Picard Lecture 1: written out

1 The geometric Picard group

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

X/k is a K3 surface; k is a number field embedded in C. Its geometric Picard lattice:

• Curves on X^{al} modulo linear equivalence.
• For K3 surfaces $\Rightarrow$ linear, algebraic and numerical equivalence agree.
• Pic^0=0.
• ρ, and more precisely the Picard lattice, is a coarse invariant.

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).

APPLIED, SOURCE-SETTLED (2026-09-14): [s01-m01] Applied the number-field, complex and geometric scope to the approved new opening; the cubic surface is smooth over the algebraic closure. Checked L:L444-449 and L:L472-480.

VERIFY-ONLY (2026-09-14), s01-c02: the discriminant tail is absent from this slide and its plan entry; no content edit needed.

OPEN (2026-09-14): [s01-m02] s01-c01 verification found "linear, algebraic and numerical", not the requested slash form. Recommendation: replace it with "linear/algebraic/numerical" in the existing equivalence bullet. Visible wording retained under the verify-only instruction. Checked V:L362 and Huybrechts, Chapter 1, Proposition 2.4, p. 12.

2026-09-14 active speaker notes (review metadata is separate):

• The embedding $k\subset\Bbb{C}$ lets us view $X_{\Bbb{C}}$ as a complex manifold. The geometric Picard group is identified with its algebraic curve classes.

AUTHOR'S CALL: [s01-m90] Which should remain: the curve-quotient display or its following prose definition? Also choose between the free-group display and the finite-free bullet. Both versions of each are preserved pending your choice.

2 The intersection pairing

no frame yet

APPLIED, s02-c03 and s02-c04: delete the two signature glosses; move the discriminant definition above the signature. Existing reveal attributes retained.

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.
• disc Pic(X^{al}) := det(D_i . D_j) in a Z-basis.
• Signature (1,rho(X^{al})-1): Hodge index theorem.
• 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 21, O or SO?, 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.

2026-09-14 comment plan, s02-c02: remove the nonzero-integer and basis-change clauses from the visible discriminant definition; retain the definition and existing spoken explanation. This is a deletion, with no new mathematical claim.

2026-09-14 active speaker notes (review metadata is separate):

• Adjunction uses $K_X=0$. The Hodge index theorem gives the signature; the discriminant does not depend on the integral basis.
• The Galois action preserves intersection numbers. Later we study the determinant of its action on the orthogonal complement.

3 Pic inside H²

Cross-reference note. The retained purpose line predates the Hodge endomorphism slide: this is the foundation for slides 15 and 25, not the only Hodge theory in the lecture.

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

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

• dim H^{1,1}(X_C)=20; 1<=rho(X^{al})<=20.
• Generic algebraic K3 surface $\Rightarrow$ rho=1.
• The degree of "difficulty" is negatively correlated with ρ(X).
• T(X): integral orthogonal complement of Pic(X^{al}) in H^2(X_C,Z).
• T(X)_Q: minimal rational sub-Hodge structure containing H^{2,0}(X_C) after complexification.

• 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.

APPLIED, SOURCE-SETTLED (2026-09-14): [s03-m01] Applied the distinction between the integral lattice and its rationalization to the new explanatory bullet. Checked Charles 2014, introduction, and Zarhin, Theorem 1.6(a).

2026-09-14 comment plan, s03-c01: move the three explanatory bullets on Lefschetz (1,1), the unique lattice of signature (3,19), and the primitive embedding to two spoken paragraphs. Preserve the displayed identification and all other content. Checked V:L381-394; Huybrechts, Chapter 1, equations (3.1)-(3.2), pp. 15-16, and Proposition 3.5, p. 17; Chapter 14, Theorem 1.1, p. 284. Primitivity follows from (3.1): the quotient embeds in the torsion-free group H^2(X_C,O_X).

2026-09-14 active speaker notes (review metadata is separate):

• Lefschetz $(1,1)$: an integral class is algebraic exactly when it has type $(1,1)$.
• $H^2(X_{\Bbb{C}},\Z)$ is the unique even unimodular lattice of signature $(3,19)$; the embedding is primitive.
• The integral lattice $T(X)$ and its rationalization are distinct: $T(X)_{\Bbb{Q}}$ is the minimal rational sub-Hodge structure whose complexification contains $H^{2,0}(X)$.
• We have some modularity results for high rank K3 surfaces, but very little is known for moderate rank.

4 Computing Pic as a Galois module

I15:L151-173; L:L472-496

"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 Picard number behave under reduction to a finite field?

What it does. States the computational goal, asks about reduction, and links the Galois module to rational points. Lecture 3 returns to the Goal box and computes the module for an actual surface.

2026-09-14 comment plan, s04-c01/c02/c03: remove both bullet lists before Goal; retain Goal and the Zariski quotation; replace the second Question bullet with the author's exact finite-field wording. Preserve the two-question structure of I15:L165-170. Put both final equations, unchanged, in one Corollary box at their existing reveal step. Leave the intended explanation under s04-m05, outside speaker notes. Checked L:L479-495 for Goal, quotation, and both final displays; Huybrechts, Chapter 18, equations (1.10)-(1.11), p. 385, for the Brauer identification over a number field.

2026-09-14 descent check: neither I15:L151-173 nor L:L472-496 states Pic(X)=Pic(X^{al})^Gal. The current draft already rationalizes this added equality; s04-c01 removes that entire bullet. Keep historical s04-m04 and the existing spoken descent caveat. The integral injection and torsion obstruction are checked in Auel-Bernardara, Remark 2.7, equation (2.1), p. 13; the rational-point case in Huybrechts, Chapter 18, equation (1.13), p. 385. No source-frame equality is silently corrected.

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

APPROVED (2026-09-14): [s04-m01] Slide 4: Pic; slide 8: the approved NS(A) gloss. Author, verbatim: "Yes, NS(A) should be used when mentioning abelian varieties, not on this slide."

APPROVED (2026-09-14): [s04-m02] Use ^{al} for varieties, fields and Galois groups. Author, verbatim: "we use ^{al} everywhere, no overlines"

APPROVED (2026-09-14): [s04-m03] Use ^{al} in Lectures 2 and 3 when those decks are worked on; cross-lecture instruction retained. Contact-sheet verdict: APPROVED; no note supplied.

APPLIED, SOURCE-SETTLED (2026-09-14): [s04-m04] Applied the rational equality to the added descent line; it is absent from both I15:L151-173 and L:L472-496. The source frames do not assert the disputed equality. Checked Auel-Bernardara, Remark 2.7, equation (2.1), and Stacks 0CDT. Historical approval request retained; no author verdict inferred.

OPEN (2026-09-14): [s04-m05] The Corollary box's label and statement await the author's explanation; the intended implication is UNVERIFIED. Recommendation: keep the two equations from L:L493-494 together under Corollary and explain their connection to the Goal orally. Author, verbatim: "The last two equations should be under a Corollary box, I will explain out loud what I mean by this".

2026-09-14 active speaker notes (review metadata is separate):

• A divisor class may require a field extension. In the finite-field examples we will count the arithmetic and geometric ranks separately.
• The Galois module matters for rational points through the Brauer-Manin obstruction. In Lecture 3 we compute it for an actual surface.
• Choose an embedding $k\subset\Bbb{C}$. Integrally, descent gives an injection into the invariant geometric Picard group; a Brauer obstruction can prevent surjectivity. Tensoring with $\Bbb{Q}$ removes that torsion obstruction. A $k$-rational point gives integral equality.

AUTHOR'S CALL: [s04-m90] Which of the two Question bullets should remain? Their formulations overlap; both are preserved pending your choice.

5 (previously 6) Picard lattice, over finite fields

APPLIED, s05-c06: add Elsenhans-Jahnel to the methods list. Checked their "Point counting on K3 surfaces and an application concerning real and complex multiplication", Section 1.2 and Algorithm 4.6: a variation of Harvey's p-adic method for double covers, implemented for degree-two K3 surfaces. The separate bullet identifies this contribution precisely.

V:L396-440; O:L616-635 and L1377-1401.

PLAN FIRST (2026-09-14), s05-c01 and s06-c01-c03: - Move the p = 89 example before the theorem to introduce P_2(t). - Keep the existing definition, normalization, example, factorization and conclusion structure. - Remove the Tate-class sentence, the general Picard-rank formula and the two-column Picard interpretation. - Combine the repeated Phi_1 factors as (t-1)^{1+4}. - Replace the columns with the requested H^2 decomposition and 10 invariant classes. - State the Tate twist explicitly: invariance is on H^2_et(X_89^{al},Q_ell(1)). - s06-c03: replace double hyphens only in the slide's visible citation text. - Preserve every existing mark ID with its subject; new example marks start at s06-m13.

Active slide content:

Verification before drafting. O:L1383-1389 prints factors [(t-1,1),(t+1,1),(t-1,4),(t^4+1,1)] and dimensions [1,1,4,4]; O:L1393 gives the complete cyclotomic rank 10. Combining the Phi_1 factors gives 5,1,4; V:L404 gives total degree 22, so the complement has dimension 12. The factors Phi_1, Phi_2 and Phi_8 divide t^8-1, while h has no cyclotomic factor. O:L622-624 and I15:L214-219 identify Frobenius-invariant dimensions with multiplicities. The four-term decomposition is on twisted cohomology over Q_ell, not a claim that H^2 is a rational Picard group. The source's split Phi_1 output is retained in this provenance; its combination on the slide is explicitly requested in s06-c02.

Primary checks: - O:L1377-1401: quartic, factors, dimensions, rank 10, Lecture 2 bound 4. - V:L402-404, L435-439: degree 22, point-counting limitation and crystalline methods. - I15:L203-219: characteristic roots and Tate multiplicities, converting the reciprocal convention. - [Ito-Ito-Koshikawa, arXiv:1809.09604v2](https://arxiv.org/pdf/1809.09604v2), Section 10.3, pp. 59-60: twisted cohomology splits into divisor classes and a complement with no root-of-unity eigenvalues.

What it does. Introduces P_2(t), then reads its cyclotomic factors as invariant cohomology. The theorem on slide 6 identifies the Picard ranks. Lecture 2 returns to this example for the obstruction.

Current marks:

APPLIED, CONFIRM REVERSAL (2026-09-14): [s06-m13] Applied s05-c01: slide 5 is now the p = 89 example; slide 6 is the theorem. This reverses the earlier order, not the decision to keep two slides. Earlier instruction (2026-09-14): "We might need to swap 5 and 6, first the theorical result, then the example." Full earlier instruction: "We might need to swap 5 and 6, first the theorical result, then the example. but we want separate slides." Latest instruction (2026-09-14): "This slide should be swapped with the next one, as we need to introduce P_2(t)". Recommendation: confirm the example-first order. Existing mark IDs follow their original subjects and are not renumbered.

APPLIED, SOURCE-SETTLED (2026-09-14): [s06-m14] Applied s06-c01 and s06-c02 on current slide 5: remove the Tate-class identification and Picard-rank count before the theorem; replace the columns by the four cohomology summands. Here $H^2=H^2_{\mathrm{et}}(X_{89}^{\mathrm{al}},\Bbb{Q}_\ell(1))$, and $P_{\Phi_k}=\ker\Phi_k(\operatorname{Frob}_{89})$, $P_h=\ker h(\operatorname{Frob}_{89})$. The Tate twist divides the untwisted eigenvalues by 89. O:L1384-1394 gives multiplicities 1,1,4,4 and the complete cyclotomic dimension 10; V:L404 gives total degree 22. Combining the two Phi_1 factors gives dimensions 5,1,4,12. Since 1,2,8 divide 8 and h has no cyclotomic factor, exactly 10 classes are fixed by Frobenius^8 on this twisted H^2. O:L622-624 and I15:L214-219 justify that these multiplicities give invariant dimensions. Ito-Ito-Koshikawa, arXiv:1809.09604v2, Section 10.3 (pp. 59-60), identifies the twisted cohomology decomposition and excludes roots of unity on the complement. This supersedes the Picard-only notation in s06-m03 and s06-m11.

Historical marks. The following verdicts use the numbering before s05-c01. Current slide 5 is the example; current slide 6 is the theorem and even-rank statement. Their IDs and verdicts remain unchanged. In particular, s06-m03 and s06-m11 describe the superseded Picard-only decomposition; s06-m14 records its replacement.

APPROVED (2026-09-14): [s06-m07] Keep historical slide 6 as the worked example. Contact-sheet verdict: APPROVED; no note supplied.

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: historical slide 5 is the theorem and historical 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."

APPROVED (2026-09-14): [s06-m02] Choose the polynomial convention for all three lectures and the Artin-Tate recall. Options: P_2 characteristic with roots q and q*zeta; P_2 reciprocal with roots 1/q and zeta/q; or separate names. Recommendation: use P_2 for det(t-Frob) and chi(t)=t^22 P_2(1/t) for the K3 reciprocal polynomial. Applied: P_2(t)=det(t-Frob); chi(t)=t^22 P_2(1/t) for the K3 reciprocal polynomial.<br />

APPROVED (2026-09-14): [s06-m03] Keep Pic(X_89^{al})_Q. Contact-sheet verdict: APPROVED; no note supplied.

APPROVED (2026-09-14): [s06-m04] Keep the even-rank fact on historical slide 5. Contact-sheet verdict: APPROVED; no note supplied.

APPROVED (2026-09-14): [s06-m05] Keep the example and its required definitions. Contact-sheet verdict: APPROVED; no note supplied.

APPROVED (2026-09-14): [s06-m06] Choose the title now that the example and theorem stay separate. Options: keep "Picard lattice, over finite fields"; or use "Example: Picard ranks at p=89". Recommendation: use "Example: Picard ranks at p=89". Author: "We should have a note that Kedlaya's Lecture will tell us more" Applied: the existing title is preserved; the Kedlaya pointer is added below the machinery citation.<br />

NEEDS APPROVAL (2026-09-14): [s06-m09] Suggestion: Keep the spoken pointer "Kedlaya's lecture will tell us more." Exact title and date remain unverified (reference-years.md, entry 20).

PROPOSED (2026-09-14): [s06-m10] Suggestion: Use "[Abbott-Kedlaya-Roe 2010; C 2015; C-Harvey-Kedlaya 2019; Pancratz-Tuitman 2015]". Author order settled by arXiv:1307.1250, title page; reference-years.md, entries 16-19. Approved slide body awaits this citation edit. Inherited source error: V:L437 prints Tuitman--Pancratz. The title page of arXiv:1307.1250 lists Sebastian Pancratz before Jan Tuitman. The visible source frame is retained pending approval of this single citation candidate.

NEEDS APPROVAL (2026-09-14): [s06-m11] Recommendation: Keep the spoken definition $P_{\zeta_k}=\ker\Phi_k(F)$, with $F$ the Tate-twisted action on rational Picard classes (cohomological Frobenius divided by 89). Dimensions 5,1,4; checked O:L1384-1396 and the displayed factors. Definition applied in notes; its placement remains open.

APPLIED, NEEDS APPROVAL (2026-09-14): [s06-m12] characteristic polynomial variable lowercased to t per the author, "let's use lower case t or x for our characteristic polynomials. in particular, avoiding u in slide 14".

2026-09-14 active speaker notes (review metadata is separate):

• The twist $(1)$ divides Frobenius eigenvalues by $89$; these are the normalized roots introduced above.
• $P_{\Phi_k}=\ker\Phi_k(\operatorname{Frob}_{89})$ and $P_h=\ker h(\operatorname{Frob}_{89})$ on this $H^2$. Their dimensions are $5,1,4,12$.
• Since $1,2,8$ divide $8$, the first three summands give $10$ classes fixed by $\operatorname{Frob}_{89}^8$. The remaining $12$ contribute none.
• "What the characteristic polynomial gives you" explains how these invariant classes determine Picard ranks.
• In practice we compute a $p$-adic approximation to the Frobenius matrix. Kedlaya's lecture will tell us more.
• We return to this quartic in Lecture 2, where the obstruction cuts the upper bound from $10$ to $4$.

AUTHOR'S CALL: [s06-m90] Keep the displayed invariant-space dimension, the following count of invariant classes, or both? Both formulations are preserved pending your choice.

6 (previously 5) What the characteristic polynomial gives you

Cross-reference note. The retained purpose line mentions definitions on slide 4. The arithmetic/geometric rank distinction is made here on slide 6; slide 4 no longer contains those definitions.

I15:L203-224; Costa-Tschinkel, arXiv:1405.2265v1, Section 2, Conjecture 2.1 and equation (8).

PLAN FIRST (2026-09-14), s05-c01-c05: - Follow the example on slide 5; retain the theorem block and full Artin-Tate display before its consequence. - Use the arithmetic lattice Pic(X_p), with X_p/F_q, throughout Artin-Tate. - Put #Br(X_p) in Q^{times 2} together with P_2(t) in the discriminant consequence. - Remove the redundant rho/Brauer/Tate-implies-Artin-Tate legend. - Display "Tate conjecture (proved)" as the single label candidate, with its approval mark. - Preserve every existing mark ID with its subject; new theorem marks start at s05-m09. - Keep spoken explanations in notes and source locators in review metadata.

Active slide content:

Answer to s05-c03. Arithmetic: Pic(X_p), with X_p/F_q. In Section 2 the source says "Let X be a smooth projective surface over" F_q, and equation (8) explicitly labels its lattice by X_{F_q}, with no algebraic closure. The group name is written as Pic here to follow the lecture's notation. Source: [Costa-Tschinkel, equation (8)](https://arxiv.org/html/1405.2265v1#S2.E8). The same paragraph says that #Br(X) is a perfect square. Slide 12 first passes to a finite extension defining all geometric classes before applying this arithmetic formula.

Formula check. Write chi(t)=t^22 P_2(1/t) for the source's reciprocal polynomial. Then lim_{t -> q} P_2(t)/(t-q)^rho = q^{22-rho} lim_{t -> 1/q} chi(t)/(1-qt)^rho. Conjecture 2.1 has the K3 exponent alpha=1 and trivial torsion denominator, giving the displayed q^{21-rho} and sign (-1)^{rho-1}. Equation (8) writes a square-class equivalent factor differing by q^2; this does not change the discriminant modulo squares. The full formula is unchanged.

Primary checks: - I15:L214-224: the two rank counts and discriminant consequence; convert reciprocal roots to characteristic roots. - V:L396-400: even geometric Picard rank for K3 surfaces over finite fields. - Costa-Tschinkel, Section 2, Conjecture 2.1 and equation (8): base field, Brauer square, full Artin-Tate formula and square class. - [Milne, 1975a article comments](https://www.jmilne.org/math/articles/1975a.html), "The condition p != 2": removal of the odd-characteristic hypothesis; Theorems 4.1 and 6.1 and the square-order input. - [Ito-Ito-Koshikawa, arXiv:1809.09604v2](https://arxiv.org/pdf/1809.09604v2), Section 1.2 and Section 6.4, Remark 6.9: Tate in all characteristics and the characteristic-2 proof repair. - Detailed earlier credit-history metadata remains in artifacts/orch/reference-years.md; it is not a spoken bibliography.

What it does. Interprets the polynomial introduced on slide 5 as arithmetic and geometric Picard ranks, then supplies the arithmetic discriminant used after base extension on slide 12, six slides later. The two rank conventions were introduced on slide 4.

Current marks:

APPLIED, CONFIRM REVERSAL (2026-09-14): [s05-m09] Applied s05-c01: slide 5 is now the p = 89 example; slide 6 is the theorem. This reverses the earlier order, not the decision to keep two slides. Earlier instruction (2026-09-14): "We might need to swap 5 and 6, first the theorical result, then the example." Full earlier instruction: "We might need to swap 5 and 6, first the theorical result, then the example. but we want separate slides." Latest instruction (2026-09-14): "This slide should be swapped with the next one, as we need to introduce P_2(t)". Recommendation: confirm the example-first order. Existing mark IDs follow their original subjects and are not renumbered.

PROPOSED LABEL (2026-09-14): [s05-m10] Applied s05-c05 on current slide 6. Author: "Theorem (many people) -> Tate Conjecture (now proved in long series of papers), can we perhaps do better here?" Recommendation: use "Tate conjecture (proved)". The draft displays this candidate for confirmation. It names the result and its status without repeating the proof history. Source: I15:L214-219; Ito-Ito-Koshikawa, arXiv:1809.09604v2, Section 1.2 and Section 6.4 (Remark 6.9), for all characteristics. This supersedes the label in s05-m05; its spoken Kuga-Satake reminder remains.

APPLIED, SOURCE-SETTLED (2026-09-14): [s05-m11] Applied s05-c02, s05-c03 and s05-c04 on current slide 6. Artin-Tate returns the arithmetic discriminant $\operatorname{disc}(\operatorname{Pic}(X_p))$, with $X_p/\Bbb{F}_q$, not the geometric discriminant. Costa-Tschinkel, arXiv:1405.2265v1, Section 2, Conjecture 2.1 and equation (8), explicitly puts $X_{\mathbb{F}_q}$ on the left of (8); the setup says "Let X be a smooth projective surface over" $\mathbb{F}_q$. The source's lattice notation is rendered as Pic for this K3 lecture. The consequence includes the square Brauer order; the redundant legend is removed. The full formula keeps $q^{21-\rho}$ after converting the reciprocal polynomial. Milne's 1975a article page, "The condition p != 2", removes the characteristic restriction and explains the square-order input. Slide 12 must first pass to an extension for the geometric discriminant. Bibliographic locators belong in this review metadata and the plan, not the spoken notes.

Historical marks. The following verdicts use the numbering before s05-c01. Current slide 5 is the example; current slide 6 is the theorem and even-rank statement. Their IDs and verdicts remain unchanged. s05-m09 explicitly reverses the earlier order; s05-m10 replaces the label candidate in s05-m05; s05-m11 moves source locators out of spoken notes.

APPROVED (2026-09-14): [s05-m01] slides 5 and 6 are swapped: historical slide 5 is the theorem and historical 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"

APPROVED (2026-09-14): [s05-m02] Choose the polynomial convention for all three lectures and the Artin-Tate recall. Options: P_2 characteristic with roots q and q*zeta; P_2 reciprocal with roots 1/q and zeta/q; or separate names. Recommendation: use P_2 for det(t-Frob) and chi(t)=t^22 P_2(1/t) for the K3 reciprocal polynomial. Applied: P_2(t)=det(t-Frob); chi(t)=t^22 P_2(1/t) for the K3 reciprocal polynomial.<br />

APPROVED (2026-09-14): [s05-m03] Choose which definitions accompany the example and which content historical slide 5 may refer forward to. Options: restore zeta, counting-range and Tate-kernel displays; keep the example alone; or add only its required definitions. Recommendation: keep the example and required definitions, then rewrite the reference in historical slide 5 to match. Author: "Here is also where, we should note that the rank tehre must be even." Applied: the example keeps its required definitions; historical slide 5 states the even-rank fact.<br />

APPROVED (2026-09-14): [s05-m04] Remove the visible Costa-Tschinkel label; keep the formula unchanged and its provenance in the speaker notes. Author, verbatim: "we should remove [Costa-Tschinkel, Conj. 2.1], that is Artin--Tate formula for K3 surfaces"

APPROVED (2026-09-14): [s05-m05] Theorem (many people). Credit history and Kuga-Satake in the speaker notes; supersedes the visible credit correction in e384285. Author, verbatim: "We should not specify whod id what on the slide, the point is that it was many people, and I should say in the speaker notes that Kuga--Satake plays a crucial role"

REFUSED (2026-09-14): [s05-m06] Refuse the citation strip on behalf of the author: s05-m04 and s05-m05 remove historical slide 5 attributions. Tate 1966, Milne 1975 and Liu-Lorenzini-Raynaud 2005, corr. 2018 stay in the speaker notes. The perfect-square Brauer order stays visible. Author on s05-m04, verbatim: "we should remove [Costa-Tschinkel, Conj. 2.1], that is Artin--Tate formula for K3 surfaces" Author on s05-m05, verbatim: "We should not specify whod id what on the slide, the point is that it was many people, and I should say in the speaker notes that Kuga--Satake plays a crucial role"

APPLIED, SOURCE-SETTLED (2026-09-14): [s05-m07] Applied the missing X_p, P_2 and abelian Picard-rank definitions to the approved new theorem expansion. Checked I15:L203-224 and Milne, Abelian Varieties, Section 17. The full Artin-Tate formula, credits policy and arithmetic exponent are unchanged.

APPLIED, NEEDS APPROVAL (2026-09-14): [s05-m08] characteristic polynomial variable lowercased to t per the author, "let's use lower case t or x for our characteristic polynomials. in particular, avoiding u in slide 14".

2026-09-14 active speaker notes (review metadata is separate):

• $P_2(t)$ is the characteristic polynomial introduced in "Picard lattice, over finite fields".
• $X_p$ is over $\Bbb{F}_q$. The multiplicity at $q$ is the arithmetic rank; summing over $q\zeta$ gives the geometric rank.
• Several papers proved the Tate conjecture. The Kuga-Satake construction plays a crucial role.
• In the K3 formula, $\rho$ is the arithmetic rank $\rho(X_p)$. Artin-Tate gives the discriminant of $\operatorname{Pic}(X_p)$ over $\Bbb{F}_q$.
• The Brauer order is a square, so the polynomial and the known sign and power of $q$ give the discriminant modulo squares.
• For the geometric discriminant, pass first to an extension defining every divisor class and use that extension's Frobenius polynomial, as in "Let's apply it to a K3 surface with a $\Bbb{Z}/5$ automorphism".

14 Why the reduction rank is even

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.

• $X_p/\Bbb{F}_q$ K3; $P_2(t)=\det(t-\operatorname{Frob}\mid H^2)$

• $h\in\Bbb{Q}[t]$: no cyclotomic factor
• $|z|=1$ $\Rightarrow$ $\operatorname{conj}(z)=z^{-1}$
• Real roots: $+1,-1$, already in $\Phi_1,\Phi_2$
• 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.

AUTHOR'S CALL: [s14-m03] Recommendation: Keep parity on 14, Hodge endomorphisms on 15, Charles on 16, rational curves on 17, jump definitions on 18, and the reunited SO proof on 15 before Charles. No uniquely forced delivery choice. The candidate follows GPT 6 astra so each frame has one main idea. GPT 6 astra: Use parity, Hodge endomorphisms, then a separate jump-definition frame. GPT 5.6 sol: Use two frames; put eta, Pi_jump and gamma with parity; move sharpness to Charles. Source: O:L616-635; I15:L523-527; V:L559-576.

NEEDS APPROVAL: [s14-m05] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "Why the reduction rank is even" | "$X_p/\Bbb{F}_q$ K3; $P_2(t)=\det(t-\operatorname{Frob}\mid H^2)$" | "$$q^{-22}P_2(qt)=h(t)\prod_i\Phi_{k_i}(t)^{\gamma_i}$$" | "$h\in\Bbb{Q}[t]$: no cyclotomic factor" | "$|z|=1$ $\Rightarrow$ $\operatorname{conj}(z)=z^{-1}$" | "Real roots: $+1,-1$, already in $\Phi_1,\Phi_2$" | "Roots of $h$: nonreal pairs; $\deg h$ even" | "Weil + Tate" | "$$\rho(X_p^{\mathrm{al}})=\sum_i\gamma_i\deg\Phi_{k_i}=22-\deg h\in2\Z$$" Speaker notes proposed: "Tate identifies the full cyclotomic degree with the geometric rank. The conjugate-pair argument for $h$ uses Weil and rationality before using Tate." SETTLED mathematical source: O:L616-635; V:L559-576; Deligne, Weil I, Thm. 1.6; finite-field Tate. Exact teaching arrangement still needs approval.

APPLIED, NEEDS APPROVAL (2026-09-14): [s14-m08] characteristic polynomial variable lowercased to t per the author, "let's use lower case t or x for our characteristic polynomials. in particular, avoiding u in slide 14".

2026-09-14 active speaker notes (review metadata is separate):

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

7 Reduction to finite characteristic (old 6)

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 $X_p := Z(f \bmod p) \subset \mathbf{P}^3_{\Bbb{F}_p}$, 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}. APPROVED (2026-09-14): [s07-m01] Keep the even-rank fact on slide 6 and the parity use on slide 7. Contact-sheet verdict: APPROVED; no note supplied.

NEEDS APPROVAL (2026-09-14): [s07-m02] Recommendation: Add "homogeneous of degree 4" after "$f\in\Bbb{Z}[x,y,z,w]$". The projective-space correction to $\mathbf{P}^3_{\Bbb{F}_p}$ is applied under s07-c01; only the degree condition remains open. Both setup paragraphs, the theorem, Goal and closing paragraphs retain the source structure. Checked O:L1144-1165 and Stacks 01NF.

2026-09-14 comment plan, s07-c01: replace $\mathbf{P}^3(\Bbb{F}_p)$ by $\mathbf{P}^3_{\Bbb{F}_p}$ in the second setup paragraph. O:L1147 supplies the old notation; [Stacks, Definition 27.13.2, tag 01NF](https://stacks.math.columbia.edu/tag/01NF) supplies the base-scheme notation. No other mathematical or structural change.

2026-09-14 active speaker notes (review metadata is separate):

• $\rho(X_p^{\mathrm{al}})$ is always even for a K3 surface over a finite field. An odd geometric rank therefore cannot equal the reduction rank.
• Take $f$ homogeneous of degree four. At a good prime, $X_p$ is the smooth quartic surface over $\Bbb{F}_p$, not merely its set of rational points.

8 Pic plays the role of End(A)

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" ↭ "compute the endomorphism algebra of an abelian variety".
• We also stratify moduli of abelian varieties via $\operatorname{End}(A)$.
• rho(A):=rank(Pic(A)/Pic^0(A)), also after base change.
• A an abelian surface, char k!=2 $\Rightarrow$ rho(Kum(A)^{al})=rho(A^{al})+16.
• $\operatorname{Kum}(A)$ is the K3 surface obtained by resolving $A/\{\pm1\}$.

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.

APPROVED (2026-09-14): [s08-m01] Pic(A)/Pic^0(A) = NS(A); state the Rosati-fixed display for (Pic(A)/Pic^0(A))_Q. This is the sole visible NS gloss in the three lectures. Author: "yes, for Abelian varieties we can also have Pic(A)/Pic^0(A) = NS(A), just to help the reader"

APPLIED, SOURCE-SETTLED (2026-09-14): [s08-m02] Applied the quotient definition of rho(A), including after base change. Checked Milne, Abelian Varieties, Section 17 and Proposition 17.2; V:L590. The approved NS gloss is unchanged.

APPLIED, SOURCE-SETTLED (2026-09-14): [s08-m03] Applied geometric base change to the new Kummer bridge. The sixteen exceptional curves are geometric; this is not an arithmetic rank identity. Checked I15:L512-518; the source uses algebraic closures.

NEEDS APPROVAL (2026-09-14): [s08-m04] Recommendation: Prepend "Fix a polarization on A" to the Rosati display. The omitted polarization is inherited from V:L366-369 and L:L452-455; retain that display pending the author. Checked Milne, Section 17, Proposition 17.2, which fixes a polarization and works over an algebraically closed field. The notes already supply the polarization; current slide 19 does so visibly.

2026-09-14 comment plan, s08-c01/c02: remove the elliptic-reduction bullet, the current replacement for the already-absent slide-number navigation sentence. Replace "is the same kind of question as" by the literal double-headed wave arrow ↭ (U+21AD). Add the moduli bullet immediately below, in the same reveal group. Keep the Rosati block, Kummer bridge and spoken notes. The analogy follows V:L366-369 and L:L452-455. Endomorphism loci are checked in [Milne, Introduction to Shimura Varieties, Theorem 8.17, p. 88](https://www.jmilne.org/math/xnotes/svi.pdf#page=88); the stratification idiom appears in [Kohel-Shieh, On Sato-Tate distributions, extremal traces, and real multiplication in genus 2, Introduction, p. 2, and Section 6, pp. 13-14](https://www.i2m.univ-amu.fr/perso/david.kohel/pub/st_g2.pdf#page=2).

2026-09-14 active speaker notes (review metadata is separate):

• For an abelian variety the quotient by $\operatorname{Pic}^0(A)$ is essential. A polarization supplies the Rosati involution.
• Computing endomorphisms of elliptic curves by reduction gives the comparison we will use for Picard lattices.
• Fix a polarization $\lambda$ on $A$. Rosati is defined using $\lambda$; the symmetric endomorphisms identify with $(\operatorname{Pic}(A)/\operatorname{Pic}^0(A))_{\Bbb{Q}}$ via $\lambda^{-1}\phi_L$. Over the ground field use a polarization defined there and rational descent.

AUTHOR'S CALL: [s08-m90] Keep the opening Pic/End analogy, the later comparison of the two computation tasks, or both? Both formulations are preserved pending your choice.

9 How to distinguish an elliptic curve with CM from one without? (old 1)

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 (a_3 = −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.

REFUSED (2026-09-14): [s09-m01] retitle and add End(E_p) table. Author: "We do not need teh table at the top, it should just be said with a simple or statemnt. ie. End = Q or Q(sqrt(-d)) (aka CM). This table is already in slide 10 implicitly" The refusal applies to the table; the earlier retitle suggestion remains open.

NEEDS APPROVAL: [s09-m02] Suggestion: Title: "How to distinguish an elliptic curve with CM from one without?" Keep the one-line End dichotomy. Source: author's retitle request and s09-m01 table refusal. The proposed title is now visible for review; approval remains open.

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.

NEEDS APPROVAL (2026-09-14): [s09-m03] UNVERIFIED: the exact probability-one and unit-constant asymptotic claims. Recommendation: Replace the last source item by the deterministic ordinary-prime test $\Bbb{Q}(\operatorname{Frob}_p)\not\simeq\Bbb{Q}(\operatorname{Frob}_q)\Rightarrow\operatorname{End}_{\Bbb{Q}}E^{\mathrm{al}}=\Bbb{Q}$, followed by the separately labelled Lang-Trotter heuristic $\#\{p\leq B:p\text{ good},a_p=0\}\sim C_E\sqrt B/\log B$. Include $E/\Bbb{Q}$ and good primes in the setup. The probability and intersection language is inherited from C22:L256-258, so the body awaits the author. Checked Akbary-Parks, Conjecture 1.1; direct counts for 11.a2 at 13 and 17 give the same field Q(i), disproving a finite-pair certainty, not a specified limiting law.

APPLIED, NEEDS APPROVAL (2026-09-14): [s09-m04] characteristic polynomial variable lowercased to t per the author, "let's use lower case t or x for our characteristic polynomials. in particular, avoiding u in slide 14".

NEEDS APPROVAL (2026-09-14): [s09-m05] s09-c01 title candidates: "Detecting CM by reduction"; "CM and reduction"; "CM or no CM". Recommendation: "Detecting CM by reduction". Keep the current title until the author chooses. Checked C22:L233-265, especially the CM/non-CM dichotomy at L238-239 and reduction criterion at L245-258. Author: "Let's brainstorm a better title."

2026-09-14 comment plan, s09-c01/c02: keep the question title; add the three candidates only in s09-m05, outside speaker notes. Remove the forced line break between "with prob. 1;" and "and we expect" in the final source item. Preserve both clauses in one continuous line of HTML. C22:L256-258 contains the inherited wording and line break. The exact probability-one and unit-constant asymptotic claims remain UNVERIFIED, with the existing recommendation in s09-m03 still open. [Akbary-Parks, Conjecture 1.1, p. 1](https://www.cs.uleth.ca/~akbary/Akbary-Parks-revised.pdf#page=1) states a conjectural prime count with a curve-dependent constant.

2026-09-14 active speaker notes (review metadata is separate):

• All primes considered are primes of good reduction. Here $a_p=p+1-\#E(\Bbb{F}_p)$ and $c_p(t)=t^2-a_pt+p$.
• At an ordinary prime, $\operatorname{End}_{\Bbb{Q}}E_p^{\mathrm{al}}\simeq\Bbb{Q}[t]/(c_p(t))$; at a supersingular prime it is a quaternion algebra.
• There are many ways to distinguish the two cases; for example, we could compute the $j$-invariant.
• By looking at the number of points mod $p$, we may deduce the endomorphism algebra of an elliptic curve.
• $\operatorname{End}_{\Bbb{Q}}$ means endomorphisms tensored with $\Bbb{Q}$. If two ordinary good reductions have non-isomorphic quadratic endomorphism fields, then the geometric endomorphism algebra in characteristic zero is $\Bbb{Q}$. For a fixed non-CM curve, Lang-Trotter predicts a count $C_E\sqrt B/\log B$ for good primes with $a_p=0$; the constant depends on the curve.

10 Two examples: 11.a2 and 27.a2 (old 2)

C22:L268-289

PLAN FIRST (2026-09-14), s10-c01: retitle only; preserve the two source blocks. Leave "universal" as the recommendation in new s10-m03, without changing the title to that wording.

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 a_2 = −2 makes 11.a2 supersingular at 2, so End over F_2^{al} is the quaternion algebra B_{2,∞} and only End over F_2 itself is Q(√−1); slide 9's superscript is the algebraic closure. The author took the p = 13 pairing (a_13 = 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. APPROVED (2026-09-14): [s10-m01] Keep "quaternion algebra". Contact-sheet verdict: APPROVED; no note supplied.

2026-09-14 active speaker notes (review metadata is separate):

• Two non-isomorphic quadratic fields both contain the characteristic-zero endomorphism algebra, so that algebra is $\Bbb{Q}$.
• For 11.a2, $a_3=-1$ and $a_{13}=4$; both reductions are ordinary. At $2$, $a_2=-2$ and the geometric endomorphism algebra is quaternionic.
• The CM example is like testing primality: the existence of Carmichael numbers does not stop us from applying Fermat's test.
• The second displayed equation has LMFDB label 27.a3. Its complex multiplication field is Q(sqrt(-3)).

NEEDS APPROVAL (2026-09-14): [s10-m02] Recommendation: Change the second label and the title suffix to 27.a3, and link to https://www.lmfdb.org/EllipticCurve/Q/27/a/3; keep y^2+y=x^3-7. C22:L277 itself gives the mismatched 27.a2 label, so the visible transcription awaits the author. Checked the official LMFDB 27.a3 equation [0,0,1,0,-7].

PROPOSED (2026-09-14): [s10-m03] Recommendation: Use "Two universal examples: non-CM and CM" as the title, with "universal" referring to the characteristic-zero dichotomy. C22:L246-261 distinguishes CM and non-CM, and L268-289 supplies the two examples. The wording remains an author decision under s10-c01; no claim of identical reduction behavior within each class is added.

11 Improving upper bounds: two specializations (old 8)

V:L466-477; O:L1167-1182

PLAN FIRST (2026-09-14), s11-c01/c02/c03 and s12-c01: retain the specialization display, hypothesis/conclusion structure and closing question. Credit the combined method to Kloosterman and van Luijk, with the specifically requested real em dash. Replace only the historical sentence; keep the surprise spoken. Citation labels name authors only. The expressly requested "(in 2005)" is a historical date, recording the first arXiv posting, not a bibliographic label.

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.

NEEDS APPROVAL: [s11-m05] Recommendation: Keep the finite-index explanation spoken: for $L\subset M$ of index $n$, $\operatorname{disc}L=n^2\operatorname{disc}M$. Both review jurors preferred a visible lemma; this recommendation retains the source frame structure and asks the author about placement. Checked Kloosterman, arXiv:math/0502439, Proposition 4.2, p. 6, and the basis-change determinant identity.

Spoken: Imagine taking until 2005 to write down a generic K3 surface over Q and prove it has geometric Picard rank one.

Deck (governs). Keep the specialization display and the existing two-prime criterion. End on "Does this always work?" Slide 27 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. APPROVED (2026-09-14): [s11-m01] Keep the primes p and q. Author, verbatim: "p and q is the natural story" APPROVED (2026-09-14): [s11-m02] Keep the historical sentence; add the spoken remark below. Author, verbatim: "that is true. he wrote a paper about it! We should put on the speaker notes, imagine, taking this long to try to prove that there are generic K3 surfaces over Q" REFUSED (2026-09-14): [s11-m03] Replace the closing failure warning with "Does this always work?" Slide 27 answers it; no added conclusion or reveal after the hypotheses. Author, verbatim: "We know that, but we should not foreshadow it right! We should isntead aks does this always work"

APPLIED (2026-09-14): [s11-m04] s11-c02/c03 replace the historical sentence with "van Luijk (in 2005) proved rho(X^{al}) = 1 for explicit K3 surfaces X/Q." The posting year fits the requested historical parenthesis; the journal year stays in the notes. This supersedes the earlier two-date citation proposal. Source: arXiv:math/0506416 submission history, 21 June 2005; published introduction and Theorem 3.1, Algebra & Number Theory 1 (2007), 1-15. Earlier existence results were ineffective.

Attribution checked: van Luijk, published Remark 3.2, pp. 8-9, credits Kloosterman's Artin-Tate refinement; Kloosterman, arXiv:math/0502439, Proposition 4.2 and Remark 4.5, pp. 6-7, credits van Luijk for the two-prime discriminant comparison. These are separate papers and complementary contributions, not a jointly authored theorem. The visible combined credit does not transfer van Luijk's explicit rank-one examples to Kloosterman.

APPLIED, NEEDS APPROVAL (2026-09-14): [s11-m06] Suggestion: Use the corrected explanation: "#Br is a square; the sign and q-power are known factors." Applied to plan rationale and notes; finite-index placement remains s11-m05. Source: Costa-Tschinkel, Conj. 2.1; redteam-astra.md, F06.

APPLIED, NEEDS APPROVAL (2026-09-14): [s11-m07] Suggestion: Retain the existing corrections "in Q^times/(Q^times)^2" and "rho(X^{al}) < 2r". Source frames V:L466-477 and O:L1167-1182 omitted the square class and wrote Pic < 2r; van Luijk 2007, Remark 3.2, pp. 8-9, supports the correction. No new body edit. Corrected transcription: these already-present deviations are retained and remain marked for the author, rather than silently represented as the literal V/O frame.

2026-09-14 active speaker notes (review metadata is separate):

• Imagine taking until 2005 to write down a generic K3 surface over Q and prove it has geometric Picard rank one.
• The same comparison as for elliptic curves, now using discriminants instead of endomorphism algebras.
• Compare square classes: the ratio of the discriminants must be a nonsquare.
• There are algorithms in principle: Charles; Poonen-Testa-van Luijk; Hassett-Kresch-Tschinkel; Shioda; Lairez-Sertoz.
• $\#\operatorname{Br}$ is a square; the sign and $q$-power are known factors. An equal-rank specialization identifies $L=\operatorname{Pic}(X^{\mathrm{al}})$ with a finite-index sublattice of $M=\operatorname{Pic}(X_p^{\mathrm{al}})$. If $[M:L]=n$, then $\operatorname{disc}L=n^2\operatorname{disc}M$.
• These were the first explicit examples over Q with geometric rank one; earlier existence arguments were ineffective. The paper was posted in 2005 and published in 2007.
• Van Luijk supplies the two-prime discriminant comparison; Kloosterman computes the discriminant square classes using Artin-Tate.

Provenance: corrected transcription, retaining the already marked mathematical deviations from the author frame.

NEEDS APPROVAL: [s11-m08] Years exception: use the posting year 2005; switch to the publication year 2007 if preferred. Exact candidate: "van Luijk (2005): first explicit K3 surfaces $X/\Bbb{Q}$ with $\rho(X^{\mathrm{al}})=1$." The added "first" makes the historical point visible; the surprise stays in the existing speaker note. Checked arXiv:math/0506416, submitted 21 June 2005, and Algebra & Number Theory 1 (2007), 1-15, introduction. This is a paper date; Elsenhans and Jahnel's later introduction dates the construction to 2004.

12 Let's apply it to a K3 surface with a Z/5 automorphism (old 9)

Historical allocation, superseded by s27-c01 and the current content map: Cross-reference note. The order-five callback is in "When every prime overshoots" (27), as the retained What it does. line now states.

new

PLAN FIRST (2026-09-14), s12-c01/c02: retain the equation, table, Artin-Tate recall and square-class legend. Remove the four specified bullets and replace the two conclusion bullets by the single ratio-and-bound bullet below. Keep fragment groups 0/1/2, with only the legend in group 1. Put both field extensions and the rank equality in the speaker notes. Visible citations name authors only, with no year or theorem/proposition number; this is a deck-wide handoff rule.

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 Picard lattice over the base field, up to squares; after the extension explained in the notes, that square class is the disc column above.

Speaker-only justification: the disc column gives geometric discriminant square-class representatives, disc Pic(X_p^{al}) modulo rational squares. 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_{11^30} and F_{13^4} respectively. There rho(X_{F_{p^r}})=rho(X_p^{al})=18 for (p,r)=(11,30),(13,4), which is what lets Artin-Tate over the base field supply the geometric discriminant.

Visible conclusion:

• $-55/-85 = 11/17 \notin (\Bbb{Q}^{\times})^2$, hence $\rho(X^{\mathrm{al}})\leq17$.
• Symplectic order-5 action $\Rightarrow$ rho(X^{al}) >= 17 [Garbagnati-Sarti].

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 When every prime overshoots (slide 26) 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) P_2(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 6 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 6 now sources the full formula to Costa-Tschinkel, Conjecture 2.1.

APPROVED (2026-09-14): [s12-m06] Keep disc Pic in the Artin-Tate recall. Author, verbatim: "Let's use Pic, not NS" CROSS-LECTURE: [s12-m01] Suggestion: Use P_2(t)=det(t-Frob); chi(t)=t^22 P_2(1/t) for K3 surfaces in all three lectures. Settled by s05-m02 and s06-m02; retain this CROSS-LECTURE tag as the follow-up. No Lecture 2 or 3 edits here. APPROVED (2026-09-14): [s12-m02] Verified: the order-5 action is symplectic; rho(X^{al}) >= 17 [Garbagnati-Sarti 2007, Prop. 1.1]. Author, verbatim: "is it symplectic? if so we should reference the paper"

• 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: sigma(x:y:z:w) = (x:zeta*y:zeta^2*z:zeta^4*w), zeta of order 5.
• Each monomial has weight 2 mod 5; det(sigma) = zeta^2.
• The residue 2-form is fixed: det(sigma)/zeta^2 = 1.
• Proposition 1.1: 16 coinvariant classes plus an invariant polarization; rho >= 17. NEEDS APPROVAL: [s12-m03] Suggestion: Heading: "Theorem (Artin-Tate)". The formula was already established on slide 6; no repeated "a theorem here". Source: I15:L203-224. Restored the source attribution as the visible review baseline; heading approval remains open. APPROVED (2026-09-14): [s12-m04] Reveal theorem, extension and conclusion at 0/1/2, in the present order. Contact-sheet verdict: APPROVED; no note supplied. APPROVED (2026-09-14): [s12-m05] Let's apply it to a K3 surface with a Z/5 automorphism. Author, verbatim: "The title should be, let's apply it to a K3 surface with a Z/5 automorphism"

APPLIED (2026-09-14): [s12-m07] s12-c01 supersedes the dated visible citation proposal. Keep "Theorem (Artin-Tate)" and [Garbagnati-Sarti] visible. Full references remain in the notes and provenance. Checked I15:L203-224 and Garbagnati-Sarti, arXiv:math/0603742, Proposition 1.1, p. 3.

APPLIED, SOURCE-SETTLED (2026-09-14): [s12-m08] Applied square-class heading and legend; retained -55 and -85. Checked NSranks/nsranks k3.ipynb:496-498 and independently powered the stored Frobenius polynomials in order5_3.data. These values are not asserted to be Gram determinants.

APPLIED, NEEDS APPROVAL (2026-09-14): [s12-m09] characteristic polynomial variable lowercased to t per the author, "let's use lower case t or x for our characteristic polynomials. in particular, avoiding u in slide 14".

2026-09-14 active speaker notes (review metadata is separate):

• The prime-field ranks are $1$ at $11$ and $5$ at $13$. Extend to $\Bbb{F}_{11^{30}}$ and $\Bbb{F}_{13^4}$, respectively $\Rightarrow$ $\rho(X_{\Bbb{F}_{p^r}})=\rho(X_p^{\mathrm{al}})=18$ for $(p,r)=(11,30),(13,4)$. Apply Artin-Tate to Frobenius to the powers $30$ and $4$, when all divisor classes are defined, to obtain the geometric discriminant square classes.
• The ratio is $11/17$, a nonsquare. The upper bound $17$ and the symplectic lower bound $17$ give $\rho(X^{\mathrm{al}})=17$.
• Garbagnati-Sarti, "Symplectic automorphisms of prime order on K3 surfaces", Journal of Algebra 318 (2007), 323-350, arXiv:math/0603742, Proposition 1.1.
• Take $\sigma(x:y:z:w)=(x:\zeta y:\zeta^2z:\zeta^4w)$, with $\zeta$ of order $5$. Every monomial has weight $2$ mod $5$, and $\det(\sigma)=\zeta^2$. The residue 2-form is fixed because $\det(\sigma)/\zeta^2=1$.
• There are $16$ coinvariant classes and an invariant polarization, giving $\rho\geq17$.
• This is the Artin-Tate formula from Tate 1966 and Milne 1975. Apply it after extending the field so every divisor class is defined.

APPLIED, SOURCE-SETTLED (2026-09-14): [s12-m10] Applied the extension-field subscript in the new Artin-Tate explanation. Prime-field ranks remain 1 and 5; geometric ranks are 18 after degrees 30 and 4. Checked and independently factored NSranks/data/17/order5_3.data, rows 11 and 13.

s12-c02 verification (2026-09-14): independently factored and powered the polynomials in NSranks/data/17/order5_3.data:2-3. The cyclotomic orders are 1,2,5,10,30 at 11 and 1,2,4 at 13; ranks are 1/5 over the prime fields and 18 geometrically. The Artin-Tate square classes after extensions of degrees 30/4 are -55/-85. Notebook locator: NSranks/nsranks k3.ipynb:493-498. The extension-field requirement is explicit in Kloosterman, arXiv:math/0502439, p. 7, immediately after Conjecture 4.6. This justification now belongs to notes only.

13 Torsion-free cokernel (old 10)

V:L479-498

APPLIED, s13-c01: rechecked the published Theorem 1.4 and Remarks 1.5(a), pp. 1028-1029. The visible theorem is correct with X/Q and good p>2. The first explicit degree-two rank-one credit is supported by the 2008 ANTS paper, introduction and Corollary 30; the first claim refers to those examples, not the later specialization theorem. Visible credit names authors only; dates and locators remain in notes.

PLAN FIRST (2026-09-14), s13-c01 and s12-c01: preserve the theorem, rank-equality consequence, three-step procedure and practical limitation. Add the missing visible setup "$X/\Bbb{Q}$ K3; $p>2$ a prime of good reduction". Add one short bullet crediting Elsenhans-Jahnel's first explicit degree-two rank-one examples. Keep the 2008 examples distinct from their 2011 specialization theorem in notes and provenance. The theorem label remains authors only.

For a K3 surface X/Q and a prime p>2 of good reduction, the geometric specialization map has torsion-free cokernel. The previous shorthand omitted the field and good-reduction hypotheses. The published theorem is more general: a proper smooth scheme over a mixed-characteristic discrete valuation ring with perfect residue field and ramification degree e<p-1. Over Q, e=1. Source: Elsenhans-Jahnel, "The Picard group of a K3 surface and its reduction modulo p", published Theorem 1.4 and Remarks 1.5(a), p. 1028; Approach 1.6, pp. 1028-1029. This published statement supersedes the older arXiv formulation over Z in the previous mark.

• Elsenhans-Jahnel: explicit $X/\Bbb{Q}$ of degree $2$ with $\rho(X^{\mathrm{al}})=1$ using a single prime.

Historical source: Elsenhans-Jahnel, "K3 surfaces of Picard rank one and degree two", ANTS VIII (2008), pp. 212-225, abstract and introduction (author PDF p. 1). The introduction contrasts the earlier degree-four examples with their degree-two construction. The same authors also gave the 2008 companion "K3 surfaces of Picard rank one which are double covers of the projective plane". Thus the first-explicit degree-two credit is theirs, but it is not a first supplied by the 2011 paper. "Generic" here means geometric Picard rank one.

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.

APPROVED (2026-09-14): [s13-m04] Keep the refinement folded into section 1.2; keep slide 13. Contact-sheet verdict: APPROVED; no note supplied. APPROVED (2026-09-14): [s13-m01] Keep slide 13 in section 1.2. Contact-sheet verdict: APPROVED; no note supplied. APPROVED (2026-09-14): [s13-m02] Keep "Theorem (Elsenhans-Jahnel)"; open with "The specialization map". Contact-sheet verdict: APPROVED; no note supplied. NEEDS APPROVAL: [s13-m03] Suggestion: Introduction: "For rho(X_p^{al}) = 2, the Elsenhans-Jahnel approach:" followed by the existing three steps. Source: V:L479-498. Restored the V:L489 source introduction as the visible baseline. The shorter recommendation above remains open; both jurors are preserved in PLAN-remaining.md D09.

SUPERSEDED (2026-09-14): [s13-m05] s12-c01 requires the authors-only label "Theorem (Elsenhans-Jahnel)". Full locator: "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. The old dated/numbered label is withdrawn.

2026-09-14 active speaker notes (review metadata is separate):

• The cokernel is torsion-free, so equality of ranks forces the specialization map to be an isomorphism of lattices.
• Integral information descends. An effective divisor of the predicted degree can be tested with Groebner bases.
• Here $X$ is a K3 surface over $\Bbb{Q}$ and $p>2$ is a prime of good reduction. The published theorem allows ramification degree $e<p-1$; over $\Bbb{Q}$, $e=1$.
• For reduction rank two, the Elsenhans-Jahnel approach computes the lattice, bounds a possible effective lift, and checks it with Groebner bases.
• Elsenhans and Jahnel had already given the first explicit degree-two examples over Q with geometric Picard rank one in 2008. Their torsion-free specialization theorem was published in 2011. Here "generic" means geometric Picard rank one.

APPLIED (2026-09-14): [s13-m06] s13-c01 authorizes the checked correction: add "$X/\Bbb{Q}$ K3; $p>2$ a prime of good reduction" before the theorem. Published Theorem 1.4 and Remarks 1.5(a), p. 1028, allow e<p-1 and specialize to e=1 over Q. The previous arXiv-based provenance was too narrow; the slide does not claim the result at arbitrary ramified places.

APPLIED (2026-09-15): [s13-m07] s13-c01: retain the first explicit degree-two K3 example over Q with geometric Picard rank one. The visible credit names only Elsenhans-Jahnel, per the standing citation rule. The ANTS VIII paper (2008), introduction and Corollary 30, constructs the examples after recalling the earlier degree-four examples. The separate torsion-free specialization theorem is Theorem 1.4 and Remarks 1.5(a) in the 2011 paper. Here generic means geometric Picard rank one.

AUTHOR'S CALL: [s13-m90] Keep the odd-prime restriction in the opening scope, in the theorem conclusion, or both? Both formulations are preserved pending your choice.

15 Endomorphisms of the transcendental Hodge structure

Order note. Applied ORDER-astra.md section 4. Historical arrangement prose and exact candidate quotations below retain their original wording; the draft allocation and heading give the current location.

2026-09-14 comments-E plan, recorded before draft edits (s15-c01--s15-c04):

1. Keep the title, T and E displays, and one proof line visible initially. Qualify T as the minimal rational sub-Hodge structure of H^2 whose complexification contains H^{2,0}. Explain the kernel/image argument and the faithful action on the one-dimensional H^{2,0} in spoken notes; record every proof step and primary-source locator in artifacts/orch/comments-E.md. 2. Add class "fragment" and data-fragment-index="0" to the theorem and the following dimension list. Change the existing SO-block fragment to index 1. Thus click 1 reveals Zarhin and d,m; click 2 reveals the existing Frobenius argument. No SO-block relocation in this pass. 3. Remove the visible X/k number-field bullet and both standalone definitions. Use exactly "$E$: a totally real field or a totally imaginary quadratic extension of one, i.e., a CM field" in the theorem. Preserve the number-field setting in spoken notes for the Frobenius block and later specialization slides. 4. Preserve all mark IDs and the settled attribution: Zarhin for the field classification, Charles for the specialization cases and eta. Visible citations contain authors only; full references belong in notes. Refresh this slide's candidate record without changing its ID. 5. Verify the parsed draft's reveal groups, requested deletions, proof and theorem text, stable IDs, and agreement with this plan entry. Append each verbatim comment to artifacts/deck-decisions.md; report completion only in artifacts/orch/comments-E.md. No other slides, database writes, publication, delegation, or git commands.

Source locator: I15:L523-527 (source introduction); Zarhin, "Hodge groups of K3 surfaces", J. reine angew. Math. 341 (1983), Thms. 1.4.1 (p. 205), 1.5.1 (p. 206), 1.6(a) and proof 1.6.1 (p. 207); van Geemen, "Real multiplication on K3 surfaces and Kuga Satake varieties", Michigan Math. J. 56 (2008), 375-399, Secs. 1.3, 1.5, 1.7-1.8, 2.1, 2.4, Thm. 2.8 and Lem. 3.2; Charles, "On the Picard number of K3 surfaces over number fields", Algebra & Number Theory 8 (2014), 1-17, Thm. 1 (p. 3), Prop. 15 (pp. 8-9), Lem. 16 (p. 9); Ito-Ito-Koshikawa, arXiv:1809.09604.

Current allocation: definitions, Zarhin and the full SO block are reunited on slide 15 before Charles on slide 16. Historical draft arrangement, after the plan above. Comment completion is recorded only in artifacts/orch/comments-E.md.

Initial view:

$$T:=T(X)_{\Bbb{Q}}=c_1(\operatorname{Pic}(X^{\mathrm{al}}))_{\Bbb{Q}}^{\perp}\subset H^2(X_{\Bbb{C}},\Bbb{Q})$$

Cross-reference note. [s18-m90] Define T_ell here by the characteristic-zero geometric Pic orthogonal complement and comparison. The existing SO setup on this same slide then sets V:=T_ell(1). The old O/SO placement is superseded.

$$E:=\operatorname{End}_{\mathrm{Hdg}}(T)=\{a\in\operatorname{End}_{\Bbb{Q}}(T):a_{\Bbb{C}}(T^{i,j})\subset T^{i,j}\}$$

$T$ minimal rational sub-Hodge structure of $H^2$ with $H^{2,0}\subset T_{\Bbb{C}}$ $\Rightarrow$ $(0\neq\alpha\in E\Rightarrow\alpha(H^{2,0})=H^{2,0}\Rightarrow\operatorname{im}\alpha=T\Rightarrow\alpha^{-1}\in E)$

Click 1, data-fragment-index="0": the theorem and its following dimension list.

> **Theorem (Zarhin)** > > $E$: a totally real field or a totally imaginary quadratic extension of one, i.e., a CM field

• $d:=[E:\Bbb{Q}]$, $m:=\dim_E T$; $dm=22-\rho(X^{\mathrm{al}})$
• $E$ totally real $\Rightarrow m\geq3$ [van Geemen]

What it does. Makes E a field before using dim_E T; separates the classification from the initial proof. The number-field setting is spoken, since the Frobenius block and later arithmetic slides still require it. Both standalone definitions are removed; the CM definition is in Zarhin's statement. Attribution is settled: Zarhin for classification, Charles for specialization and eta. Visible citations contain authors only.

Spoken (active notes, without review metadata):

• We are still working with a projective K3 surface $X/k$, with $k\subset\Bbb{C}$ a number field. For reduction, $p$ is a finite place of good reduction and $\ell$ differs from its residue characteristic.
• $T$ is the smallest rational sub-Hodge structure of $H^2(X_{\Bbb{C}},\Bbb{Q})$ whose complexification contains $H^{2,0}(X_{\Bbb{C}})$. That line has complex dimension one. Endomorphisms preserve the Hodge decomposition; their kernels and images are rational sub-Hodge structures.
• If $\alpha$ kills $H^{2,0}$, minimality gives $\ker\alpha=T$, hence $\alpha=0$. Otherwise its image contains that line, so minimality gives $\operatorname{im}\alpha=T$. Finite dimension gives $\ker\alpha=0$; the inverse also preserves the Hodge decomposition.
• Restriction to $H^{2,0}$ embeds $E$ into $\operatorname{End}_{\Bbb{C}}(H^{2,0})=\Bbb{C}$. Thus the division algebra is commutative, and finite dimensionality over $\Bbb{Q}$ makes it a number field. $E=\Bbb{Q}$ means no real or complex multiplication.
• Zarhin proves simplicity and fieldhood in "Hodge groups of K3 surfaces", J. reine angew. Math. 341 (1983), Theorem 1.6(a) and proof 1.6.1, p. 207; the classification is Theorem 1.5.1, p. 206. Van Geemen explains the Hodge structures in "Real multiplication on K3 surfaces and Kuga Satake varieties", Michigan Math. J. 56 (2008), 375-399, Sections 1.3, 1.5, 1.7-1.8 and 2.1; Lemma 3.2 gives $m\geq3$ in the totally real case.

Review records, with their stable IDs:

APPLIED (2026-09-14): [s14-m09] The existing invertibility line is CORRECT with its stated minimality, but the first implication uses the kernel argument. Clarified the ambient H^2. Exact line: $T$ minimal rational sub-Hodge structure of $H^2$ with $H^{2,0}\subset T_{\Bbb{C}}$ $\Rightarrow$ $(0\neq\alpha\in E\Rightarrow\alpha(H^{2,0})=H^{2,0}\Rightarrow\operatorname{im}\alpha=T\Rightarrow\alpha^{-1}\in E)$. If alpha kills H^{2,0}, its kernel is a rational sub-Hodge structure containing that line after complexification; minimality gives alpha=0. Otherwise the image contains H^{2,0}, hence equals T; finite dimension gives ker alpha=0 and the inverse is Hodge. Restriction to the one-dimensional H^{2,0} embeds E into C, proving commutativity. Source: Zarhin, Thm. 1.6(a), proof 1.6.1, p. 207; van Geemen, Secs. 1.3, 1.5, 1.7-1.8. Full check: artifacts/orch/comments-E.md, s15-c01. The stable mark ID is retained.

SETTLED (2026-09-14): [s14-m11] Zarhin owns the classification of E as totally real or CM; Charles owns the specialization cases and eta. Source: Zarhin, Thms. 1.5.1 and 1.6(a); Charles, Thm. 1 and Prop. 15. This supersedes the earlier attribution question; no new verdict is requested.

The author requires the full SO block rejoined with the Hodge definitions on slide 15, before Charles on slide 16. The earlier split-authorization claim is superseded.

NEEDS APPROVAL: [s14-m06] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Author comments s15-c01--s15-c04 are incorporated; other arrangement decisions retain this stable ID. Allocation note: Current allocation: the complete original candidate is reunited on slide 15 before Charles on slide 16. Definitions, invertibility, Zarhin, dimensions, the Frobenius/SO block and all original notes are retained. The new T_ell definition is covered by s18-m90. Historical quotations retain their original wording and IDs. Exact candidate text: "Endomorphisms of the transcendental Hodge structure" | "$$T:=T(X)_{\Bbb{Q}}=c_1(\operatorname{Pic}(X^{\mathrm{al}}))_{\Bbb{Q}}^{\perp}\subset H^2(X_{\Bbb{C}},\Bbb{Q})$$" | "$$E:=\operatorname{End}_{\mathrm{Hdg}}(T)=\{a\in\operatorname{End}_{\Bbb{Q}}(T):a_{\Bbb{C}}(T^{i,j})\subset T^{i,j}\}$$" | "$T$ minimal rational sub-Hodge structure of $H^2$ with $H^{2,0}\subset T_{\Bbb{C}}$ $\Rightarrow$ $(0\neq\alpha\in E\Rightarrow\alpha(H^{2,0})=H^{2,0}\Rightarrow\operatorname{im}\alpha=T\Rightarrow\alpha^{-1}\in E)$" | "Theorem (Zarhin)" | "$E$: a totally real field or a totally imaginary quadratic extension of one, i.e., a CM field" | "$d:=[E:\Bbb{Q}]$, $m:=\dim_E T$; $dm=22-\rho(X^{\mathrm{al}})$" | "$E$ totally real $\Rightarrow m\geq3$ [van Geemen]" | "$E$ totally real; $V:=T_\ell(1)$; $g=\operatorname{Frob}_p^a$ in connected monodromy" | "$$V\otimes\Bbb{Q}_\ell^{\mathrm{al}}=\bigoplus_{\sigma:E\hookrightarrow\Bbb{Q}_\ell^{\mathrm{al}}}V_\sigma,\quad\dim V_\sigma=m,\quad g|V_\sigma\in SO(V_\sigma)$$" | "$$m\text{ odd}\Rightarrow\dim\ker(g-1)\geq d\Rightarrow\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+d$$" Speaker notes: "We are still working with a projective K3 surface $X/k$, with $k\subset\Bbb{C}$ a number field. For reduction, $p$ is a finite place of good reduction and $\ell$ differs from its residue characteristic." | "$T$ is the smallest rational sub-Hodge structure of $H^2(X_{\Bbb{C}},\Bbb{Q})$ whose complexification contains $H^{2,0}(X_{\Bbb{C}})$. That line has complex dimension one. Endomorphisms preserve the Hodge decomposition; their kernels and images are rational sub-Hodge structures." | "If $\alpha$ kills $H^{2,0}$, minimality gives $\ker\alpha=T$, hence $\alpha=0$. Otherwise its image contains that line, so minimality gives $\operatorname{im}\alpha=T$. Finite dimension gives $\ker\alpha=0$; the inverse also preserves the Hodge decomposition." | "Restriction to $H^{2,0}$ embeds $E$ into $\operatorname{End}_{\Bbb{C}}(H^{2,0})=\Bbb{C}$. Thus the division algebra is commutative, and finite dimensionality over $\Bbb{Q}$ makes it a number field. $E=\Bbb{Q}$ means no real or complex multiplication." | "Zarhin proves simplicity and fieldhood in "Hodge groups of K3 surfaces", J. reine angew. Math. 341 (1983), Theorem 1.6(a) and proof 1.6.1, p. 207; the classification is Theorem 1.5.1, p. 206. Van Geemen explains the Hodge structures in "Real multiplication on K3 surfaces and Kuga Satake varieties", Michigan Math. J. 56 (2008), 375-399, Sections 1.3, 1.5, 1.7-1.8 and 2.1; Lemma 3.2 gives $m\geq3$ in the totally real case." | "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." | "These new cyclotomic roots are removed from $h$. Commutation with $E$ does not force every eigenvalue orbit to have size $d$." | "Zarhin classifies the Hodge endomorphism field. Charles computes the minimum increase of the geometric Picard rank under specialization in "On the Picard number of K3 surfaces over number fields", Algebra & Number Theory 8 (2014), 1-17, Theorem 1, p. 3; Proposition 15 and Lemma 16, pp. 8-9, give the fixed-space argument. The Tate theorem holds in every residue characteristic; see Ito-Ito-Koshikawa, arXiv:1809.09604." Checked sources: I15:L523-527; Zarhin, Hodge groups of K3 surfaces (1983), Thms. 1.4.1 (p. 205), 1.5.1 (p. 206), 1.6(a) and proof 1.6.1 (p. 207); van Geemen, Real multiplication on K3 surfaces and Kuga Satake varieties, Secs. 1.3, 1.5, 1.7-1.8, 2.1, 2.4, Thm. 2.8 and Lem. 3.2; Charles, On the Picard number of K3 surfaces over number fields (2014), Thm. 1 (p. 3), Prop. 15 (pp. 8-9) and Lem. 16 (p. 9).

Cross-reference note. s27-c01 rejoins this complete Frobenius block with endomorphisms before Charles. Index 1 adds it after Zarhin and dimensions; earlier content remains visible.

Order note. Applied ORDER-astra.md section 4. Historical arrangement prose and exact candidate quotations below retain their original wording; the draft allocation and heading give the current location.

Spoken justification for the reunited SO block: suppose $E$ is totally real. Choose $a>0$ with $g=\operatorname{Frob}_p^a$ in connected monodromy. The Tate twist divides the original Frobenius eigenvalues by the residue-field size.

Spoken: This is the forced-eigenspace assertion of Charles, Proposition 15(2), with Lemma 16 for connected monodromy. Before taking the power, these eigenvalues are roots of unity; Tate supplies the new geometric divisor classes. The new cyclotomic roots are removed from $h(t)$ of slide 14. Commutation with $E$ does not force every eigenvalue orbit to have size $d$.

Click 2, retained data-fragment-index="1": add the complete existing Frobenius block; keep initial content and group 0 visible.

• $E$ totally real; $V:=T(1)\otimes\Bbb{Q}_\ell$; $g=\operatorname{Frob}_p^a$ in connected monodromy

$$V\otimes\Bbb{Q}_\ell^{\mathrm{al}}=\bigoplus_{\sigma:E\hookrightarrow\Bbb{Q}_\ell^{\mathrm{al}}}V_\sigma,\quad\dim V_\sigma=m,\quad g|V_\sigma\in SO(V_\sigma)$$

$$m\text{ odd}\Rightarrow\dim\ker(g-1)\geq d\Rightarrow\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+d$$

What it does. Supplies the SO-block reason before Charles states the minimum.

• 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.
• These new cyclotomic roots are removed from $h$. Commutation with $E$ does not force every eigenvalue orbit to have size $d$.
• Zarhin classifies the Hodge endomorphism field. Charles computes the minimum increase of the geometric Picard rank under specialization in "On the Picard number of K3 surfaces over number fields", Algebra & Number Theory 8 (2014), 1-17, Theorem 1, p. 3; Proposition 15 and Lemma 16, pp. 8-9, give the fixed-space argument. The Tate theorem holds in every residue characteristic; see Ito-Ito-Koshikawa, arXiv:1809.09604. NEEDS APPROVAL: [s14-m04] Recommendation: Keep the qualified SO-block statement and the odd-m condition on slide 15, before Charles. GPT 6 astra: Use embedding blocks after a suitable Frobenius power; reject fixed root-orbit sizes. GPT 5.6 sol: Use the same qualified block decomposition; reject fixed root-orbit sizes. Source: Zarhin 1983, Thms. 1.5.1, 1.6(a), 2.2.1; van Geemen 2008, Lem. 3.2; Charles 2014, Prop. 15 and Lem. 16.

16 Computing $\rho(X^{\mathrm{al}})$

Cross-reference note. The retained What it does. line describes the original combined Charles slide. The minimum is here; the prime pairs are now on slide 25.

Order note. Applied ORDER-astra.md section 4. Historical arrangement prose and exact candidate quotations below retain their original wording; the draft allocation and heading give the current location.

• $X/k$ projective K3; $k\subset\Bbb{C}$ a number field; $p$ a finite place of good reduction.
• $r:=\rho(X^{\mathrm{al}})$, $r_p:=\rho(X_p^{\mathrm{al}})$; $E,T,d,m$ as on slide 15.

> **Theorem (Charles; using finite-field Tate as recalled earlier).** At every good place, > > $$\boxed{r_p\geq\begin{cases}r&\text{if }E\text{ is CM or }m\text{ is even},\\r+d&\text{if }E\text{ is totally real and }m\text{ is odd}.\end{cases}}$$ > > Equality holds at infinitely many good places. After a suitable finite extension of $k$, equality holds at a set of places of density $1$.

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

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

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

Theorem (Charles)

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

With density $1$,

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

Cross-reference note. The minimum and prime-pair discriminants are together in "Computing rho(X^{al})" (slide 16).

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

Cross-reference note. [s22-m90] The historical sentence "Further, assume that we are in the second case" refers to Charles's theorem in "Computing rho(X^{al})": E is totally real and m is odd. "The equality" means both reduction ranks equal rho(X^{al})+d. The active sentence and notes now name that theorem explicitly. Historical candidate quotations in s22-m04 and s22-m07 retain their original wording.

NEEDS APPROVAL: [s22-m04] Recommendation: Keep the source sentence "Further, assume that we are in the second case, then exist infinitely many pairs (p,q) such that the equality holds and" before the geometric discriminant display; say ordinary in the notes. Source: I15:L539-543; Charles 2014, Prop. 18.

NEEDS APPROVAL: [s22-m05] Recommendation: Recall T=T(X)_Q from slide 15; use algebraic divisor classes in its definition. SETTLED: the complement is algebraic Pic, not topological line bundles. Source: Charles 2014, introduction. Retained correction: T is the orthogonal complement of algebraic Pic in rational H^2, not topological line bundles. Source: Charles 2014, introduction. The repaired body was already present before this pass.

NEEDS APPROVAL: [s22-m06] Recommendation: Keep geometric discriminants and the conjunction "totally real and m odd". SETTLED by the theorem: both geometric base changes and AND are necessary corrections to I15:L533-541. Source: Charles 2014, Thm. 1 and Prop. 18. Content anchors: the conjunction is in "Computing rho(X^{al})" (slide 16); the geometric discriminants are in "Two primes at the minimum" (slide 25). This single ID covers both parts.

NEEDS APPROVAL: [s22-m07] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "Computing $\rho(X^{\mathrm{al}})$" | "$T=T(X)_{\Bbb{Q}}$; $E=\operatorname{End}_{\mathrm{Hdg}}(T)$; $d=[E:\Bbb{Q}]$; $m=\dim_E T$" | "Theorem (Charles)" | "$$\rho(X_p^{\mathrm{al}})\geq\begin{cases}\rho(X^{\mathrm{al}})&\text{if }E\text{ is CM or }m\text{ is even,}\\\rho(X^{\mathrm{al}})+d&\text{if }E\text{ is totally real and }m\text{ is odd.}\end{cases}$$" | "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" | "$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$" Speaker notes proposed: "Charles computes the minimum and proves its attainment. Density one is over a suitable finite extension, not necessarily over the original field." | "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." | "The primes are good; the pairs can be chosen ordinary, with both ranks equal to $\rho(X^{\mathrm{al}})+d$. The minimum $\eta$ is zero in the first case and $d$ in the second." SETTLED mathematical source: I15:L523-550; Charles 2014, Thm. 1 and Prop. 18. Exact teaching arrangement still needs approval. Allocation note: Allocation after the move: this single ID still covers the complete original candidate. Charles's minimum and equality are on slide 16; its original prime-pair sentence and discriminants are on slide 25. The implicit second-case reference is flagged by s22-m90.

2026-09-14 active speaker notes (review metadata is separate):

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

Provenance: corrected transcription, retaining the already marked mathematical deviations from the author frame.

27 A real multiplication example

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
• $X$: minimal resolution of

• $6$ lines; $15=\binom{6}{2}$ nodes; $15$ exceptional $(-2)$-curves
• $H,E_{ij}$: $16$ independent classes
• $\rho(X^{\mathrm{al}})=16$
• RM: $E=\Bbb{Q}(\sqrt{2})$; $\dim_E T=(22-16)/2=3$
• $\eta=2$; $\rho(X_p^{\mathrm{al}})\geq18$ at every good prime
• Rank-$18$ pair at $p=17$, $q=23$, unequal square classes, certified RM $\Rightarrow$ $\rho(X^{\mathrm{al}})\leq16$ Without certified real multiplication, one could only prove $\rho(X^{\mathrm{al}})\leq17$.

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 $\Bbb{Q}(\sqrt2)$ into six lines; no three meet. $H$ is the pullback of a general line. With the fifteen exceptional curves its Gram matrix is $\operatorname{diag}(2,-2,\ldots,-2)$, 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. Point counts at the good primes $p=17$ and $q=23$ give both geometric ranks $18$; Artin-Tate gives the unequal square classes after extending the residue fields to define all divisor classes. Specialization gives $\rho(X^{\mathrm{al}})\leq18$; the unequal square classes exclude $18$, so $\rho(X^{\mathrm{al}})\leq17$. Certified quadratic RM makes $2$ divide $22-\rho(X^{\mathrm{al}})$, so $\rho(X^{\mathrm{al}})$ is even and at most $16$. The sixteen independent classes $H,E_{ij}$ give $\rho(X^{\mathrm{al}})\geq16$, hence equality. Point counts alone do not prove characteristic-zero rank $16$.

Spoken: Further integral generators satisfy $2D_i=H+\sum_{j\neq i}E_{ij}$. For $w^2=\prod_i l_i$, the quintic $\prod_{j\neq i}l_j-l_i^5=0$ splits into $w=\pm l_i^3$. 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.

AUTHOR'S CALL: [s24-m01] Recommendation: Keep the certified-RM application after the known sixteen classes on slide 27. The authorized running order keeps the certified bound on slide 26, before the example on slide 27. The equation, nodes, rank and field remain unchanged. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Elsenhans-Jahnel 2014, proof of Thm. 6.6; Charles 2014, Prop. 23. The authorized running order keeps the complete certified bound on slide 26 before this example on slide 27; the earlier KEEP/DROP placement alternative is superseded.

AUTHOR'S CALL: [s24-m02] Recommendation: Keep the certified-RM application after the known sixteen classes on slide 27. The authorized running order keeps the certified bound on slide 26, before the example on slide 27. The equation, nodes, rank and field remain unchanged. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Elsenhans-Jahnel 2014, proof of Thm. 6.6; Charles 2014, Prop. 23. The authorized running order keeps the complete certified bound on slide 26 before this example on slide 27; the earlier KEEP/DROP placement alternative is superseded.

NEEDS APPROVAL: [s24-m03] Recommendation: Keep the equation and the short geometry, rank, RM and certified-bound bullets shown in the candidate. Source: Elsenhans-Jahnel 2014, Thms. 5.12 and 6.6. The authorized running order keeps the complete certified bound on slide 26 before this example on slide 27; the earlier KEEP/DROP placement alternative is superseded.

AUTHOR'S CALL: [s24-m04] Recommendation: Show the equation and credit first; reveal geometry at 0, rank and RM at 1, and the certified bound at 2. GPT 6 astra: Use geometry, then rank/RM, then the method application. GPT 5.6 sol: Use three groups: equation/source, geometry/rank, then RM and the certified conclusion. Source: Elsenhans-Jahnel 2014, Thm. 6.6. The authorized running order keeps the complete certified bound on slide 26 before this example on slide 27; the earlier KEEP/DROP placement alternative is superseded.

APPROVED (2026-09-14): [s24-m05] title "A real multiplication example"; correct the intended word "multiplication" from the author's typo. Author, verbatim: "Regarding Slide : "A surface where that happens", The title should be "A real multiplaction example", we should credit Elsenhans and Jahnel, We should explain that 15 = 6 choose 2. I think Elsenhans--Jahnel even tell us the shape of the extra cycles. I do not understand the questions about that slide"

APPROVED (2026-09-14): [s24-m06] credit Elsenhans-Jahnel 2014, Theorems 5.12 and 6.6; this is X^(2,1). Explain 15 = 6 choose 2 nodes and 15 exceptional (-2)-curves. Author, verbatim: "Regarding Slide : "A surface where that happens", The title should be "A real multiplaction example", we should credit Elsenhans and Jahnel, We should explain that 15 = 6 choose 2. I think Elsenhans--Jahnel even tell us the shape of the extra cycles. I do not understand the questions about that slide"

NEEDS APPROVAL: [s24-m07] Recommendation: Keep the split-quintic integral generators in notes; leave the two reduction-only representatives explicitly unverified. No clear answer for explicit representatives of the two new classes at 83 was found. GPT 6 astra: The split quintics complete the same rank-16 integral lattice; the reduction-only shapes are unverified. GPT 5.6 sol: Give no description of the two additional reduction classes; their representatives are unverified. Source: Elsenhans-Jahnel, period integration, Rem. 4.6; 2-adic point counting, Lem. 3.11. UNVERIFIED: explicit representatives in Pic(X_83^{al})_Q / sp(Pic(X^{al})_Q). Checked EJ-RM Theorem 6.6 and family definition; period integration Remark 4.6; 2-adic point counting Lemma 3.11, equation (11). These passages supply no representatives for the two quotient directions. The split component has class D_i+2H; D_i is an integral saturation generator. The author can supply another exact locator.

DECIDED, SUPERSEDED (2026-09-14): [s24-m08] Superseded by the explicit approved Elsenhans-Jahnel credit in s24-m06. Retain this ID and its history; no separate credit decision remains.

NEEDS APPROVAL: [s24-m09] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "A real multiplication example" | "Elsenhans-Jahnel" | "$X$: minimal resolution of" | "$$w^2=(-y^2/8+yz-z^2)(7x^2/8+5xz+7z^2)(2x^2+3xy+y^2)$$" | "$6$ lines; $15=\binom{6}{2}$ nodes; $15$ exceptional $(-2)$-curves" | "$H,E_{ij}$: $16$ independent classes" | "$\rho(X^{\mathrm{al}})=16$" | "RM: $E=\Bbb{Q}(\sqrt{2})$; $\dim_E T=(22-16)/2=3$" | "$\eta=2$; $\rho(X_p^{\mathrm{al}})\geq18$ at every good prime" | "Rank-$18$ pair, unequal square classes, certified RM $\Rightarrow$ $\rho(X^{\mathrm{al}})\leq16$" Speaker notes proposed: "This is $X^{(2,1)}$ in Elsenhans-Jahnel 2014. The three quadratics split over $\Bbb{Q}(\sqrt2)$ into six lines; no three meet. $H$ is the pullback of a general line. With the fifteen exceptional curves its Gram matrix is $\operatorname{diag}(2,-2,\ldots,-2)$, determinant $-65536$." | "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." | "Further integral generators satisfy $2D_i=H+\sum_{j\neq i}E_{ij}$. For $w^2=\prod_i l_i$, the quintic $\prod_{j\neq i}l_j-l_i^5=0$ splits into $w=\pm l_i^3$; a split component has class $D_i+2H$. The full characteristic-zero lattice has index thirty-two over the displayed sublattice and discriminant $-64$. These generators add no rational rank and do not identify the two new reduction classes." | "At eighty-three there are two additional divisor-class directions. We have not identified explicit curves representing them." SETTLED mathematical source: 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. Exact teaching arrangement still needs approval. SETTLED author instructions, s24-m05 and s24-m06: title "A real multiplication example"; credit Elsenhans and Jahnel; explain "15 = 6 choose 2". Approval is requested for the additional content and arrangement. Allocation note: Allocation after the move: the complete original example and its unchanged reveals are on slide 27, after the certified bound on slide 26.

Count: 22 candidate content bodies plus the unchanged title. Existing mark IDs retain their original subject association. No Lecture 2 or 3 changes.

2026-09-14 active speaker notes (review metadata is separate):

• This is $X^{(2,1)}$ in Elsenhans-Jahnel 2014. The three quadratics split over $\Bbb{Q}(\sqrt2)$ into six lines; no three meet. $H$ is the pullback of a general line. With the fifteen exceptional curves its Gram matrix is $\operatorname{diag}(2,-2,\ldots,-2)$, determinant $-65536$.
• 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. Point counts at the good primes $p=17$ and $q=23$ give both geometric ranks $18$; Artin-Tate gives the unequal square classes after extending the residue fields to define all divisor classes. Specialization gives $\rho(X^{\mathrm{al}})\leq18$; the unequal square classes exclude $18$, so $\rho(X^{\mathrm{al}})\leq17$. Certified quadratic RM makes $2$ divide $22-\rho(X^{\mathrm{al}})$, so $\rho(X^{\mathrm{al}})$ is even and at most $16$. The sixteen independent classes $H,E_{ij}$ give $\rho(X^{\mathrm{al}})\geq16$, hence equality. Point counts alone do not prove characteristic-zero rank $16$.
• Further integral generators satisfy $2D_i=H+\sum_{j\neq i}E_{ij}$. For $w^2=\prod_i l_i$, the quintic $\prod_{j\neq i}l_j-l_i^5=0$ splits into $w=\pm l_i^3$; a split component has class $D_i+2H$. The full characteristic-zero lattice has index thirty-two over the displayed sublattice and discriminant $-64$. These generators add no rational rank and do not identify the two new reduction classes.
• At eighty-three there are two additional divisor-class directions. We have not identified explicit curves representing them.

Citation references (speaker notes; s12-c01): Elsenhans-Jahnel, "Examples of K3 surfaces with real multiplication" (2014), arXiv:1402.4555v2, Theorems 5.12 and 6.6.

AUTHOR'S CALL: [s24-m90] Within s24-m01, keep the opening placement recommendation, the later placement note, or both? Their wording differs; both are preserved pending your choice.

AUTHOR'S CALL: [s23-m01] Recommendation: KEEP the interpretation on slide 26; Charles's minimum is on slide 16, the prime pairs are on slide 25, and the RM example follows on slide 27. The earlier DROP alternative is superseded by the authorized running order; the certified bound stays on slide 26. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Charles 2014, Thm. 1, Props. 18 and 23.

AUTHOR'S CALL: [s23-m02] Recommendation: Keep the certified quadratic-RM box, including both rank-18 reductions and unequal geometric square classes. GPT 6 astra: Show the elementary 18 -> 17 -> 16 argument. GPT 5.6 sol: Show the degree-sensitive subtraction, then its quadratic instance. Source: Charles 2014, Prop. 23.

NEEDS APPROVAL: [s23-m03] Recommendation: Keep "eta: forced minimum; an individual reduction can exceed it" in the notes. SETTLED: Charles proves the minimum and its attainment. Source: Charles 2014, Thm. 1.

APPROVED (2026-09-14): [s23-m04] matching lower bound on slide 12, via s12-m02; symplectic order-5 action and a polarization, Garbagnati-Sarti, Proposition 1.1. Author, verbatim: "is it symplectic? if so we should reference the paper"

NEEDS APPROVAL: [s23-m05] Recommendation: Keep the two cases with their actual two-prime bounds; reserve the broad discussion for the notes. Source: Charles 2014, Remark 19.

AUTHOR'S CALL: [s23-m06] Recommendation: Show the parity case and rank-17 example first; reveal nontrivial RM at 0 and the certified bound at 1. Source: Charles 2014, Thm. 1 and Prop. 23.

AUTHOR'S CALL: [s23-m07] Recommendation: KEEP the interpretation on slide 26; Charles's minimum is on slide 16, the prime pairs are on slide 25, and the RM example follows on slide 27. The earlier DROP alternative is superseded by the authorized running order; the certified bound stays on slide 26. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Charles 2014, Thm. 1, Props. 18 and 23.

NEEDS APPROVAL: [s23-m08] Recommendation: Keep "Charles 2014" visible; cite van Geemen 2008, Lem. 3.2, after Zarhin 1983, in the notes. Source: reference-years, entries 23, 43, 46.

NEEDS APPROVAL: [s23-m09] Recommendation: Keep "eta: forced minimum; an individual reduction can exceed it" in the notes. SETTLED: Charles proves the minimum and its attainment. Source: Charles 2014, Thm. 1.

NEEDS APPROVAL: [s23-m10] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "When every prime overshoots" | "$r:=\rho(X^{\mathrm{al}})$; $d:=[E:\Bbb{Q}]$; $m:=\dim_E T$" | "$E=\Bbb{Q}$, $m$ odd $\Rightarrow$ $\eta=1$; van Luijk succeeds [Charles] $$\text{Order-5 example:}\quad17\leq\rho(X^{\mathrm{al}})<18$$" | "$$\text{Order-5 example:}\quad17\leq\rho(X^{\mathrm{al}})<18$$" | "$E$ totally real, $E\neq\Bbb{Q}$, $m$ odd $\Rightarrow$ $\eta=d\geq2$ $$\min_p\rho(X_p^{\mathrm{al}})=r+d;\qquad\text{two-prime upper bound: }r+d-1$$" | "$$\min_p\rho(X_p^{\mathrm{al}})=r+d;\qquad\text{two-prime upper bound: }r+d-1$$" | "Certified quadratic RM" | "$$F\hookrightarrow E,\quad[F:\Bbb{Q}]=2;\qquad\rho(X_p^{\mathrm{al}})=\rho(X_q^{\mathrm{al}})=18$$" | "$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$" | "$$\rho(X^{\mathrm{al}})\leq17,\quad\rho(X^{\mathrm{al}})\text{ even}\quad\Rightarrow\quad\rho(X^{\mathrm{al}})\leq16$$" Speaker notes proposed: "$\eta$ 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." | "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." | "For certified quadratic RM, $2$ divides $22-\rho$, so $\rho$ is even. The unequal rank-eighteen discriminants exclude eighteen, hence give at most sixteen. This is the elementary quadratic case of Charles, Proposition 23." | "A projective Kummer surface has transcendental dimension at most five. Nontrivial totally real multiplication requires $dm\geq2\cdot3=6$ by van Geemen 2008, Lemma 3.2, after Zarhin 1983. Kummer surfaces avoid this obstruction even when their determinant character is nontrivial." SETTLED mathematical source: Charles 2014, Thm. 1, Remark 19 and Prop. 23; Garbagnati-Sarti 2007, Prop. 1.1. Exact teaching arrangement still needs approval.

2026-09-14 active speaker notes (review metadata is separate):

• $\eta$ 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.
• 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.
• For certified quadratic RM, $2$ divides $22-\rho$, so $\rho$ is even. The unequal rank-eighteen discriminants exclude eighteen, hence give at most sixteen. This is the elementary quadratic case of Charles, Proposition 23.
• A projective Kummer surface has transcendental dimension at most five. Nontrivial totally real multiplication requires $dm\geq2\cdot3=6$ by van Geemen 2008, Lemma 3.2, after Zarhin 1983. Kummer surfaces avoid this obstruction even when their determinant character is nontrivial.

AUTHOR'S CALL: [s22-m01] Recommendation: KEEP the interpretation on slide 26; Charles's minimum is on slide 16, the prime pairs are on slide 25, and the RM example follows on slide 27. The earlier DROP alternative is superseded by the authorized running order; the certified bound stays on slide 26. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Charles 2014, Thm. 1, Props. 18 and 23.

AUTHOR'S CALL: [s22-m02] Recommendation: KEEP the interpretation on slide 26; Charles's minimum is on slide 16, the prime pairs are on slide 25, and the RM example follows on slide 27. The earlier DROP alternative is superseded by the authorized running order; the certified bound stays on slide 26. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Charles 2014, Thm. 1, Props. 18 and 23.

NEEDS APPROVAL: [s22-m03] Recommendation: KEEP the interpretation on slide 26; Charles's minimum is on slide 16, the prime pairs are on slide 25, and the RM example follows on slide 27. The earlier DROP alternative is superseded by the authorized running order; the certified bound stays on slide 26. GPT 6 astra: KEEP the interpretation between Charles and the RM example. GPT 5.6 sol: KEEP the interpretation; the earlier absorption proposal is withdrawn. Source: Charles 2014, Thm. 1, Props. 18 and 23.

17 Infinitely many rational curves

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 $\rho(X^{\mathrm{al}})\leq\rho(X_p^{\mathrm{al}})$.Can we use the inequality to our advantage?

Theorem (Li-Liedtke)

If there are infinitely many $p$ primes such that

then $X^{\mathrm{al}}$ contains infinitely many rational curves.

Theorem (Bogomolov-Zarhin)

The set $\{p:\rho(X_p^{\mathrm{al}})\neq22\}$ has positive density (density 1 after finite extension).

What it does. Uses odd Picard rank to produce infinitely many integral rational curves. The general lifting theorem is also used in the later half-density application.

APPROVED (2026-09-14): [s15-m01] Historical approval, superseded by s17-c02: generalize the odd-rank statement to the two verified cases; label "Theorem (Li-Liedtke; C-Elsenhans-Jahnel)". Li-Liedtke supplies odd rank; Costa-Elsenhans-Jahnel supplies even rank, no real or complex multiplication, and a nontrivial jump character. Author, verbatim: "We should write the "Corollary (Li-Liedtke)" more generically, so we can use it immediately when we show the density is at least 1/2. We can add our names to it also. In particular, this should help with the delivery in slide "We can explain the 1/2", and now the cororllary is obvious"

REFUSED (2026-09-14), CHECK FAILED: [s15-m02] the proposed universal Kummer-to-SO assertion is false on the transcendental representation. Costa-Elsenhans-Jahnel 2020, Example 2.36(b): rank-18 Kummer surfaces from quadratic-conjugate elliptic factors have a nontrivial jump character. No universal assertion added. Author, verbatim: "I am also unsure what is the purpose of Slide 17, in particular given Slide 16, some of teh questions are already answered in the previous slide. On slide 15, not sure we should write "The jump criterion on the slides that immediately follow is a consequence of that one fact.". Intead, we should point, there is an easy way to explain some jumps, O vs SO . And maybe there one should point out that for kummer varieties we always land in SO (check this for me please)"

APPLIED (2026-09-15): [s15-m03] s17-c02: the final box is "Corollary (Li-Liedtke)". $\rho(X^{\mathrm{al}})$ odd $\Rightarrow$ infinitely many integral rational curves on $X^{\mathrm{al}}$. Jun Li and Christian Liedtke, "Rational curves on K3 surfaces", Inventiones Mathematicae 188 (2012), 713-727; arXiv:1012.3777, introduction and Theorem 3.3. The introduction says integral; the proof produces integral rational curves of arbitrarily large degree.

NEEDS APPROVAL: [s15-m05] Recommendation: Keep the source Bogomolov-Zarhin box; speak the credit "Positive density: Joshi-Rajan; density one after finite extension: Bogomolov-Zarhin." Source: Bogomolov-Zarhin 2009, Thm. 0.1 and following note.

NEEDS APPROVAL: [s15-m07] Review the current candidate below; s17-c02 settles the odd-rank corollary and its required cross-references. The title choice remains open under s15-m08. Exact candidate text: "K3 surfaces" | "So far we have been trying to improve the inequality $\rho(X^{\mathrm{al}})\leq\rho(X_p^{\mathrm{al}})$." | "Can we use the inequality to our advantage?" | "Theorem (Li-Liedtke)" | "If there are infinitely many $p$ primes such that" | "$$\rho(X^{\mathrm{al}})<\rho(X_p^{\mathrm{al}})\text{ and }\rho(X_p^{\mathrm{al}})\neq22,$$" | "then $X^{\mathrm{al}}$ contains infinitely many rational curves." | "Theorem (Bogomolov-Zarhin)" | "The set $\{p:\rho(X_p^{\mathrm{al}})\neq22\}$ has positive density (density 1 after finite extension)." | "Corollary (Li-Liedtke)" | "$\rho(X^{\mathrm{al}})$ odd $\Rightarrow$ infinitely many integral rational curves on $X^{\mathrm{al}}$."

Active speaker notes:

• 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.
• Odd geometric rank and even reduction rank force an increase at every good prime. Infinitely many of these reductions are non-supersingular, so the lifting theorem gives the corollary.
• Jun Li and Christian Liedtke, "Rational curves on K3 surfaces", Inventiones Mathematicae 188 (2012), 713-727; arXiv:1012.3777, introduction and Theorem 3.3. The introduction says integral; the proof produces integral rational curves of arbitrarily large degree.

NEEDS APPROVAL: [s15-m08] s17-c03. Title candidates: "Rational curves"; "Infinitely many rational curves"; "What do rank jumps give us?". Recommendation: "Rational curves". The current title remains until the author chooses.

18 Jumping Picard ranks

Cross-reference note. [s14-m90] The historical speaker note said "Charles will compute this minimum and prove its attainment." Applied s17-c01: the active note now points back to Charles on slide 16. Its case formula identifies eta as 0 or d; the quotation is retained as history.

Cross-reference note. The retained What it does. line describes the earlier combined proposal. Charles is on slide 16; this slide names the excess and asks the jump questions. The active spoken reference now points back to Charles on slide 16; s14-m90 retains its earlier finding.

Order note. Applied ORDER-astra.md section 4. Historical arrangement prose and exact candidate quotations below retain their original wording; the draft allocation and heading give the current location.

Source locator: no frame yet for this arrangement; I15:L523-550; V:L559-576; Charles, "On the Picard number of K3 surfaces over number fields", Algebra & Number Theory 8 (2014), Thm. 1 (p. 3), Prop. 15(2) (p. 8), Lem. 16 (p. 9) and proof of Thm. 1 (pp. 9-10).

PROPOSED (2026-09-14): [s14-m10] Charles on slide 16 before eta on slide 18; author, verbatim: "Slide 15, "Endomorphisms of the transcendental Hodge structure", we should add one proof line "alpha must preserve H^2,0, as T is minimal, and thus invertible". We should make clear that this is a theorem of charles. Maybe have that in slide 16, and then eta definition is natraul". The quoted "Maybe" is historical. The later reorder instruction authorizes the separate Charles and eta slides specified by ORDER-astra.md section 4.

Define the minimum excess, now computed by the theorem:

Consider

Is this set infinite? What is its density?

What about, for $X/\Bbb{Q}$,

What it does. States Charles's sharp bound before naming its minimum excess, then asks which primes exceed that minimum.

Spoken: Charles's printed Theorem 1(2) states the lower bound in residue characteristic at least $5$. The all-good-places formulation above uses the finite-field Tate theorem recalled earlier and the same proof. Density one is asserted after finite extension, not necessarily over $k$. The lecture's symbol $\eta$ names the minimum that the theorem computes. Odd $r$ forces an increase but does not imply $\eta=1$; $E=\Bbb{Q}$ and odd $m$ does. Over a number field count places by norm; the displayed $\gamma$ counts rational primes.

Content accounting, APPLIED: slide 16 gives the sharp minimum and equality; slide 18 keeps $\eta$, $\Pi_{\mathrm{jump}}$, the infinitude/density questions and $\gamma$; slide 15 gives the reunited SO proof. Slide 25 gives infinitely many ordinary pairs $(p,q)$ with $r_p=r_q=r+d$ and distinct geometric $\operatorname{disc}\operatorname{Pic}$ square classes. Slides 26-27 retain the interpretation, certified RM bound and example.

Dependency check: slide 23 supplies the determinant bound; the general lifting theorem on slide 17 supplies rational curves on slide 24; the odd-rank corollary is separate. The $\eta=0$ and $\gamma$ notation comes from slide 18. Slides 26-28 receive $E,d,m$ from slide 15 and the attained minimum from slide 16.

Earlier candidate marks retain their stable IDs and historical quotations. The authorized order supersedes postponing sharpness: Charles is on slide 16, the unchanged jump definitions on slide 18, and the reunited proof on slide 15.

NEEDS APPROVAL: [s14-m01] Recommendation: Keep the displayed minimum definition before Pi_jump on slide 18; recall sharpness from slide 16. Source: Charles 2014, Thm. 1.

NEEDS APPROVAL: [s14-m02] Recommendation: Keep the source questions and displays on slide 18. Source: V:L559-576.

NEEDS APPROVAL: [s14-m07] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "Jumping Picard ranks" | "$$\eta(X^{\mathrm{al}}):=\min_{p\text{ good}}\bigl(\rho(X_p^{\mathrm{al}})-\rho(X^{\mathrm{al}})\bigr)$$" | "Consider" | "$$\Pi_{\mathrm{jump}}(X):=\{p\text{ good}:\rho(X_p^{\mathrm{al}})>\rho(X^{\mathrm{al}})+\eta(X^{\mathrm{al}})\}$$" | "Is this set infinite? What is its density?" | "What about" | "$$X/\Bbb{Q}:\quad\gamma(X,B):=\frac{\#\{p\leq B:p\in\Pi_{\mathrm{jump}}(X)\}}{\#\{p\leq B:p\text{ prime}\}}\quad\text{as }B\rightarrow\infty\quad ?$$" Speaker notes proposed: "$\eta$ is the minimum excess. Odd characteristic-zero rank forces an increase but does not imply $\eta=1$. Which primes exceed the minimum? Is that set infinite? What is its density?" | "Charles will compute this minimum and prove its attainment. The counting function here uses rational primes; over a number field, count places by norm." | "Charles's theorem identifies the forced minimum: zero in the CM or even-dimensional case, and the endomorphism-field degree in the totally real odd-dimensional case." SETTLED mathematical source: V:L559-576; Charles 2014, Thm. 1. Exact teaching arrangement still needs approval. Allocation note: Allocation after the move: this candidate is on slide 18. The active Charles note points back to slide 16; the earlier finding is retained in s14-m90; Charles is now on slide 16.

2026-09-14 active speaker notes (review metadata is separate):

• $\eta$ is the minimum excess. Odd characteristic-zero rank forces an increase but does not imply $\eta=1$. Which primes exceed the minimum? Is that set infinite? What is its density?
• Charles's theorem in "Computing $\rho(X^{\mathrm{al}})$" computes this minimum and proves its attainment. The counting function here uses rational primes; over a number field, count places by norm.
• Charles's theorem identifies the forced minimum: zero in the CM or even-dimensional case, and the endomorphism-field degree in the totally real odd-dimensional case.

20 Product of elliptic curves

Order note. Applied ORDER-astra.md section 4. Historical arrangement prose and exact candidate quotations below retain their original wording; the draft allocation and heading give the current location.

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.

• $A=E_1\times E_2$; $E_i/\Bbb{Q}$
• $X=\operatorname{Kum}(E_1\times E_2)$

• $X$ | $\rho(X^{\mathrm{al}})$ | $\gamma(X,B)$, predicted | What is known
• square of CM | 20 | $1/2$ | $1/2+o(1)$, CM theory
• square of non-CM | 19 | $\sim c_X/\sqrt{B}$ | infinitely many [Elkies]
• CM times CM | 18 | $1/4$ | $1/4+o(1)$, CM theory
• CM times non-CM | 18 | $\sim c_X/\sqrt{B}$ | infinitely many [Charles]
• non-CM times non-CM | 18 | $\sim c_X/\sqrt{B}$ | infinitely many [Charles]
• Product rows: geometrically non-isogenous factors
• Non-CM rates: Lang-Trotter heuristics; per-prime scale $1/\sqrt{p}$

Remark

$p\in\Pi_{\mathrm{jump}}(X)$ depends uniquely on the pair $(a_{E_1}(p),a_{E_2}(p))$.

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

Spoken: NS, End and Hom are over $\Bbb{Q}^{\mathrm{al}}$; $\dagger$ is Rosati for the product polarization.

Spoken: Recall "Pic plays the role of $\operatorname{End}(A)$": $\rho(X^{\mathrm{al}})=16+\rho(A^{\mathrm{al}})$. At common good primes of odd residue characteristic, $\rho(X_p^{\mathrm{al}})=16+\rho(A_p^{\mathrm{al}})$.

Spoken: Use the definitions in "Jumping Picard ranks" also for $A$, with $\rho(A^{\mathrm{al}})=\operatorname{rk}(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))$. Compare $A$ and $X$ on the same common good odd primes. The two rank identities give equal rank excesses and equal minima over this prime set, hence $\Pi_{\mathrm{jump}}(X)=\Pi_{\mathrm{jump}}(A)$ on this set.

Spoken: All isogenies are geometric. For $A^{\mathrm{al}}\sim E^2$, a jump occurs exactly when $E$ has supersingular reduction. For geometrically non-isogenous factors, a jump occurs exactly when $E_{1,p}^{\mathrm{al}}\sim E_{2,p}^{\mathrm{al}}$.

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 $B$, $c(\log\log B)\log B/B<\gamma(X,B)<C\log B/B^{1/4}$. 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.

NEEDS APPROVAL: [s16-m02] Recommendation: Keep the combined transfer, geometric product formula and comparison table as this candidate synthesis. GPT 6 astra: The table is a synthesis of source frames and needs explicit approval. GPT 5.6 sol: The table is a synthesis of source frames and needs explicit approval. Source: I15:L298-352, L512-518; V:L577-596. SYNTHESIS, not a transcription: this combines I15:L298-352 and L512-518 with V:L577-596 into the current four-column table. The report describing five columns refers to an older version. Recommendation remains to keep this single four-column candidate; no structural approval is inferred. Allocation note: Allocation after the move: the original combined subject spans the dictionary on slide 19 and product/frequencies on slide 20; no content approval is inferred from the move.

NEEDS APPROVAL: [s16-m03] Recommendation: Keep geometric Hom and End; read the product rows as geometrically non-isogenous factors. Source: I15:L298-326; Skorobogatov-Zarhin, Sec. 1, eq. (10).

NEEDS APPROVAL: [s16-m04] Recommendation: Keep Charles in both product rows and Elkies for infinitude; put publication years, the 1991 bounds and their GRH qualification in notes. GPT 6 astra: Correct the mixed row; omit the original-field density-zero footer unless its separate source is supplied. GPT 5.6 sol: Correct the mixed row; distinguish infinitude from an unproved asymptotic. Source: Charles 2018, Thm. 1.1; Elkies 1991, Thms. A-B.

NEEDS APPROVAL: [s16-m06] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "Product of elliptic curves" | "$X=\operatorname{Km}(A)$; $A/\Bbb{Q}$ an abelian surface" | "$\rho(A^{\mathrm{al}}):=\operatorname{rk}(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))$" | "$\rho(X^{\mathrm{al}})=16+\rho(A^{\mathrm{al}})$" | "$\rho(X_p^{\mathrm{al}})=16+\rho(A_p^{\mathrm{al}})$; $p>2$ good" | "$\eta(X^{\mathrm{al}})=\eta(A^{\mathrm{al}})=\rho(A^{\mathrm{al}})\bmod2$" | "$\Pi_{\mathrm{jump}}(X)=\Pi_{\mathrm{jump}}(A)$" | "Fix a polarization on $A$; $\dagger$ the Rosati involution" | "$$(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))_{\Bbb{Q}}\simeq\{\phi\in\operatorname{End}(A^{\mathrm{al}})_{\Bbb{Q}}:\phi^\dagger=\phi\}$$" | "$A=E_1\times E_2$; $E_i/\Bbb{Q}$" | "$$\rho(X^{\mathrm{al}})=18+\operatorname{rk}\operatorname{Hom}(E_1^{\mathrm{al}},E_2^{\mathrm{al}})$$" | "$X$" | "$\rho(X^{\mathrm{al}})$" | "$\gamma(X,B)$, predicted" | "What is known" | "square of CM" | "20" | "$1/2$" | "$1/2+o(1)$, CM theory" | "square of non-CM" | "19" | "$\sim c_X/\sqrt{B}$" | "infinitely many [Elkies]" | "CM times CM" | "18" | "$1/4$" | "$1/4+o(1)$, CM theory" | "CM times non-CM" | "18" | "$\sim c_X/\sqrt{B}$" | "infinitely many [Charles]" | "non-CM times non-CM" | "18" | "$\sim c_X/\sqrt{B}$" | "infinitely many [Charles]" | "Product rows: geometrically non-isogenous factors" | "Non-CM rates: Lang-Trotter heuristics; per-prime scale $1/\sqrt{p}$" | "Remark" | "$p\in\Pi_{\mathrm{jump}}(X)$ depends uniquely on the pair $(a_{E_1}(p),a_{E_2}(p))$." Speaker notes proposed: "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." | "For a fixed non-CM square and sufficiently large $B$, $c(\log\log B)\log B/B<\gamma(X,B)<C\log B/B^{1/4}$. 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." | "For a CM square, the good unramified jump primes are exactly the primes inert in the CM field. Their density is one half." SETTLED mathematical source: I15:L298-352, L512-518; V:L577-596; Charles 2018, Thm. 1.1; Elkies 1991, Thms. A-B. Exact teaching arrangement still needs approval. Allocation note: Allocation after the move: this single ID still covers the complete original candidate, with its dictionary and Rosati display on slide 19 and product formula, table and trace-pair remark on slide 20.

2026-09-14 active speaker notes (review metadata is separate):

• NS, End and Hom are over $\Bbb{Q}^{\mathrm{al}}$; $\dagger$ is Rosati for the product polarization.
• Recall "Pic plays the role of $\operatorname{End}(A)$": $\rho(X^{\mathrm{al}})=16+\rho(A^{\mathrm{al}})$. At common good primes of odd residue characteristic, $\rho(X_p^{\mathrm{al}})=16+\rho(A_p^{\mathrm{al}})$.
• Use the definitions in "Jumping Picard ranks" also for $A$, with $\rho(A^{\mathrm{al}})=\operatorname{rk}(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))$. Compare $A$ and $X$ on the same common good odd primes. The two rank identities give equal rank excesses and equal minima over this prime set, hence $\Pi_{\mathrm{jump}}(X)=\Pi_{\mathrm{jump}}(A)$ on this set.
• All isogenies are geometric. For $A^{\mathrm{al}}\sim E^2$, a jump occurs exactly when $E$ has supersingular reduction. For geometrically non-isogenous factors, a jump occurs exactly when $E_{1,p}^{\mathrm{al}}\sim E_{2,p}^{\mathrm{al}}$.
• 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.
• For a fixed non-CM square and sufficiently large $B$, $c(\log\log B)\log B/B<\gamma(X,B)<C\log B/B^{1/4}$. 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.
• For a CM square, the good unramified jump primes are exactly the primes inert in the CM field. Their density is one half.

Provenance: SYNTHESIS of I15:L298-352, L512-518 and V:L577-596; four-column arrangement remains open.

Citation references (speaker notes; s12-c01): Elkies, "The existence of infinitely many supersingular primes for every elliptic curve over Q", Inventiones Mathematicae 89 (1987), 561-567. Charles, "Frobenius distribution for pairs of elliptic curves and exceptional isogenies", Duke Mathematical Journal 167 (2018), 2039-2072, Theorem 1.1; arXiv:1411.2914. The quantitative bounds use Elkies, "Distribution of supersingular primes", Asterisque 198-200 (1991), 127-132, Theorems A and B.

23 Discriminant of a K3 surface

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.

• $V:=T(1)\otimes\Bbb{Q}_\ell$; cup-product pairing

• $\det\tau=1\Longleftrightarrow\operatorname{im}\tau\subset SO(V)$
• $\det\tau\neq1$ $\Rightarrow$ nontrivial quadratic character
• An easy way to explain some jumps: $O$ vs $SO$.
• $\rho(X^{\mathrm{al}})$ even; $\varphi:=\operatorname{Frob}_p|V$; $\det\varphi=-1$

• $X/\Bbb{Q}$ quartic K3
• $D_X:=\Delta_{H^2}(X)\in\Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2$: determinant-character square class
• $D_X\in\Z\setminus\{0\}$ a representative; $p$ good, $p\nmid2D_X$

Theorem (Deligne; Suh)

The functional equation of the Frobenius action on $H^2(X)$ has the plus sign if and only if $D_X$ is square mod $p$.

• $\operatorname{Gal}(\Bbb{Q}^{\mathrm{al}}/\Bbb{Q})$ fixes $\operatorname{Pic}(X^{\mathrm{al}})$ $\Rightarrow$ $\Delta_{\operatorname{Pic}}(X)=1$

Corollary

• $V:=T(1)\otimes\Bbb{Q}_\ell$
• $X/\Bbb{Q}$ K3; $r:=\rho(X^{\mathrm{al}})$ even; $\eta(X^{\mathrm{al}})=0$
• $d_X:=\Delta_{H^2}(X)\Delta_{\operatorname{Pic}}(X)$ modulo squares; $d_X\in\Z\setminus\{0\}$

Theorem (C-Elsenhans-Jahnel)

Corollary

• $d_X$ nonsquare $\Rightarrow$ $L=\Bbb{Q}(\sqrt{d_X})$, $[L:\Bbb{Q}]=2$
• $p$ good, inert in $L$ $\Rightarrow p\in\Pi_{\mathrm{jump}}(X)$, up to finitely many primes
• $\displaystyle\liminf_{B\rightarrow\infty}\gamma(X,B)\geq1/2$
• $E=\Bbb{Q}$ $\Rightarrow$ infinitely many integral rational curves on $X^{\mathrm{al}}$
• Example: Costa-Tschinkel

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 $H^2(1)$. It is neither the Picard intersection discriminant nor an unspecified equation discriminant.

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

Spoken: $\Delta_{\operatorname{Pic}}$ is the square class of the Picard representation determinant. Galois fixing every geometric class makes it one; an integral descent equality is unnecessary.

NEEDS APPROVAL: [s20-m01] Recommendation: Use "Theorem (Deligne; C-Elsenhans-Jahnel)"; keep Suh in the notes. GPT 6 astra: Credit Deligne visibly for the projective sign theorem; Suh belongs in broader notes. GPT 5.6 sol: Keep the split credit Deligne-Suh for the sign and C-Elsenhans-Jahnel for the character. Source: CEJ 2020, Prop. 2.1.

NEEDS APPROVAL: [s20-m02] Recommendation: Define D_X as the cohomological determinant square class before its residue symbol; choose a nonzero integer representative. Source: CEJ 2020, Def. 2.4, Thm. 2.15.

AUTHOR'S CALL: [s20-m03] Recommendation: Reveal the trivial Picard-representation premise with the second theorem. Source: I15:L626-646.

AUTHOR'S CALL: [s20-m04] Recommendation: Keep the cohomological D_X and the explicit Galois-fixed Picard premise on slide 23. GPT 6 astra: Credit Deligne visibly for the projective sign theorem; Suh belongs in broader notes. GPT 5.6 sol: Keep the split credit Deligne-Suh for the sign and C-Elsenhans-Jahnel for the character. Source: CEJ 2020, Def. 2.4 and Thm. 2.15.

NEEDS APPROVAL: [s20-m05] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "Discriminant of a K3 surface" | "$X/\Bbb{Q}$ quartic K3" | "$D_X:=\Delta_{H^2}(X)\in\Bbb{Q}^{\times}/(\Bbb{Q}^{\times})^2$: determinant-character square class" | "$D_X\in\Z\setminus\{0\}$ a representative; $p$ good, $p\nmid2D_X$" | "Theorem (Deligne; C-Elsenhans-Jahnel)" | "The functional equation of the Frobenius action on $H^2(X)$ has the plus sign if and only if $D_X$ is square mod $p$." | "$$\varepsilon_p=\det(-\operatorname{Frob}_p\mid H^2_{\mathrm{et}}(X^{\mathrm{al}},\Bbb{Q}_\ell(1)))=\left(\frac{D_X}{p}\right)$$" | "$\operatorname{Gal}(\Bbb{Q}^{\mathrm{al}}/\Bbb{Q})$ fixes $\operatorname{Pic}(X^{\mathrm{al}})$ $\Rightarrow$ $\Delta_{\operatorname{Pic}}(X)=1$" | "Theorem (C-Elsenhans-Jahnel)" | "$$\rho(X^{\mathrm{al}})=2r,\quad\left(\frac{D_X}{p}\right)=-1\quad\Rightarrow\quad\rho(X_p^{\mathrm{al}})\geq2r+2$$" Speaker notes proposed: "$D_X$ represents the quadratic extension cut out by the determinant on $H^2(1)$. It is neither the Picard intersection discriminant nor an unspecified equation discriminant." | "Dimension twenty-two gives $\det(-\operatorname{Frob})=\det(\operatorname{Frob})$. C-Elsenhans-Jahnel, Proposition 2.1, attributes the projective sign statement to Deligne; Suh treats the proper nonprojective extension." | "$\Delta_{\operatorname{Pic}}$ is the square class of the Picard representation determinant. Galois fixing every geometric class makes it one; an integral descent equality is unnecessary." SETTLED mathematical source: I15:L626-646; C-Elsenhans-Jahnel 2020, Prop. 2.1, Def. 2.4 and Thm. 2.15. Exact teaching arrangement still needs approval.

2026-09-14 active speaker notes (review metadata is separate):

• $D_X$ represents the quadratic extension cut out by the determinant on $H^2(1)$. It is neither the Picard intersection discriminant nor an unspecified equation discriminant.
• Dimension twenty-two gives $\det(-\operatorname{Frob})=\det(\operatorname{Frob})$. C-Elsenhans-Jahnel, Proposition 2.1, attributes the projective sign statement to Deligne; Suh treats the proper nonprojective extension.
• $\Delta_{\operatorname{Pic}}$ is the square class of the Picard representation determinant. Galois fixing every geometric class makes it one; an integral descent equality is unnecessary.

Citation references (speaker notes; s12-c01): Costa-Elsenhans-Jahnel, "On the distribution of the Picard ranks of the reductions of a K3 surface" (2020), arXiv:1610.07823v3, Proposition 2.1, including the attribution to Deligne.

AUTHOR'S CALL: [s20-m90] Keep the prose plus-sign criterion, the determinant/Legendre-symbol display, or both? The display also identifies the determinant; both formulations are preserved pending your choice.

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.

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 $H^2(A)(1)$. Removing all divisor classes gives determinant equal to the algebraic determinant. For $A=(y^2=x^3-x)^2$, complex conjugation on the CM field gives a nontrivial character.

NEEDS APPROVAL: [s18-m02] Recommendation: Keep "Forced excess can survive determinant one" spoken on slide 21. GPT 6 astra: Keep the forced-excess qualification in notes. GPT 5.6 sol: Keep the forced-excess qualification in notes. Source: Charles 2014, Prop. 15.

APPROVED (2026-09-14): [s18-m03] replace the quoted closing sentence with "An easy way to explain some jumps: O vs SO." The author calls it slide 15; the exact sentence is on slide 21 (historical slide 18). Apply at the text location; preserve the author's quoted numbers. Author, verbatim: "I am also unsure what is the purpose of Slide 17, in particular given Slide 16, some of teh questions are already answered in the previous slide. On slide 15, not sure we should write "The jump criterion on the slides that immediately follow is a consequence of that one fact.". Intead, we should point, there is an easy way to explain some jumps, O vs SO . And maybe there one should point out that for kummer varieties we always land in SO (check this for me please)"

AUTHOR'S CALL: [s18-m04] Recommendation: Show the representation first, then the determinant-one and nontrivial-character alternatives, then the approved closing line. Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s18-m05] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "O or SO?" | "$V:=T_\ell(1)$; cup-product pairing" | "$$\tau:\operatorname{Gal}(k^{\mathrm{al}}/k)\longrightarrow O(V)$$" | "$\det\tau=1\Longleftrightarrow\operatorname{im}\tau\subset SO(V)$" | "$\det\tau\neq1$ $\Rightarrow$ nontrivial quadratic character" | "An easy way to explain some jumps: $O$ vs $SO$." Speaker notes proposed: "The Tate twist makes the pairing orthogonal. The determinant detects a quotient of order two, not all components of monodromy." | "Forced excess can survive determinant one: quadratic RM with $\dim_E T=3$ still forces two Tate classes." | "For Kummer surfaces the universal SO assertion is false. Serre, Lectures on N_X(p), Section 8.5.6.4, removes one polarization from $H^2(A)(1)$. Removing all divisor classes gives determinant equal to the algebraic determinant. For $A=(y^2=x^3-x)^2$, complex conjugation on the CM field gives a nontrivial character." SETTLED mathematical source: C-Elsenhans-Jahnel 2020, Prop. 2.13; Serre, Sec. 8.5.6.4. Exact teaching arrangement still needs approval. SETTLED author instruction, s18-m03: "An easy way to explain some jumps: O vs SO." Approval is requested for the additional content and arrangement.

2026-09-14 active speaker notes (review metadata is separate):

• The Tate twist makes the pairing orthogonal. The determinant detects a quotient of order two, not all components of monodromy.
• Forced excess can survive determinant one: quadratic RM with $\dim_E T=3$ still forces two Tate classes.
• For Kummer surfaces the universal SO assertion is false. Serre, Lectures on N_X(p), Section 8.5.6.4, removes one polarization from $H^2(A)(1)$. Removing all divisor classes gives determinant equal to the algebraic determinant. For $A=(y^2=x^3-x)^2$, complex conjugation on the CM field gives a nontrivial character.

Spoken: For $X=\operatorname{Kum}(A)$, the determinant on $V$ equals the determinant on $(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))_{\Bbb{Q}_\ell}$. It can be nontrivial. - For $X=\operatorname{Kum}(A)$, the determinant on $V$ equals the determinant on $(\operatorname{Pic}(A^{\mathrm{al}})/\operatorname{Pic}^0(A^{\mathrm{al}}))_{\Bbb{Q}_\ell}$. It can be nontrivial.

NEEDS APPROVAL: [s15-m04] Recommendation: Keep "An easy way to explain some jumps: O vs SO." on slide 21 only. GPT 6 astra: Keep the approved line at its actual location on the O/SO frame. GPT 5.6 sol: Keep the O/SO frame and its approved transition; the earlier report proposed a second pointer. Source: author decision on s18-m03.

NEEDS APPROVAL: [s15-m06] Recommendation: Keep the qualified Kummer determinant explanation in the notes. GPT 6 astra: The universal Kummer-to-SO claim is false; a swapping lift need not have a fixed spectrum. GPT 5.6 sol: The final jury agrees the universal claim is false; its earlier fixed-spectrum argument is superseded. Source: Serre, Lectures on N_X(p), Sec. 8.5.6.4; CEJ 2020, Ex. 2.36.

C-Elsenhans-Jahnel 2020, Prop. 2.13

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

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

Spoken: Remove the reciprocal pairs other than $\pm1$. 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.

AUTHOR'S CALL: [s19-m01] Recommendation: Keep slide 22 as the four-step proof of the two new Tate classes. GPT 6 astra: KEEP the elementary determinant proof. GPT 5.6 sol: KEEP the elementary determinant proof. Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s19-m02] Recommendation: Keep the even-rank hypothesis in the setup and the bound rho(X_p^al) >= rho(X^al)+2 in the final step. GPT 6 astra: Put the even-rank hypothesis in the setup. GPT 5.6 sol: Say "In this even-rank case" in the final step. Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s19-m03] Recommendation: Keep "Orthogonality $\Rightarrow$ lambda and lambda^{-1}, with equal multiplicities." GPT 6 astra: State reciprocal pairing explicitly; absolute value one alone is insufficient. GPT 5.6 sol: Rely on the preceding orthogonal representation; omit the repeated word. Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m04] Recommendation: Show reciprocal pairing first; reveal minus one, plus one and Tate at 0/1/2. Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m05] Recommendation: Keep slide 22 as the four-step proof of the two new Tate classes. GPT 6 astra: KEEP the elementary determinant proof. GPT 5.6 sol: KEEP the elementary determinant proof. Source: CEJ 2020, Prop. 2.13.

NEEDS APPROVAL: [s19-m06] Recommendation: Keep the quoted candidate wording, formulas and arrangement on this slide. Exact candidate text: "What $\det=-1$ costs you" | "$\rho(X^{\mathrm{al}})$ even; $\varphi:=\operatorname{Frob}_p|T_\ell(1)$; $\det\varphi=-1$" | "Orthogonality $\Rightarrow$ $\lambda$ and $\lambda^{-1}$, with equal multiplicities." | "Other pairs: determinant $+1$; multiplicity of $-1$ odd." | "$\dim T_\ell(1)$ even $\Rightarrow$ multiplicity of $+1$ odd." | "Tate: $+1,-1$ give two new geometric divisor classes. $$\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+2$$" | "$$\rho(X_p^{\mathrm{al}})\geq\rho(X^{\mathrm{al}})+2$$" Speaker notes proposed: "Remove the reciprocal pairs other than $\pm1$. Determinant minus one makes the multiplicity of minus one odd. Even dimension then makes the multiplicity of plus one odd." | "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." SETTLED mathematical source: C-Elsenhans-Jahnel 2020, Prop. 2.13. Exact teaching arrangement still needs approval.

2026-09-14 active speaker notes (review metadata is separate):

• Remove the reciprocal pairs other than $\pm1$. Determinant minus one makes the multiplicity of minus one odd. Even dimension then makes the multiplicity of plus one odd.
• 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.

AUTHOR'S CALL: [s18-m01] Recommendation: Keep the separate four-step determinant proof on slide 22. GPT 6 astra: KEEP the proof; it explains the increase by two. GPT 5.6 sol: KEEP the proof; the earlier drop recommendation is superseded. Source: CEJ 2020, Prop. 2.13.

AUTHOR'S CALL: [s19-m90] Which proof-placement recommendation should remain: s18-m01 or s19-m01? Their wording differs; both are preserved pending your choice.

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.

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

Cross-reference note. The retained purpose line predates s17-c02. This application uses the general Li-Liedtke lifting theorem in "K3 surfaces" (slide 17), with the separate E=Q hypothesis stated here; it does not use the odd-rank corollary.

Spoken: Inert primes in $L$ have density one half. The assumption $\eta=0$ makes the rank increase a jump. For $E=\Bbb{Q}$, Costa-Elsenhans-Jahnel supply infinitely many non-supersingular inert reductions, so the Li-Liedtke lifting theorem in "K3 surfaces" applies.

Spoken: The rational-curves conclusion requires $E=\Bbb{Q}$. 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.

NEEDS APPROVAL: [s21-m01] Recommendation: Keep the title "We can explain the 1/2". Source: V:L672-702.

APPROVED (2026-09-14): [s21-m02] Invoke the general lifting theorem on slide 17 for rational curves, with no real or complex multiplication. The odd-rank corollary is not used here. The density bound remains unchanged. Costa-Elsenhans-Jahnel 2020, Corollary 2.16 and Theorem 3.1. Author, verbatim: "We should write the "Corollary (Li-Liedtke)" more generically, so we can use it immediately when we show the density is at least 1/2. We can add our names to it also. In particular, this should help with the delivery in slide "We can explain the 1/2", and now the cororllary is obvious"

NEEDS APPROVAL: [s21-m03] Use $L=\Bbb{Q}(\sqrt{d_X})$ and the inert-prime implication on the half-density slide (24); retain $\eta=0$ and the extra $E=\Bbb{Q}$ rational-curve hypothesis. The latter uses the general lifting theorem on slide 17, with Costa-Elsenhans-Jahnel, Theorem 3.1 and Lemma 3.3. It does not use the odd-rank corollary.

NEEDS APPROVAL: [s21-m04] Recommendation: Keep one factorization, visibly credited to Costa-Tschinkel; identify the 2014 paper and Example 3.3 in notes. E=Q for this example remains unverified. GPT 6 astra: Remove the orphan integer from the candidate. GPT 5.6 sol: Identify its source example before retaining the integer, or remove it. Source: V:L697; CEJ 2020, Ex. 2.37. SYNTHESIS: one identified example is selected from the three source rows in CEJ, Example 2.6.11 of arXiv:1610.07823 (published Example 2.37); V:L697-699. This does not certify E=Q for this surface; do not instantiate the rational-curve branch with an unverified endomorphism field.

NEEDS APPROVAL: [s21-m05] Review the existing half-density candidate with its updated lifting-theorem reference. Exact candidate text: "We can explain the $1/2$" | "Theorem (C-Elsenhans-Jahnel)" | "$$p\text{ good},\ p\nmid2d_X\Rightarrow\quad(\det(\operatorname{Frob}_p\mid T_\ell(1))=\left(\frac{d_X}{p}\right)=-1\Rightarrow\rho(X_p^{\mathrm{al}})\geq r+2)$$" | "Corollary" | "$d_X$ nonsquare $\Rightarrow$ $L=\Bbb{Q}(\sqrt{d_X})$, $[L:\Bbb{Q}]=2$" | "$p$ good, inert in $L$ $\Rightarrow p\in\Pi_{\mathrm{jump}}(X)$, up to finitely many primes" | "$\displaystyle\liminf_{B\rightarrow\infty}\gamma(X,B)\geq1/2$" | "$E=\Bbb{Q}$ $\Rightarrow$ infinitely many integral rational curves on $X^{\mathrm{al}}$" | "Example: Costa-Tschinkel" | "$$d_X=-1\cdot5\cdot151\cdot22490817357414371041\cdot387308497430\allowbreak 149337233666\allowbreak 358807996260\allowbreak 780875056740\allowbreak 850984213276\allowbreak 970343278935\allowbreak 342068889706\allowbreak 146733313789$$"

2026-09-14 active speaker notes (review metadata is separate):

• Inert primes in $L$ have density one half. The assumption $\eta=0$ makes the rank increase a jump. For $E=\Bbb{Q}$, Costa-Elsenhans-Jahnel supply infinitely many non-supersingular inert reductions, so the Li-Liedtke lifting theorem in "K3 surfaces" applies.
• The rational-curves conclusion requires $E=\Bbb{Q}$. It is not certified for the numerical example. The negative sign alone proves nonsquareness.
• 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.

Provenance: SYNTHESIS selecting the first of V:L697-699, identified in CEJ Example 2.37. The half-density application uses the general lifting theorem on slide 17; s17-c02 replaces the earlier generic corollary by the odd-rank statement.

Citation references (speaker notes; s12-c01): Costa-Tschinkel, "Variation of Neron-Severi ranks of reductions of K3 surfaces" (2014), arXiv:1405.2265, Example 3.3. The displayed factorization is Costa-Elsenhans-Jahnel, "On the distribution of the Picard ranks of the reductions of a K3 surface" (2020), arXiv:1610.07823v3, Example 2.37.