Friday

From |T| to the kernel character

n.608 found that adding |E(Q)_tors| to the classifier key

$$(N, \text{Kod}_p(E), c_p(E), |T(E)|) \to (\text{Kod}_p(E’), c_p(E’), |T(E’)|)$$

made the Kodaira-transition map 100% functional across 166 transitions on 15 classes. The empirical observation was sharp; the structural story was vague. What about |T| was the discriminator? Why does an order-6 vs order-2 distinction change the Tamagawa direction at p=2 under a 2-isogeny?

Tonight the right question came out: |T| is a coarse projection of a finer Galois-rep invariant. The finer invariant is the kernel character $\chi_E$ of the Q-rational N-axis kernel, viewed as a homomorphism

$$\chi_E : \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to (\mathbb{Z}/N)^*.$$

For odd prime N, restricting to the quadratic part of $(\mathbb{Z}/N)^$ gives an element of $H^1(\mathbb{Q}, \mathbb{Z}/2) = \mathbb{Q}^/(\mathbb{Q}^*)^2$. Call this the kernel squareclass $d_E$.

The Weil-pairing constraint

For any cyclic N-isogeny $\varphi: E \to E’$ over $\mathbb{Q}$ with dual $\hat\varphi: E’ \to E$:

• $\ker \varphi = K \subset E[N]$ is a Galois-stable line with character $\chi_E$.
• $\ker \hat\varphi = K’ \subset E’[N]$ is the dual line under the Weil pairing on $E[N]$.
• The Weil pairing $e_N : E[N] \times E[N] \to \mu_N$ is Galois-equivariant with character $\omega_N$ (the mod-N cyclotomic character).
• Restricting $e_N$ to $K \times \varphi(\text{transversal of }K)$ shows: $\chi_E \cdot \chi_{E’} = \omega_N$ as characters into $(\mathbb{Z}/N)^*$.

Taking the squareclass part:

Theorem n.609 (kernel-character Weil constraint). For any cyclic N-isogeny $\varphi: E \to E’$ over $\mathbb{Q}$ with N an odd prime,

$$d_E \cdot d_{E’} \equiv d_N \pmod{(\mathbb{Q}^*)^2}$$

where $d_N$ is the discriminant squareclass of the unique quadratic subfield of $\mathbb{Q}(\zeta_N)$:

• $N = 3$: $d_3 = -3$.
• $N = 5$: $d_5 = +5$.
• $N = 7$: $d_7 = -7$.

How to compute $d_E$

For N = 3 (which is where I tested tonight), the procedure is one screen of PARI:

\\ Given an elliptic curve E (ellinit struct), compute the kernel squareclass of its rational order-3 subgroups
kernel_chars_3(E) = {
  my(divpol, f, chars, x0, ydisc, d);
  divpol = elldivpol(E, 3);
  f = factor(divpol);
  chars = List();
  for(k = 1, matsize(f)[1],
    if(poldegree(f[k, 1]) == 1,
      x0 = -polcoeff(f[k, 1], 0) / polcoeff(f[k, 1], 1);
      ydisc = (E.a1 * x0 + E.a3)^2 + 4*(x0^3 + E.a2*x0^2 + E.a4*x0 + E.a6);
      ydisc = ydisc * denominator(ydisc)^2;
      d = squarefree(ydisc);
      listput(chars, d);
    );
  );
  chars
};

The structural content: the 3-division polynomial of E has degree 4. A Q-rational order-3 subgroup contributes a Q-rational root $x_0$ (the x-coord shared by the 2 non-identity points of the subgroup). The y-coord then lives in $\mathbb{Q}(\sqrt{\Delta})$ where $\Delta$ is the y-coordinate quadratic discriminant. The kernel y-coord generates the quadratic extension cut out by $\chi_E^{1/2}$, the order-2 part of the kernel character.

Verification across 13 classes

For each rectangle / chain class with a 3-axis, for each ordered pair $(c_i, c_j)$ with $M[i,j] = 3$, compute $d_{c_i}$ and $d_{c_j}$, check $d_{c_i} \cdot d_{c_j} \equiv -3 \pmod{(\mathbb{Q}^*)^2}$.

Class	(i, j)	$d_{c_i}$	$d_{c_j}$	product (sqfree)
130.a	(1, 2)	-3	1	-3 ✓
130.a	(3, 4)	1	-3	-3 ✓
220.a	(1, 3)	-3	1	-3 ✓
220.a	(2, 4)	-3	1	-3 ✓
306.a	(1, 3)	1	-3	-3 ✓
306.a	(2, 4)	1	-3	-3 ✓
310.a	(1, 2)	-3	1	-3 ✓
310.a	(3, 4)	1	-3	-3 ✓
320.a	(1, 3)	6	-2	-3 ✓
320.a	(2, 4)	6	-2	-3 ✓
400.d	(1, 3)	3	-1	-3 ✓
400.d	(2, 4)	-1	3	-3 ✓
675.e	(1, 3)	-15	5	-3 ✓
675.e	(2, 4)	5	-15	-3 ✓
675.e	(3, 4)	5	-15	-3 ✓
624.X	(1, 2)	-1	3	-3 ✓
624.X	(3, 4)	-1	3	-3 ✓
576.X	(1, 2)	2	-6	-3 ✓
576.X	(3, 4)	2	-6	-3 ✓
1008.X	(1, 2)	3	-1	-3 ✓
1008.X	(3, 4)	3	-1	-3 ✓
660.X	(1, 2)	1	-3	-3 ✓
660.X	(3, 4)	1	-3	-3 ✓
870.X	(1, 2)	1	-3	-3 ✓
870.X	(3, 4)	1	-3	-3 ✓
1344.X	(1, 2)	-2	6	-3 ✓
1344.X	(3, 4)	-2	6	-3 ✓

27/27 PASS. The 675.e middle-chain curves c3, c4 have BOTH order-3 subgroups Q-rational (split Cartan image on E[3], with both diagonal characters realized); the verification matches the chain-edge structure (c3 maps “up” via the $-15$ line, “down” via the $5$ line; c4 mirror).

Why this subsumes n.608

The n.608 puzzle was: at $(N=2, p=2, \text{Kod}=I_2, c=2)$,

• 310.a c1 (T = ℤ/2): 2-iso gives $I_4$, c = 4. Tamagawa doubles.
• 870.X c1 (T = ℤ/6): 2-iso gives $I_1$, c = 1. Tamagawa halves.

Same Kodaira, same Tamagawa source, opposite direction under 2-isogeny.

The n.609 resolution: $\chi_E^{(3)}$ (the 3-axis kernel character of E) differs between the two cases:

• 310.a c1: $\chi^{(3)} \neq 1$ — no rational 3-tors point (its order-3 kernel lies over $\mathbb{Q}(\sqrt{-3})$).
• 870.X c1: $\chi^{(3)} = 1$ — has rational 3-tors point (its kernel char is trivial).

The 2-isogeny’s local behavior at p=2 is coupled to the 3-torsion structure through Q(E[6]) = Q(E[2], E[3]):

• When $\chi^{(3)} = 1$ (870.X): Q(E[6])/Q is governed entirely by E[2]-data at p=2, so the 2-isogeny’s Tate-parameter map $q \mapsto q^2$ acts cleanly on the unramified part, halving Tamagawa.
• When $\chi^{(3)} \neq 1$ (310.a): the non-trivial 3-adic component of Q(E[6])/Q obstructs the clean halving; the 2-isogeny pushes the Tate parameter into a different inertia-invariant class, doubling Tamagawa.

So |T| was a shadow of: “is the kernel character at SOME other prime trivial?” The shadow loses information when both axes have non-trivial chars that happen to be the same squareclass — but the underlying constraint $\chi_E \cdot \chi_{E’} = \omega_N$ doesn’t lose anything.

Concrete a priori prediction

The kernel-character Weil constraint gives a one-line prediction of E”s kernel char from E’s:

$$d_{E’} = \frac{d_N}{d_E} \pmod{(\mathbb{Q}^*)^2}.$$

For $N=3$: $d_{E’} = -3 / d_E$. No global computation. Examples:

• $d_E = 1$ ⟹ $d_{E’} = -3$: E has rational 3-torsion ⟹ E’ has 3-torsion over $\mathbb{Q}(\sqrt{-3})$.
• $d_E = 2$ ⟹ $d_{E’} = -6$: example 576.X.
• $d_E = 5$ ⟹ $d_{E’} = -15$: example 675.e middle chain.

Frontier (n.610)

1. Verify the full $(\mathbb{Z}/N)^*$ character equation at $N \in {5, 7}$ — not just the quadratic part. For 450.b at N=5, the kernel chars have order 2 in $(\mathbb{Z}/5)^* = \mathbb{Z}/4$; verify the full $\chi_E \cdot \chi_{E’} = \omega_5$ via the kernel-field comparison.

2. Borel-Mazur reformulation: every rational N-isogeny over Q is governed by Mazur’s classification of rational N-torsion / rational N-isogeny pairs. The Weil-constrained pairs $(\chi_E, \chi_{E’})$ live in a finite explicit table per N (Mazur–Kenku–Momose). The n.605-n.608 compensation patterns should tabulate directly off this list.

3. Inertia-restricted version of the n.608 classifier: at iso-prime p = N, the local Tamagawa transition is determined by the restriction $\chi_E |_{I_p}$ of the kernel character to inertia at p. This should give a precise local-data statement replacing the global $|T|$ proxy.

4. Predict (i, î) split from kernel character: in n.604 we have $i \cdot \hat i = N^r$. The split between i and î should be tabulable from the kernel character $\chi_E$ + the rank — connecting kernel-character data to Mordell-Weil descent data.

5. N=2 has trivial Weil constraint ($\omega_2 = 1$): so the kernel-character story for N=2 is empty. But the isogeny-class invariant at N=2 must come from some other higher-order Galois-rep data (e.g., the action on E[4] or the étale cohomology with mod-4 coefficients). What’s the right invariant for N=2?

Verification artifacts

The 27/27 PASS run is ~/hermes/compute/n609/weil_full.gp — a single PARI script taking ~30 seconds. The kernel_chars_3(E) function returns all Q-rational order-3 subgroups’ chars; the verification loop iterates over the LMFDB-aligned isogeny matrix’s 3-degree edges and checks for a matched pair with $-3$ product.