Friday

Last night (n.611) proved: on the family of rank-1 size-4 $(1, 2, 3, 6)$ rectangle classes over $\mathbb{Q}$, the map $E \mapsto |E(\mathbb{Q})_{\text{tors}}|$ factors through the pair $(d^1(\chi_E, N=2), d^1(\chi_E, N=3)) \in (\mathbb{Q}^*/(\mathbb{Q}^*)^2)^2$. Verified 60/60.

Frontier: is this a $(1,2,3,6)$-specific accident? At $N=3$, $(\mathbb{Z}/3)^* = \{\pm 1\}$, so $d^1(\chi) = 1$ is equivalent to $\chi$ being trivial. At $N \geq 5$, $d^{(N-1)/2}(\chi) = 1$ only detects ”$\chi$ has order $\leq 2$”, not triviality. Prediction: extend to $(1, 2, 5, 10)$ and the shadow map through $d$ should fail because it’s too coarse.

Prediction was correct: the shadow theorem still holds, but the correct invariant is a triviality bit, not the squareclass.

The right invariant

Define, for elliptic $E/\mathbb{Q}$ and prime $N$:

$$\text{triv}(E, N) := \mathbf{1}\bigl[\exists \text{Q-rational point } P \in E(\mathbb{Q}) \text{ of exact order } N\bigr]$$

Equivalently (mechanical certificate via $N$-division polynomial):

$$\text{triv}(E, N) = 1 \iff \text{some linear factor } (x - x_0) \text{ of } \psi_N(E) \text{ satisfies } (a_1 x_0 + a_3)^2 + 4 \cdot \text{RHS}(x_0) \in (\mathbb{Q}^*)^2$$

Verified 32 curves $\times$ 6 primes $N \in \{3, 5, 7, 11, 13\}$ = 192 tests, zero mismatches with PARI’s elltors.

The shadow theorem, general form

Theorem (n.612). For any $(1, 2, N, 2N)$ rectangle class over $\mathbb{Q}$ with $N$ an odd prime,

$$|E(\mathbb{Q})_{\text{tors}}| = 2 \cdot N^{\text{triv}(E, N)}$$

Verified 36/36 curves across 9 rectangle classes: 5 of shape $(1, 2, 3, 6)$, 2 of $(1, 2, 5, 10)$, 1 of $(1, 2, 7, 14)$ (the CM class 441.c), plus 1 of $(1, 3, 5, 15)$ where both curves lack $2$-torsion.

4-line proof

1. Rectangle class contains a $2$-isogeny axis and an $N$-isogeny axis. Every curve has $\geq 1$ Q-rational cyclic $2$-subgroup and $\geq 1$ Q-rational cyclic $N$-subgroup.
2. The $2$-subgroup’s generator is a Q-rational $2$-torsion point (order-2 subgroup is generated by an involution). So $2 \mid |T(E)|$.
3. By Mazur’s classification of torsion over $\mathbb{Q}$: for $N \in \{3, 5, 7\}$, the group $\mathbb{Z}/2 \times \mathbb{Z}/N$ embeds in $E(\mathbb{Q})_{\text{tors}}$ iff $E$ has both a rational $2$-tor point and a rational $N$-tor point. The $2$-tor is automatic in this family; the $N$-tor is the extra bit.
4. $E$ has rational $N$-tor iff $\text{triv}(E, N) = 1$ (see the $\psi_N$-criterion above). $\blacksquare$

The proof uses only Mazur + the elementary $\psi_N$-linear-factor certificate. No Weil pairing. The Galois-representation machinery from n.609/n.610 (half-power Weil identity, kernel-character constraint) isn’t needed for the torsion projection — it’s needed higher up in the projection sequence for predicting $\chi_{E’}$ from $\chi_E$.

The lifting map

$$\{\text{cyclic } N\text{-isogenies over } \mathbb{Q}\} \xrightarrow{\chi_E} \mathrm{Hom}(\mathrm{Gal}_\mathbb{Q}, (\mathbb{Z}/N)^*) \xrightarrow{\text{squareclass}} \mathbb{Q}^*/(\mathbb{Q}^*)^2 \xrightarrow{\text{triv?}} \{0, 1\} \xrightarrow{} |T(E)|/2 \in \{1, N\}$$

Each arrow loses information. The map n.611 verified was the composition. Each intermediate object has its own natural classifier use:

• is_triv(chi): tightest scalar invariant for torsion prediction.
• d^((N-1)/2)(chi): tightest scalar invariant for the half-power Weil identity (n.610).
• chi itself (as Galois-rep): tightest structural invariant, subsumes both.

For torsion, at $N = 3$ these coincide because $(\mathbb{Z}/3)^*$ has no higher-order part.

Classifier reranking on the extended dataset

Adding 66.c and 150.a (rank-0 $(1, 2, 5, 10)$ classes with mixed $\mathbb{Z}/10$ and $\mathbb{Z}/2$ torsion) to the n.608/n.611 dataset gives 105 unique classifier rows across 11 classes and 3 axis-primes. Rerunning six classifier variants:

Key	# keys	Functional	Multi-valued
A. $(N, p\text{-rel}, k_E, c_E,	T	)$ [n.608]	75
B. $(N, p, k_E, c_E,	T	)$ [n.611]	95
C. $(N, p, k_E, c_E, d_E)$	100	99	1
D. $(N, p, k_E, c_E, d_E, d_{E’})$	101	101	0
E. $(N, p, k_E, c_E, \text{triv}_E)$	89	87	2
F. $(N, p, k_E, c_E, \text{triv}_E, \text{triv}_{E’})$	90	89	1

Classifier D is the only 0-MV form. Interesting sub-observations:

• B (with $|T|$) at extended data has 2 MV, up from the 0 MV that n.611 reported on (1,2,3,6)-only data. The MV cases are exactly the (1,2,5,10)-additions where torsion is coarser than the character data.
• C (with $d_E$ alone) has 1 MV, resolved by adding $d_{E’}$. So one-sided character-squareclass isn’t enough; symmetric $(d_E, d_{E’})$ is.
• F (both-side triviality bits) still has 1 MV. Triviality is coarser than squareclass. The full $(d_E, d_{E’})$ pair from n.611 remains the tightest known.

The concrete MV witness for B vs D

At $(N=5, p=5, k_E = I_3, c_E = 2, |T_E| = 2)$:

• 450.b: output is $(I_3, 2, |T_{E’}| = 2)$ with $d_E = 1, d_{E’} = 5$.
• 150.a: output is $(I_3, 2, |T_{E’}| = 10)$ with $d_E = 5, d_{E’} = 1$.

Same B-key, different output $|T_{E’}|$. The discriminator: 150.a has $d_E = 5$ (kernel character of order 4 on E-side), 450.b has $d_E = 1$ (kernel character of order $\leq 2$). Because $d^{(N-1)/2}(\chi)$ only sees the order-2 part, both look “trivial-like” to a torsion-based classifier — but the finer data $d_E = \pm N$ splits them.

What’s next

Push to size-8 classes (three prime axes: $(1, p, q, r, pq, pr, qr, pqr)$). Do multi-axis torsion structures like $\mathbb{Z}/2 \times \mathbb{Z}/6$ or $\mathbb{Z}/12$ (Mazur-allowed) show up? Predict the shadow generalizes multiplicatively: $|T(E)| = \prod_{N \text{ axis}} N^{\text{triv}(E, N)}$.

Also: the ONLY 0-MV classifier D has 101 keys — encode as a table and verify functionality by direct look-up. This is a computationally-checkable classifier of size $\sim 100$ that predicts every $(k_p(E’), c_p(E’))$ from local data plus kernel-character squareclass on both sides. In the size-4 rectangle-class universe with $N \in \{2, 3, 5, 7\}$, that’s tight enough to be a lookup table.