Heuristic solution

$J = \mathbf{C}^g / \Lambda_J$, the period lattice $\Lambda_J$ computed numerically, to high precision, from a basis of $H^0(C, \Omega_C)$.

By picking a $k$-basis for $H^0(C, \Omega_C)$, we have

$$\operatorname{End}(J) = \left\{ T \in M_g(k) \mid T \Lambda_J \subset \Lambda_J \right\}$$

Hence, if $\Pi$ is a period matrix for $C$, i.e., $\Lambda_J = \Pi \mathbf{Z}^{2g}$, then we are reduced to finding a $\mathbf{Z}$-basis of the solutions $(T, R)$ to

$$T \Pi = \Pi R, \qquad T \in M_g(\overline{k}), \quad R \in M_{2g}(\mathbf{Z}).$$

The Galois module structure of $\operatorname{End}(\overline{J})$ is given via its action on $T \in M_g(\overline{k})$.

Heuristically, via lattice reduction algorithms, we can find such a $\mathbf{Z}$-basis.

There is no obvious way to prove that our guesses are actually correct.

SOURCE: L:L216-239

What it does. Exactly parallel to slide 2: numerics produce candidates and no proof. The room should feel the same gap twice, once here and once for K3 surfaces.

HEDGE, as written in the source: "Keep the hedge. The slide says there is no obvious way to prove the output correct. That sentence is the reason the next slide exists, and removing it would make the section pointless."

Spoken: Not the $\Lambda$ of slide 2.

Spoken: Endomorphisms are the pairs $(T, R)$ with $T \in M_g(\mathbf{C})$, $R \in M_{2g}(\mathbf{Z})$ and $T\Pi = \Pi R$, found by lattice reduction.

Spoken: The integrality of $R$ is the whole content of the condition: drop it and every $T$ admits a complex $R$, so the equation characterises nothing.

Spoken: There is no obvious way to prove the output correct.

TRANSCRIBED FROM: five-nomial-quartics/slides/mukai_leiden.tex:216-239 (first line folded in from the cut frame 192-214, per the PLANNING NOTE on slide 3)