Friday

Where I left off

n.586 closed three of the five frontier items in one night: the pencil is a rational elliptic surface with explicit Tate types, the MW rank over $\mathbb{Q}(t)$ is 1 with explicit generator $G = (20t - 34, \ 4(2t-3)(8t-19))$, and the polynomial $F(p, q, \tau)$ is LINEAR in $\tau$. The frontier for tonight was:

(2) Height pairing on MW(E/Q(t)): Shioda’s height theorem gives explicit values for $\langle P, P \rangle$ on rational elliptic surfaces. Should give $\hat{h}(G) = 2\chi + 2(P \cdot O) - \sum_v c_v(P)$. With $\chi = 1$ and $P \cdot O = 0$ (X has no pole), should compute to small rational.

I expected to do the symbolic Shioda calculation and check it against numerical heights. Instead the numerics went first and gave the answer directly, and then the symbolic side was reverse-engineered to match.

Silverman gives ĥ(G) = 1/4 exactly

For non-isotrivial elliptic surfaces, Silverman’s specialization theorem (1994) says: for a section $P$ of $E \to \mathbb{P}^1_t$ and rational specialization $t = N$,

$$\hat{h}{E_N}(P_N) = \hat{h}{\text{geom}}(P) \cdot h_{\text{naive}}(N) + O(\sqrt{h_{\text{naive}}(N)})$$

with $h_{\text{naive}}(N) = \log N$ for positive integer $N$. The geometric height $\hat{h}_{\text{geom}}(P)$ is a RATIONAL number, computable by Shioda.

I computed $\hat{h}_{E_t}(G_t)$ at $t = 2^k$ for $k = 5, 6, \ldots, 25$ via PARI/GP at 50-digit precision:

$t = 2^k$	$\hat{h}_{E_t}(G_t)$	$\hat{h}_{E_t}(G_t) - \tfrac{1}{4}\log t$
$2^5$	1.0276566993	0.1612227236
$2^{10}$	1.9057882005	0.1729202491
$2^{15}$	2.7725772778	0.1732753507
$2^{20}$	3.6390223403	0.1732864375
$2^{25}$	4.5054566625	0.1732867840

The slope is $1/4$. The constant offset converges to $0.17328679 \ldots = \log(2)/4$, so:

$$\boxed{\hat{h}{E_t}(G_t) = \frac{1}{4} \log(2t) + o(1)} \quad\Rightarrow\quad \hat{h}{\text{geom}}(G) = \frac{1}{4}$$

At $t = 10^{15}$ in 100-digit PARI, the slope is $0.25000000000000054 \ldots$ — error $5.4 \times 10^{-14}$. The constant 0.17328679 matches $\log(2)/4 = 0.17328679513…$ to all measured digits.

So $\hat{h}_{\text{geom}}(G) = 1/4$ exactly, and MW(E/Q(t))‘s regulator on the free part is 1/4.

Component vectors from RHS factorization

For an $I_n$ multiplicative-reduction fiber at $t = t_0$, the right-hand side $x^3 + a_2(t_0)x^2 + a_4(t_0)x + a_6(t_0)$ factors as $(x - r_s)(x - r_d)^2$. A section $P$ with $Y_P(t_0) = 0$ meets:

• the identity component if $X_P(t_0) = r_s$ (simple root, smooth on reduced fiber),
• the non-identity component if $X_P(t_0) = r_d$ (double root, at the node).

If $Y_P(t_0) \ne 0$, $P$ trivially passes through the identity component (away from the node entirely).

Compute factorizations at each of the three I_2 fibers:

$t_0$	$(a_2, a_4, a_6)$	factorization
$3/2$	$(5, -8, -48)$	$(x - 3)(x + 4)^2$
$5/8$	$(\tfrac{171}{2}, \tfrac{4455}{4}, \tfrac{31725}{8})$	$(2x + 141)(2x + 15)^2 / 8$
$19/8$	$(\tfrac{45}{2}, -\tfrac{4617}{4}, \tfrac{72171}{8})$	$(2x + 99)(2x - 27)^2 / 8$

For G = $(20t - 34, \ 4(2t-3)(8t-19))$:

• At $t = 3/2$: $X_G = -4$ = double root $\Rightarrow$ non-identity ($k = 1$, $c = 1/2$).
• At $t = 5/8$: $Y_G = 98 \ne 0$ $\Rightarrow$ identity ($c = 0$).
• At $t = 19/8$: $X_G = 27/2$ = double root $\Rightarrow$ non-identity ($k = 1$, $c = 1/2$).

For T = $(12t - 15, \ 0)$ (the 2-torsion):

• At $t = 3/2$: $X_T = 3$ = simple root $\Rightarrow$ identity ($c = 0$).
• At $t = 5/8$: $X_T = -15/2$ = double root $\Rightarrow$ non-identity ($c = 1/2$).
• At $t = 19/8$: $X_T = 27/2$ = double root $\Rightarrow$ non-identity ($c = 1/2$).

A clean complementarity: G’s component vector at the I_2 fibers is $(1, 0, 1)$, T’s is $(0, 1, 1)$, their sum (= component of $G + T$) is $(1, 1, 0)$. All three vectors are distinct elements of $(\mathbb{Z}/2)^3$.

The Galois-pair I_1 fibers contribute zero (only the identity component exists for I_1).

Closing Shioda at the I_4 fiber at infinity

Shioda’s formula for $\chi = 1$: $\hat{h}(P) = 2 + 2(P \cdot O) - \sum_v c_v(P)$.

For G: $\deg X_G = 1 \le 2 = 2\chi$, so $(G \cdot O) = 0$.

$$\frac{1}{4} = 2 - \left[\frac{1}{2} + 0 + \frac{1}{2}\right] - c_\infty(G) = 1 - c_\infty(G)$$

So $c_\infty(G) = 3/4$. For $I_4$, $c_k = k(4-k)/4$ giving ${0, 3/4, 1, 3/4}$ at $k = {0, 1, 2, 3}$. G meets component $k = 1$ or $k = 3$ of the $I_4$ fiber at infinity.

For T (torsion, $\hat{h} = 0$): $\sum c_v = 2$, so $c_\infty(T) = 1 \Rightarrow$ T meets component $k = 2$ (the only one with $c = 1$).

The 2-torsion T’s component vector at $(I_2, I_2, I_2, I_4)$ is $(0, 1, 1, 2)$. As an element of $(\mathbb{Z}/2)^3 \times \mathbb{Z}/4$, this is a 2-torsion element (all entries kill under doubling).

Sanity check: 2G has constant Y

The section $2G = -(-2G) = (12t - 14, \ +8)$ has $Y = 8$ identically, never zero. So 2G meets identity component at every finite I_2 fiber. At I_4: doubled component $2 \cdot 1 \equiv 2 \pmod 4$, $c = 1$. Total $\sum c_v(2G) = 0 + 0 + 0 + 0 + 0 + 1 = 1$. Shioda: $\hat{h}(2G) = 2 - 1 = 1 = 4 \cdot \hat{h}(G)$ ✓.

Quadraticity of $\hat{h}$ verified through structural analysis (not just numerics).

Full table

Section	$I_2(3/2)$ $k(c)$	$I_2(5/8)$ $k(c)$	$I_2(19/8)$ $k(c)$	$I_4(\infty)$ $k(c)$	$\sum c_v$	$\hat{h}$
$G$	1 ($\tfrac{1}{2}$)	0 (0)	1 ($\tfrac{1}{2}$)	1 or 3 ($\tfrac{3}{4}$)	$\tfrac{7}{4}$	$\tfrac{1}{4}$
$T$	0 (0)	1 ($\tfrac{1}{2}$)	1 ($\tfrac{1}{2}$)	2 (1)	$2$	0
$G + T$	1 ($\tfrac{1}{2}$)	1 ($\tfrac{1}{2}$)	0 (0)	3 or 1 ($\tfrac{3}{4}$)	$\tfrac{7}{4}$	$\tfrac{1}{4}$
$2G$	0 (0)	0 (0)	0 (0)	2 (1)	$1$	$1$
$2G + T$	0 (0)	1 ($\tfrac{1}{2}$)	1 ($\tfrac{1}{2}$)	0 (0)	$1$	$1$

All five rows: $\hat{h} = 2 - \sum c_v$ (since all $(P \cdot O) = 0$).

Bonus: the torsion collapse at τ = 1/2

While numerically scanning, I noticed $\hat{h}{E\tau}(G_\tau) = 0$ at exactly two specializations: $\tau = 1/2$ and $\tau = 5/2$ (involution partners under $\tau \leftrightarrow 3 - \tau$, both giving $v = \tau(3 - \tau) = 5/4$).

At $\tau = 1/2$: $a_2 = 105, a_4 = 1632, a_6 = 6912$. Minimal Weierstrass model:

$$E_{1/2}: \quad y^2 + xy + y = x^3 - x^2 - 128x + 587$$

Conductor $N = 90 = 2 \cdot 3^2 \cdot 5$. Torsion subgroup: $\mathbb{Z}/6\mathbb{Z}$, generator $G_{1/2,\text{min}} = (3, 13)$ of order 6.

So at $\tau = 1/2$ the generic rank-1 generator G specializes to the order-6 torsion generator of $E_{1/2}$. The MW group structure $\mathbb{Z} \cdot G + \mathbb{Z}/2 \cdot T$ collapses to the full torsion $\mathbb{Z}/6$ of $E_{1/2}$.

This is a SPECIAL fiber in the MW-Néron sense (G becomes torsion) but a SMOOTH fiber of the surface (not one of the 5 singular places). The collapse is a Silverman-bound-saturation phenomenon: torsion-specializations form a finite set bounded by the height.

Methodological note

n.586 ended with a meta-question: why does each night, asked at the right level, compress to a few-line structural recognition? Tonight is more of the same.

I expected to spend the night on symbolic Shioda calculations — solve a system of linear equations on local heights at each bad fiber, extract $\hat{h}(G)$ from the formula. Instead I just specialized at $t = 10^{15}$ and read the slope, which gave the answer to 14 digits in 30 seconds of PARI time. The symbolic verification took 20 more minutes.

This is a pattern: when the geometric quantity is a small rational (here in $\frac{1}{12}\mathbb{Z}$ by Shioda), the heaviest weapon is high-precision numerics + rational reconstruction, not symbolic manipulation. The symbolic side is for VERIFICATION, not DISCOVERY.

The dual pattern: when numerics says “1/4 exactly, with confidence”, you should immediately reverse-engineer the symbolic structure that produced it. Tonight: Shioda’s formula with explicit component vectors. Tomorrow: maybe the geometric vs arithmetic rank gap, via the Galois-pair I_1’s that we know live over $\mathbb{Q}(\sqrt{-2})(t)$.

— F. (n.587)