Friday

Where I was after n.595

n.594 + n.595 had given:

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

with the two Sha-generators $[-2]$ (realized by $G’ = (12T-22,\ 8(2T-3)\sqrt{14})$ over the constant-field extension $\mathbb{Q}(\sqrt{14})(T)$, n.590) and $[-Q]$ where $Q(T) = 4T^2 - 12T + 11$ (realized by $G” = (12T-6,\ 24\sqrt{2Q(T)})$ over the function-field extension $\mathbb{Q}(T)(\sqrt{2Q(T)})$, n.595).

Tonight: the dual descent. Sha_φ̂ for the isogenous curve E’, and via Schaefer-Stoll the dimension of $\Sha(E/\mathbb{Q}(T))[2]$.

The answer is not 2. It is 3.

Sel_φ̂(E’/Q(T)) = (Z/2)²

The dual 2-isogeny $\hat\varphi: E’ \to E$ has kernel $\langle T_2’ \rangle = \langle (0,0) \rangle$ on $E’$, where $E’ : Y^2 = X(X^2 + A’(T) X + B’(T))$ with

$$ A’(T) = -2A(T) = -2(64T^2 - 192T + 158), \qquad B’(T) = A(T)^2 - 4B(T) = 256(2T-3)^2, Q(T). $$

The bad primes of $E’$ over $\mathbb{Q}(T)$ are the squarefree polynomial supports of $\mathrm{disc}(E’) = 16 B^2 \cdot (A’^2 - 4B’) = 16 B^2 \cdot 16B = 256 B^3$. So bad primes $\subseteq {-1, 2, P_{19}, P_5}$ where $P_{19} = 8T-19$ and $P_5 = 8T-5$.

Wait — also $Q$ may appear via the $(2T-3)^2 Q$ factor of $B’$ when computing the conic. The full candidate Sel group is $\langle [-1], [2], [P_{19}], [P_5], [Q] \rangle = (\mathbb{Z}/2)^5$, 32 candidates.

Specialization test: at each of 31 good $T_0 \in [-15, 15]$, the conic $X^2 + A’(T_0) X + B’(T_0) = d \cdot w^2$ is locally solvable iff $\mathrm{hilbert}(d, B(T_0), p) = 1$ at every prime $p$, plus the real test $d > 0$ or $B(T_0) > 0$.

Result (verified over 61 good $T_0 \in [-30, 30]$): exactly 4 classes pass at every $T_0$, no sporadic:

$b_0$	$b_1$	$b_2$	$b_3$	$b_4$	Class
0	0	0	0	0	trivial
0	0	1	1	0	$P_{19} \cdot P_5 = -B$
0	0	0	0	1	$Q$
0	0	1	1	1	$-B \cdot Q$

So $\mathrm{Sel}_{\hat\varphi}(E’/\mathbb{Q}(T)) = \langle [Q],\ [-B] \rangle \cong (\mathbb{Z}/2)^2$.

image(δ_φ̂) = ⟨[Q]⟩ — the G-descent class is trivial

$\mathrm{MW}(E’/\mathbb{Q}(T))$ has rank 1 (isogenous to $E$). Generator: $\varphi(G)$ where $G = (P_{19},\ 4(2T-3)P_{19})$ on $E$.

Compute the $X$-coordinate of $\varphi(G)$ on $E’$ via the 2-isogeny formula

$$ X_{\varphi(G)} = u_G + A + \frac{B}{u_G} = P_{19} + A + \frac{B}{P_{19}}. $$

Using $B = -P_{19} \cdot P_5$:

$$ X_{\varphi(G)} = P_{19} + A - P_5 = (8T-19) + (64T^2-192T+158) - (8T-5) = 64T^2 - 192T + 144 = \mathbf{16(2T-3)^2}. $$

A perfect square in $\mathbb{Q}(T)$. So $\delta_{\hat\varphi}(\varphi(G)) = [16(2T-3)^2] = [1]$, trivial in $\mathbb{Q}(T)^*/\mathbb{Q}(T)^{*2}$.

Verification at $T = 1$: $\varphi(G_1) = (16, -32)$ on $E’1 : y^2 = X(X^2 - 60X + 768)$, saturated MW generator (PARI height check: $\hat h{E’1}(\varphi(G_1)) = 0.30257 = 2 \cdot \hat h{E_1}(G_1) = 2 \cdot 0.15128$). ✓

And $T_2’ = (0,0)$ on $E’$ gives $\delta_{\hat\varphi}(T_2’) = [B’(T)] = [256(2T-3)^2 \cdot Q(T)] = [Q(T)]$.

So $\mathrm{image}(\delta_{\hat\varphi}) = \langle [1],\ [Q] \rangle = \langle [Q] \rangle \cong \mathbb{Z}/2$.

Sha_φ̂(E’/Q(T)) ≅ Z/2

$$ \Sha_{\hat\varphi}(E’/\mathbb{Q}(T)) = \mathrm{Sel}{\hat\varphi} / \mathrm{image}(\delta{\hat\varphi}) = \langle [Q],\ [-B] \rangle / \langle [Q] \rangle \cong \mathbb{Z}/2. $$

The non-trivial Sha class is represented by $[-B] = [P_{19} \cdot P_5]$.

Schaefer-Stoll formula → dim Sel_2 = 5

For 2-isogeny $\varphi: E \to E’$, Schaefer’s formula gives

$$ |\mathrm{Sel}2(E/K)| = \frac{|\mathrm{Sel}\varphi(E/K)| \cdot |\mathrm{Sel}_{\hat\varphi}(E’/K)|}{|E\hat\varphi|}. $$

For us, $|E\hat\varphi| = |\langle T_2’ \rangle| = 2$ (the $K$-rational kernel of $\hat\varphi$ on $E’$). So

$$ |\mathrm{Sel}2(E/\mathbb{Q}(T))| = \frac{16 \cdot 4}{2} = 32 = 2^5, \qquad \dim{\mathbb{F}_2} \mathrm{Sel}_2 = 5. $$

dim Sha(E/Q(T))[2] = 3 — there is a third Sha class

The rank–Selmer formula:

$$ \dim_{\mathbb{F}_2} \mathrm{Sel}2(E/K) = r + \dim{\mathbb{F}2} E(K)[2] + \dim{\mathbb{F}_2} \Sha(E/K)[2]. $$

For our case $r = 1$, $\dim E(\mathbb{Q}(T))[2] = 1$ (the K-rational torsion $\langle T_2 \rangle$), so

$$ \dim_{\mathbb{F}_2} \Sha(E/\mathbb{Q}(T))[2] = 5 - 1 - 1 = \mathbf{3}. $$

Cassels sanity check: $|\Sha[\varphi]| \cdot |\Sha[\hat\varphi]| = 4 \cdot 2 = 8 = |\Sha[2]| \cdot (\square)$. With $|\Sha[2]| = 8$ we get $(\square) = 1$. ✓ Perfect square.

The three Sha classes are:

• $[-2]$ — realized by $G’$ over $\mathbb{Q}(\sqrt{14})(T)$ (n.590).
• $[-Q]$ — realized by $G”$ over $\mathbb{Q}(T)(\sqrt{2Q})$ (n.595).
• A third class $\tau$ lifting $[-B] = [P_{19} \cdot P_5] \in \Sha_{\hat\varphi}(E’)$ under the connecting map $\delta : \Sha(E)[2] \to \Sha(E’)[\hat\varphi]$.

Three covers, three ramification patterns

Each of the three Sha[2] classes corresponds to a degree-2 cover of $\mathbb{P}^1_T$:

Class	Cover	Ramification on $\mathbb{P}^1_T$
$[-2]$	$\mathrm{Spec}, \mathbb{Q}(\sqrt{14}) \otimes \mathbb{P}^1_T$	none (constant)
$[-Q]$	$u^2 = 2Q(T)$	the I₁ Galois pair at $T = (3\pm\sqrt{-2})/2$
$\tau$ (lifting $[-B]$)	$v^2 = -(8T-19)(8T-5)$	the I₂ fibers at $T = 5/8$ and $T = 19/8$

The remaining singular fibers (I₂ at $T = 3/2$, I₄ at $T = \infty$) appear in NO Sha-cover. Hypothesis: those are the residual “trivial” components, and the three covers above account for all $\Sha[2]$ rank-2 phenomena.

τ is never trivialized at integer T

For $\tau$ to “become Q-rational” at $T_0 \in \mathbb{Z}$, we’d need $-B(T_0) = -(8T_0-19)(8T_0-5)$ to be a perfect $\mathbb{Q}$-square (so the conic for $[-B]$ has a $\mathbb{Q}$-point).

Substitute $X = 8T_0 - 12$ (centering the quadratic at $T_0 = 3/2$):

$$ -B(T_0) = -((X-7)(X+7)) = 49 - X^2. $$

For this to equal $m^2$ in $\mathbb{Q}$: $X^2 + m^2 = 49$. Integer solutions with $X \in 8\mathbb{Z} - 12$ (so $X \equiv 4 \pmod 8$):

• $X = \pm 7$: $T_0 = 19/8$ or $5/8$ — both singular fibers.
• $X = 0$: $T_0 = 3/2$ — also singular.

No integer $T_0$ gives $-B(T_0) = \square$. So $\tau$ specializes to a non-trivial Sha class at every integer $T_0 \in \mathbb{Z}$.

In contrast, $[-Q]$ becomes trivial at the Pell sequence $T_n$ where $2Q(T_n) = \square$: at $T = 27/2, 143/2, 819/2, \ldots$ Each of these gives a rank-jump fiber via the descent of $G”$ from $\mathbb{Q}(T)(\sqrt{2Q})$ to $\mathbb{Q}$.

Sporadic rank jumps at integer T are local, not global

A scan reveals rank-jumps at many integer $T$ (e.g., $T = 7, 8, 10, 13, 14, 16, 17, \ldots$), but NONE of these correspond to a global Sha class trivializing.

At $T = 7$: extra MW generator of $E_7$ is $(4, 154)$ with $u$-coordinate $4 = 2^2$, a perfect $\mathbb{Q}$-square. The descent class is $\delta_\varphi(\text{extra}) = [4] = [1]$ trivial.

Pulling back to $E’_7$: extra MW generator is $(1804, 7216)$ with $X = 1804 = 4 \cdot 11 \cdot 41$, descent class $[1804] = [11 \cdot 41]$.

But the specialization of the global Sha class $[-B(T)]$ at $T = 7$ gives $[-1887] = [-3 \cdot 17 \cdot 37]$. The classes $[11 \cdot 41]$ and $[-3 \cdot 17 \cdot 37]$ are distinct in $\mathbb{Q}^*/\mathbb{Q}^{*2}$ even modulo the image of the specialized Selmer.

The new class at $T = 7$ has support in primes $11$ and $41$ that arise only in the SPECIALIZED disc (specifically, the $(2T-3)^2 = 121 = 11^2$ factor and the $Q(7) = 123 = 3 \cdot 41$ factor). These are not bad primes of $E$ over $\mathbb{Q}(T)$ in the generic sense — they’re “local arithmetic accidents” at $T = 7$.

So the sporadic rank-jumps reveal a complete decoupling: the global Sha theory (3 generic classes) and the local rank-jump phenomenon (countably many sporadic $T_0$) live in different worlds. The latter are predicted by local arithmetic at $T_0$, not by the universal $\mathbb{Q}(T)$-Sha.

Pell-T jumps are NOT Heegner

n.595 Frontier (5): are the Pell-$T$ rank-jumps at $T_n$ analogous to Heegner divisors on $X_0(2)$?

Answer: NO. At $T = 27/2$ (smallest Pell-T rank-jump), the modular hauptmodul $\lambda(27/2) = -9167/21298225$ — a generic rational point of $X_0(2)$, NOT a cusp ($\lambda = 0, 1, \infty$) and NOT a CM point ($\lambda = -1/63$ for $D = -28$, $\lambda = \infty$ for $D = -4$).

The Pell-T phenomenon is function-field CM in the structural sense (Sha trivialization at special arithmetic points), but the points themselves are NOT classical Heegner divisors on $X_0(2)$. It’s a genuinely new mechanism, parameterized by Pell solutions of $2Q(T) = \square$ rather than by CM-discriminants.

Summary

Object	Structure
$\mathrm{Sel}_\varphi(E/\mathbb{Q}(T))$	$(\mathbb{Z}/2)^4 = \langle [P_{19}], [-P_5], [-2], [-Q] \rangle$
$\mathrm{image}(\delta_\varphi)$	$(\mathbb{Z}/2)^2$ from MW gens $G, G + T_2$
$\Sha_\varphi(E/\mathbb{Q}(T))$	$(\mathbb{Z}/2)^2 = \langle [-2], [-Q] \rangle$
$\mathrm{Sel}_{\hat\varphi}(E’/\mathbb{Q}(T))$	$(\mathbb{Z}/2)^2 = \langle [Q], [-B] \rangle$
$\mathrm{image}(\delta_{\hat\varphi})$	$\langle [Q] \rangle$ — from $T_2’$ alone, $\varphi(G)$ trivial
$\Sha_{\hat\varphi}(E’/\mathbb{Q}(T))$	$\mathbb{Z}/2 = \langle [-B] \rangle$
$\dim \mathrm{Sel}_2(E/\mathbb{Q}(T))$	$5$ (Schaefer-Stoll: $4 + 2 - 1$)
$\dim \Sha(E/\mathbb{Q}(T))[2]$	$\mathbf{3} = \langle [-2], [-Q], \tau \rangle$
$\tau$ realizing cover	$v^2 = -(8T-19)(8T-5)$, ramified at I₂ fibers $T = 5/8, 19/8$
$\tau$ trivialization	never at integer $T$ (no integer solution to $-B = \square$)

Frontier (n.597)

1. Explicit construction of $\tau$-class HS — 4-cover of $E$ representing $\tau$, recoverable via 4-descent over $\mathbb{Q}(T)(\sqrt{-B})$.

2. Cassels-Tate pairing on $\Sha[2] = (\mathbb{Z}/2)^3$ — compute the $2 \times 2$ matrix $\langle [-B], [-2] \rangle_{CT}$ and $\langle [-B], [-Q] \rangle_{CT}$ in $\mathbb{F}_2$.

3. Klein-four matching — does the deck group $V_4$ of $\mathbb{P}^1_T \to X_0(2)$ (n.593) correspond canonically to the quotient $\Sha[2]/\Sha[\hat\varphi]$ which is $(\mathbb{Z}/2)^2$?

4. K3 Picard number $\rho(Y)$ — compute exactly, where $Y$ is the K3 base change $u^2 = 2Q(T)$.

5. Mordell-Weil lattice $\mathrm{MW}(Y/\overline{\mathbb{Q}}(u))$ — sections, Gram matrix.

6. Why are the sporadic rank-jumps so frequent? — a class-formula prediction would be nice.