How to distinguish between the two types?
non-CM CM
$\operatorname{End}_{\mathbf{Q}} E^{\mathrm{al}} = \mathbf{Q}$
$\operatorname{End}_{\mathbf{Q}} E^{\mathrm{al}} = \mathbf{Q}(\sqrt{-d})$
SOURCE: C22:L233
SECTION 1.1, Two primes determine the answer. Two slides. A deduction from reduction data that settles a characteristic-zero question, before any K3 appears.
EDITORIAL (from the written-out lecture): drop the Sato-Tate figures that sit on this frame; they belong to a different talk.
CORRECTION kept: the frame's second item reads $a_p \neq 0$; the deck's criterion is $p \nmid a_p$ (deck-decisions.md, 2026-09-13, minima-w1x).
The frame's tabular carries su2.pdf and nu1.pdf in its second row; the EDITORIAL above removes them, so the table keeps only the two endomorphism algebras.
Frame note: there are many ways, for example we could have simply computed the $j$-invariant and call it a day.
Frame note: thus by just looking at the number of points mod p, we may deduce the endomorphism algebra of an elliptic curve.
Spoken: For an elliptic curve,
Spoken: $\operatorname{End}_{\mathbf{Q}}(E^{\mathrm{al}}) \dashrightarrow \operatorname{End}_{\mathbf{Q}}(E_p^{\mathrm{al}}),$
Spoken: and the right-hand side is $\mathbf{Q}[T]/(c_p(T))$ when $p \nmid a_p$, a quaternion algebra otherwise.
Spoken: So the reduction sees more endomorphisms than the curve has, never fewer.
TRANSCRIBED FROM: frobenious-dist/frobenius-dist-ctnt.tex:233-265
Examples: 11.a2 and 27.a2
$E: y^2 + y = x^3 - x^2 - 10x - 20$ (LMFDB label: 11.a2 )
$\operatorname{End}_{\mathbf{Q}} E_3^{\mathrm{al}} \simeq \mathbf{Q}(\sqrt{-11})$
$\operatorname{End}_{\mathbf{Q}} E_{13}^{\mathrm{al}} \simeq \mathbf{Q}(\sqrt{-1})$
$\Rightarrow \operatorname{End}_{\mathbf{Q}} E^{\mathrm{al}} = \mathbf{Q}$
$E: y^2 + y = x^3 - 7$ (LMFDB label: 27.a2 )
$p = 2 \bmod 3 \Rightarrow a_p = 0 \Rightarrow \operatorname{End}_{\mathbf{Q}} E_p^{\mathrm{al}}$ is a Quaternion algebra
$p = 1 \bmod 3 \Rightarrow \operatorname{End}_{\mathbf{Q}} E_p^{\mathrm{al}} \simeq \mathbf{Q}(\sqrt{-3})$
$\leadsto \operatorname{End}_{\mathbf{Q}} E^{\mathrm{al}} = \mathbf{Q}(\sqrt{-3})$
SOURCE: C22:L268
What it does. Establishes the move the whole lecture repeats: compare structures at two primes, not dimensions.
PLANNING NOTE: 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.
CORRECTION kept: the frame uses $p = 2$ ($\mathbf{Q}(\sqrt{-1})$) and $p = 3$ ($\mathbf{Q}(\sqrt{-11})$); this deck uses $p = 3$ and $p = 13$ (deck-decisions.md, 2026-09-13, minima-w1x).
Frame note: this is like testing primality, existence of Carmichael number's doesn't stop us from applying Fermat's test.
Spoken: Two quadratic fields, not isomorphic, both containing the characteristic-zero algebra. So that algebra is $\mathbf{Q}$.
Spoken: Then 27.a2 for contrast: quaternionic at $p \equiv 2 \ (3)$, $\mathbf{Q}(\sqrt{-3})$ at $p \equiv 1 \ (3)$, and the curve does have CM.
TRANSCRIBED FROM: frobenious-dist/frobenius-dist-ctnt.tex:268-289
Picard lattice
A key geometric invariant for an algebraic K3 surface is its Picard lattice
$$\operatorname{Pic}(X) = \operatorname{NS}(X) \simeq \mathbf{Z}^{\rho}, \qquad \rho(X) := \operatorname{rank} \operatorname{Pic}(X)$$
Over $\mathbf{Q}^{\mathrm{al}}$, we have
$$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \simeq H^{1,1}(X_{\mathbf{C}}) \cap H^2(X_{\mathbf{C}}, \mathbf{Z}) \subset H^2(X_{\mathbf{C}}, \mathbf{Z}) \simeq (-E_8)^2 \oplus U^3 \simeq \mathbf{Z}^{22}$$
and $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \in \{1, 2, \dots, 20\}$. For a generic K3 surface we have $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 1$.
The degree of "difficulty" is negatively correlated with $\rho(X)$
$$H^2(X_{\mathbf{C}}, \mathbf{Q}) \simeq \operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}})_{\mathbf{Q}} \oplus T(X)_{\mathbf{Q}}$$
The "new and interesting" Galois representations arise from $T(X)$.
SOURCE: V:L358, L381 compressed
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.
SECTION 1.2, The same strategy for K3 surfaces: van Luijk. Seven slides. Carry the move to Picard lattices and state the one foundation the later lectures need.
EDITORIAL (from the written-out lecture): two frames in the source both titled "Picard lattice"; they compress to this one. Keep the transcendental lattice, because 1.5 is entirely about it.
The compression drops what the second frame repeats verbatim from the first (the $\mathbf{Q}^{\mathrm{al}}$ display, the range of $\rho$, the generic line), keeping one copy of each.
Spoken: Geometrically it describes the algebraic cycles on $X$ under linear/algebraic/numerical equivalency.
Spoken: curves in $X$ up to your favorite equivalency relation.
Spoken: Plays a similar role as $\operatorname{End}(A)$ for an abelian variety $A$: $\operatorname{NS}(A)_{\mathbf{Q}} \simeq \{\phi \in \operatorname{End}(A)_{\mathbf{Q}} : \phi^{\dagger} = \phi\}$, where $\dagger$ denotes the Rosati involution.
Spoken: for X = P2 we have Z, for X = P1xP1 we have Z2 and for a cubic surface we have Z7.
Spoken: when comes to state Theorems about K3 surfaces, we usually state them for K3 surfaces with a certain lattice or with a certain rank, just like for abelian varieties.
Spoken: This one of the key features for K3 surfaces, and what makes them interesting.
Spoken: we have some modularity results for high rank K3 surfaces, but very little is known for moderate rank.
Spoken: The transcendental lattice $T(X)$ is the orthogonal complement of $\operatorname{Pic}(X_{\mathbf{C}})$ in $H^2(X,\mathbf{Z})$.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:358-394
EDITORIAL: the two vantage.tex frames (358-380 and 381-395) are compressed into this slide; kept out of view, as Spoken lines above: "Geometrically it describes the algebraic cycles on X under linear/algebraic/numerical equivalency" and the End(A) / NS(A)_Q Rosati display (vantage.tex:362, 367-369), which returns on slide 16. Restoring them overflows the frame by about 150px; splitting the slide back into two is the alternative and needs the author.
Picard lattice, over finite fields
Over $\mathbf{F}_p^{\mathrm{al}}$ we have $\rho(\overline{X}_p) \in \{2, 4, \dots, 22\}$ (Over $\mathbf{Q}^{\mathrm{al}}$ it was $\{1, 2, \dots, 20\}$)
The Hasse-Weil zeta function $Z_X(t)$ plays a key role for the computation of $\rho(X_{\mathbf{F}_{p^n}})$
$$Z_X(t) := \exp\left( \sum_{m=1}^{\infty} \frac{\# X(\mathbf{F}_{p^m})}{m} t^m \right) = \frac{1}{(1-t) \, P_2(t) \, (1-p^2 t)}$$
where $P_2(t) = \det(1 - t \operatorname{Frob} \mid H^2_{\mathrm{et}}(\overline{X}_p, \mathbf{Q}_{\ell})) \in \mathbf{Z}[t]$ and $\deg P_2 = 22$.
One may deduce $Z_X(t)$ by naively computing $\# X(\mathbf{F}_{p^m})$ for $m \leq 11$.
From $P_2(t)$ we may deduce $\rho(X_{\mathbf{F}_{p^n}})$ for any $n$, via Tate conjecture:
$$\operatorname{Pic}(X_p)_{\mathbf{Q}_{\ell}} = \ker( \operatorname{Frob}_p - p \cdot \operatorname{id} \mid H^2_{\mathrm{et}}(\overline{X}_p, \mathbf{Q}_{\ell}))$$
Tate conjecture is a theorem for K3 surfaces over finite fields. [Charles, Madapusi, Kim-Madapusi]
For $p > 7$ computing $Z_X(t)$ by naive point counting is not practical. Instead, one relies in a infrastructure of methods in crystalline cohomology [Abbott-Kedlaya-Roe, C, C-Harvey-Kedlaya, Tuitman-Pancratz]
SOURCE: V:L396
$P_2(T)$ is the name this lecture uses throughout: the same object slide 5 states Tate's theorem for, the same one Lecture 2 normalises as $P_2(pT)$, and, up to the reciprocal normalisation $\det(1 - T\operatorname{Frob})$, the same one the sources write $\chi$, $\chi_1$ or $L(t)$.
Four cyclotomic factors can give rank 10; the count is of degrees.
NOTATION FLAG: the frame writes this polynomial $\chi(t) = \det(1 - t\operatorname{Frob})$, and the slide 5 frame writes the same object $P_2(X,T) := \det(1 - T\operatorname{Frob})$. Transcription keeps the author's definition and only renames $\chi$ to $P_2$. That reverses the deck's earlier $\det(T - \operatorname{Frob})$ rewrite, so slide 5 now reads $\operatorname{ord}_{T=1/q}$, as the frame does. Lecture 2's "$P_2(pT)$" normalisation line needs the matching check.
The last line comes from the immediately following frame (V:L421-440), which the deck's third paragraph was drawn from; its earlier lines repeat this frame verbatim.
Spoken: surfaces attaning the maximum value are known as supersingular.
Spoken: Kuga-Satake construction plays a key role in its proofs.
Spoken: in practice, we will compute a p-adic approximation of the frob matrix instead of counting points.
Spoken: recent advances in how to use this infrastructure will play a key role in the later part of the talk.
Spoken: Write $P_2(T) = \det(T - \operatorname{Frob} \mid H^2)$, whose roots have absolute value $q$; the normalised polynomial is $P_2(qT)$, whose roots have absolute value 1.
Spoken: The Tate conjecture is a theorem for K3 surfaces over finite fields, so $\rho(\overline{X}_p)$ is the number of roots of $P_2(qT)$ that are roots of unity, counted with multiplicity: the sum of the degrees of its cyclotomic factors, not the number of factors.
Spoken: Point counting is impractical beyond very small $p$; the machinery that makes it practical is cited, not taught.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:396-419 (last line from 421-440)
What the characteristic polynomial gives you
Theorem (Tate) [Tate], [Charles], [Madapusi Pera], [Maulik], [Kim-Madapusi Pera]
Let $X/\mathbf{F}_q$, where $q = p^n$, be an abelian surface or a K3 surface. Then:
$\rho(X_p) = \operatorname{ord}_{T = 1/q} P_2(T)$
$\rho(\overline{X}_p) = \sum_{\zeta} \operatorname{ord}_{T = \zeta/q} P_2(T)$, where $\zeta$ runs over all roots of unity.
(Artin-Tate, a theorem here) $P_2(T) \leadsto \operatorname{disc}(\operatorname{NS}(X_p)) \bmod \mathbf{Q}^{\times 2}$
SOURCE: I15:L213
Attribution: [Madapusi Pera], not [Pera]. Characteristic 2 is Kim-Madapusi Pera, arXiv:1512.02540; the earlier finite-height cases in odd characteristic are Nygaard-Ogus.
Artin-Tate is a theorem for K3 and abelian surfaces, not a conjecture: by Milne, Tate is equivalent to Artin-Tate here. Say it that way.
Equivalently to passing to an extension, use an appropriate power of Frobenius.
What it does. Closes the only silent dependency in Lecture 1: every use of a discriminant up to now has assumed the audience already knows where discriminants come from.
Say out loud: the distinction between the rank over the field of definition and the rank over its algebraic closure is one nothing earlier in the lecture has drawn.
Artin-Tate is what supplies the discriminant, and the discriminant is what van Luijk needs two slides later. Without this slide, slide 9's two discriminants arrive unexplained.
Artin-Tate is quoted and used here, never unpacked: no slide in this deck states a formula for it. It is black-boxed exactly the way Berthelot-Ogus-Raynaud is black-boxed in Lecture 2.
CORRECTION kept: the frame's hypothesis reads "$X$ an abelian surface or a K3 surface then:"; the deck's line is the one on the slide (deck-decisions.md, 2026-09-13, minima-uv3.10). Frame label reads "Theorem (Tate Conjecture) [Tate], [Charles], [Pera] and [Maulik]" and the third item "(Artin-Tate Conjecture)"; both carry the note-flagged corrections above.
The frame opens with the zeta function and the definition $P_2(X,T) := \det(1 - T\operatorname{Frob} \mid H^2)$, "which have reciprocal roots of absolute value $q$". Slide 4 already carries both, so this slide keeps only the theorem block.
Spoken: $Z_{X_p}(T) := \exp\left(\sum_{i=1}^{\infty} \frac{\# X_p(\mathbf{F}_{q^i})}{i} T^i\right) = \frac{1}{(1-T) P_2(X,T) (1-q^2 T)}$, where $P_2(X,T) := \det(1 - T\operatorname{Frob}_p \mid H^2) \in \mathbf{Z}[t]$, which have reciprocal roots of absolute value $q$.
Spoken: The discriminant is that of $\operatorname{NS}$ over the base field. Van Luijk needs $\operatorname{disc}\operatorname{Pic}(\overline{X}_p)$: the two are guaranteed to agree mod squares when $\rho(X_p) = \rho(\overline{X}_p)$, so first pass to a finite extension over which all divisor classes are defined.
Spoken: Both ranks come from the same polynomial: $\rho(X_p)$ and $\rho(\overline{X}_p)$ are two different counts of its roots, the rank over the field of definition and the rank over its algebraic closure.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:203-224
Reduction to finite characteristic
Take $f \in \mathbf{Z}[x,y,z,w]$ and $X := Z(f) \subset \mathbf{P}^3_{\mathbf{Q}}$.
We may consider the surface $X_p := Z(f \bmod{p}) \subset \mathbf{P}^3(\mathbf{F}_p)$.
Theorem
If $X$ and $X_p$ are smooth then the specialization map is injective
$$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p) \quad \text{and} \quad \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p).$$
Goal
For a given $f$ and $p$, improve the inequality $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p)$.
Parity reasons might already force the inequality to not be sharp.
Endomorphisms of the transcendental lattice can complicate things even further.
SOURCE: O:L1144
Spoken: Also: $\rho(\overline{X}_p)$ is always even for a K3 over a finite field.
Spoken: That parity fact does more work in this course than anything else on this slide.
TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1144-1165
Computing the Picard lattice over $\mathbf{Q}^{\mathrm{al}}$
Computing $\rho(X_{\mathbf{Q}^{\mathrm{al}}})$ is in principle, solved.
[Charles, Poonen-Testa-van Luijk, Hassett-Kresch-Tschinkel, Shioda, Lairez-Sertöz]
These algorithms are not practical.
Usually rely on searching for explicit generators for the Picard lattice.
We do not know how to do that efficiently.
Various ad hoc methods exist to improve the inequality above.
SOURCE: V:L442
EDITORIAL (from the written-out lecture): compressed to a citation line. One slide, no survey.
The frame's specialization sentence and display ("To terminate such a search, one makes use of the specialization being injective ... for a prime of good reduction.") are the ones slide 6 already carries, so they are not repeated here.
Spoken: The first 2 algorithms rely on searching for explicit generators for the Picard, the 3rd passes this problem to the Kuga-Satake variety of a K3 surface.
Spoken: Shioda worked it out for Elliptic surfaces and surfaces with extra automorphisms.
Spoken: Lairez-Sertöz give a numerical approach to the problem.
Spoken: To terminate such a search, one makes use of the specialization being injective, $\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p)$ and $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p)$, for a prime of good reduction.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:442-464
Improving upper bounds: two specializations
$$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p) \quad \text{and} \quad \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p)$$
van Luijk
If $p$ and $q$ are two primes of good reduction, and
$$\begin{gathered} \rho(\overline{X}_p) = \rho(\overline{X}_q) = 2r, \\ \operatorname{disc} \operatorname{Pic}(\overline{X}_p) \neq \operatorname{disc} \operatorname{Pic}(\overline{X}_q). \end{gathered}$$
then
$$\rho(X_{\mathbf{Q}^{\mathrm{al}}}) < 2r.$$
van Luijk, used this technique with $r = 1$, to provide the first known examples of K3 surfaces over $\mathbf{Q}$ such that $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 1$
Under the right conditions we know that this method will not succeed to give a tight upper bound (parity + endomorphisms of the transcendental lattice).
SOURCE: V:L466
What it does. The same move as slide 2, now with discriminants in place of endomorphism algebras. The parallel should be said out loud.
EDITORIAL (from the written-out lecture): fold in from the Lecture 2 copy only its statement of when the method fails. That is the small line at the bottom of the slide.
The bottom line is that statement, verbatim from the Lecture 2 copy of this frame, nyc-jnts.tex:1180-1181.
CORRECTION kept: both source frames conclude "$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) < 2r$"; the object bounded is the rank, so the slide reads $\rho$.
Frame note: we already applied this method to abelian surfaces earlier in the talk.
Spoken: The comparison is of square classes: the test is whether the ratio of the two discriminants is a square, not whether the computed representatives differ.
Spoken: The method fails when the two discriminants agree modulo squares: equal ranks and equal discriminant classes leave the bound untouched.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:466-477 (closing line from frobenious-dist/nyc-jnts.tex:1167-1182)
The quartic, worked
$$X : x^3 z + 3x^2 y^2 + 5xw^3 + y^3 w + 3yz^3 - 5z^2 w^2 = 0 \ \subset \ \mathbf{P}^3$$
A quartic with an automorphism of order 5. Two primes, read off the recorded data:
$p$ $\rho(\overline{X}_p)$ disc 11 18 $-55$ 13 18 $-85$
The disc column is the geometric discriminant $\operatorname{disc}\operatorname{Pic}(\overline{X}_p)$. The base-field ranks are 1 at 11 and 5 at 13, so Artin-Tate is applied over $\mathbf{F}_{11^{30}}$ and $\mathbf{F}_{13^4}$, where $\rho(X_p) = \rho(\overline{X}_p) = 18$.
Equal ranks, and the ratio $-55 : -85$ is $11/17$, not a square. Van Luijk gives $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) < 18$, hence $\leq 17$.
NEW SLIDE: newly written, not reused from an existing talk.
The equality of the base-field and geometric ranks holds only after the extension named on the slide; that is the hypothesis under which Artin-Tate's discriminant is the one van Luijk wants.
What it does. Runs the criterion once, with numbers. Slide 8 states it and never does it; this room will want to watch it happen. It also plants the surface that slide 24 comes back to.
PLANNING NOTE: Provenance and two caveats. Surface from NSranks/data/polynomials.m:1286-1290, annotated "rank 17", citing arXiv:math/0603742v2. 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.
Spoken: Over the prime field the base-field ranks are 1 at 11 and 5 at 13.
Spoken: so the two classes in $\mathbf{Q}^{\times}/(\mathbf{Q}^{\times})^2$ differ.
TRANSCRIBED FROM: no source, restyled
Torsion-free cokernel
Elsenhans-Jahnel showed that the specialization map
$$\operatorname{Pic}(X_{\mathbf{Q}^{\mathrm{al}}}) \hookrightarrow \operatorname{Pic}(\overline{X}_p)$$
has torsion-free cokernel for $p \neq 2$.
Thus, if $\rho(\overline{X}_p) = \rho(X_{\mathbf{Q}^{\mathrm{al}}})$ every invertible sheaf lifts.
For example, if $\rho(\overline{X}_p) = 2$, Elsenhans-Jahnel approach is
compute $\operatorname{Pic}(\overline{X}_p)$
estimate the degree of a hypothetical effective divisor of the lift
use Gröbner bases to verify that such a divisor does or does not exist
This approach is only practical if one can compute $\operatorname{Pic}(\overline{X}_p)$ and if the obtained estimates are low.
SOURCE: V:L479
SECTION 1.3, Refinements: Elsenhans-Jahnel. Two slides, both pointers. The second one is load-bearing: it is how you compute the character in 1.5.
Spoken: so integral and not merely rational information descends.
Spoken: With a Gröbner-basis test for whether a divisor class is realised.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:479-498
The sign in the functional equation
Let $X$ be a quartic K3 surface over $\mathbf{Q}$ and $D_X$ be its discriminant.
Theorem (Costa-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 $p$.
SOURCE: I15:L629, split from the Discriminant frame
What it does. Turns an abstract character into something readable from point counts at one prime. 1.5 depends on this slide, which is why 1.3 is not purely decorative.
EDITORIAL (from the written-out lecture): the 2015 slide labels this [Elsenhans-Jahnel]; it is yours.
PANEL FLAG: the panel attributes this theorem to Elsenhans-Jahnel, Duke Math. J. 164 (2015), building on Saito, and holds that the 2015 label was correct; on its reading Costa-Elsenhans-Jahnel is the jump-character paper used in section 1.5. You stated it is Costa-Elsenhans-Jahnel. The attribution on the slide is unchanged pending your decision.
NOTE: slide 20's box states the same theorem with the frame's label "Theorem (Elsenhans-Jahnel)"; the two labels differ pending minima-pln.
Spoken: The sign-to-determinant dictionary is Deligne and Suh.
Spoken: $D_X = \Delta_{H^2}(X)$, the discriminant of the frame's opening line.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:626-646, first block only, as the SOURCE note's split directs; the frame's two later blocks are the 1.5 material.
Jumping Picard ranks
Theorem (Charles)
We have
$$\rho(\overline{X}) + \eta(\overline{X}) \leq \rho(\overline{X}_p)$$
for some $\eta(\overline{X}) \geq 0$. Equality occurs infinitely often (density 1 after some finite extension).
Consider
$$\Pi_{\mathrm{jump}}(X) := \lbrace\, p : \rho(\overline{X}_p) > \rho(\overline{X}) + \eta(\overline{X}) \,\rbrace$$
Is this set infinite? What is its density?
What about
$$\gamma(X,B) := \frac{\#\lbrace\, p \leq B : p \in \Pi_{\mathrm{jump}}(X) \,\rbrace}{\#\lbrace\, p \leq B \,\rbrace} \quad \text{as } B \rightarrow \infty \quad ?$$
SOURCE: V:L559
What it does. Note that the jump set is defined net of eta. Jumping and forced excess are two different things from the first slide of the section, and keeping them apart is the whole point of the 1.4/1.5 split.
SECTION 1.4, Jumping, and choosing your primes. Six slides. Reduction ranks jump often, and for Kummer surfaces you can say exactly when. The bound is still attainable; the primes just stop being free.
Spoken: $\eta(\overline{X})$ is the forced excess: the amount by which every good reduction must exceed $\rho(\overline{X})$.
Spoken: and its counting function $\gamma(X,B)$, the proportion of primes $p \leq B$ lying in $\Pi_{\mathrm{jump}}(X)$.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:559-576
Kummer surface
If $X \simeq \operatorname{Kummer}(A)$, where $A$ is an abelian surface, then
$$\rho(\overline{X}) = 16 + \rho(\overline{A})$$
SOURCE: I15:L512
Setup only.
Spoken: $X \simeq \operatorname{Km}(A)$ for $A$ an abelian surface.
Omitted from the frame: the second display, $\mathrm{ST}_X = \mathrm{ST}_A/\lbrace \pm 1 \rbrace \subset \mathrm{SO}_5 \cong \mathrm{USp}_4/\lbrace \pm 1 \rbrace$, which is Sato-Tate and is dropped for the same reason as the Sato-Tate figures on slide 1.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:512-518
Product of elliptic curves
Let $X \simeq \operatorname{Km}(E_1 \times E_2)$ and $E_i$ elliptic curve over $\mathbf{Q}$.
$$\rho(X) = 18 + \operatorname{rk}(\operatorname{Hom}(E_1,E_2)) = 18 + \begin{cases} 0 & E_1 \not\sim E_2;\\ \operatorname{rk}(\operatorname{End}(E_1)) & E_1 \sim E_2 \end{cases}$$
$X$ $\rho(\overline{X})$ $\gamma(X,B)$ square of CM 20 $1/2 + o(1)$ CM theory square of non-CM 19 $\tfrac{c(\log\log B)(\log B)}{B} < \cdot < \tfrac{C \log B}{B^{1/4}}$ ‡ [Elkies] CM times CM 18 $1/4 + o(1)$ CM theory CM times non-CM 18 ? non-CM times non-CM 18 infinitely many jump primes [Charles]
‡ the lower bound is conditional on ERH, with an absolute implied constant $c > 0$; the upper bound is unconditional. In the last row Charles gives infinitely many jump primes, not a positive proportion: for a geometrically non-isogenous non-CM pair over $\mathbf{Q}$, $\gamma(X,B) \to 0$.
Remark
$p \in \Pi_{\mathrm{jump}}(X)$ depends uniquely on the pair $(a_{E_1}(p), a_{E_2}(p))$
SOURCE: I15:L298
Entries are the source's own, from the first of two successive frames of K3workshop.tex, with $E_1, E_2$ over Q throughout.
Spoken: which is 18 when they are not isogenous, and $18 + \operatorname{rk}\operatorname{End}(E_1)$ when they are.
The last row and the dagger line carry the deck's correction to the frame's $\gg 0$ entry; kept, in the frame's structure.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:298-326
The simplest case
Let $X \simeq \operatorname{Km}(E_1 \times E_2)$ and $E_i$ elliptic curve over $\mathbf{Q}$.
$$\rho(X) = 18 + \operatorname{rk}(\operatorname{Hom}(E_1,E_2)) = 18 + \begin{cases} 0 & E_1 \not\sim E_2;\\ \operatorname{rk}(\operatorname{End}(E_1)) & E_1 \sim E_2 \end{cases}$$
$X$ $\rho(\overline{X})$ $P(p \in \Pi_{\mathrm{jump}}(X))$ $\gamma(X,B)$ square of CM 20 $1/2$ $1/2$ square of non-CM 19 $\sim 1/\sqrt{p}$ † $c/\sqrt{B}$ † CM times CM 18 $1/4$ $1/4$ CM times non-CM 18 $\sim 1/\sqrt{p}$ † $c/\sqrt{B}$ † non-CM times non-CM 18 $\sim 1/\sqrt{p}$ † $c/\sqrt{B}$ †
† conjectural, the Lang-Trotter heuristic. The CM rows are unconditional. No row is 1.
Remark
$p \in \Pi_{\mathrm{jump}}(X)$ depends uniquely on the pair $(a_{E_1}(p), a_{E_2}(p))$
SOURCE: I15:L328
What it does. This is where the section's real claim lives, and the table already makes it. Jumping is frequent, sometimes very frequent, and never universal.
Entries are the source's own, from the second of the two successive frames of K3workshop.tex; the source sets the 1/sqrt(p) and c/sqrt(B) entries in red, i.e. as the conjectural ones, which is what the dagger carries.
Same remark as the previous slide: whether p is a jump prime depends only on the pair (a_{E_1}(p), a_{E_2}(p)).
Spoken: The same table with the per-prime probability column added:
Spoken: Every one of these surfaces has attainable bounds.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:328-352
Jumping Picard ranks for Kummer surfaces
Let $X \simeq \operatorname{Km}(A) := \widetilde{A/\pm}$ be a Kummer surface, where $A$ is an abelian surface.
We have
$\rho(\overline{X}) = \rho(\overline{A}) + 16$
$\rho(\overline{X}_p) = \rho(\overline{A}_p) + 16$
$\eta(\overline{X}) = \eta(\overline{A}) = (\rho(\overline{A}) \bmod 2)$
Thus, $\Pi_{\mathrm{jump}}(X) = \Pi_{\mathrm{jump}}(A)$
Moreover
$$\operatorname{Pic}(A_k)/\operatorname{Pic}^0(A_k) \simeq \operatorname{NS}(A_k)_{\mathbf{Q}} \simeq \lbrace \phi \in \operatorname{End}(A_k)_{\mathbf{Q}} : \phi^{\dagger} = \phi \rbrace,$$
where $\dagger$ denotes the Rosati involution and
$\rho(\overline{A}_p) \geq 4 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ an elliptic curve
$\rho(\overline{A}_p) = 6 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ a supersingular elliptic curve
SOURCE: V:L577
Spoken: and $\operatorname{NS}(A)_{\mathbf{Q}}$ is the Rosati-fixed part of $\operatorname{End}(A)_{\mathbf{Q}}$.
Spoken: So the K3 question becomes a question about the abelian surface.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:577-596
Jumping Picard ranks for Kummer surfaces
$\rho(\overline{A}_p) \geq 4 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ an elliptic curve
$\rho(\overline{A}_p) = 6 \Longleftrightarrow \overline{A}_p \sim E^2$, $E$ a supersingular elliptic curve
If $A \sim E^2$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $p$ is supersingular for $E$.
This is related to the Lang-Trotter conjecture.
It states that $p$ should be supersingular with probability proportional to $1/\sqrt{p}$.
Elkies has shown that there are infinitely many supersingular primes for $E/\mathbf{Q}$.
If $A \sim E_1 \times E_2$ with $E_1 \not\sim E_2$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $E_1 \sim E_2$ over $\overline{\mathbf{F}_p}$.
Charles has shown that there are also infinitely many such primes.
If $\operatorname{End}(\overline{A}) = \mathbf{Z}$, then $p \in \Pi_{\mathrm{jump}}(A)$ iff $\overline{A}_p \sim E^2$.
What do you think it should happen in this case?
SOURCE: V:L597
The 1/sqrt(p) is a heuristic probability, not a natural density. The CM case, density 1/2, is what the CM rows of the table on slides 14-15 record.
What it does. Exact control of which primes jump, for a whole family, ending on something unknown. Closes 1.4 on a question rather than a result. This is the slide the section exists for.
PLANNING NOTE: Default taken. The question to the room stays. Under the purpose you named, an open case put to the audience is the content, not a loose end.
Spoken: Three cases:
Spoken: For CM $E$ they are the inert primes, density $1/2$.
Spoken: Case 3: does anyone here know?
Omitted from the frame: its closing line, "Let's do some numerical experiments for some non Kummer surfaces!", which hands over to experiments this lecture does not run; slide 18 opens section 1.5 instead.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:597-630
O or SO?
Galois acts on the Tate-twisted $\ell$-adic realisation $T_{\ell}(1)$, where the cup product is an orthogonal pairing.
$$\tau : \operatorname{Gal}(\overline{K}/K) \longrightarrow O(T_{\ell}(1)).$$
There is no reason for the image to sit inside $SO(T_{\ell}(1))$.
When it does not, $\det \tau$ is a non-trivial quadratic character and the image meets both determinant components of $O(T_{\ell}(1))$.
The determinant records only a quotient of order two; the monodromy group itself can have more components than two.
The jump criterion on the slides that immediately follow is a consequence of that one fact.
NEW SLIDE: newly written, not reused from an existing talk.
What it does. Gives the room one picture to hold for the rest of the lecture. The paper says the same thing about itself: it describes the effects of det tau being non-trivial.
Galois does not act on the integral Betti lattice T; and untwisted etale cohomology preserves the cup product only up to the cyclotomic multiplier, i.e. through a similitude representation. The Tate twist is what makes tau orthogonal.
The Sato-Tate notation from your ICERM deck (ST_X inside O_{22-rho(X)}, ST_X-zero inside SO_{22-rho(Xbar)}, at K3workshop.tex:503) is a parenthetical only: nothing that follows uses it.
Spoken: The forced excess of the later slides is not.
Spoken: It is a separate mechanism, due to Charles, and it survives passing to the finite extension that kills $\det\tau$: quadratic RM with $\dim_E T = 3$ still forces excess 2, and $SO_3 \times SO_3$ has determinant $+1$ and still forces two eigenvalues 1.
TRANSCRIBED FROM: no source, restyled
What $\det = -1$ costs you
Let $\varphi = \operatorname{Frob}_p$ on $T(1)$, and suppose $\det \varphi = -1$.
Every eigenvalue has absolute value 1, so those other than $\pm 1$ come in conjugate pairs $\lbrace z, \bar z \rbrace$ with $z \bar z = 1$.
Those pairs contribute $+1$ to the determinant, so some eigenvalue must be $-1$.
If $\operatorname{rk}\operatorname{Pic} X_{\overline{K}}$ is even , so is $\dim T = 22 - \operatorname{rk}\operatorname{Pic} X_{\overline{K}}$; then one more eigenvalue is forced, and it is $+1$.
Both are roots of unity, so both are new Tate classes. The rank jumps by at least 2 .
NEW SLIDE: newly written, not reused from an existing talk.
The even-rank hypothesis is needed: in odd dimension determinant -1 forces no +1 eigenvalue. It is the same hypothesis slide 20 states as rho(Xbar) = 2r.
What it does. Explains the 2, which is otherwise an unmotivated constant. It is the only argument in Lecture 1 given in full, and it fits on one slide because it is genuinely four lines.
TRANSCRIBED FROM: no source, restyled
Discriminant of a K3 surface
Let $X$ be a quartic K3 surface over $\mathbf{Q}$ and $D_X$ be its discriminant.
Theorem (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 $p$.
Theorem
If $\rho(\overline{X}) = 2r$, then $\rho(\overline{X}_p) \geq 2r + 2$ at every prime $p$ of good reduction with $p \nmid 2 D_X$ at which $D_X$ is not a square mod $p$.
Assume $\operatorname{NS}(X/k) = \operatorname{NS}(X/\bar k)$, so the Picard representation is trivial, $\Delta_{\operatorname{Pic}} = 1$, and a single discriminant $D_X$ does both jobs.
SOURCE: I15:L626
The parity fact is about the geometric rank, hence rho(Xbar) = 2r. The primes dividing 2 D_X are excluded so that the quadratic character chi_{D_X} is defined and non-zero at p: at p | D_X the residue is 0 = 0^2 and the character vanishes, so "not a square mod p" cannot hold there and the criterion says nothing.
PLANNING NOTE: Default taken, as you instructed. The simplified form is on the slide. Every worked example in the paper satisfies it, at discr_sing_v7e.tex:2607-2610, so it costs no generality in practice. The general statement, with Delta_{H^2} times Delta_{Pic} in place of D_X, belongs here in the speaker notes or in a footnote.
Spoken: the source frame's opening line is restored as the lead, and the deck's hypothesis sentence is kept below the boxes because the simplified D_X form depends on it.
Spoken: the frame's third box is the Corollary "If $D_X$ is not a square, $\rho(X) = \rho(\overline{X}) = 2r$ and $\eta(\overline{X}) = 0$, then $\lbrace p : p \text{ inert in } \mathbf{Q}(\sqrt{D_X}) \rbrace \subset \Pi_{\mathrm{jump}}(X)$, $\liminf \gamma(X,B) \geq 1/2$, and $\overline{X}$ has infinitely many rational curves." The deck carries it on the next slide, from the vantage.tex frame.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:626-646
We can explain the $1/2$
Theorem (C, C-Elsenhans-Jahnel)
If $\rho(\overline{X}) = \min_p \rho(\overline{X}_p)$, then there is $d_X \in \mathbf{Z}$ such that:
$$\bigl\lbrace p > 2 : p \text{ inert in } \mathbf{Q}(\sqrt{d_X}) \bigr\rbrace \subset \Pi_{\mathrm{jump}}(X).$$
$d_X$ represents the quadratic character $p \mapsto \det(\operatorname{Frob}_p | T(X)(1)) \in \lbrace \pm 1 \rbrace$.
Corollary
If $d_X$ is not a square:
$\liminf_{B \rightarrow \infty} \gamma(X,B) \geq 1/2$
$\overline{X}$ has infinitely many rational curves.
$d_{X_3} = -1 \cdot 5 \cdot 151 \cdot 22490817357414371041 \cdot 387308497430149337233666358807996260780875056740850984213276970343278935342068889706146733313789$
SOURCE: V:L672
The determinant argument gives an increase relative to rho, not relative to rho + eta, which is why the source's hypothesis eta = 0 has to be in force before the jump set is the right target. D_X nonsquare is what makes Q(sqrt(D_X)) a nontrivial quadratic extension, without which Chebotarev gives no density 1/2.
What it does. The payoff of 17 and 18 made arithmetic, and the size of d_X is the point. Half of all primes are useless for this surface, and here is the integer that says which half.
The source prints this factorization at \tiny, running to hundreds of digits; the slide keeps it at that size.
Spoken: the frame's hypothesis $\rho(\overline{X}) = \min_p \rho(\overline{X}_p)$ is the source's own way of saying $\eta(\overline{X}) = 0$, so it replaces the deck's paragraph-form hypothesis.
Spoken: the inert primes of $\mathbf{Q}(\sqrt{D_X})$ are jump primes, by Chebotarev they have density one half.
Spoken: the source frame prints three factorizations, $d_{X_3}$, $d_{X_4}$ and $d_{X_5}$; the deck shows only $d_{X_3}$.
Spoken: d is well defined up to squares; d is the discriminant of the quadratic extension, over which the Galois representation associated to the transcendental lattice has det = 1.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:672-702
K3 surfaces
So far we have been trying to improve the inequality $\rho(\overline{X}) \leq \rho(\overline{X}_p)$.
Theorem (Li-Liedtke)
If there are infinitely many $p$ primes such that
$$\rho(\overline{X}) < \rho(\overline{X}_p) \text{ and } \rho(\overline{X}_p) \neq 22,$$
then $\overline{X}$ contains infinitely many rational curves.
Theorem (Bogomolov-Zarhin)
The set $\lbrace p : \rho(\overline{X}_p) \neq 22 \rbrace$ has positive density (density 1 after finite extension).
Corollary (Li-Liedtke)
If $\rho(\overline{X})$ is odd, then $\overline{X}$ contains infinitely many rational curves.
SOURCE: V:L533
What it does. Stops the section being purely about obstacles. A forced jump is also a tool, and this is the slide that says so.
Spoken: What the jumping buys:
Spoken: For odd geometric Picard rank the reduction rank exceeds the characteristic-zero rank at every good prime, by parity.
Spoken: Membership in $\Pi_{\mathrm{jump}}$, defined net of $\eta$, does not follow.
Spoken: The general statement is now a theorem of Chen, Gounelas and Liedtke; the non-trivial jump character gives an easy proof in a special case.
TRANSCRIBED FROM: frobenious-dist/vantage.tex:533-557
Computing $\rho(\overline{X})$
Let $T_X$ be the orthogonal complement of $\operatorname{NS}(X_{\mathbf{C}}^{\mathrm{top}})$ in $H^2(X_{\mathbf{C}}^{\mathrm{top}}, \mathbf{Q})$.
Theorem (Charles)
$$\rho(\overline{X}_p) \geq \begin{cases} \rho(\overline{X}) & \text{if } E_X \text{ is CM or } \dim_{E_X}(T_X) \text{ is even,} \\ \rho(\overline{X}) + [E_X:\mathbf{Q}] & \text{if } E_X \text{ is totally real and } \dim_{E_X}(T_X) \text{ is odd.} \end{cases}$$
Further, assume that we are in the second case, then exist infinitely many pairs $(p, q)$ such that the equality holds and
$$\operatorname{disc}(\operatorname{NS}(X_p)) \not\equiv \operatorname{disc}(\operatorname{NS}(X_q)) \bmod \mathbf{Q}^{\times 2}$$
SOURCE: I15:L523
Once [E_X:Q] is at least 2 the excess is [E_X:Q], and two primes can cut only one below it, so the old "second case" wording had it backwards.
CORRECTION: the source frame writes the second case as "$E_X$ is a totally real or $\dim_{E_X}(T_X)$ is odd"; the slide keeps the corrected "and", which is what makes the two cases exclusive.
Spoken: Charles: van Luijk's method computes $\rho(\overline{X})$ unless $E_X$ is totally real with $E_X \neq \mathbf{Q}$ and $\dim_{E_X} T_X$ odd.
Spoken: The case it survives is $E_X = \mathbf{Q}$ with $\dim T_X$ odd: excess 1, and infinitely many pairs achieve equality with mismatched discriminants.
TRANSCRIBED FROM: K3workshop/K3workshop.tex:523-550
When every prime overshoots
Forced excess is a different question, and there are exactly two cases with $\eta(\overline{X}) > 0$, so that every good reduction has rank strictly above $\rho(\overline{X})$:
$E_X = \mathbf{Q}$ with $\dim T_X$ odd: $\eta = 1$, van Luijk survives;
$E_X$ totally real, $E_X \neq \mathbf{Q}$, $\dim_{E_X} T_X$ odd: excess $[E_X:\mathbf{Q}] \geq 2$, a genuine wall.
And these are the only two, by a theorem of Charles.
Only the second is a wall. In case 1 parity hands the excess straight back: the quartic on slide 9 has $\rho(\overline{X}) = 17$, every good prime overshoots to 18 or more, and cutting 18 to below 18 gives the upper bound 17.
Case 2 is different. The excess is $[E_X:\mathbf{Q}] \geq 2$, nothing tells you it is exactly that, and no pair of primes cuts far enough on reduction data alone .
NEW SLIDE: newly written, not reused from an existing talk.
If E_X is totally real of odd degree at least 3 with dim_{E_X} T odd, then rho(Xbar) is odd and the excess is already at least 3; that is why odd rank alone does not land you in case 1.
Cutting 18 to below 18 is an upper bound of 17 only; smaller ranks are not excluded by that argument, which is why the lower bound has to be named.
What it does. Draws the last distinction of the lecture, and it is a fine one: forced jumping is not the same as a lost cause. Slide 9's surface jumps at every prime and is still solved. The next slide is one that is not.
SECTION 1.5, Why the excess is forced. Eight slides, and the destination. Why jumping happens at all, how often, what it buys you, and the two cases where no prime escapes it.
Spoken: Everything so far has been about which primes jump.
Spoken: Odd $\rho(\overline{X})$ alone does not put you in case 1.
Spoken: Note $\eta$ is a forced minimum, not the excess at a given prime: an individual reduction can overshoot by more, and by slide 12 equality holds infinitely often.
Spoken: The matching lower bound 17 comes from the symplectic order-5 action together with a polarization.
Spoken: With certified RM the subtraction improves: certified quadratic RM plus two rank-18 reductions with mismatched discriminants gives $\rho \leq 16$.
Spoken: Kummer surfaces are never here, by van Geemen's bound $\dim_E T \geq 3$ for totally real $E$: that forces $\operatorname{rk} T \geq 6$, while $\operatorname{rk} T(\operatorname{Km}(A)) \leq 5$. That is why 1.4 was safe.
TRANSCRIBED FROM: no source, restyled
A surface where that happens
$$X : w^2 = (-y^2/8 + yz - z^2)(7x^2/8 + 5xz + 7z^2)(2x^2 + 3xy + y^2)$$
$X$ is the minimal desingularization of this double cover: the branch locus is a union of six lines, so the cover is singular at their 15 intersection points.
A double cover of $\mathbf{P}^2$, $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 16$, RM by $\mathbf{Q}(\sqrt{2})$. The known classes are the polarization and the 15 exceptional curves.
Case 2 exactly: $(22 - 16)/2 = 3$, odd. Reduction will overshoot at every good prime, and unaugmented reduction, van Luijk's two-prime argument included, stalls here; only the certified RM of slide 24 cuts further.
NEW SLIDE: newly written, not reused from an existing talk.
SOURCE: saard PDF p.35
What it does. Hands over. Tomorrow this same surface comes back, and the new method gets closer without getting there.
PLANNING NOTE: Source not yet available. This slide exists only in the two obstruction decks, as PDF. It needs padicperiods/slides/.
TRANSCRIBED FROM: no source, restyled