobenius polynomials to geometry"), frobenious-dist/frobenius-dist-ctnt.tex ("Geometric invariants from counting points"), and the Frobenius Distributions talks in the same directory.

[C] The block was cut from this lecture's own source. O:L335-468 is four frames on exactly this question, and none of it survived into the deck:

[C] One detail the plan currently hides: the algorithm works on U = P^n minus X, the affine complement, not on H^2(X) directly, and deduces the H^2 factor from Q(t) = det(1 - q^{-1} t Frob | H^{n,dagger}(U)). If slide 10 is to "say something" rather than wave, that is the something. The cost is recorded too, at O:L1431: "Computing the zeta function or the Frobenius approximation for a prime takes several hours, thus trying to use 2 primes", which is also the honest answer to why slides 15 and 16 run one prime at a time.

[C] Cross-lecture: Lecture 1's s05.html sends the room to Kedlaya's lecture for the same machinery. Two pointers to two different talks is fine, but they should not contradict each other about whose treatment is the reference.

Deck (governs). The slide makes these four statements four short bullets.

What it does. The boulders say to indicate how the inputs are obtained without teaching the engines. This is that slide, and it also sets up the next one.

11 Why finite precision still proves something

new

An approximate Frobenius gives an approximate eigenspace, so one might expect only heuristic conclusions. The asymmetry saves it: π(C) ≠ 0 is an open condition. Establishing it to finite precision establishes it, full stop. So every dimension the method removes is removed rigorously, and the result is a genuine upper bound at any N.

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

Deck (governs). The first paragraph is four bullets and the monotonicity claim is the final line.

What it does. Converts the method from evidence into proof in the room's mind. Without it the examples read as experiments.

12 Abelian surface

O:L1267–1290

A = Jac(y² = 4x⁵ − 36x⁴ + 56x³ − 76x² + 44x − 23), at p = 31 and N = 3. Frobenius on H¹_dR(A/Q_p) is printed in full: Frob | H¹_dR(A/Q_p) ≡ [ 31·482 31·284 16241 3075 ] [ 31·386 31·886 2644 12126 ] [ 31·284 31·659 6336 9750 ] [ 31·194 31·876 27408 10841 ] (mod 31³)

with L(t) = det(1 − t Frob | H¹) = 1 − 3t + 14t² − 93t³ + 961t⁴; L(t) is the H¹ polynomial, the H² one is still P₂(T). From it one deduces Frobenius on H²_dR(A/Q_p) and det(1 − t·31⁻¹ Frob | H²_dR) = (t − 1)² (31t⁴ + 48t³ + 43t² + 48t + 31)/31,

whence ρ(A_F31al) = 2. The basis of H¹ respects the Hodge filtration, so the induced basis of H² does too. PANEL FLAG. The panel recomputed this example and got #C(F₃₁) = 42 and L(t) = 1 + 10t + 68t² + 310t³ + 961t⁴ for this curve, which disagrees with the L(t) shown here (both are valid Weil polynomials for p = 31). The slide faithfully reproduces nyc-jnts.tex:1267–1290, which displays this curve, this L(t), this Frobenius matrix and this H² determinant. So if the discrepancy is real, either the curve or the L-polynomial in the original deck came from a different example. Nothing here has been changed and nothing has been guessed; it awaits the author.

[C] DECIDED (jury, 2026-09-15). The matrix and displayed L(t) agree; the printed genus 2 equation is the single wrong item, giving #C(F_31) = 42 where the polynomial needs 29. Replace only the equation with y^2 = x^5 - 2x^4 + 4x^3 - 4x^2 + 3x - 1, LMFDB 1104.a.17664.1, with the point count verified. Matching Euler factors does not verify the basis, and the same misprint is in Costa-Sertoz Example 5.1, so the author should issue an erratum.

What it does. The only example whose whole computation fits on a slide, because an abelian surface has an H¹ to work from. A K3 would need the 22-by-22 matrix on H², which is why the K3 examples are transcripts.

13 Abelian surface, continued

O:L1291–1315

The basis of H¹ respects the filtration, so the induced basis of H² does too. Compute the eigenvectors; one of them fails to lie in F¹. The bound drops from 2 to 1, so the geometric Néron–Severi rank is 1.

The endomorphism statement needs one more step, and it is the standard one. Under the Rosati involution attached to the polarization, NS(A)_Q is the set of Rosati-fixed elements of End(A)_Q; so a Rosati-fixed part of dimension 1 leaves only Q, and the classification of endomorphism algebras of abelian surfaces then rules out everything but End(A_Qal)_Q = Q. Hence End(A_Qal) = Z.

One line notes that this is crystalline_obstruction in Sage; the K3 runs below are the same call.

Deck (governs). The slide prints the two eigenvectors modulo 31², identifies the last coordinate as projection to H²/F¹, and closes by saying van Luijk would have succeeded with a second prime.

What it does. The payoff. The room watches a dimension get removed by hand, once, and then trusts the transcripts.

14 K3 surface

O:L1377–1401

X = Z(y⁴ − x³z + yz³ + zw³ + w⁴) at p = 89, precision 3. The transcript reports ρ(X̄89) = 10, from the cyclotomic factors (t−1), (t+1), (t−1)⁴, (t⁴+1) of p⁻²²P₂(pT): four factors, degrees 1 + 1 + 4 + 4 = 10.

The distinct cyclotomic primary pieces have dimensions 5, 1 and 4: the (t−1) and (t−1)⁴ entries of the transcript are the polarization and the primitive part of the same +1-eigenspace, not two separate eigenspaces. Write Ti for the i-th such Frobenius-stable piece of Tate classes and Li = ker πi ⊆ Ti, the largest Frobenius-stable subspace of Ti killed by the obstruction map; dim Li is an upper bound for how much of that piece can lift, not an identification of lifted classes. The transcript prints dim Ti against dim Li, factor by factor, and at finite precision those entries themselves need only bound that dimension. Applying the obstruction to each factor separately, as on slide 9, gives ∑ dim Li = 4, so ρ(X_Qal) ≤ 4, and four lines in z = 0 give equality.

Closing line, and it reaches back to Lecture 1: previous approaches would not have used p = 89.

>>[minima-uv3.219] AUTHOR (2026-09-15): the slide 14: "K3 surface", should be improved to explain what what is going on, we are basically looking at each eigenspace individually, and say for which ones we found obsctruction, and for which ones we didn't

[C] The per-eigenspace account, read off the numbers already on the slide. Grouping the transcript's four factors into the three distinct eigenspaces, as the plan's own paragraph says to:

[C] That is the shape the comment asks for, and it is also the sentence the slide is missing: the polarization can never be obstructed, since it lifts by construction, so the only question on the +1 eigenspace is its primitive part.

[C] Bears on open item 4. This example is evidence in the slide 19 dispute. The plan's slide 19 says the obstruction "removes at most one dimension per invariant piece", and the panel disputes it by pointing at this slide. The numbers here settle the factual half: the +1 eigenspace loses exactly one dimension, while the -1 and zeta_8 eigenspaces lose all of theirs. So one-per-piece is the behaviour on the +1 piece only, not a general bound. Whether that changes the conclusion on slide 19 is the author's call, but the per-eigenspace presentation requested here will make the tension visible to the room.

Deck (governs). The transcript itself prints dim T_i = [1, 1, 4, 4] and dim L_i = [1, 0, 3, 0], followed by the five conclusions above.

What it does. The K3 example where the obstruction improves the reduction bound from 10 to the sharp value 4.

15 Quartic surface

O:L1402–1423

The same surface at p = 31. No obstruction is found. The bound is ρ ≤ 4 and stays there.

>>[minima-uv3.268] AUTHOR (2026-09-15): yes, but many slides are examples, that are a bit repetitive. so the hard part is to not make them repetitive, and try to get some content in between

[C] Scope: slides 14, 15 and 16, the three runs of the same call. Said in a discussion of how long the lecture takes, but it is a content instruction, not a timing one: repetitive slides are cheap in minutes and expensive in attention, so the fix is not to cut them but to make each one carry something the previous did not.

[C] The content between them is already being asked for elsewhere in this pass, and this comment says why it matters. The author's slide 17 comment moves each example's lesson onto its own slide: 14 gets "the obstruction improves the bound to the sharp value", 15 gets "a prime that says nothing is not a prime that says no", 16 gets "looking harder cannot help at this prime". Those three lines are exactly the content between the examples, and they are what stops the run reading as one slide shown three times.

[C] Two further sources of differentiation already available, neither of which needs new material. Slide 14 is the only one that gets a per-eigenspace breakdown, which the author asked for separately, so it is visibly the detailed one. Slide 16 carries the quintic remark, that a smooth quintic in P^3 has p_g = 4 and is not a K3, so the method is not K3-specific; that is genuinely new information arriving at the third example rather than a repeat.

[C] The risk to name when drafting: with the Sage transcripts gone and all three rewritten as mathematical statements in the same shape, they will look MORE alike, not less, unless each is given its own closing line first. The no-Sage-blocks instruction and this one have to be applied together.

Deck (governs). The visible slide says that working modulo p⁵ finds no obstruction and gives "some extra confidence" that equality might hold; it does not call the bound sharp.

What it does. The honest middle case, and the reason the method is stated per-prime. A prime that says nothing is not a prime that says no.

[C] DECIDED (jury, 2026-09-15). Slide 14 proves rho(X^al) = 4 for this quartic via four lines in z = 0, and slide 15 is the same surface with reduction bound 4, so the bound at 31 is sharp. Make slide 15's closing line say the obstruction finds nothing because the bound is already exact, not because the attempt was inconclusive. Dissent: one juror preferred leaving 15 at "a prime that says nothing is not a prime that says no" and keeping sharpness on 17.

16 Quintic surface

O:L1428–1458

A quintic with complex multiplication by Q(ζ₅), at p = 23 and p = 29. A smooth quintic in P³ has p_g = 4 and is not a K3; the method is not K3-specific, it applies to smooth surfaces in P³ generally, which is why it can be run here. Say that out loud, or the room spends the slide wondering what a quintic is doing in a K3 course.

At 29 the method fails to certify ρ = 1 even at infinite precision; what it does reach, at sufficiently high precision, is ρ ≤ 3, down from the reduction bound 5, and no further precision can improve that bound.

Deck (governs). The slide prints both Sage transcripts, including the sharp result 1 at p = 23 and the limiting result 3 at p = 29, then states the CM field.

What it does. Categorically stronger than slide 15. Not "we did not look hard enough" but "looking harder cannot help at this prime". This is the slide that makes the closing question a question rather than a to-do.

17 What the three examples say

new

Three outcomes, and they are different in kind. At p = 89 the obstruction improves the bound from 10 to 4 and the result is sharp. At p = 31 the reduction bound is already the sharp value 4, so the obstruction has nothing left to remove: no obstruction found is not an inconclusive attempt here. At p = 29 on the quintic the obstruction does fire, improving 5 to 3, but 3 is not sharp and no amount of precision will bring it down 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.

>>[minima-uv3.220] AUTHOR (2026-09-15): Slide 17 needs to be reworked, these lessons shoould already be in the each slide, and here be recalled, otherwise it is hard to recall

[C] Where each lesson currently lives: in the plan's "What it does." paragraphs, which are notes about the slides, not lines on them. Slide 14's is "the obstruction improves the reduction bound from 10 to the sharp value 4"; slide 15's is "a prime that says nothing is not a prime that says no"; slide 16's is "not 'we did not look hard enough' but 'looking harder cannot help at this prime'". None of the three is visible to the room, so slide 17 is the first and only time each is said. The comment asks to put each on its own slide as a closing line, leaving 17 as a three-line recall of things already heard.

[C] One conflict this surfaces, and it has to be resolved before the lesson can be moved. Slide 17 asserts "At p = 31 the reduction bound is already the sharp value 4, so the obstruction has nothing left to remove: no obstruction found is not an inconclusive attempt here." Slide 15 deliberately declines to say that: its Deck line records that the visible slide claims only "some extra confidence" that equality might hold and "does not call the bound sharp". The two cannot both be the closing line of slide 15. The sharper reading is defensible, since slide 14 gives rho(X_Q^{al}) = 4 for the same surface and slide 15 reports rho <= 4 at p = 31, so there is genuinely nothing left to remove; but that is a change to what slide 15 claims, and it is the author's to make.

[C] With the lessons distributed, slide 17's own work shrinks to the sentence the examples cannot say individually: only the third outcome is a statement about the method rather than about an attempt, which is what turns the closing slide's question into a question rather than a to-do.

>>[minima-uv3.221] AUTHOR (2026-09-15): And the two questions should show up: Does this always work? No and then the otehr One, and maybe the first question should show up earlier

[C] The two questions, and who answers them. "Does this always work?" is answered No, and slide 19 is the No: the theoretical example has RM by Q(sqrt 2), every good prime overshoots, and the obstruction cannot recover the 2 dimensions it would need. The second question is slide 20's, currently "Is there a way to get a sharp upper bound on the Picard number of a K3 surface?" on the plan and the narrower "Is there a prime for which the bound will be tight?" on the deck.

[C] Where the first question could go earlier. The natural place is straight after slide 14, the one unambiguous success: 14 succeeds, the question is asked, 15 stalls, 16 hits a wall at a prime, 17 recalls the three, 19 answers No for good, 20 asks what is left. Asking it at 17 instead means the room has already sat through both the stall and the wall without being told what they were evidence for.

[C] This also unpicks open item 5 and slide 20's panel flag. Separating the two questions gives each its own scope: the first is about this obstruction method and is answered No by the RM example; the second is the open one and is the only place the wording matters. The panel objects that the unrestricted second question is settled, since Poonen-Testa-van Luijk give unconditional algorithms for the geometric Neron-Severi group over a number field. With the questions split, that objection lands only on the second, and the RM example stops looking like a counterexample to it.

Deck (governs). The three outcomes are three bullets; the final question is the closing claim.

What it does. Stops the three examples reading as a list. Without it the room takes the quintic as discouragement rather than as the precise shape of what is unknown.

18 What is being computed, exactly

O:L1462–1480, second bullet

What the computation returns is not Pic(X)_Qp, exactly or approximately. The exact obstruction kernel is a containing space, Pic(X)_Qp ⊆ ker π ⊆ T_ev ∩ F¹_Qp,

with T_ev the eventual Tate space of slide 8, which is the object that lives over Q_p; the geometric Picard group itself appears only after extending scalars.

and what is computed is an approximation to it. An approximate matrix has its own exact kernel, which need not contain the true one nor sit inside the filtration: y and y + p^N x agree mod p^N with neither kernel containing the other. What the computation certifies is a bound on dim ker π, by slide 11's open condition, not the displayed containments for the approximate kernel.

Two different things separate that containing space from the answer, and only one of them is precision. Finite-precision error shrinks as N grows. The other does not: even at infinite precision ker π can stay strictly larger than the characteristic-zero Picard span, because the construction supplies no rational structure. That is also why primes do not combine: comparing the lattices at two primes needs Pic(X̄_p)_Q inside each, and separate Q_p-spaces do not give it. Raising N does not touch that.

Deck (governs). The visible slide retains the source's three Picard-space bullets and the claim that raising N does not recover the missing rational structure; the exact-kernel qualifications are in the speaker notes.

What it does. The distinction between rational classes and their p-adic span, which your plan asks to be kept visible. It is also why this method and Lecture 1's cannot simply be added together.

19 Theoretical example

saard PDF pp.35–37

The sextic is a product of three conics, so the double cover is singular at the 15 points where its six branch lines meet; those points are not divisor classes. Take X to be the minimal resolution of the double cover, a K3. The known sublattice is then the polarization together with the 15 exceptional curves above the 15 singular points, which gives ρ ≥ 16. At p = 83 its characteristic polynomial is χ₁ = (t−1)¹⁰(t+1)⁶, the factor of p⁻²²P₂(pT) coming from that rank-16 sublattice, and the reduction bound leaves ρ = 16, 17 or 18 a priori. The two extra classes at 83 span the single Q-irreducible Frobenius piece t² + 1, a two-dimensional piece that cannot supply exactly one extra rational divisor class, so they arrive together or not at all: 17 is ruled out and ρ = 16 or 18. In fact ρ(X_Qal) = 16, and X has RM by Q(√2).

This is exactly the wall from Lecture 1. RM by a field of degree 2 with (22 − 16)/2 = 3 odd puts it in the second of the two cases where every good prime overshoots. The obstruction respects the Galois-invariant decomposition, meaning the decomposition of H² into the Frobenius-stable pieces cut out by the Q-irreducible factors of P₂(T), and removes at most one dimension per invariant piece, so it cannot recover the 2 it would need. PANEL FLAG. The panel holds the one-dimension-per-piece claim to be false, and to contradict slide 14: there the drop from rank 10 to 4 requires the 4-dimensional (t⁴+1) piece to lose all 4 dimensions, since t⁴+1 splits over Q₈₉; on a Frobenius-stable piece that is Q_p-irreducible the method removes all of it or none of it. The panel also reads the author's own closing line below as saying the obstacle is the precision of Frobenius, not a structural bound. The author asserted the one-dimension mechanism, so the text is unchanged pending the author's decision.

The slide's own closing line: "Given a good enough approximation to Frobp we will obstruct 2 extra cycles. However, we don't yet know how to compute such approximation!"

Applied. Deck decision of 2026-09-13: over Q the t² + 1 block is one Galois orbit, so both extra classes go and the bound at 83 is 16; over Q(√2), Frob² acts as −1 and one stable kernel line survives, so the best bound is at least 17.

>>[minima-uv3.222] AUTHOR (2026-09-15): The "Theoretical example" i s nwo in slide1, so we should say that clearly, and with this technique, we could still prove that it works, if we didn't know RM. And we should make clear, that if we base change it wouldn't work

[C] Reading "slide1" as Lecture 1. The same surface is Lecture 1's s27.html, "A real multiplication example", at deck position 18, credited to Elsenhans-Jahnel as X^(2,1): six lines, 15 = 6 choose 2 nodes, 15 exceptional (-2)-curves, rho(X^{al}) = 16, RM by E = Q(sqrt 2) with dim_E T = 3, hence eta = 2 and rho(X_p^{al}) >= 18 at every good prime. This lecture re-derives the setup from scratch instead of naming it, which is what the comment asks to fix: say it is yesterday's surface and skip the reconstruction.

[C] The two claims are already decided in this plan, in the "Applied. Deck decision of 2026-09-13" line above; the comment asks for them to be visible rather than buried. Spelled out:

[C] That contrast is the real content of the slide: the method is not field-invariant, and it is strongest exactly where the Galois action is most irreducible. It also explains the closing quote, "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!", which is about precision over the base field, not about the base change.

[C] Bears on open item 4 and the panel flag below. If the slide is rebuilt around this two-field contrast, the old "removes at most one dimension per invariant piece" paragraph is no longer load-bearing and the panel's objection to it becomes moot rather than needing adjudication.

Deck (governs). The visible sixth bullet carries that decided two-field comparison, and the author's closing sentence remains verbatim in the highlighted box.

What it does. One concrete surface carries both lectures: Lecture 1 says why every prime overshoots here, Lecture 2 says why the obstruction cannot fix it, and the slide says exactly what is missing. This is the recruitment slide, and it is already written.

20 The question

O:L1462–1480

Is there a way to get a sharp upper bound on the Picard number of a K3 surface?

Open, with the RM case above as the known obstacle. Hopeful over Q. Not a claim that the method always works, and not a claim that it cannot be made to. PANEL FLAG. The panel holds the unrestricted question to be settled: unconditional algorithms to compute the geometric Néron–Severi group of a K3 surface over a number field exist (Poonen–Testa–van Luijk, Theorem 8.38, math.mit.edu/~poonen/papers/compute_ns.pdf). As the panel reads it, the question needs scoping to this particular obstruction method and to a stated base field, and the preceding real-multiplication example is not a failure of the obstruction method. The wording is the author's own, so it is unchanged pending the author's decision.

>>[minima-uv3.223] AUTHOR (2026-09-15): the last slide should frame more questions. And agian with answers referecing back to the previous examples. And also there was a chatgpt log, that gave a idealized theorem. Something about the galois group being cyclic

[C] The log is found: artifacts/chatgpt-picard-rank/transcript.md in the root checkout, 2154 lines, exported from the share link https://chatgpt.com/share/6aa5d676-e1a4-83ea-9e15-20cd1cc4fe9b (which never fetched; it renders client-side). It is already summarised in eleven numbered points in artifacts/decisions/decisions-brief.md.

[C] The cyclic statement is point 7 of that summary, the transcript's global criterion in the odd RM case: "There is a sharp good place iff H contains an element of order d", and for X/Q this is equivalent to "E/Q is cyclic and G_K acts on E through its full Galois group". Point 8 turns that into the arithmetic realization question, and point 10 is the transcript's own proposed closing question for a minicourse: "Can a K3 surface over Q have noncyclic totally real multiplication and odd dimension over its full endomorphism field?" The supporting local statement is point 5, the orbit lemma B_v(X) >= rho + d - t_v with t_v the order of the Frobenius action on E, which is also what makes the base-change failure on slide 19 exact: over Q(sqrt 2), t_v = 1 everywhere, so B_v >= 17 > 16 at every good place.

[C] A recorded rule points the other way, and the author's comment may be overriding it deliberately. Point 11 of the summary says the transcript labels these as "proposed new theorems supported by the preceding arguments", not established results, and artifacts/decisions/jury-fable.md concluded on that basis that the reduction to noncyclic RM "cannot go on a slide under the no-new-mathematics rule". Putting the cyclic criterion on the last slide is exactly that. It can be done honestly by marking it as a conjecture or as "what one would like to be true", which is what "idealized theorem" suggests, but the no-new-mathematics rule should be stated as consciously relaxed rather than quietly broken.

[C] Questions with their answers, each pointing back at an example already shown:

[C] That arrangement also does the work open item 5 asks for: the narrow question ("is there a prime for which the bound is tight?") is the first three rows, all answered by examples, and the broad question is the last row, which is the only one still open and the only place the panel's Poonen-Testa-van Luijk objection applies.

>>[minima-uv3.226] AUTHOR (2026-09-15): we are not going to cite the source, we can ask a question, when does this work? and then ask about what conditions are sufficient? and give some bullet point of most likely necessary statments or something like that

[C] DECIDED, and it supersedes the four options put to him. No citation, no attribution, and no theorem claimed. The slide becomes two questions with candidate conditions under them:

[C] The bullets are candidates, phrased as what would have to be true, not as results. That is what keeps the no-new-mathematics rule intact while still putting the content in front of the room: nothing is asserted, so nothing needs a proof or a source. The earlier jury ruling that the criterion "cannot go on a slide" was about asserting it; it does not bite on a question.

[C] Do not attribute these to the transcript, to Costa-Sertoz, or to anyone. The author's instruction is explicit. Whoever drafts this must also check each bullet against the sources before it goes up, since "most likely necessary" is the author's own hedge and the drafting must not harden it into a claim.

Deck (governs). The visible question is the narrower "Is there a prime for which the bound will be tight?", followed by the RM counterexample and the stated hope for K3 surfaces and abelian threefolds over Q.

What it does. Ends the lecture, and the break falls here.