Theoretical example

$X : w^2 = (-y^2/8 + yz - z^2)(7x^2/8 + 5xz + 7z^2)(2x^2 + 3xy + y^2)$

The real multiplication example from Lecture 1: $X$ is the minimal resolution of this double cover of $\mathbf{P}^2$; the sextic is a product of three conics, singular at 15 points.
Known sublattice: the polarization and the 15 exceptional curves, so $\rho(X^{\mathrm{al}})\geq16$.
At $p = 83$ it has $\chi_1 = (t-1)^{10}(t+1)^6$, and the reduction bound leaves $\rho(X^{\mathrm{al}})=16,17$ or $18$.
The two extra classes span the piece $t^2+1$, irreducible over $\mathbf{Q}_{83}$, so 17 is out: $\rho(X^{\mathrm{al}})=16$ or $18$. In fact $\rho(X_{\mathbf{Q}^{\mathrm{al}}})=16$, and $X$ has RM by $\mathbf{Q}(\sqrt{2})$.
This is the wall from Lecture 1: RM by a field of degree 2 with $(22 - 16)/2 = 3$ odd, so every good prime overshoots.
Over $\mathbf{Q}$, a nonzero obstruction on the $t^2+1$ piece would remove both extra dimensions and prove $\rho(X_{\mathbf{Q}^{\mathrm{al}}})=16$, without knowing RM.
After base change to $\mathbf{Q}(\sqrt{2})$, the residue degree at $83$ is $2$. Using only $F^2$, where $F=83^{-1}\operatorname{Frob}_{83}$, gives $F^2=-1$ on the extra piece. A kernel line is Frobenius-stable and survives: the bound is at least $17$.

"Given a good enough approximation to $\operatorname{Frob}_p$ we will obstruct 2 extra cycles. However, we don't yet know how to compute such approximation!"

NEW SLIDE: newly written, not reused from an existing talk.

SOURCE: saard PDF p.35-37

What it does. One concrete surface carries both lectures: Lecture 1 says why every prime overshoots here, Lecture 2 says why the obstruction cannot fix it, and the slide says exactly what is missing. This is the recruitment slide, and it is already written.

The quoted line is the source slide's own closing sentence and is reproduced word for word.

Why the wall is over Q(sqrt 2) and not over Q: at 83 the two extra classes span the block t^2 + 1, one Galois orbit of two conjugate Frobenius eigenlines; the obstruction is Galois-equivariant, so if one eigenvector fails to lift so does its conjugate, and the exact bound at 83 is 16 (Costa-Sertoz, Example 5.9). Over Q(sqrt 2) the residue field at 83 is F_{83^2}, Frobenius is squared and acts as -1 on the block, so the obstruction kernel is a Frobenius-stable line and survives: 17. At every good place where the RM is defined, Frobenius commutes with the RM and one RM eigenline is kept, so at least 17. What is missing over Q is the Frobenius matrix on the resolution.

Spoken: Let $X$ be the minimal resolution of this double cover of $\mathbf{P}^2$: the sextic is a product of three conics, so the cover is singular at the 15 points where its six branch lines meet, and those points are not divisor classes.

Spoken: The known sublattice is the polarization together with the 15 exceptional curves above those 15 singular points, giving $\rho \geq 16$.

Spoken: At $p = 83$ that sublattice has $\chi_1 = (t-1)^{10}(t+1)^6$, the factor of $p^{-22}P_2(pT)$ it accounts for, and the reduction bound leaves $\rho = 16, 17$ or $18$ a priori.

Spoken: The two extra classes at 83 span the single $\mathbf{Q}$-irreducible piece $t^2 + 1$, which cannot supply exactly one extra rational class, so they come together or not at all: 17 is out and $\rho = 16$ or $18$.

Spoken: This is exactly the wall from Lecture 1. RM by a field of degree 2 with $(22 - 16)/2 = 3$ odd puts it in the second of the two cases where every good prime overshoots.

TRANSCRIBED FROM: no source, restyled