What is being computed, exactly

Can we combine both approaches?

At the moment we only certify an upper bound for $\dim_{\mathbf{Q}_p}L$, where $L$ is the largest $\operatorname{Frob}_p$-stable subspace of $T_{\mathrm{ev}}\cap F^1$.
To combine several primes we need at least $\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}$, to be able to use $\operatorname{sp}\bigl(\operatorname{Pic}(X^{\mathrm{al}})_{\mathbf{Q}}\bigr)=\operatorname{Pic}(X_p^{\mathrm{al}})_{\mathbf{Q}}\cap F^1_{K'}$ in its full strength.
At the moment we are only using $\operatorname{sp}\bigl(\operatorname{Pic}(X^{\mathrm{al}})\bigr)\otimes_{\mathbf{Z}}K'\subseteq L\otimes_{\mathbf{Q}_p}K'\subseteq\bigl(T_{\mathrm{ev}}\otimes_{\mathbf{Q}_p}K'\bigr)\cap F^1_{K'}$

Raising $N$ does not touch the missing rational structure.

We would also love to combine this with methods over $\mathbf{C}$.

SOURCE: O:L1462-1480, second bullet

What it does. The distinction between rational classes and their p-adic span, which your plan asks to be kept visible. It is also why this method and Lecture 1's cannot simply be added together.

Concretely: y and y + p^N x agree mod p^N, and neither of their kernels contains the other, which is why the certified output is a dimension bound rather than a containment for the approximate kernel.

EDITORIAL: the closing claim line "Raising N does not touch the missing rational structure." is deck-written, not on the frame.

Spoken: What the computation returns is not $\operatorname{Pic}(X)_{\mathbf{Q}_p}$.

Spoken: Write $L$ for the largest Frobenius-stable subspace of $T_{\mathrm{ev}}$ contained in $F^1_{\mathbf{Q}_p}$, the slide-14 object taken over all pieces at once.

Spoken: The exact $L$ is a containing space, $\operatorname{Pic}(X)_{\mathbf{Q}_p} \subseteq L \subseteq T_{\mathrm{ev}} \cap F^1_{\mathbf{Q}_p},$ with $T_{\mathrm{ev}}$ the eventual Tate space of slide 8, which is where the $\mathbf{Q}_p$-statement lives; the geometric Picard group itself only appears after extending scalars.

Spoken: and what is computed is an approximation to it: the exact kernel of an approximate matrix need not contain the true kernel nor lie in the filtration.

Spoken: What the computation certifies is a bound on $\dim L$, not those containments for the approximate kernel.

Spoken: Two things separate it from the answer, and only one is precision.

Spoken: Finite-precision error shrinks as $N$ grows.

Spoken: The other does not: even at infinite precision $L$ can stay strictly larger than the characteristic-zero Picard span, because the construction supplies no rational structure.

Spoken: Comparing lattices across primes needs $\operatorname{Pic}(\overline{X}_p)_{\mathbf{Q}}$ inside each, which separate $\mathbf{Q}_p$-spaces do not give.

TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1462-1480 (second item)