$$\rho(X_{\mathbf{Q}^{\mathrm{al}}}) \leq \rho(\overline{X}_p) \quad \text{for every good } p.$$
Only the second defeats the unaugmented two-prime argument.
Theorem (Charles)
$$\rho(X_p^{\mathrm{al}})\geq\begin{cases}\rho(X^{\mathrm{al}})&\text{if }E\text{ is CM or }m\text{ is even,}\\\rho(X^{\mathrm{al}})+d&\text{if }E\text{ is totally real and }m\text{ is odd,}\end{cases}$$
Equality occurs infinitely often (density $1$ after some finite extension).
If $E$ is totally real and $m$ is odd, infinitely many good ordinary prime pairs $(p,q)$ satisfy $\rho(X_p^{\mathrm{al}})=\rho(X_q^{\mathrm{al}})=\rho(X^{\mathrm{al}})+d$ and
$$\operatorname{disc}\operatorname{Pic}(X_p^{\mathrm{al}})\not\equiv\operatorname{disc}\operatorname{Pic}(X_q^{\mathrm{al}})\bmod(\Bbb{Q}^{\times})^2$$
It is very hard to prove RM!
We are now not only asking to find algebraic cycles in $X$ but also in $X \times X$.
NEW SLIDE: newly written, not reused from an existing talk.
What it does. Re-enters after the break, and re-draws the one distinction the whole lecture rests on: a prime that overshoots is a nuisance, a surface where every prime overshoots is a wall.
As on Lecture 1 slide 24: a totally real E_X of odd degree at least 3 with dim_{E_X} T_X odd also has odd geometric rank, and its excess is already at least 3, so odd rank by itself does not put you in the survivable case.
SECTION 2.1, From reduction to a lifting problem. Four slides. Pose the geometric question before any cohomological test, and re-enter after the break without assuming momentum.
THE CEILING SET FOR THIS LECTURE: Berthelot-Ogus-Raynaud is quoted and used, not unpacked. What stays defined is what the method computes with: the filtration F^1, Frobenius on H^2_dR, and the projection into H^2/F^1. Every slide is written to that level.
Spoken: and Lecture 1 closed by separating two things.
Spoken: so half your primes are wasted, but the bound is still attainable at the others.
Spoken: there are exactly two cases in which every good prime overshoots.
Spoken: Odd geometric rank alone is not the first case.
Spoken: Every good prime overshoots in both; only the second defeats the unaugmented two-prime argument.
TRANSCRIBED FROM: no source, restyled