A running example inspired by Klein-Mukai

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

This example started at a workshop at ICERM about thinking about K3 surfaces on the LMFDB.

It is the $ψ=-1/4$ fiber of the pencil $F_1L_3$, which has generic rank 19, thus $\operatorname{rank}\operatorname{Pic}X^{al} \geq 19$.
Matching upper bounds can be deduced by positive characteristic methods: Lectures 1 and 2 give $\operatorname{rank}\operatorname{Pic}X^{al} \leq 19$.
The pencil has a symplectic $\mathbf{Z}/7\mathbf{Z}$ action. Its coinvariant lattice $\Omega_7=(H^2(X,\mathbf{Z})^{\mathbf{Z}/7\mathbf{Z}})^\perp$ has rank $18$ and determinant $7^3$.
$\operatorname{Pic}(X^{al})=\langle4\rangle\oplus\Omega_7$, of determinant $4\cdot7^3=1372$ and saturation index $1$.

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.