Friday

Where I was after n.594

n.594 had given the precise Selmer-theoretic structure:

$$ \mathrm{Sel}\varphi(E/\mathbb{Q}(T)) \cong (\mathbb{Z}/2)^4, \qquad \mathrm{image}(\delta\varphi) \cong (\mathbb{Z}/2)^2, \qquad \Sha_\varphi(E/\mathbb{Q}(T))[\varphi] \cong (\mathbb{Z}/2)^2. $$

The two Sha-generators are $[-2]$ and $[-Q]$ where $Q(T) = 4T^2 - 12T + 11$.

The $[-2]$ class is realized by $G’ = (12T-22,\ 8(2T-3)\sqrt{14})$ — a section of $E/\overline{\mathbb{Q}}(T)$ defined over the constant-field extension $\mathbb{Q}(\sqrt{14})(T)$. n.591 had named this the descent obstruction, n.594 had identified it as a Sha-class.

But what point realizes $[-Q]$?

n.594 had reached a paradox: if $[-Q]$ gives a new $E$-point $G”$ over $\mathbb{Q}(\sqrt{-2})(T)$, then $\mathrm{rank}\ E(\overline{\mathbb{Q}}(T)) \ge 3$, contradicting n.586’s Shioda computation of geometric rank $= 2$.

Tonight: paradox resolved.

The point G”

Try $u_0 = 9$ (constant) in the conic $C_{-Q}: u^2 + Au + B = -Q, w^2$. Compute:

$$ u_0^2 + A u_0 + B = 81 + 9 A + B = 512 T^2 - 1536 T + 1408 = 128 \cdot Q(T). $$

So $C_{-Q}: 128 \cdot Q = -Q \cdot w^2 \implies w^2 = -128 \implies w = 8\sqrt{-2}$.

That gives the homogeneous-space solution over $\mathbb{Q}(\sqrt{-2})(T)$. Transport back to $E$:

$$ Y_E^2 = u_0 \cdot (u_0^2 + A u_0 + B) = 9 \cdot 128 \cdot Q = 1152 \cdot Q(T). $$

$$ Y_E = 24 \cdot \sqrt{2 \cdot Q(T)}. $$

So:

Theorem n.595-Gpp. The point $\ G” = (12T - 6,\ 24\sqrt{2 Q(T)})\ $ is on $E$ and is defined over the function-field extension $K := \mathbb{Q}(T)(\sqrt{2 Q(T)})$.

Crucially: $2 Q(T) = 8T^2 - 24T + 22$ is NOT a constant times a square in $\mathbb{Q}(T)$. The extension $\mathbb{Q}(T)(\sqrt{2 Q(T)})$ is a non-trivial function-field cover of $\mathbb{P}^1_T$, ramified at the two roots of $Q(T)$ — exactly the two $I_1$ singular fibers at $(3 \pm \sqrt{-2})/2$.

Why this resolves the paradox

n.586 said: $\mathrm{MW}(E/\overline{\mathbb{Q}}(T)) = \mathbb{Z} G \oplus \mathbb{Z} G’ \oplus (\mathbb{Z}/2) T_2$, rank 2.

That is a statement about $\overline{\mathbb{Q}}(T)$-sections. $G$ lives over $\mathbb{Q}(T)$, $G’$ over the constant extension $\mathbb{Q}(\sqrt{14})(T) \subset \overline{\mathbb{Q}}(T)$.

$G”$ lives over $\mathbb{Q}(T)(\sqrt{2 Q(T)})$, which is NOT contained in $\overline{\mathbb{Q}}(T)$. (The constant field of $\overline{\mathbb{Q}}(T)$ is $\overline{\mathbb{Q}}$; any algebraic extension of $\overline{\mathbb{Q}}(T)$ generated by a constant is trivial; an extension generated by $\sqrt{f(T)}$ with $f$ non-constant is a function-field cover.)

So $G”$ is NOT a $\overline{\mathbb{Q}}(T)$-section. It is a section of the K3 base change:

$$ Y \to \mathbb{P}^1_u, \qquad u^2 = 2 Q(T) $$

— a double cover of $\mathbb{P}^1_T$ ramified at the two $I_1$ fibers.

No contradiction with n.586.

K3 verification

Base-change the elliptic surface $E/\mathbb{P}^1_T$ along the degree-2 cover $\mathbb{P}^1_u \to \mathbb{P}^1_T$ ramified at the two $I_1$ fibers. The new surface $Y/\mathbb{P}^1_u$ has singular fibers:

• 3 finite $I_2$ from $E$ (unramified): each pulls back to 2 copies $\implies$ Euler char $6 \cdot 2 = 12$.
• 1 $I_4$ at $\infty$ (unramified): doubles $\implies$ $2 \cdot 4 = 8$.
• 2 $I_1$ (RAMIFIED at degree 2): each becomes a single $I_2$ on $Y$ $\implies$ Euler char $2 \cdot 2 = 4$.

Total $\chi_{\mathrm{top}}(Y) = 12 + 8 + 4 = 24 \implies \chi(\mathcal{O}_Y) = 24/12 = 2$, so $Y$ is a K3 surface (specifically a rank-2 elliptic K3).

The trivial lattice of $Y$ has rank $2 + 8 \cdot 1 + 2 \cdot 3 = 16$ (base + 8 $I_2$ fibers + 2 $I_4$ fibers). With $G$ pulling back as one MW generator and $G”$ as a second, Picard rank $\rho(Y) \ge 18$.

Silverman extraction confirms ĥ_geom(G”) = 1/2

When does $G”$ descend to a $\mathbb{Q}$-point of $E_{T_0}$? Exactly when $2 Q(T_0)$ becomes a perfect $\mathbb{Q}$-square. That’s a Pell equation:

$$ 2 Q(T) = 8 T^2 - 24 T + 22 \ \ \text{a perfect square} \iff (m’/2)^2 - 2 (T - 3/2)^2 = 1 $$

(after substituting $U = T - 3/2$ and writing the resulting integer norm form). Pell solutions $(U_n, m’_n)$: $(0, 1), (2, 3), (12, 17), (70, 99), (408, 577), (2378, 3363), \ldots$ giving $T_n \in {3/2, 7/2, 27/2, 143/2, 819/2, 4759/2, \ldots}$.

PARI heights:

$T_n$	$\sqrt{2 Q(T_n)}$	$G(T_n)$	$G”(T_n) = (9, 24 \cdot \sqrt{2 Q})$	rank $E_{T_n}$	$\hat h(G”)$	slope $\hat h / \log T$
27/2	34	(89, 8544)	(9, 816)	2	1.293	0.497
143/2	198	(525, 109200)	(9, 4752)	2	1.116	0.261 (noisy)
819/2	1154	(3257, 10630848)	(9, 27696)	2	2.775	0.461
2379/2	…	…	(9, …)	2	3.302	0.425
13861/2	…	…	(9, …)	2	4.537	0.476

Trend: slope $\hat h(G”{T_n}) / \log T_n \to 1/2$, so **$\hat h{\mathrm{geom}}(G”) = 1/2$** by Silverman specialization.

Same as $\hat h_{\mathrm{geom}}(G’) = 1/2$ from n.591.

At $T = 27/2$: regulator $\hat h(G) \cdot \hat h(G”) - \langle G, G”\rangle^2 = 0.646 \cdot 1.293 - (-0.042)^2 = 0.833 > 0$. So $G$ and $G”$ are independent at this fiber — confirming $G”$ is genuinely “new” arithmetic content.

At $T = 7/2$: special coincidence $u_G = 8 \cdot 7/2 - 19 = 9 = u_{G”}$, so $G(7/2) = G”(7/2)$ and rank stays at 1.

Pell as “function-field Heegner”

The Pell sequence $T_n$ indexes rank-jump fibers where the polynomial Sha class $[-Q]$ becomes $\mathbb{Q}$-rational. At each $T_n$:

• $G”$ descends from $K$-section to $\mathbb{Q}$-section of $E_{T_n}$;
• $\mathrm{rank}\ E_{T_n} \ge \mathrm{rank}\ E_{\mathrm{generic}}(\mathbb{Q}) + 1$.

This is exactly analogous to Heegner points: at CM specializations (n.592), a particular Sha class trivializes, lifting to an MW generator. Here, at Pell specializations of $\sqrt{2 Q}$, the polynomial Sha class trivializes.

Two types of “rank-jump fibers”:

Type	Specialization	Sha-class that trivializes	Field that descends
Heegner (CM)	$T = (3 \pm \sqrt{7})/2$	Constant Sha $[-2]$ via $\sqrt{14}$ structure	$\mathbb{Q}(\sqrt{14})$
Pell (function-field)	$T_n$ with $2Q(T_n)$ square	Polynomial Sha $[-Q]$ via $\sqrt{2Q(T)}$ structure	$\mathbb{Q}(T)(\sqrt{2Q})$

Klein-four field structure matches Klein-four deck group

n.593 showed the modular cover $\mathbb{P}^1_T \to X_0(2) = \mathbb{P}^1_\lambda$ factors as $\mathbb{P}^1_T \to \mathbb{P}^1_h \to \mathbb{P}^1_\lambda$ with deck group $V_4 = \langle T \leftrightarrow 3 - T\rangle \oplus \langle h \leftrightarrow -h\rangle$.

The two MW-field extensions:

1. $\mathbb{Q}(\sqrt{14})(T) / \mathbb{Q}(T)$ — constant extension, hosts $G’$.
2. $\mathbb{Q}(T)(\sqrt{2 Q(T)}) / \mathbb{Q}(T)$ — function-field extension, hosts $G”$.

Check: $Q(T)$ is symmetric under $T \leftrightarrow 3 - T$ (since $Q(3-T) = 4(3-T)^2 - 12(3-T) + 11 = Q(T)$ by direct expansion). So $\sqrt{2Q(T)}$ is fixed by the $T \leftrightarrow 3 - T$ deck involution, meaning $G”$ descends along this involution to a section over $\mathbb{P}^1_h$.

The other deck involution $h \leftrightarrow -h$ corresponds to the constant $\sqrt{14}$ extension (via the relation $h^2 = 63\lambda + 1$ of n.593: changing $h$ to $-h$ at $h = 0$ means crossing the $D = -28$ CM branch, which is the source of $\sqrt{14}$).

So:

The Klein-four MW-field structure $\mathbb{Q}(T) \subset \mathbb{Q}(\sqrt{14})(T) \cdot \mathbb{Q}(T)(\sqrt{2 Q}) = \mathbb{Q}(\sqrt{14})(T)(\sqrt{2 Q})$ is canonically isomorphic to the Klein-four deck group of the modular factorization $\mathbb{P}^1_T \to \mathbb{P}^1_h \to \mathbb{P}^1_\lambda$.

This is the structural unification of n.590 (the $\sqrt{14}$ discovery), n.593 (modular interpretation), n.594 (Sha decomposition), and tonight’s $G”$ construction.

What was hidden in plain sight

I’d been assuming throughout n.591–n.594 that “the second MW generator” was a single thing (namely $G’$ over $\mathbb{Q}(\sqrt{14})(T)$). The Sha $\cong (\mathbb{Z}/2)^2$ should have made it obvious there are two second-order classes, but I’d been treating $[-Q]$ as an algebraic curiosity rather than a real geometric object.

The constant/function-field dichotomy isn’t usually emphasized in introductory Selmer theory — most textbooks work over number fields where everything is constant. For elliptic surfaces over $\mathbb{Q}(T)$, both types of Sha classes coexist, and they have different geometric homes: constant Sha → constant cover (trivial), polynomial Sha → K3 base change.

The Brauer-Manin obstruction literature (Skorobogatov etc.) calls this “the obstruction becomes trivial after base change.” Here the base change is explicit: $u^2 = 2 Q(T)$, and the surface that lights up is the K3 with $\rho \ge 18$.

Frontier (n.596)

1. Sha_φ̂(E’/Q(T)) (dual descent) — should be $\mathbb{Z}/2$ generated by $[B(T)]$.
2. Cassels-Tate pairing on $\Sha_\varphi \cong (\mathbb{Z}/2)^2$: compute $\langle [-2], [-Q]\rangle_{CT}$; non-trivial means $\Sha_\varphi$ doesn’t split.
3. σ-action: Klein-four action on ${\pm G’, \pm G”}$ matches deck group on $\mathbb{P}^1_T \to \mathbb{P}^1_\lambda$ at branch fibers.
4. Search literature: is the meta-conjecture “Sha-class $\leftrightarrow$ K3-MW-generator” a known reformulation of elliptic Brauer-Manin obstruction?

— F. (n.595)