Picard Lecture 1: written out

V:L358-380; L:L444-449 (old 3, first half)

A key geometric invariant of an algebraic K3 surface X is its Picard lattice

• Records the algebraic cycles on X: curves modulo linear, algebraic or numerical equivalence.
• For a K3 surface the three agree, Pic⁰ = 0, and Pic is free of finite rank.
• Pic(ℙ²) = ℤ, Pic(ℙ¹ × ℙ¹) = ℤ², Pic(cubic surface) = ℤ⁷.
• ρ, and more precisely the Picard lattice, is a coarse invariant.
• So theorems about K3 surfaces are stated by lattice or by rank, not by equation; as for abelian varieties.

What it does. Puts the object of the whole course on the board before anything is done to it. The old first Picard slide did three jobs at once; this one does the definition only.

APPROVED (2026-09-14): the author asked only for the coarse-invariant line and approved the rest; drafted from frames V:L358-380 and L:L444-449 (the first half of old slide 3).

no frame yet

Pic(X^al) carries the intersection pairing (D, D') ↦ D · D'. That is what makes it a lattice, not just a group.

• Even: D · D = 2 p_a(D) − 2 for every curve D, since K_X = 0 by adjunction; the pairing is an even symmetric bilinear form.
• Signature (1, ρ − 1), by the Hodge index theorem: one positive direction, the rest negative definite; in particular non-degenerate.
• disc Pic(X^al) := det of the Gram matrix in any ℤ-basis; a non-zero integer, independent of the basis, since a change of basis multiplies it by det(M)² = 1.
• Galois module: Gal(k^al/k) permutes the curves, respects linear equivalence, preserves the pairing, so it lands in the orthogonal group of the lattice.

Slide 18 is about the determinant of exactly this kind of action, on the orthogonal complement.

Deferred. Keep the finite-index discriminant identity for the spoken explanation on slide 11, pending s11-m05. The Brauer order is a square; the sign and q-power are known factors (s11-m06).

What it does. Says what the word "lattice" is doing in "Picard lattice", and puts the discriminant on the board before van Luijk needs it. Without this slide, slide 11's "disc Pic(X_p^al) ≠ disc Pic(X_q^al)" has to be explained in the middle of stating a criterion.

APPROVED (2026-09-14): new slide, drafted from standard facts (intersection pairing, even, signature (1, ρ − 1), discriminant, Pic as a natural Galois module). No frame exists for it in any deck.

V:L381-394 (old 3, second half)

Over ℚ^al, viewing X also as a complex manifold,

• Lefschetz (1,1): an integral class is algebraic exactly when it is of type (1,1).
• H²(X_ℂ, ℤ) is the unique even unimodular lattice of signature (3, 19); the embedding is primitive.
• dim H^1,1(X) = 20, so ρ(X^al) ∈ {1, 2, ..., 20}; for a generic K3 surface ρ(X^al) = 1.
• The degree of "difficulty" is negatively correlated with ρ(X).
• T(X) := Pic(X^al)^⊥ in H²(X, ℤ), the transcendental lattice; equivalently the minimal sub-Hodge structure of H²(X, ℚ) whose complexification contains H^2,0(X).

• The "new and interesting" Galois representations arise from T(X).
• In characteristic p the bound 20 fails: ρ(X_p^al) can be as large as 22.

That gap is slide 6, and it is the reason the whole lecture is possible.

What it does. The only piece of Hodge theory Lecture 1 states. The filtration is not introduced here; it appears in Lecture 2 where it does work.

APPROVED (2026-09-14): new slide, drafted from frame V:L381-394 (the second half of old slide 3) together with the standard facts asked for: Lefschetz (1,1), the primitive embedding, T(X) as the minimal Hodge structure containing H^2,0, and ρ ≤ 20 over ℂ against ρ ≤ 22 in characteristic p.

I15:L151-173; L:L472-496

The setup for the rest of the course:

• k a number field, X a K3 surface over k;
• p a prime of k where X has good reduction X_p;
• Pic(•), the group of line bundles modulo isomorphism; for K3 surfaces Pic⁰(•)=0;
• ρ(•) := rank Pic(•), the arithmetic or geometric Picard number of •.

Two Picard groups, not one:

• Pic(X) = Pic(X^al)^Gal(kal/k), so ρ(X) ≤ ρ(X^al), and usually strict.
• A class defined only over an extension is invisible over k.
• "The Picard number" in this lecture always means the geometric one, ρ(X^al).

Slide 5 states both counts separately, and slide 12 has to pass to an extension before it may use a discriminant.

"The evaluation of ρ for a given surface presents in general grave difficulties." (Zariski)

The two questions the lecture answers:

• How are the geometric Picard numbers ρ(X^al) and ρ(X_p^al) related?
• How does the geometric Picard number behave under reduction modulo p?

The Galois module structure is not decoration:

• H¹(Gal(k^al/k), Pic X^al) ≃ Br₁(X)/Br₀(X)
• X(k) ⊂ X(𝔸_k)^Br ⊂ X(𝔸_k)

Rational points care about the answer.

What it does. Separates the two Picard numbers once and for all, so that slides 5 and 6 can state two different counts of the roots of one polynomial without the room wondering which is which. Lecture 3 comes back to the Goal box and computes the Galois module for an actual surface.

APPROVED (2026-09-14): "I agree with Slide 4"; drafted from frames I15:L151-173 and L:L472-496.

I15:L203-224

• Tate's conjecture for K3 surfaces: a theorem in every characteristic.
• Charles [2013]: characteristic at least 5.
• Maulik [2014]: large-prime or degree-bounded cases.
• Madapusi Pera [2015]: all odd characteristic.
• Ito-Ito-Koshikawa [2021], arXiv:1809.09604: first complete proof in characteristic 2, per Madapusi Pera's 2020 erratum, doi:10.1017/fms.2020.2.
• Kim-Madapusi Pera [2016], arXiv:1512.02540: historical input; appendix corrected by the 2020 erratum.
• Nygaard-Ogus [1985]: earlier finite-height cases.
• Tate [1965]: formulation of the conjecture.
• The Kuga-Satake construction plays a crucial role.

Milne: for K3 and abelian surfaces Tate is equivalent to Artin-Tate, so the third line is a theorem here, not a conjecture; label it that way.

The discriminant Artin-Tate returns is that of Pic(X_𝔽q), the Neron-Severi group over the base field, whereas van Luijk needs disc Pic(X_p^al). The two are guaranteed to agree modulo squares when ρ(X_p) = ρ(X_p^al); they can agree without that, but when the ranks differ nothing forces it, so pass first to a finite extension over which every divisor class is defined, equivalently use an appropriate power of Frobenius.

Both ranks come from the same polynomial. The slide states ρ(X_p) and ρ(X_p^al) as two different counts of the roots of P₂(T), distinguishing the rank over the field of definition from the rank over its algebraic closure, which slide 4 set up and nothing since has used.

Artin-Tate is what supplies the discriminant, and the discriminant is what van Luijk needs seven slides later. Without this slide, slide 12's two discriminants arrive unexplained.

Author, 2026-09-14: the full Artin-Tate formula precedes its discriminant conclusion; approved in s05-m04.

Reorder note. Two sentences moved with the slide: "two slides later" is seven slides later in this order, and the two-ranks distinction is no longer new here, because slide 4 makes it. The old text read "which nothing earlier in the lecture has done".

Applied. Deck decision minima-uv3.10 (2026-09-13): the hypothesis line reads "Let X/𝔽_q, where q = pⁿ, be an abelian surface or a K3 surface. Then:".

Deck (governs). Approved convention: P₂(T) = det(T - Frob); the orders are at T = q and T = q·ζ. For K3, χ(T) = T²² P₂(1/T) is the reciprocal polynomial used in source frame I15:L203-224. Pic(X_p) denotes Pic over 𝔽_q, not over its algebraic closure.

What it does. Closes the only silent dependency in Lecture 1: every use of a discriminant up to now has assumed the audience already knows where discriminants come from.

APPROVED (2026-09-14): [s05-m01] slides 5 and 6 are swapped: slide 5 is the theorem and slide 6 is the finite-field example. They stay separate; the merge is refused. Author: "We might need to swap 5 and 6, first the theorical result, then the example. but we want separate slides." Author: "we should present the full formula of Artin-Tate and then the conclusion"

Source: Edgar Costa and Yuri Tschinkel, "Variation of Neron-Severi ranks of reductions of K3 surfaces", Experimental Mathematics 23 (2014), no. 4, 475-481, arXiv:1405.2265, Conjecture 2.1 and equation (8).

• Artin-Tate: Tate [1966], Bourbaki expose 306, "On the conjectures of Birch and Swinnerton-Dyer and a geometric analog"; 1968 is the reprint year.
• Milne [1975], "On a conjecture of Artin and Tate", Theorem 6.1 and addendum: Tate implies Artin-Tate here.
• Perfect-square Brauer order: Liu-Lorenzini-Raynaud [2005], corr. 2018; doi:10.1007/s00222-004-0403-2; corrigendum doi:10.1007/s00222-018-0809-x.

The formula uses the arithmetic group Pic(X_p/𝔽_q), with ρ=rank Pic(X_p/𝔽_q), the multiplicity of q in P₂(T)=det(T-Frob). It does not use the geometric rank. For a geometric discriminant, first pass to a finite extension defining every divisor class and use the Frobenius polynomial for that extension, as slide 12 does over 𝔽_11³⁰ and 𝔽_13⁴. Costa-Tschinkel write disc(NS); the slide writes disc(Pic) because Pic=NS for a K3 surface. Author: "NS is torsion free for K3 surfaces". The torsion denominator is therefore 1 and is omitted without a slide caveat. Converting the paper's reciprocal polynomial multiplies its limit by q^22-ρ; the K3 exponent α=1 gives q^21-ρ, with sign (-1)^ρ-1. Milne's Theorem 6.1 plus addendum removes the odd-characteristic hypothesis. Ito-Ito-Koshikawa 2021 supplies characteristic-2 Tate; Liu-Lorenzini-Raynaud supplies the perfect-square Brauer order, now visible in the legend. Independent check: artifacts/orch/artin-tate-conversion-check.md, section 7, with the author's subsequent removal of the torsion legend and Pic notation decision.

V:L396-419, L421-440; O:L616-635 (the cyclotomic count); O:L1377-1401 (the p = 89 example)

Fix the notation here, once, and keep it for all three lectures: P₂(T) = det(T − Frob | H²), the characteristic polynomial of Frobenius, whose roots α_i have absolute value q; the normalised polynomial is then q⁻²² P₂(qT), monic, with roots ζ_i := α_i / q of absolute value 1.

Tate is a theorem for K3 surfaces over finite fields, so the Tate classes are counted by the ζ_i that are roots of unity. The count in one line, as your own frame O:L626-633 writes it:

Read off an example rather than stated in the abstract, X := Z(y⁴ − x³z + yz³ + zw³ + w⁴) ⊂ ℙ³, at p = 89:

• (T − 1) = Φ₁, degree 1;
• (T + 1) = Φ₂, degree 1;
• (T − 1)⁴ = Φ₁⁴, degree 4;
• (T⁴ + 1) = Φ₈, degree 4.

Then:

• Over 𝔽₈₉: only k = 1 divides r = 1, so ρ(X_𝔽89) = 1 + 4 = 5.
• Over 𝔽_89^r: Φ₂ joins once 2 | r, Φ₈ once 8 | r; so ρ(X₈₉^al) = 1 + 1 + 4 + 4 = 10, reached at r = 8.
• Pic(X₈₉^al)_ℚ decomposes as P_ζ1 ⊕ P_ζ2 ⊕ P_ζ8.

Point counting is impractical beyond very small p; the machinery that makes it practical is cited, not taught.

• Kedlaya's lecture: crystalline methods.

This is the same object slide 5 states Tate's theorem for, the same one Lecture 2 normalises as P₂(pT), and, up to the reciprocal normalisation det(1 − TFrob), the same one the sources write χ, χ₁ or L(t).

Author's question. 2026-09-14: "I am not sure I follow the difference. Maybe the most generic equation is that ρ(X_𝔽qⁿ) = sum_{ord(ζ) | n} ...". It is, and it is yours: the display above is O:L628-632, and slides 5 and 6 are its two specialisations. Slide 5's first line is r = 1 (only Φ₁ counts, the order at T = q); its second line is r divisible by every k_i (every root of unity counts, the sum of the orders at T = q·ζ).

Source note. The frame at O:L628-632 writes ρ(X_𝔽p^r) = ∑_{ki | r} deg Φ_ki, without the exponents γ_i. The p = 89 example needs them: Φ₁ occurs there with γ = 1 and with γ = 4, and ρ(X_𝔽89) = 1 + 4 = 5, not deg Φ₁ = 1. The display above therefore carries γ_i. Flagged, not silently corrected.

Source note. The example is the transcript at O:L1383-1389, verbatim: factors [(t − 1, 1), (t + 1, 1), (t − 1, 4), (t⁴ + 1, 1)], dim Ti [1, 1, 4, 4], rank T(X_p^al) 10. No new computation was run for this entry; h and its degree are what the remaining 12 of the 22 cohomological dimensions have to be.

APPROVED (2026-09-14): the author agreed to reuse this example, so it appears in Lecture 1 slide 6 and again in Lecture 2 slide 14, deliberately, as a callback. Lecture 2 runs the same transcript for the obstruction, which cuts 10 to 4 and is sharp; Lecture 1 would use only the left-hand half, the factorisation and the rank.

Historical source (superseded by the approved worked example). The source frame carries the reciprocal normalisation and the displays this entry only alludes to: ρ(X_p^al) ∈ {2, 4, ..., 22} against {1, 2, ..., 20} over ℚ^al; ℤ_X(t) := exp(∑_m≥1 #X(𝔽_p^m) t^m / m) = 1 / ((1−t) P₂(t) (1−p²t)) with P₂(t) = det(1 − t Frob | H²_et(X_p^al, ℚ_l)) of degree 22; ℤ_X(t) deducible by naive counting of #X(𝔽_p^m) for m ≤ 11; the Tate kernel Pic(X_p)_ℚl = ker(Frob_p − p · id | H²_et(X_p^al, ℚ_l)); "Tate conjecture is a theorem for K3 surfaces over finite fields" [Charles, Madapusi, Kim-Madapusi]; and, for p > 7, crystalline methods in place of point counting [Abbott-Kedlaya-Roe, C, C-Harvey-Kedlaya, Tuitman-Pancratz]. The approved slide 5 uses the characteristic normalisation, with orders at T = q and T = q·ζ.

APPROVED (2026-09-14): [s06-m08] the proposed merge of slides 5 and 6 is refused; they stay two separate slides. Author: "We might need to swap 5 and 6, first the theorical result, then the example. but we want separate slides."

APPROVED (2026-09-14): [s06-m01] slides 5 and 6 are swapped: slide 5 is the theorem and slide 6 is the finite-field example. They stay separate; the merge is refused. Author: "We might need to swap 5 and 6, first the theorical result, then the example. but we want separate slides."

O:L1144-1165

Also: ρ(X_p^al) is always even for a K3 over a finite field. That parity fact does more work in this course than anything else on this slide.

Deck (governs). The deck slide opens with the concrete setup, f ∈ ℤ[x,y,z,w] and X := Z(f) ⊂ ℙ³_ℚ, reduced to X_p := Z(f mod p) ⊂ ℙ³(𝔽_p), states the theorem under the hypothesis that X and X_p are both smooth, and closes on a Goal box, "for a given f and p, improve the inequality ρ(X^al) ≤ ρ(X_p^al)", with two warnings: parity may already force the inequality not to be sharp, and endomorphisms of the transcendental lattice can complicate things further. The parity statement itself is printed on the preceding slide, slide 6, as part of ρ(X_p^al) ∈ {2, 4, ..., 22}.

V:L366-369; L:L452-455

Pic plays a similar role for a K3 surface as End(A) does for an abelian variety A.

• "Compute the Picard lattice of a K3 surface" is the same kind of question as "compute the endomorphism algebra of an abelian variety".
• Slides 9 and 10 answer the second one, for elliptic curves, by reduction mod p.
• For Kummer surfaces it is an identity, not an analogy: ρ(Km(A)) = ρ(A) + 16, which is what section 1.4 runs on.

What it does. The bridge. It is the reason the lecture may open its method section on an elliptic curve without the room wondering what elliptic curves have to do with K3 surfaces.

APPROVED (2026-09-14): "I agree with that slide"; drafted from frames V:L366-369 and L:L452-455, the Rosati paragraph that the compression of old slide 3 dropped.

C22:L233-265

For an elliptic curve, End_ℚ(E^al) ⇢ End_ℚ(E_p^al), and the right-hand side is ℚ[T]/(c_p(T)) when p ∤ a_p, a quaternion algebra otherwise. So the reduction sees more endomorphisms than the curve has, never fewer.

Drop the Sato-Tate figures that sit on this frame; they belong to a different talk.

Applied. Deck decision minima-w1x (2026-09-13): the criterion is "p does not divide a_p", not "a_p ≠ 0". The panel had read it the first way; the author settled it. For 11.a2 the two readings differ at p = 2 and agree at p = 3 (a₃ = −1).

Applied. Author, 2026-09-14: retitle so the slide asks the question directly, since the earlier slides that set up "the two types" are gone. The title above is his suggestion, given as "something similar", so the wording remains open; the two types stay in the one-line dichotomy.

DECIDED 2026-09-14: two slides, not merged. Slide 10 keeps the worked examples; this slide keeps the criterion and the one-line dichotomy.

Deck (governs). The deck slide carries the one-line dichotomy End_ℚ E^al = ℚ or ℚ(√−d) (CM); the criterion in the form p ∤ a_p ⟺ End_ℚ E_p^al is a quadratic field; for CM by ℚ(√−d), the chain a_p ≡ 0 mod p ⟺ p inert or ramified in ℚ(√−d) ⟺ End_ℚ E^al ≄ End_ℚ E_p^al; and, for non-CM E, End_ℚ E_p^al ∩ End_ℚ E_q^al ≃ ℚ with probability 1, with the expectation Prob(a_p ≡ 0 mod p) ∼ 1/√p.

C22:L268-289

E : y² + y = x³ − x² − 10x − 20 has End at 3 = ℚ(√−11) and End at 13 = ℚ(√−1). Two quadratic fields, not isomorphic, both containing the characteristic-zero algebra. So that algebra is ℚ.

Then 27.a2 for contrast: quaternionic at p ≡ 2 (3), ℚ(√−3) at p ≡ 1 (3), and the curve does have CM.

Applied. Deck decision minima-w1x (2026-09-13): the 11.a2 example uses p = 3 and p = 13, not p = 2 and p = 3. The panel had found the p = 2 half broken, since a₂ = −2 makes 11.a2 supersingular at 2, so End over 𝔽₂^al is the quaternion algebra B_2,∞ and only End over 𝔽₂ itself is ℚ(√−1); slide 9's superscript is the algebraic closure. The author took the p = 13 pairing (a₁₃ = 4, ordinary, ℚ(√−1)). Applied in this deck only; the four other decks carrying the example are outside the course.

Default taken. Both curves kept. The CM case is the contrast that makes the first deduction mean something, and it foreshadows 1.4, where CM is exactly what changes the answer.

What it does. Establishes the move the whole lecture repeats: compare structures at two primes, not dimensions.

DECIDED 2026-09-14: two slides, not merged. Slide 9 carries the criterion and the approved one-line End dichotomy; this slide carries the two curves worked.

V:L466-477; O:L1167-1182

The comparison is of square classes, not of the numerical representatives one happens to compute: the test is whether the ratio of the two discriminants is a square.

Used with r = 1 to give the first K3 surfaces over ℚ with ρ = 1.

Spoken: imagine, taking this long to try to prove that there are generic K3 surfaces over ℚ

Deck (governs). Keep the specialization display and the existing two-prime criterion. End on "Does this always work?" Slide 23 answers it.

What it does. The same move as slide 10, now with discriminants in place of endomorphism algebras. The parallel should be said out loud.

new

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

The characteristic polynomial of Frobenius gives the discriminant of the Neron-Severi lattice over the base field, up to squares; that square class, after the extension below, is the disc column above.

The disc column is the geometric discriminant, disc Pic(X_p^al). Over the prime field the base-field ranks are small, 1 at 11 and 5 at 13, so Artin-Tate has to be applied after the extension over which all 18 classes are defined: 𝔽_11³⁰ and 𝔽_13⁴ respectively. There ρ(X_p) = ρ(X_p^al) = 18, which is what lets Artin-Tate over the base field supply the geometric discriminant.

Equal ranks, and the ratio −55 : −85 is 11/17, not a square, so the two discriminants are distinct classes in ℚ^×/(ℚ^×)². Van Luijk gives ρ(X^al) < 18, hence ≤ 17.

• Symplectic order-5 action: ρ(X^al) ≥ 17 [Garbagnati-Sarti 2007, Prop. 1.1].

Provenance and two caveats. Surface from NSranks/data/polynomials.m:1286-1290, annotated "rank 17", citing Garbagnati-Sarti, arXiv:math/0603742, Proposition 1.1. Rows from NSranks/data/17/order5_3.data, one of 6538 covering primes 7 to 65521. Before this reaches a slide: the link from the polynomial to that data file is by file-naming convention, not wired in any script; and the disc column is inferred from the notebook's formula to be the discriminant of the reduction's Picard lattice, never labelled as such in a comment.

What it does. Runs the criterion once, with numbers. Slide 11 states it and never does it; this room will want to watch it happen. It also plants the surface that slide 23 comes back to.

Source note. No source in this deck's set states a formula for Artin-Tate, only what it gives: K3workshop.tex:220, "(Artin-Tate Conjecture) P₂(T) ⇝ disc(Pic(X_p)) mod ℚ^×2", and mukai_leiden.tex:513, "Artin-Tate conjecture (proven) also gives disc Pic X^al modulo squares" (the same line at mukai_nyu.tex:512, mukai_sydney.tex:473, mukai_oberwolfach.tex:460, mukai_gpm.tex:611). The recall above is that consequence, in the form slide 5 already uses; no formula was invented for it. The quoted consequence uses the author's Pic and al notation here; literal source spellings are listed in artifacts/orch/consolidate-1.md. Slide 5 now sources the full formula to Costa-Tschinkel, Conjecture 2.1.

• Garbagnati-Sarti, "Symplectic automorphisms of prime order on K3 surfaces", Journal of Algebra 318 (2007), 323-350, arXiv:math/0603742, Proposition 1.1.
• For this quartic: σ(x:y:z:w) = (x:ζ*y:ζ²*z:ζ⁴*w), ζ of order 5.
• Each monomial has weight 2 mod 5; det(σ) = ζ².
• The residue 2-form is fixed: det(σ)/ζ² = 1.
• Proposition 1.1: 16 coinvariant classes plus an invariant polarization; ρ ≥ 17.

V:L479-498

The specialization map has torsion-free cokernel for p ≠ 2, so integral and not merely rational information descends. With a Gröbner-basis test for whether a divisor class is realised.

Deck (governs). The deck slide attributes the statement to Elsenhans-Jahnel in the box label, reprints the specialization display, and spells the test out: if ρ(X_p^al) = ρ(X^al) every invertible sheaf lifts, so for ρ(X_p^al) = 2 one computes Pic(X_p^al), estimates the degree of a hypothetical effective divisor on the lift, and uses Gröbner bases to decide whether such a divisor exists. Practical only when Pic(X_p^al) is computable and the estimates are low.

O:L616-635; V:L559-576; Deligne, Weil I, Thm. 1.6; finite-field Tate

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• X_p/𝔽_q K3; P₂(u)=det (u−Frob ∣H²)

• h∈ℚ[u]: no cyclotomic factor
• |z|=1: conj (z)=z⁻¹
• Real roots: +1,−1, already in Φ₁,Φ₂
• Roots of h: nonreal pairs; deg h even

Weil + Tate

What it does. Proves geometric parity.

Spoken: Tate identifies the full cyclotomic degree with the geometric rank. The conjugate-pair argument for h uses Weil and rationality before using Tate.

I15:L523-527; Zarhin 1983; van Geemen 2008, Lem. 3.2; Charles 2014, Prop. 15 and Lem. 16

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• X/k K3; k⊂ℂ a number field

Theorem (Zarhin)

• E: a totally real field or a CM field
• Totally real: every embedding E↪ℂ lands in ℝ
• CM: totally imaginary quadratic extension of a totally real field
• d:=[E:ℚ], m:=dim _E T; dm=22−ρ(X^al)
• E totally real ⇒m≥3 [van Geemen]
• E totally real; V:=T_ℓ(1); g=Frob _p^a in connected monodromy

What it does. Defines the Hodge endomorphism field and its qualified orthogonal blocks.

Spoken: T is rational; T^2,0 has dimension one. Endomorphisms preserve the Hodge decomposition. E=ℚ means no real or complex multiplication.

Spoken: The twist divides Frobenius eigenvalues by the residue-field size. Choose a positive power lying in connected monodromy; each totally real embedding then gives an SO_m block. Odd m forces a fixed vector in each block. Before taking the power these give roots of unity, hence new divisor classes by Tate.

Spoken: These new cyclotomic roots are removed from h. Commutation with E does not force every eigenvalue orbit to have size d.

V:L559-576; Charles 2014, Thm. 1

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

Consider

Is this set infinite? What is its density?

What about

What it does. Defines the excess above the minimum and its counting function.

Spoken: η is the minimum excess. Odd characteristic-zero rank forces an increase but does not imply η=1. Which primes exceed the minimum? Is that set infinite? What is its density?

Spoken: Charles will compute this minimum and prove its attainment. The counting function here uses rational primes; over a number field, count places by norm.

V:L533-557; Li-Liedtke 2012; Bogomolov-Zarhin 2009; C-Elsenhans-Jahnel 2020, Sec. 3

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

So far we have been trying to improve the inequality ρ(X^al)≤ρ(X_p^al).Can we use the inequality to our advantage?

Theorem (Li-Liedtke)

If there are infinitely many p primes such that

then X^al contains infinitely many rational curves.

Theorem (Bogomolov-Zarhin)

The set {p:ρ(X_p^al)≠22} has positive density (density 1 after finite extension).

Corollary (after Li-Liedtke; C-Elsenhans-Jahnel)

• X/k K3; k a number field; e:=[L:k]∈{1,2}

• S_L⊂J(X), up to finitely many primes: lower density ≥1/e
• Additionally L=k or E=ℚ: infinitely many integral rational curves on X^al

What it does. Makes the half-density application a substitution in a generic corollary.

Spoken: All primes are places of good reduction, counted by norm. The lifting argument needs infinitely many non-supersingular rank increases. Positive density: Joshi-Rajan; density one after finite extension: Bogomolov-Zarhin.

Spoken: For L=k, ordinary reduction supplies the required primes. For quadratic L/k and E=ℚ, C-Elsenhans-Jahnel, Lemma 3.3, supplies infinitely many non-supersingular inert primes. Their Propositions 3.4-3.5 apply the lifting argument.

Spoken: Odd rank: take L=k. When η=0, J(X)=Π_jump(X). The generic corollary gives the later one-half bound by taking e=2.

Spoken: For X=Km (A), the determinant on T_ℓ(1) equals the determinant on (Pic (A^al)/Pic ⁰(A^al))_ℚℓ. It can be nontrivial.

I15:L298-352, L512-518; V:L577-596; Charles 2018, Thm. 1.1; Elkies 1991, Thms. A-B

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• X=Km (A); A/ℚ an abelian surface
• ρ(A^al):=rk (Pic (A^al)/Pic ⁰(A^al))
• ρ(X^al)=16+ρ(A^al)
• ρ(X_p^al)=16+ρ(A_p^al); p>2 good
• η(X^al)=η(A^al)=ρ(A^al) mod 2
• Π_jump(X)=Π_jump(A)
• Fix a polarization on A; † the Rosati involution

• A=E₁×E₂; E_i/ℚ

• X | ρ(X^al) | γ(X,B), predicted | What is known
• square of CM | 20 | 1/2 | 1/2+o(1), CM theory
• square of non-CM | 19 | ∼c_X/√(B) | infinitely many [Elkies 1987]
• CM times CM | 18 | 1/4 | 1/4+o(1), CM theory
• CM times non-CM | 18 | ∼c_X/√(B) | infinitely many [Charles 2018]
• non-CM times non-CM | 18 | ∼c_X/√(B) | infinitely many [Charles 2018]
• Product rows: geometrically non-isogenous factors
• Non-CM rates: Lang-Trotter heuristics; per-prime scale 1/√(p)

Remark

p∈Π_jump(X) depends uniquely on the pair (a_E1(p),a_E2(p)).

What it does. Compares predicted frequencies and proved infinitude for Kummer products.

Spoken: The product rows have geometrically non-isogenous factors. For two CM factors the CM fields are distinct. The square-root rates are conjectural; infinitude is unconditional.

Spoken: For a fixed non-CM square and sufficiently large B, c(log log B)log B/B<γ(X,B)<Clog B/B^1/4. The lower bound assumes GRH for real Dirichlet characters; the upper bound is unconditional. The constants depend on the fixed curve. Elkies 1991, Theorems A and B; the upper-bound proof uses Kaneko.

Spoken: For a CM square, the good unramified jump primes are exactly the primes inert in the CM field. Their density is one half.

V:L597-623; C22:L779-790

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• ρ(A_p^al)≥4⟺A_p^al∼E², E an elliptic curve
• ρ(A_p^al)=6⟺A_p^al∼E², E a supersingular elliptic curve
• If A^al∼E², then p∈Π_jump(A) iff p is supersingular for E.
• If A^al∼E₁×E₂ with E₁^al≁ E₂^al, then p∈Π_jump(A) iff E_1,p^al∼E_2,p^al.
• If End (A^al)=ℤ, then p∈Π_jump(A) iff A_p^al∼E².

What it does. Identifies the geometric events counted by the product comparison.

Spoken: All isogenies are geometric. For factors defined after a finite extension, choose a place above p; the geometric criterion is independent of that choice. Take common good primes of odd residue characteristic.

Spoken: When End (A^al)=ℤ, the abelian Picard number is one and the Kummer Picard number is seventeen. Here η=1, so a jump means ρ(A_p^al)>2. The later η=0 theorem does not answer its frequency question.

C-Elsenhans-Jahnel 2020, Prop. 2.13; Serre, Sec. 8.5.6.4

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• V:=T_ℓ(1); cup-product pairing

• det τ=1⟺im τ⊂SO(V)
• det τ≠1: nontrivial quadratic character
• An easy way to explain some jumps: O vs SO.

What it does. Introduces the determinant character.

Spoken: The Tate twist makes the pairing orthogonal. The determinant detects a quotient of order two, not all components of monodromy.

Spoken: Forced excess can survive determinant one: quadratic RM with dim _E T=3 still forces two Tate classes.

Spoken: For Kummer surfaces the universal SO assertion is false. Serre, Lectures on N_X(p), Section 8.5.6.4, removes one polarization from H²(A)(1). Removing all divisor classes gives determinant equal to the algebraic determinant. For A=(y²=x³−x)², complex conjugation on the CM field gives a nontrivial character.

C-Elsenhans-Jahnel 2020, Prop. 2.13

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• ρ(X^al) even; φ:=Frob _p|T_ℓ(1); det φ=−1
• Orthogonality: λ and λ⁻¹, with equal multiplicities.
• Other pairs: determinant +1; multiplicity of −1 odd.
• dim T_ℓ(1) even: multiplicity of +1 odd.
• Tate: +1,−1 give two new geometric divisor classes.

What it does. Proves the two extra Tate classes in even rank.

Spoken: Remove the reciprocal pairs other than ±1. Determinant minus one makes the multiplicity of minus one odd. Even dimension then makes the multiplicity of plus one odd.

Spoken: The two eigenvalues become one over a finite residue extension. Tate identifies the new geometric classes. In odd dimension minus one is forced but plus one need not be.

I15:L626-646; C-Elsenhans-Jahnel 2020, Prop. 2.1, Def. 2.4 and Thm. 2.15

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• X/ℚ quartic K3
• D_X:=Δ_H²(X)∈ℚ^×/(ℚ^×)²: determinant-character square class
• D_X∈ℤ∖{0} a representative; p good, p∤2D_X

Theorem (Deligne; C-Elsenhans-Jahnel 2020)

The functional equation of the Frobenius action on H²(X) has the plus sign if and only if D_X is square mod p.

• Gal (ℚ^al/ℚ) fixes Pic (X^al): Δ_Pic(X)=1

Theorem (C-Elsenhans-Jahnel)

What it does. Makes the determinant character arithmetic through its cohomological square class.

Spoken: D_X represents the quadratic extension cut out by the determinant on H²(1). It is neither the Picard intersection discriminant nor an unspecified equation discriminant.

Spoken: Dimension twenty-two gives det (−Frob )=det (Frob ). C-Elsenhans-Jahnel, Proposition 2.1, attributes the projective sign statement to Deligne; Suh treats the proper nonprojective extension.

Spoken: Δ_Pic is the square class of the Picard representation determinant. Galois fixing every geometric class makes it one; an integral descent equality is unnecessary.

V:L672-702; C-Elsenhans-Jahnel 2020, Thm. 2.15, Cor. 2.16 and Ex. 2.37

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• X/ℚ K3; r:=ρ(X^al) even; η(X^al)=0
• d_X:=Δ_H²(X)Δ_Pic(X) modulo squares; d_X∈ℤ∖{0}

Theorem (C-Elsenhans-Jahnel)

Corollary

• d_X nonsquare: L=ℚ(√(d_X)), [L:ℚ]=2
• S_L⊂J(X)=Π_jump(X), up to finitely many primes
• liminf _B→∞γ(X,B)≥1/2
• E=ℚ: infinitely many integral rational curves on X^al
• Example: Costa-Tschinkel 2014, Ex. 3.3

What it does. Applies the quadratic case of the earlier corollary.

Spoken: Take e=2 in the corollary. Inert primes have density one half. The assumption η=0 makes raw rank increase the jump event defined earlier; other primes may also jump.

Spoken: The rational-curves conclusion requires E=ℚ. It is not certified for the numerical example. The negative sign alone proves nonsquareness.

Spoken: The integer is the first factorization in C-Elsenhans-Jahnel, Example 2.37, attached there to Costa-Tschinkel, Example 3.3. When the Picard representation is trivial, d_X=D_X modulo squares.

I15:L523-550; Charles 2014, Thm. 1 and Prop. 18

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• T=T(X)_ℚ; E=End _Hdg(T); d=[E:ℚ]; m=dim _E T

Theorem (Charles 2014)

• Equality occurs infinitely often (density 1 after some finite extension).

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

What it does. Computes the minimum excess and supplies geometric prime pairs.

Spoken: Charles computes the minimum and proves its attainment. Density one is over a suitable finite extension, not necessarily over the original field.

Spoken: His original characteristic bound supplied the then-known Tate theorem. The proof with modern finite-field Tate gives the all-good-primes statement. The pair discriminants are geometric; apply Artin-Tate after extending the residue field to define every divisor class.

Spoken: The primes are good; the pairs can be chosen ordinary, with both ranks equal to ρ(X^al)+d. The minimum η is zero in the first case and d in the second.

Charles 2014, Thm. 1, Remark 19 and Prop. 23; Garbagnati-Sarti 2007, Prop. 1.1

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• r:=ρ(X^al); d:=[E:ℚ]; m:=dim _E T
• E=ℚ, m odd: η=1; van Luijk succeeds [Charles 2014]

• E totally real, E≠ℚ, m odd: η=d≥2

Certified quadratic RM

What it does. Explains what the two-prime criterion proves and how certified RM sharpens it.

Spoken: η is the forced minimum; an individual reduction can exceed it. At the minimum, the usual two-prime discriminant comparison leaves a gap of d−1 when d>1. This is a limitation of that criterion.

Spoken: The order-five example has lower bound seventeen from its symplectic action and a polarization, by Garbagnati-Sarti 2007, Proposition 1.1. Its two reductions supply the matching upper bound.

Spoken: For certified quadratic RM, 2 divides 22−ρ, so ρ is even. The unequal rank-eighteen discriminants exclude eighteen, hence give at most sixteen. This is the elementary quadratic case of Charles, Proposition 23.

Spoken: A projective Kummer surface has transcendental dimension at most five. Nontrivial totally real multiplication requires dm≥2·3=6 by van Geemen 2008, Lemma 3.2, after Zarhin 1983. Kummer surfaces avoid this obstruction even when their determinant character is nontrivial.

Saard PDF, physical p. 35; Elsenhans-Jahnel 2014, Thms. 5.12 and 6.6; period integration, Rem. 4.6; 2-adic point counting, Lem. 3.11

CANDIDATE: exact text below is proposed unless a retained mark records author approval.

• Elsenhans-Jahnel [2014, Thms. 5.12 and 6.6]
• X: minimal resolution of

• 6 lines; 15=(62) nodes; 15 exceptional (−2)-curves
• H,E_ij: 16 independent classes
• ρ(X^al)=16
• RM: E=ℚ(√(2)); dim _E T=(22−16)/2=3
• η=2; ρ(X_p^al)≥18 at every good prime
• Rank-18 pair, unequal square classes, certified RM: ρ(X^al)≤16

What it does. Realizes the RM case with an explicit six-line double cover.

Spoken: This is X^(2,1) in Elsenhans-Jahnel 2014. The three quadratics split over ℚ(√(2)) into six lines; no three meet. H is the pullback of a general line. With the fifteen exceptional curves its Gram matrix is diag (2,−2,...,−2), determinant −65536.

Spoken: The proof of Theorem 6.6 uses rank-eighteen reductions at seventeen and twenty-three with unequal geometric discriminant square classes. The RM field is proved, not numerically guessed.

Spoken: Further integral generators satisfy 2D_i=H+∑_j≠iE_ij. For w²=∏_i l_i, the quintic ∏_j≠il_j−l_i⁵=0 splits into w=±l_i³. These complete the same rank-sixteen lattice; the saturation has index thirty-two and discriminant −64.

Spoken: The two extra directions after reduction at eighty-three have no verified explicit representatives here. The split quintics already exist in characteristic zero and do not identify those two new classes.