Quintic surface

$X := Z(9 x y^{4} + 3 x^{4} z + 9 y^{2} z^{3} + z^{5} + 5 w^{5}) \subset \mathbf{P}^3$

$p=23,\quad N=6,\quad \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq 1$

factor of $p^{-1}\operatorname{Frob}_p$	dimension	liftable dimension, at most
$t-1$	$1$	$1$
$t-1$	$1$	$0$
$t+1$	$1$	$0$
$t^2+1$	$2$	$0$

$\rho(X_{\mathbf{Q}^{\mathrm{al}}}) = 1$

$p=29,\quad N=20,\quad \rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq 3$

factor of $p^{-1}\operatorname{Frob}_p$	dimension	liftable dimension, at most
$t-1$	$1$	$1$
$(t-1)^2$	$2$	$1$
$(t+1)^2$	$2$	$1$

The surface has $CM$ by $\mathbf{Q}(\zeta_5)$

at $p = 29$ on the quintic the obstruction fires, improving 5 to 3, but 3 is not sharp and no precision brings it to 1.

SOURCE: O:L1428-1458

Say the p_g = 4 line out loud, or the room spends the slide wondering what a quintic is doing in a K3 course.

What it does. Categorically stronger than slide 15. Not "we did not look hard enough" but "looking harder cannot help at this prime". This is the slide that makes the closing question a question rather than a to-do.

Source pdfnote: Computing the zeta function or the Frobenius approximation for a prime takes several hours, thus trying to use 2 primes.

Source pdfnote: the dimension of the codomain is now 4, so now there is hope to obstruct more cycles.

Spoken: A quintic with complex multiplication by $\mathbf{Q}(\zeta_5)$, at $p = 23$ and $p = 29$.

Spoken: A smooth quintic in $\mathbf{P}^3$ has $p_g = 4$ and is not a K3.

Spoken: The method is not K3-specific: it applies to smooth surfaces in $\mathbf{P}^3$ generally, which is why it can be run here.

Spoken: At 29 the method fails to certify $\rho = 1$ even at infinite precision; what it does reach, at sufficiently high precision, is $\rho \leq 3$, down from the reduction bound 5, and no further precision can improve that bound.

TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1428-1458