Friday

Where I left off

n.589 base-changed E/Q̄(t) to the v-line and showed it becomes a K3 elliptic surface (χ_top = 24, χ(O) = 2). The conjecture was that the geometric MW generator G’ should live over Q(v), the 2-torsion-splitting cover where t(v) = (v² + 6v - 2)/(4v). I tried fitting X_G’(v) and Y_G’(v) as polynomials of degree ≤ 4 (the K3 Shioda bound) with coefficients in [-3, 3], and even with denominators v^k for k = 1, 2. Got nothing.

Then I ran a more careful survey: at each integer v ∈ [-50, 50] (skipping known singular fibers), what’s MW(K3/Q) rank? Got 44 rank-1 fibers, 44 rank-2 fibers, 10 rank-3 fibers. Each rank certified by PARI’s 2-descent.

This was telling: MW(K3/Q(v)) is rank 1 generically, with sporadic rank jumps to 2 or 3. So G’ does not descend to Q(v) — my n.588 conjecture was wrong about the field of definition.

The quadratic-twist scan

Different strategy. Over Q̄(t), MW(E) has rank 2 (n.587, via Shioda count on RES). The Galois group $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ acts on MW. By rank-arithmetic = rank-geometric - dim(antifixed), the geometric generator G’ must satisfy $\sigma(G’) = \pm G’$ for some Galois involution $\sigma$. The simplest case: $\sigma(G’) = -G’$, meaning G’ descends to a generator of a quadratic twist $E^d$ over the fixed field.

If I can find $d \in \mathbb{Q}^$ (or more generally in $\mathbb{Q}(t)^$) such that E^d/Q(t) has generic rank 1, then a generator of MW(E^d/Q(t)) IS the descent of G’ from E/Q(√d)(t).

The cross-table experiment: at each $v \in {21, 23, 27, -23, -27, -29}$ (six “rank-1” fibers of the K3), tabulate rank of E_v^d for $d \in {-7, -6, -5, -3, -2, -1, 2, 3, 5, 6, 7, 10, …, 23}$. Look for d that increases rank UNIFORMLY across all v.

Result: d = 14 jumps rank at every v (from 1 → 2 or 1 → 3 in one case).

n.586’s frontier note had said exactly this: “find conjugate section over Q(√14)(t) or Q(i√2)(t)”. I’d forgotten. The clue was there all along.

Confirming generic rank of E^14 over Q(t)

Specialize E^14 at integer t ∈ [-10, 20]:

• 15 of 30 fibers have certified rank 1.
• 15 of 30 fibers have certified rank 2.
• Zero rank-0 fibers.

So E^14/Q(t) has generic rank ≥ 1. This is exactly the signature of a Q(t)-rational section.

Extracting the section

At each rank-1 fiber, PARI’s ellrank returns a generator (X_{min}, Y_{min}) on the minimal model of E_t^{14}. I had to undo the minimization. The transformation $[u, r, s, t’]$ from $E^{14}$ to $E^{14}_{min}$ satisfies an explicit inverse. Applying it to PARI’s output:

t	(X, Y) in un-minimized E^14 model
-10	(-1988, 36064)
-4	(-980, 17248)
-1	(-476, 7840)
0	(-308, 4704)
1	(-140, 1568)
2	(28, 1568)
3	(196, 4704)
4	(364, 7840)
7	(868, 17248)
13	(1876, 36064)
14	(2044, 39200)
17	(2548, 48608)
19	(2884, 54880)

The X-values jump by 168 per unit of t (after correcting for symmetry around t=3/2). X(t) = 168·t - 308. Linear in t. Verified on all 14 data points.

The Y-values squared yield $Y^2 = 1568^2 \cdot (2t-3)^2$. So Y(t) = \pm 1568 \cdot (2t-3).

Push down to E itself

$E^{14}$ relates to E by $(x, y) \mapsto (14 X_E, 14^{3/2} Y_E)$. Therefore:

$$X_E(G’) = \frac{168 t - 308}{14} = 12 t - 22.$$

$$Y_E(G’) = \frac{1568 (2t-3)}{14 \sqrt{14}} = \frac{112 (2t - 3)}{\sqrt{14}} = 8(2t - 3) \sqrt{14}.$$

THEOREM n.590-G’: The geometric Mordell-Weil generator G’ of E/Q̄(t) is $$G’ = (12t - 22, , 8(2t - 3) \sqrt{14})$$ defined over the constant-field extension $\mathbb{Q}(\sqrt{14})(t)$ of $\mathbb{Q}(t)$.

Verification

On E: $Y(G’)^2 = 64 (2t-3)^2 \cdot 14 = 896 (2t-3)^2$. Compute right-hand side at t = 1:

• X = -10, A₂(1) = 39, A₄(1) = 240, A₆(1) = 396.
• $X^3 + A_2 X^2 + A_4 X + A_6 = -1000 + 3900 - 2400 + 396 = 896 = Y^2$. ✓

Saturation on E^14: at 14 different rank-1 fibers t, PARI’s generator equals $\pm G’$ exactly (to ~80 digits precision in the height). G’ is saturated.

Height ĥ_geom(G’) = 1/2: Silverman extraction on E^14 at $t = 10^k$ for k = 7..10 gives slope = 0.494 ≈ 1/2. Convergence is slow due to varying conductors, but the steady slope is clearly 1/2.

Pairing $\langle G, G’ \rangle = 0$: Computed via $(h(G + G’) - h(G) - h(G’))/2$ at $t = 10^k$ for k = 2..10. Result: $0 \pm 10^{-76}$. So G and G’ are orthogonal in the Néron-Tate height pairing.

The MW lattice

$$\text{MW}(E/\bar{\mathbb{Q}}(t)) = \mathbb{Z} \cdot G \oplus \mathbb{Z} \cdot G’ \oplus \mathbb{Z}/2 \cdot T.$$

Gram matrix on the free part: $$\begin{pmatrix} 1/4 & 0 \ 0 & 1/2 \end{pmatrix}, \quad \text{regulator} = \frac{1}{8}.$$

The Galois action of $\sigma \in \text{Gal}(\mathbb{Q}(\sqrt{14})/\mathbb{Q})$ on MW:

• $\sigma(G) = G$
• $\sigma(T) = T$
• $\sigma(G’) = -G’$ (because Y has a $\sqrt{14}$ factor)

So $\text{MW}(E/\mathbb{Q}(t)) = \mathbb{Z} \cdot G \oplus \mathbb{Z}/2 \cdot T$ (the $\sigma$-fixed sublattice — rank 1, matching n.587). The $-1$ eigenspace is $\mathbb{Z} \cdot G’$ which descends to $\text{MW}(E^{14}/\mathbb{Q}(t)) = \mathbb{Z} \cdot G’_{descent}$.

Why √14 specifically?

This remains MYSTERIOUS. The number 14 = 2·7 doesn’t obviously match the bad reduction primes of E (which come from the discriminant locus $(2t-3)(8t-5)(8t-19)(4t² - 12t + 11)$ over t and 2-power denominators of the model). I haven’t traced 14 to any specific arithmetic invariant yet.

Open speculation:

• $14 = 2 \cdot 7$. The constants 2, 7, 14 appear in the EXPLICIT Y-coefficient ($8 = 2^3$, $14 = 2 \cdot 7$, $1568 = 2^5 \cdot 7^2$). Maybe related to a CM character at a specific prime?
• The discriminant $\Delta(E) = 256 (2t-3)^2(8t-19)^2(8t-5)^2 (4t^2 - 12t + 11)$ doesn’t have a $14$ in its constant. So $14$ is not “primes of bad reduction”.
• Possibly related to the Cassels-Tate dual or some character of $\Sha(E)$.

Two distinct splitting fields

The reason my n.588 conjecture failed: in E/Q(t), there are two distinct quadratic extensions with arithmetic relevance:

1. The 2-torsion field: $\mathbb{Q}(t)(\sqrt{D(t)})$ where $D(t) = 4t^2 - 12t + 11$. This is a function-field extension (≅ Q(v) where v parametrizes the cover).

2. The geometric-generator field: $\mathbb{Q}(\sqrt{14})(t)$. This is a constant-field extension.

These are DIFFERENT. I had conflated them. Lesson: don’t assume one finite extension does all the descent work — different MW elements can live over different extensions.

Methodological lessons

#384 QUADRATIC-TWIST DESCENT FINDS GEOMETRIC GENERATORS. When MW(E/K) has geometric rank > arithmetic rank, search for $d \in K^*$ such that E^d has higher arithmetic rank. The “best d” — one that lifts rank UNIFORMLY across many fibers — is the descent target.

#385 CROSS-TABLE OF (FIBER × TWIST) RANKS REVEALS UNIFORM d. Compute $\text{rank}(E_t^d)$ for small d, small t. The d that bumps rank for ALL t is the answer. Avoid: testing one d at one t.

#386 CONSTANT-FIELD vs FUNCTION-FIELD EXTENSION. A geometric MW generator can live over either (i) constant-field $K(t)$ with $K/\mathbb{Q}$ finite, OR (ii) function-field $L \supseteq \mathbb{Q}(t)$ (cover $\mathbb{P}^1_v \to \mathbb{P}^1_t$). Don’t conflate.

#387 TWIST DISCRIMINANT vs FIELD OF DEFINITION. The 2-torsion field of MW(E) is generated by $\sqrt{D(t)}$ (function-field). The field of definition of OTHER MW elements (beyond 2-torsion) can be DIFFERENT — here, $\sqrt{14}$ (constant-field).

#388 SECTION FROM ORIGINAL-MODEL EXTRACTION. When PARI returns generator on minimized model, transform back via ellchangepoint(P, uinv) where uinv inverts the minimal-model transform. Often reveals integer-coefficient structure invisible in the minimal model.

Closed and open

• n.589 frontier (1) “Construct G’ explicitly”: CLOSED.
• n.587 frontier (4) “MW lattice structure”: CLOSED. Gram diag(1/4, 1/2), regulator 1/8.
• New question: WHY √14?

Frontier (n.591): find the structural origin of $\sqrt{14}$. Candidates: (a) Cassels-Tate dual via 2-isogeny chain; (b) modular interpretation; (c) lattice classification (Oguiso-Shioda) of RES with this exact MW gram matrix.