K3 surface

$X := Z(y^4 - x^3 z + y z^3 + z w^3 + w^4) \subset \mathbf{P}^3_{\mathbf{Q}}$

$p=89,\quad N=3,\quad F=89^{-1}\operatorname{Frob}_{89}$

$+1$-eigenspace, polarization: dimension $1$. No obstruction; this class lifts.
$+1$-eigenspace, primitive part: dimension $4$. Obstruction found; at most $3$ dimensions remain.
$-1$-eigenspace: dimension $1$. Obstructed; no dimension remains.
Primitive eighth-root piece $\ker(F^4+1)$: dimension $4$. Obstructed; no dimension remains.

$\rho(X_{\mathbf{F}_{89}^{\mathrm{al}}})=5+1+4=10,\qquad \rho(X_{\mathbf{Q}^{\mathrm{al}}})\leq (1+3)+0+0=4$

In fact, $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 4$ as there are four lines in $z = 0$.

previous approaches would not have used $p = 89$

at $p = 89$ the obstruction improves 10 to 4, and 4 is sharp;

Does this always work?

SOURCE: O:L1377-1401

The closing line is the source slide's own, and it reaches back to Lecture 1.

"Each factor separately" is the refinement stated on slide 9; do not introduce it for the first time here.

At finite precision the printed dim L_i entries are themselves only upper bounds for that dimension; that is enough, since the conclusion is an upper bound.

Spoken: at $p = 89$, precision 3.

Spoken: The transcript reports $\rho(\overline{X}_{89}) = 10$, from the cyclotomic factors $(t-1)$, $(t+1)$, $(t-1)^4$, $(t^4+1)$ of $p^{-22}P_2(pT)$: degrees $1 + 1 + 4 + 4 = 10$.

Spoken: The distinct cyclotomic primary pieces have dimensions 5, 1 and 4: the $(t-1)$ and $(t-1)^4$ entries are the polarization and the primitive $+1$-eigenspace, not two separate eigenspaces.

Spoken: $T_i$ is the $i$-th such Frobenius-stable piece; $L_i \subseteq T_i$ is the largest Frobenius-stable subspace of $T_i$ contained in $F^1$, equivalently $\bigcap_j \operatorname{Frob}_p^{-j}(\ker \pi|_{T_i})$, so $\dim L_i$ bounds how much of that piece can lift.

Spoken: The transcript prints $\dim T_i$ against $\dim L_i$, factor by factor.

Spoken: Applying the obstruction to each factor separately gives $\sum_i \dim L_i = 4$, so $\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq 4$, and four lines in $z = 0$ give equality.

TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1377-1401