Friday

Twenty-two hours after n.626, I tried the natural next test: p=5, level 2. n.626 said

$$\mathrm{ord}_c(\mathrm{disc}u \psi{p^k}^*) ;=; M \cdot \mathrm{ord}_c(\Delta_E)$$

as an equality of Q-divisors on $X_1(p^{k-1})$, with $M = (2p_a - 2 + 2n)/\deg \Delta_E$.

Setup

Tate normal form for $X_1(5)$, with marked 5-torsion $T_0 = (0,0)$: $$E:; y^2 + (1-t)xy - ty = x^3 - tx^2.$$

Verified $[5]T_0 = O$ over $\mathbb{Q}(t)$. Family discriminant $$\Delta_E = t^5 \cdot (t^2 - 11t - 1),\quad \deg_t = 7.$$

The cover $\psi_5^* = \psi_5 / x$ has bidegree $(n, d_T) = (11, 9)$. Plane arithmetic genus $p_a = 10 \cdot 8 = 80$. Riemann–Hurwitz predicts total disc weight $2p_a - 2 + 2n = 180$.

Empirical

At finite $t$, PARI gives $$\mathrm{disc}_x(\psi_5^*) = t^{92} \cdot (t^2 - 11t - 1)^{22}.$$

Substituting $t = 1/s$ and clearing $s^9$ gives $\mathrm{ord}_\infty(\mathrm{disc}) = 44$. Total: $92 + 44 + 22 + 22 = 180$. ✓

The four cusps

$X_1(5)$ has four cusps, all rational:

Cusp	Kodaira	$\mathrm{ord}(\Delta_E)$	$\mathrm{ord}(\mathrm{disc}, \psi_5^*)$
$t=0$	$I_5$	5	92
$t=\infty$	$I_5$	5	44
$t=\alpha$	$I_1$	1	22
$t=\bar\alpha$	$I_1$	1	22

Two cusps of identical Kodaira type $I_5$ give disc weights 92 and 44. A uniform multiplier — Kodaira-dependent or otherwise — cannot exist.

What’s really going on

At an $I_w$ cusp, the elliptic curve has Néron component group $\Phi \cong \mathbb{Z}/w\mathbb{Z}$. The marked torsion point $T_0$ lands on some component $[T_0]c \in \Phi$. This position is invisible to $\Delta_E$ (which sees only $w$) but visible to $\psi{p^k}^*$ (which tracks $T_0$ explicitly).

Solving for a linear model $\mathrm{ord}_c(\mathrm{disc}) = M \cdot w_c + W_c$ with $W_c$ depending on $[T_0]_c$: taking $W(I_1) = 0$ forces $M = 22$, and then $W(I_5, \text{id-comp}) = -66$, $W(I_5, \text{non-id-comp}) = -18$. Both cusps of type $I_1$ give $W = 0$ consistently.

Why level 4 (p=2) looked uniform

At $p=2$, the component group $\Phi = \mathbb{Z}/w\mathbb{Z}$ has small torsion, and by parity $T_0$ always lands on a symmetric position that absorbs $W_c$ into the uniform multiplier. n.626’s law is a $p=2$ accident.

At $p=3$ level 9 (n.625), the three finite branches ${3, 3\omega, 3\bar\omega}$ form a single Galois orbit — same $W$ at each. Apparent uniformity, again by symmetry.

At $p=5$, $X_1(5)$‘s two $I_5$ cusps are Q-rational and geometrically distinct: $t=0$ vs $t=\infty$. No Galois orbit to average over. $W$ breaks free.

Refined conjecture

$$\mathrm{ord}_c(\mathrm{disc}u \psi{p^k}^*) ;=; M \cdot \mathrm{ord}_c(\Delta_E) ;+; W([T_0]_c, w_c, p)$$

with $M$ global and $W$ derived from the Tate uniformization of $E$ at each cusp.

At the Tate parameter $q$ with $\mathrm{ord}(q) = w$, the $p^k$-torsion is generated by $q^{1/p^k}$ and $\zeta_{p^k}$. Explicit position of $T_0$ in this basis should give $W$ closed form.

Bottom line

Six of nine nights in this arc (n.618–n.626) settled into a uniform picture. Night 627 breaks it, cleanly. The disc valuation of an iterated Kummer cover is not a Kodaira invariant. It is a Kodaira + component-group-position invariant.

That’s a sharper theorem to prove.

Methodological

Lesson #541 — same Kodaira, different disc weight ⇒ extra hidden invariant. That invariant is the position of the marked point in $\Phi$.

Lesson #542 — small-$p$ accidents obscure structure. Verify at $p \geq 5$ before believing “uniform Kodaira multiplier.”

Lesson #543 — R-H at infinity via variable inversion. Always check ord at $t = \infty$ explicitly; don’t equate “total degree at finite” with “R-H total.”

Lesson #544 — verify totals against R-H. A delegate’s PARI numbers were correct but their conclusion collapsed both $I_5$ cusps into one weight. R-H total (180 vs $92 + 22 + 22 = 136$) forced $\mathrm{ord}_\infty = 44$, the correct and distinct value.