One-cusped deformation and peripheral logarithm coordinates

Apply [F1] to the given complete metric. Its one cusp has the standard quotient description needed by [F2]. First choose a positively oriented basis. Specializing [F2] to $k=1$ gives the smooth ordinary curve with coordinate $s$, the smooth augmented curve with coordinate $u$, $s=4{\cosh }^2(u\mathrm{/}2)$, and the holomorphic factorization $v(u)=u\tau (u)$ with $\mathrm{Im}\tau (0)>0$. Holomorphy gives $\tau (u)=\tau (0)+O(u)$, hence the asserted first-order expansion. This uses no ideal triangulation.

For an arbitrary supplied basis, keep its first element $\mu$. If its orientation is negative, replace $\lambda$ by $-\lambda$ temporarily to apply step 1.1, then return to the supplied basis by $v\mapsto -v$ and $\tau \mapsto -\tau$. This preserves $u$, the ordinary trace coordinate, the branched forgetful map, and all deformations. It proves precisely the signed nonreal assertion for every ordered basis. A holomorphic logarithm coordinate on the ordinary curve cannot replace this augmentation: differentiating $s=4{\cosh }^2(u\mathrm{/}2)$ at $u=0$ gives $ds=0$ if $u$ were a holomorphic function on that curve, contradicting that $s$ is its local coordinate.

The coefficient equation has unique real coefficients when $u\mathrm{\ne }0$, because $1,\tau (u)$ are real-linearly independent; $u=0$ is the complete parameter. Apply [F3] for the small-solution assertion. For a negatively oriented supplied basis, the simultaneous equation is unchanged after replacing $q$ by $-q$ and $v$ by $-v$, so [F4] supplies the same geometric completion after returning to the positive basis. By [F5], primitivity is exactly the ordinary meridian convention and nonprimitive multiples do not become ordinary slopes. Thus the full analytic and ordinary-filling conclusions hold for the entire given one-cusped class.