From periods to explicit geometry | Computing Picard Lattices of K3 Surfaces

The other direction

Two lectures of upper bounds. Reduction gave $\rho \leq \rho(\overline{X}_p)$, the obstruction cut that down, and both stop at a number.
A number is not a Picard lattice: it says nothing about the intersection form, nothing about the Galois action, and exhibits no curve.

Today: produce the classes. A lower bound is a curve you can write down, and enough curves with their intersections is the lattice itself.

An analytic approach

Lefschetz (1,1) theorem

A homology class $\gamma \in H_2(X, \mathbf{Z})$ is in $\operatorname{Pic} \overline{X}$ if and only if $\int_\gamma \omega_X = 0$, where $\omega_X$ is the nonzero holomorphic 2-form $\omega_X$ on $X$, unique up to scaling.

Hence, if $\Pi = [\int_\gamma \omega_X]_{\gamma \in H_2(X, \mathbf{Z})} \in \mathbf{C}^{22}$ is the period vector for $\omega_X$, then we are reduced to finding a (saturated) lattice $\Lambda \subset H_2(X, \mathbf{Z})$ of solutions

$$\Pi R = 0, \qquad R \in H_2(X, \mathbf{Z}) \simeq \mathbf{Z}^{22}.$$

$\Pi$ can be computed:
- rigorously as a ball via deformation for projective hypersurfaces (Sertöz)
- heuristically for degree 2 surfaces branched over 6 lines (Elsenhans-Jahnel)
Heuristically, via lattice reduction algorithms, we can find $\Lambda \subset H_2(X, \mathbf{Z})$.
There is no obvious way to prove that our guesses are actually correct.

SOURCE: L:L524-554

Say the name out loud here. Lambda is introduced on this slide and nowhere else; the Jacobian period lattice on slide 5 is a different object and carries a subscript.

What it does. Converts geometry into numerical linear algebra in one line. This is the counterpart of Lecture 1's Pic = H^2 intersect H^{1,1}, now made computational.

Spoken: So with the period vector $\Pi = [\int_{\gamma} \omega_X]_{\gamma} \in \mathbf{C}^{22}$, finding Picard classes means finding the integer solutions of $\Pi R = 0$.

Spoken: With exact periods that kernel already is $\operatorname{Pic}\overline{X}$.

Spoken: What one has is $\Pi$ to finite precision: write $\Lambda$ for the saturated lattice produced from those approximate periods, satisfying $\Pi R \approx 0$ to the working precision.

Spoken: Whether its vectors are Picard classes is what the rest of the lecture establishes.

Spoken: Every $\Lambda$ later in this lecture is this object.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:524-554 (items 1 to 3 of the closing itemize; the Lairez-Sertoz item is slide 3)

How far numerics can be trusted

Nonetheless, given $\Pi$ as a ball, one can compute $B \gg 0$ such that
$$\operatorname{Pic}(\overline{X})_{\mid B} := \mathbf{Z} \langle \gamma \in \operatorname{Pic} \overline{X} \mid -\gamma^2_{\mathrm{prim}} < B \rangle \subseteq \Lambda \qquad \text{(Lairez-Sert\"{o}z).}$$

SOURCE: L:L550-552

Only once B exceeds 6 does the enumeration of Lambda account for every smooth rational curve of degree at most 4; the primitive class C - H of a quartic has square -6, so that is the norm the bound must in particular cover. The -2 in the formula is C^2 = 2 p_a(C) - 2 for a smooth rational C; for other curves the square is 2 p_a(C) - 2 - d^2/4. The margin at 300 digits is enormous, which is why this step is cheap, but it is a step.

PLANNING NOTE: Default taken. The Jacobian period-lattice slide from Leiden, L:L192-214, is cut. It is written for an audience meeting complex tori for the first time, and this room is not that audience. Its one working part, that the period lattice of the Jacobian, written Lambda_J so that it is not confused with the candidate Picard lattice Lambda of slide 2, is computed numerically to high precision, is folded into slide 5.

Spoken: Given $\Pi$ as a ball, one can compute $B \gg 0$ with $\operatorname{Pic}(\overline{X})|_B \subseteq \Lambda$, where $\operatorname{Pic}(\overline{X})|_B$ is generated by the classes with $-\gamma^2_{\mathrm{prim}} < B$.

Spoken: Numerics alone give candidates.

Spoken: This theorem says how much of the lattice the candidates are guaranteed to cover, which is the difference between a guess and a bounded guess.

Spoken: $B$ grows with the working precision, and the bound has to be checked, not assumed: a degree-$d$ smooth rational curve on a quartic surface ($H^2 = 4$) has primitive projection $C - \frac{d}{4}H$ of square $-(2 + d^2/4)$, which is $-6$ at $d = 4$, so degree $\leq 4$ needs $B > 6$.

Spoken: At 300 digits, $B \approx 10^{300}$.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:524-554 (last item of the closing itemize)

Our setup

Let $C$ be a nice (smooth, projective, geometrically integral) curve over $k$ of genus $g$ given by equations. Let $J$ be the Jacobian of $C$.

Goal

Given the equations of $C$, compute the endomorphism ring $\operatorname{End} \overline{J}$.

Heuristic solution

$J = \mathbf{C}^g / \Lambda_J$, the period lattice $\Lambda_J$ computed numerically, to high precision, from a basis of $H^0(C, \Omega_C)$.

By picking a $k$-basis for $H^0(C, \Omega_C)$, we have

$$\operatorname{End}(J) = \left\{ T \in M_g(k) \mid T \Lambda_J \subset \Lambda_J \right\}$$

Hence, if $\Pi$ is a period matrix for $C$, i.e., $\Lambda_J = \Pi \mathbf{Z}^{2g}$, then we are reduced to finding a $\mathbf{Z}$-basis of the solutions $(T, R)$ to

$$T \Pi = \Pi R, \qquad T \in M_g(\overline{k}), \quad R \in M_{2g}(\mathbf{Z}).$$

The Galois module structure of $\operatorname{End}(\overline{J})$ is given via its action on $T \in M_g(\overline{k})$.

Heuristically, via lattice reduction algorithms, we can find such a $\mathbf{Z}$-basis.

There is no obvious way to prove that our guesses are actually correct.

SOURCE: L:L216-239

What it does. Exactly parallel to slide 2: numerics produce candidates and no proof. The room should feel the same gap twice, once here and once for K3 surfaces.

HEDGE, as written in the source: "Keep the hedge. The slide says there is no obvious way to prove the output correct. That sentence is the reason the next slide exists, and removing it would make the section pointless."

Spoken: Not the $\Lambda$ of slide 2.

Spoken: Endomorphisms are the pairs $(T, R)$ with $T \in M_g(\mathbf{C})$, $R \in M_{2g}(\mathbf{Z})$ and $T\Pi = \Pi R$, found by lattice reduction.

Spoken: The integrality of $R$ is the whole content of the condition: drop it and every $T$ admits a complex $R$, so the equation characterises nothing.

Spoken: There is no obvious way to prove the output correct.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:216-239 (first line folded in from the cut frame 192-214, per the PLANNING NOTE on slide 3)

Representing endomorphisms via correspondences

$$\begin{aligned} \alpha_C : \ & C \xrightarrow{\ \ AJ\ \ } J \xrightarrow{\ \ \alpha\ \ } J \dashrightarrow \operatorname{Sym}^g(C) \\ & P \mapsto \{Q_1, \ldots, Q_g\} \Longleftrightarrow \alpha([P - P_0]) = \left[ \sum_{i=1}^g Q_i - P_0 \right] \end{aligned}$$

This traces out a divisor on $C \times C$, which determines $\alpha$.

This divisor is a certificate of containment for $\alpha \in \operatorname{End} \overline{J}$.

Theorem (C-Mascot-Sijsling-Voight)

We give an algorithm for nondegenerate $\alpha \in \mathrm{M}_g(\overline{k})$

$$\alpha \mapsto \begin{cases} \texttt{true} & \text{if } \alpha \in \operatorname{End} \overline{J}, \text{ and a certificate} \\ \texttt{false} & \text{if } \alpha \notin \operatorname{End} \overline{J} \end{cases}$$

By interpolation via $\alpha_C$ or by locally solving a differential equation on $C \times C$.

SOURCE: L:L241-279

Nondegeneracy is a hypothesis on the input, not a branch of the output. A degenerate alpha that does lie in End Jbar is outside the algorithm's scope; it is not returned false.

What it does. The template. Numerical candidate, then an exact object that witnesses it. Everything in 3.3 is this pattern with quadric surfaces in place of correspondences.

[figure: images/certificatew.png] The source frame sets this certificate icon, labelled alpha, next to the words "certificate of containment" and inside the theorem's true branch. The deck carries no images, so the icon is dropped and the word "certificate" carries it.

Spoken: Close the gap: realise a candidate as a divisor on $C \times C$, which is an object one can verify.

Spoken: An algorithm which, given a nondegenerate $\alpha \in M_g(\overline{\mathbf{Q}})$, returns true with a certificate if $\alpha \in \operatorname{End} \overline{J}$, and false if $\alpha \notin \operatorname{End} \overline{J}$.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:241-279

Examples

We have verified, decomposed and matched the $66158$ curves over $\mathbf{Q}$ of genus $2$ in the L-functions and modular form database LMFDB.org
The algorithm verifies that the following genus 4 curve over $\mathbf{Q}(\sqrt{3})$
$$\begin{aligned} 0 &= -8x^2 + 8xy + 17y^2 - 34xz - 2yz - 28z^2 - 10xw - 9yw - 18zw + 2w^2, \\ 0 &= 4x^3 - 6x^2 y - 6x y^2 + 12x^2 z + 6xyz + 24y^2 z - 12x z^2 - 24z^3 + 2x^2 w + 7xyw \\ &\qquad + 4y^2 w + 4xzw - 13yzw - 8z^2 w - 20x w^2 - 3z w^2 - 12w^3 \end{aligned}$$
has real multiplication by the maximal order of $\mathbf{Q}(x)/(x^4 - x^3 - 3x^2 + x + 1)$.
The first step to show that, under Langlands, it corresponds to a specific Hilbert modular form $f$, i.e., $J_{\mathbf{Q}(\sqrt{3})} \sim A_f$. We used this in a recent project, where we show that the 2-isogeny field of $A_f$ solves the inverse Galois problem for $\operatorname{PSL}_2(\mathbf{F}_{16}) \rtimes C_2 \simeq \texttt{17T7}$.
Our method works just as well for isogenies and projections.
Try it: https://github.com/edgarcosta/endomorphisms

SOURCE: L:L281-380, compressed to one

PLANNING NOTE: Default taken. The full rigorous-endomorphism-ring slide and its day/night structure are compressed into this one. They are a method talk, and 3.2 is meant to be a shape, not a method.

PLANNING NOTE: Default taken. 3.2 runs from Leiden alone, without concrete numerical data. The UNSW 2019 deck has a 600-digit period matrix and explicit candidate matrices, but its own example stops at candidates rather than a finished certificate, and your no-new-computations rule settles it: use what exists and stop where it stops. Say the word and I will request that source instead.

Spoken: Scale. 66158 genus-2 curves from the LMFDB, verified.

Spoken: Reach. A genus-4 curve over $\mathbf{Q}(\sqrt{3})$ with RM, tied to a Hilbert modular form and to 17T7 as a Galois group over $\mathbf{Q}$.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:352-380 (the Examples frame; the Rigorous Endomorphism ring frame, 281-351, stays compressed away per the PLANNING NOTE)

Picard lattice of a K3 surface

Let $X$ be a K3 surface defined over $k \subset \mathbf{C}$. We view $X$ also as a complex manifold.

$$\operatorname{Pic} \overline{X} \simeq \mathbf{Z} \langle \text{algebraic curves in } X \rangle / \langle \text{linear equivalences} \rangle \subset H_2(X, \mathbf{Z})$$

Goal

From the equations of $X$, compute $\operatorname{Pic} \overline{X} \subset H_2(X, \mathbf{Z})$ as a $\operatorname{Gal}(\bar k/k)$-module.

"The evaluation of $\rho$ for a given surface presents in general grave difficulties." -- Zariski

"New and interesting" Galois representations arise from $T(X)$:

$$H^2(X, \mathbf{Q}) \simeq \operatorname{Pic}(\overline{X})_{\mathbf{Q}} \oplus T(X)_{\mathbf{Q}}$$

Useful for studying rational points, via a potential Brauer-Manin obstruction, for $k$ a number field:

$$H^1(\operatorname{Gal}(\bar k/k), \operatorname{Pic} \overline{X}) \simeq \operatorname{Br}_1(X)/\operatorname{Br}_0(X).$$

$$X(k) \subset X(\mathbf{A}_k)^{\operatorname{Br}} \subset X(\mathbf{A}_k)$$

SOURCE: L:L472-496, compressed

The number-field hypothesis is doing work. Hochschild-Serre gives the isomorphism only once the next obstruction H^3(k, G_m) vanishes, which holds over a number field; over a general base one has an exact sequence rather than an isomorphism.

PLANNING NOTE: the Brauer-Manin clause that used to sit here has moved to slide 20, where H^1(Gal, Pic Xbar) returns as an open question. Nothing in this lecture picked it up between here and there.

What it does. Names what "computing the Picard lattice" has meant all along, and makes clear that Lectures 1 and 2 were only ever computing one number of it.

SECTION 3.3, Picard lattices of K3 surfaces. Thirteen slides, the destination of the course. Candidate classes, then curves, then a certified lattice, ending on what is still open.

Spoken: The target, stated once and properly: compute $\operatorname{Pic} \overline{X}$ as a $\operatorname{Gal}(\bar k/k)$-module, not merely its rank.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:472-496

A running example inspired by Klein-Mukai

$$X : x^{4} + xyzw + y^{3}z + yw^{3} + z^{3}w = 0 \subset \mathbf{P}^3$$

It is a fiber in a pencil that has generic rank 19, thus $\operatorname{rank}\operatorname{Pic}\overline{X} \geq 19$.
Matching upper bounds can be deduced by positive characteristic methods: Lectures 1 and 2 give $\operatorname{rank}\operatorname{Pic}\overline{X} \leq 19$.
No known explicit descriptions of $\operatorname{Pic}\overline{X}$.
Heuristically, one computes $\Lambda \simeq \mathbf{Z}^{19}$ such that
$$\Pi \Lambda \approx 0 \qquad \operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic}\overline{X}.$$
We can compute $\operatorname{Aut}\Lambda$, the isomorphism class seems to be $F_{42} \times \operatorname{PGL}(2,7)$.
No small rational curves: There are no lines, no conics, no twisted cubics.
The "smallest" non-trivial curves that appear are smooth rational quartics.
Lattice computations with $\Lambda$ predict that there are
133056
smooth rational quartics spanning $\Lambda$.

SOURCE: L:L556-607

What it does. The number 133056 is the hook. A prediction that large, from pure lattice theory, is what makes the rest of the lecture worth sitting through.

PLANNING NOTE: Default taken. The slide's \only<3-6> overlay recaps the even-rank fact, van Luijk, and the C-Sertoz p-adic step. With Lectures 1 and 2 behind us that overlay compresses to one back-reference line. This was flagged in the source map as the clearest cross-lecture edit, and it needed your approval; it is applied here and marked.

The upper bound rank Pic Xbar <= 19 is not decoration: slide 15 needs it, since saturation and a common rank-19 sublattice on their own do not exclude Picard rank 20.

PANEL FLAG: the isomorphism Aut Lambda = F_42 x PGL(2,7) and the predicted count 133056 are asserted from the author's lattice computation. The panel could not certify either from the supplied data; nothing has been changed.

What F_7 meant, now settled by the orders. #Aut Pic Xbar = 14112 and 14112 / 336 = 42, so the first factor has order 42, the Frobenius group AGL(1,7), not Z/7 (which would give 2352). Slide 17 compares 14112 against 1008, again a factor of 14. Aut here is the polarization-preserving orthogonal group O(Lambda, H); -1 is an isometry of Lambda but does not fix H, so it is not in the product.

Spoken: Then $\operatorname{Aut}\Lambda := O(\Lambda, H) \simeq F_{42} \times \operatorname{PGL}(2,7)$, of order $42 \cdot 336 = 14112$, where $F_{42} = AGL(1,7)$ is the Frobenius group of order 42 and $O(\Lambda,H)$ means the isometries fixing $H$, so $-1 \notin \operatorname{Aut}\Lambda$.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:556-607

Spoken (not on the frame): Where that rank-19 starting point came from: even rank of the reduction, van Luijk, and the C-Sertöz $p$-adic step, all of Lectures 1 and 2.

Reconstructing isolated curves from their Hodge classes

Turns out one can compute a bit more for hypersurfaces

$$\varphi \colon H_2(X, \mathbf{Z}) \times H^2_{\mathrm{dR}}(X/k) \to \mathbf{C} \qquad (\gamma, \omega) \longmapsto \int_\gamma \omega$$

Note, if $\gamma \in \operatorname{Pic} \overline{X}$, then $\frac{1}{2\pi i}\int_\gamma \omega \in \bar k$ for $\omega \in F^1 H^2_{\mathrm{dR}}(X/k)$.

Theorem (Movasati-Sertöz)

If $\gamma = [C] \in H_2(X, \mathbf{Z})$ for a curve $C \subset X$ then from $\frac{1}{2\pi i}\left(\int_\gamma \omega\right)_{\omega \in F^1}$ one can construct an ideal $I_\gamma$ such that $I(C) \subsetneq I_\gamma$.

In favorable circumstances we expect low order equations in $I_\gamma$ to span $I(C)$.
For example, smooth rational curves of degree up to 4 in K3s.

Theorem (Cifani-Pirola-Schlesinger)

For a smooth rational quartic curve $C \subset X$ we have that the equation of the quadric surface containing $C$ generates $I_{[C],2}$, i.e., $I(C)_2 = I_{[C],2}$.

Reconstructing quadric surfaces

$$X : x^{4} + xyzw + y^{3}z + yw^{3} + z^{3}w = 0 \subset \mathbf{P}^3$$

$$\operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic}\overline{X}$$

Goal

Reconstruct the quadric surfaces containing some of the 133056 smooth rational quartics in $X$ using the curve classes.

Fortunately, there is a small $\operatorname{Aut}(\Lambda)$ orbit of size 336:
$133056 = 336 + 1008 + 1176 + 3528 \cdot 3 + 4704 \cdot 3 + 7056 \cdot 9 + 14112 \cdot 3$
For each quartic curve $C \subset X$, we can compute
$$I_{[C],2} = \langle a_0 x^2 + \cdots + a_9 w^2 \rangle_{\mathbf{C}}$$
that defines a quadric surface $Q$, such that $Q \cap X = C \cup \overline{C}$. Hence, we expect an orbit of 168 quadrics each containing a pair of quartics.
We aim reconstruct the ten (algebraic!) coefficients of these quadrics.

Reconstructing quadric surfaces

Goal

Reconstruct the ten coefficients $a_i$ of these quadrics in a Galois orbit of size 168.

The minimal polynomials have large height about 9k characters, e.g.:
$x^{168} - 10014013832542203812872613924739x^{161} + 171047690745503707515328576627906817785436888130925209472262244x^{154} - 1268317331496745879603035032448157273146519836562713924560050631153969519297207668270922371313x^{147} + 23237703563539410755436556575134206593366430461423708193774287327245213403024087108979694756912313 \cdots$
Every computation must be done extremely selectively!
We are presented with same 168 degree field $L$ in 9 different ways.

Isomorphism problem

The abstract isomorphism problem feels hopeless.

Goal

Construct $\mathbf{Q}(a_k) \hookrightarrow L$, where $L = \mathbf{Q}(a_0, \dots, a_9) = \mathbf{Q}(a_0)$.

In our case, we have all the compatible embeddings

$$\sigma_i : \mathbf{Q}(a_k) \hookrightarrow L \hookrightarrow \mathbf{C}$$

Thus the isomorphism is given by the solution of the following linear system

$$\{\sigma_i(a_k)^j\}_{i, j} \cdot v = \{\sigma_i(a_0)\}_i, \qquad v \in \mathbf{Q}^{168}$$

Distinct nodes make $\{\sigma_i(a_k)^j\}$ invertible, so the solution $v \in \mathbf{Q}^{168}$ is unique; the denominators of $v$ are bounded a priori, so enough precision pins $v$ down exactly and the isomorphism is then verified exactly.

In practice, it is faster to iteratively refine the complex embeddings, as their height is smaller than theoretically possible: 4k vs 120k digits.

SOURCE: L:L694-720, compressed

PANEL FLAG: the Leiden slide says the system is "numerically stable, as it is a Vandermonde matrix". Distinct nodes give invertibility, not stability; Vandermonde matrices can be severely ill-conditioned (Pan, arXiv:1504.02118), and compatibility of the embeddings supplies no stability guarantee. The slide now states the weaker true claim: unique solution, bounded denominators, exact verification. If a condition-number estimate for these particular nodes or a certified error bound at the precision used is available, supply it and the stronger sentence can come back.

HEDGE, as written in the source: "Keep the phrase. The slide says the isomorphism problem 'feels hopeless' before the trick arrives. That reversal is the slide."

Spoken: Then: build $\mathbf{Q}(a_k) \to L$ by a Vandermonde system over compatible embeddings, with a height-reduction trick taking 120k digits down to 4k.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:694-720

Intersecting the quadric surfaces with the K3 surface

$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$

Goal

Show that $Q \cap X$ decomposes into two quartic curves.

It suffices to show that the singular locus $S$ of $Q \cap X$ consists of 10 distinct reduced points.
Hopeless to do this directly! Operations in $L$ are seriously expensive!
Linear algebra. Gröbner basis.
One needs to compute $S$ by hand, and clear denominators before that.
Working over $\mathbf{F}_p$ we find 10 distinct points.
Hence, $S$ is zero-dimensional and reduced, and $\deg S \leq 10$.
We conclude $\deg S = 10$ via Gotzmann regularity theorem, by checking that $\dim L[x,y,z,w]_{\bullet}/I_{\bullet} = 10$ for $\bullet = 6,7$, where $I$ is saturated and $V(I) = S$.

SOURCE: L:L723-753

What it does. The candidates become curves. This is the moment the numerics turn into geometry.

PANEL FLAG: that Lambda has no degree-4 class of square 0 is asserted from the author's enumeration and was not independently certifiable from the supplied data. The exclusion of (2,2) + (2,2) rests on it.

Smoothness gives the two rulings only over the algebraic closure: a smooth quadric over L need not be split. The bidegree argument and the component degrees are geometric, so Lbar is where they belong; if the two rulings are wanted over L itself, splitness over L is an extra thing to establish.

The division of labour matters. The special fibre over F_p bounds the degree from above and shows S is reduced and zero-dimensional; it cannot on its own prove that the characteristic-zero intersection splits. The equality deg S = 10 is a characteristic-zero Hilbert-function claim, checked over L in degrees 6 and 7. The exclusion of other decompositions rests on the bidegree count, the absence of lines, conics and twisted cubics from slide 9, and separately on the absence of a degree-4 square-zero class in Lambda, which is what rules out (2,2) + (2,2); primitive norm 4 is below B, so the enumeration supplies it.

Gotzmann regularity, in one clause: it bounds the regularity of a saturated ideal in terms of its Hilbert polynomial, so the Hilbert polynomial is determined by finitely many explicitly known degrees. The saturation hypothesis is not optional: adding a degree-8 equation can leave the values in degrees 6 and 7 unchanged while shrinking V(I).

Spoken: $Q$ is smooth, so $Q_{\overline{L}} \simeq \mathbf{P}^1 \times \mathbf{P}^1$ and $Q \cap X$ has bidegree $(4,4)$ over $\overline{L}$; slide 9 excludes components of degree $1, 2, 3$, leaving $(1,3) + (3,1)$ and $(2,2) + (2,2)$.

Spoken: The second is excluded by the enumeration of $\Lambda$, which has no degree-4 class of square 0: the singularity count does not separate the two, since two nodal $(2,2)$ curves also give $8 + 1 + 1 = 10$ reduced singular points.

Spoken: That leaves $(1,3) + (3,1)$, whose components meet in $1 \cdot 1 + 3 \cdot 3 = 10$ points, so it suffices that the singular locus $S$ is 10 distinct reduced points.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:723-753

Certifying $\operatorname{Pic} \overline{X} = \Lambda$

$$\begin{aligned} &Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168\\ &\Lambda_Q := \langle [C] : C \subset \sigma(Q) \cap X, \, \sigma : L \hookrightarrow \mathbf{C} \rangle \subseteq \operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic} \overline{X} \end{aligned}$$

The inclusion $\Lambda_Q \subseteq \Lambda$ is not explicit!

Nonetheless, $\operatorname{Pic} \overline{X}$ and $\Lambda$ are saturated in $H_2(X, \mathbf{Z})$.

Hence, it is sufficient to show that $\operatorname{rank} \Lambda_Q = \operatorname{rank} \Lambda = 19$.

We can do this in two ways:

Compute the intersections of these 336 curves with each other over $\mathbf{F}_p$.
Certify that these correspond to the original classes.
Showing that there are at most 66528 distinct quadrics. Can be done over $\mathbf{C}$.
This establishes a bijection between these quadric surfaces and the $168$ pairs of quartic curve classes that they correspond to.

$$\operatorname{Pic} \overline{X} = \Lambda$$

KEEP SLIDE: reused verbatim from an existing talk.

SOURCE: L:L755-791

What it does. The destination of the whole course. Note what it delivers that Lectures 1 and 2 could not: not a number, but the lattice.

Say why the upper bound is there. Saturation plus a common rank-19 sublattice does not exclude Picard rank 20, with Lambda a corank-one saturated sublattice of Pic Xbar. The bound rank Pic Xbar <= 19 closes that off: Lambda_Q sits inside both Lambda and Pic Xbar with rank 19, so the two saturated lattices have the same rational span and coincide.

PANEL FLAG: the bijection and the count of 336 curves do not by themselves give rank 19. A bijection of two sets of size 168 is compatible with the 336 classes spanning any rank up to 19, so the rank computation is a separate step, and the supplied sources do not show it being performed. State it as a requirement, not as done.

The two counts are at different scales, and the source's one-line version runs them together. 66528 = 133056/2 bounds the quadrics globally, one per residual pair of quartic classes. The bijection is the local statement: 168 embeddings sigma of L into C, 168 conjugate quadrics sigma(Q), 168 pairs of quartic classes, 336 curves. It is not a bijection between 66528 quadrics and 168 pairs.

HEDGE, as written in the source: "Verbatim, including: 'The inclusion Lambda_Q is contained in Lambda is not explicit!' Then the rescue: both Pic Xbar and Lambda are saturated in H_2(X,Z), so it suffices that rank Lambda_Q = rank Lambda = 19."

The frame closes the display Pic Xbar = Lambda with a certificate stamp image and a check mark; the deck carries no images, so the stand-in line names the file.

Spoken: With $\operatorname{rank}\operatorname{Pic}\overline{X} \leq 19$ from slide 9, all three lattices have rank 19, so $\Lambda$ and $\operatorname{Pic}\overline{X}$ are saturated with the same rational span, hence equal.

Spoken: Two routes to the rank equality: intersect the 336 curves with each other over $\mathbf{F}_p$; or, over $\mathbf{C}$, bound the quadrics arising from the 133056 classes by $66528 = 133056/2$, one per residual pair, and match the 168 conjugates $\sigma(Q)$ bijectively with the 168 pairs of quartic classes they cut out.

Spoken: Either route still owes the rank itself: an explicit certificate that $\operatorname{rank} \Lambda_Q = 19$, a nonzero $19 \times 19$ minor of the intersection matrix of the 336 classes, or a prior verification that those 336 vectors span $\Lambda$.

Spoken: With that, $\operatorname{Pic} \overline{X} = \Lambda$, with the certificate mark.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:755-791

Computing the Galois action

$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$

$Q \cap X$ decomposes into a pair of quartics over $K$ a quadratic extension of $L$.

Goal

Compute $K$ and $\operatorname{Gal}(K/\mathbf{Q})$ acting on $\Lambda_Q$.

Via the identification with the original classes we have $\frac{1}{2 \pi i} \left( \int_C \omega \right)_{\omega \in F^1} \in K^{21}$.

These can be reconstructed in the same fashion as we reconstructed $a_i$.

Unclear how to certify this step! What are the denominators of $\frac{1}{2 \pi i} \int_C \omega$?

Can one compute $K$ using geometry without Gröbner basis?

To try: For a generic hyperplane $Q \cap X \cap H$ is a degree 8 reduced scheme.
The number field $K$ is the quadratic extension where we observe two orbits.

Computing the Galois action

$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$

$Q \cap X$ decomposes into a pair of quartics over $K$ a quadratic extension of $L$.

Goal

Compute $K$ and $\operatorname{Gal}(K/\mathbf{Q})$ acting on $\Lambda_Q$.

The direct computation of $\operatorname{Gal}(K/\mathbf{Q})$ looks hopeless.

We guess that $K = F(\sqrt[14]{u})$ where $[F : \mathbf{Q}] = 24$ and $\operatorname{Gal}(F/\mathbf{Q}) = C_3 \times \operatorname{PGL}(2,7)$.
Note, $\#\operatorname{Gal}(F/\mathbf{Q})$ is 14 times smaller than $\#\operatorname{Aut} \operatorname{Pic} \overline{X}$.

Can we compute $\operatorname{Gal}(K/\mathbf{Q})$?
$\operatorname{Gal}(K/\mathbf{Q}) \overset{?}{=} \operatorname{Aut} \Lambda$?
$H^1(\operatorname{Gal}(\bar k/k), \operatorname{Pic} \overline{X}) = ?$

KEEP SLIDE: reused verbatim from an existing talk.

SOURCE: L:L821-849

What it does. Under the purpose you named, this is not an apology for an incomplete result. It is three questions handed to a room full of people who could answer them.

The radical degree is 14, not 16. The written-out lecture had it as a fourth root of a fourth root, which would give [K:Q] = 384; the Leiden source says the 14th root of u, and 14 is what makes the degrees close: 14 * 24 = 336 = 2 * 168, so K is quadratic over the degree-168 field L of slide 12, as slide 16 says.

PANEL FLAG: [F:Q] = 24 and Gal(F/Q) = C_3 x PGL(2,7), of order 1008, cannot both describe a Galois extension F/Q. The reading that makes the slide consistent is that Gal(F/Q) denotes the Galois group of the Galois closure of F/Q, of order 1008, since 1008 * 14 = 14112 = #Aut Pic Xbar is exactly the "14 times smaller" in the box. Confirm that reading before speaking to it.

Spoken: Degrees: $[K:\mathbf{Q}] = 14 \cdot 24 = 336 = 2 \cdot 168 = [K:L][L:\mathbf{Q}]$, matching slides 12 and 16.

Spoken: Three questions.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:821-849

Summary

Theorem (C-Sertöz)

The quartic surface $X : x^4 + xyzw + y^3 z + yw^3 + z^3 w = 0 \subset \mathbf{P}^3$ has $\operatorname{Pic} \overline{X} = \Lambda$, generated by quartics over a quadratic extension of $L := \mathbf{Q}(\{a_i\}_i)$.

We are still developing the method and figure out its applications/limitations.

Wanna be a Theorem (C-Sertöz)

There is a practical algorithm to compute the saturation of the lattice generated by rational curves of degree up to 4.

KEEP SLIDE: reused verbatim from an existing talk.

SOURCE: L:L851-870

CORRECTION candidate (unapproved, minima-41h / minima-rza): "The quartic surface $X : x^4 + xyzw + y^3 z + yw^3 + z^3 w = 0 \subset \mathbf{P}^3$ has $\operatorname{Pic} \overline{X} = \Lambda$, the saturation of the lattice spanned by the 336 Galois-conjugate smooth rational quartics, which are all defined over the normal closure of $K/\mathbf{Q}$."

PANEL FLAG: the index 343 is asserted from the author's computation and was not independently checkable by the panel.

Two words carrying the weight. The field: a quadratic extension of L defines the two components of a single quadric section, but the Galois conjugates of those curves are only simultaneously defined over the normal closure, which is the field the Galois action of slides 16 and 17 is computed over. The saturation: the 336 curves generate a full-rank sublattice of index 343 in Pic Xbar, so integral generation requires an explicit saturation step, which is precisely what the wanna-be theorem below asks for.

Spoken: A quadratic extension of $L = \mathbf{Q}(\lbrace a_i \rbrace)$ splits one quadric section; the whole conjugate family needs the normal closure.

Spoken: And the 336 curves span only a sublattice of index 343, so the saturation is not cosmetic.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:851-870 (the frame's closing line 868 is deck slide 19)

Do you have a challenge K3 surface for us?

What is open, in one place

Does $\operatorname{End}(\overline{A}) = \mathbf{Z}$ force $\overline{A}_p \sim E^2$ infinitely often? (1.4)
Is there a way to get a sharp upper bound on the Picard number of a K3 surface? The double cover with RM by $\mathbf{Q}(\sqrt{2})$ is the known obstacle, where the method has not yet succeeded. (2.2)
How does one compute a good enough approximation to $\operatorname{Frob}_p$ to obstruct the last two cycles? (2.2)
Can $\operatorname{Gal}(K/\mathbf{Q})$ be computed, and is it $\operatorname{Aut}\Lambda$? (3.3)
What is $H^1(\operatorname{Gal}, \operatorname{Pic}\overline{X})$? Over a number field it is $\operatorname{Br}_1(X)/\operatorname{Br}_0(X)$, and from there the algebraic Brauer-Manin obstruction. (3.3)
Is there a practical saturation algorithm for lattices generated by low-degree rational curves? (3.3)