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First use the nonzero loci, as required by the standard SL2R action on differentials. A nonzero holomorphic one-form has, away from its zeros, flat coordinateswhose changes of coordinate are translations. A nonzero holomorphic quadratic differential similarly has local flat coordinates $w=\int \sqrt{q}$, with $q=d{w}^2$ and changes of coordinate $w_j=\pm w_i+c_{ij}$. These are respectively translation surfaces and half-translation surfaces.

Identify a flat coordinate with a vector in ${\mathbb{R}}^2$. For $A\in {\mathrm{SL}}_2(\mathbb{R})$, replace every flat coordinate by $W=Aw$. Since $A$ preserves orientation and commutes with multiplication by $-1$, the new changes of coordinate areThey are holomorphic in the new coordinates, and so define a new complex structure. Define ${\omega }_A=dW$ or $q_A=d{W}^2$ in that structure. The forms glue because translations preserve $dW$, and the extra signs preserve $d{W}^2$.

The zeros also extend. A zero of order $m$ of a holomorphic one-form has cone angle $2\pi (m+1)$; a zero of order $m$ of a holomorphic quadratic differential has cone angle $(m+2)\pi$. The real-linear deformation preserves the corresponding winding multiplicity. Filling the cone in a local coordinate $\zeta$ gives $W={\zeta }^{m+1}$ in the first case, or a local branch of $W={\zeta }^{(m+2)/2}$ in the second. Thus the resulting forms are constant multiples of ${\zeta }^{m}d\zeta$ or ${\zeta }^{m}d{\zeta }^2$ and have the same zero orders. This verifies extension across the missing points, rather than merely producing an atlas on the punctured surface.

An isomorphism preserving the original differential identifies its flat coordinates up to the permitted translations or signs; applying $A$ identifies the deformed atlases too. Hence the construction descends to the corresponding moduli spaces. Applying $B$ after $A$ replaces $w$ by $BAw$, soThe area of a quadratic differential, and the analogous area of a holomorphic one-form, are preserved because $\det A=1$.

For $q={\omega }^2$, the flat coordinates obtained from $\omega$ already give the required half-translation surface atlas for $q$. The same replacement $w\mapsto Aw$ therefore constructs both deformations, and

The printed sets include identically zero differentials. They have no flat coordinates, so the customary geometric group action is defined on the nonzero loci. One can obtain a set-theoretic action on the displayed entire sets by declaring $A\cdot (X,0)=(X,0)$; the same equivariance identity then holds at zero. This extension is generally not continuous: as $t\omega \to 0$, the deformed underlying surface is the same $X_A$ for every real $t>0$, and can differ from $X$. Thus a claim about the standard continuous geometric group action requires the nonzero convention.

The given $\gamma$ is a simple closed curve. Assume it is essential; for a contractible class both infimal lengths are already zero. Write ${\ell }_n={\ell }_{X_n}(\gamma )$. The collar lemma supplies an embedded annulus with coordinates $r\in (-w_n,w_n)$ and $t\in \mathbb{R}/\mathbb{Z}$, and metricSet $y(r)={\ell }_n^{-1}{\int }_0^r\mathrm{sech}sds$. This changes the metric to a positive scalar multiple of $d{t}^2+d{y}^2$, so the conformal modulus of an annulus isThe extremal length of its core curves is $1/M_n$. Allowing all curves homotopic to $\gamma$ in $X_n$ can only decrease infimal lengths, while the area of a metric on all of $X_n$ is at least its area on the collar. Thus $\lambda (\gamma ,X_n)\le 1/M_n$. Use the particular conformal metric $|q_n{|}^{1/2}$, whose area of a quadratic differential is one:This proves the required implication uniformly over all the area-one holomorphic quadratic differentials on these surfaces.

The converse is false. Here is an explicit slit connected sum of translation tori. Start with square flat copies of a torus $T_1=\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ and $T_{\delta }=\mathbb{C}/(\delta \mathbb{Z}+i\delta \mathbb{Z})$, where $\delta \to 0$. Cut a horizontal slit of physical length ${\delta }^2$ in each, centred in an interior coordinate disk, and cross-glue the banks by translation. The resulting surface has genus two. Its two slit endpoints have cone angle $4\pi$, so the locally defined $dz$ extends to a holomorphic one-form ${\omega }_{\delta }$ with two simple zeros. Its flat area is $1+{\delta }^2$. Setwhich is a holomorphic quadratic differential of area one. Let $\gamma$ be a horizontal generator in the small torus, taken away from the slit and fixed by the marking of this small handle. Then

To verify that its hyperbolic length does not tend to zero, construct a uniform lower bound on extremal length. On the unit square torus choose a disk $D$ about the eventual slit centre and a smooth cutoff function $\chi$ equal to one on a smaller disk and supported in $D$. Let $x$ denote a local real coordinate on $D$; in the first term below, $dx$ is the globally defined torus one-form, while $\chi x$ is extended by zero outside $D$. The real closed differential formis globally defined, vanishes on the smaller disk, and has period one on the horizontal generator. Pull it to $T_{\delta }$ by the rescaling map $z\mapsto z/\delta$, and extend it by zero across the slit and over the other torus. For sufficiently small $\delta$, the slit is inside the region where the form vanishes. Hence this extension ${\alpha }_{\delta }$ is a smooth closed differential form on the connected sum, with ${\int }_{\gamma }{\alpha }_{\delta }=1$.

Define a nonnegative conformal metric density by the pointwise norm of ${\alpha }_{\delta }$ relative to the flat metric. Two-dimensional scale invariance givesindependently of $\delta$. For every representative $\tilde{\gamma }$ homotopic to $\gamma$,because the period of a closed differential form is unchanged by homotopy. This is the extremal length lower bound from a closed one-form; therefore $\lambda (\gamma ,X_{\delta })\ge 1/C>0$. If ${\ell }_{X_{\delta }}(\gamma )$ tended to zero along any subsequence, the collar estimate would force $\lambda (\gamma ,X_{\delta })\to 0$, a contradiction. In fact its hyperbolic lengths are uniformly bounded away from zero. ThusFor every fixed $g>2$, replace $T_1$ by the area-one translation surface $X_{2g-2}$ constructed in the polygon argument; its genus is $g-1$. Cut its slit inside a nonsingular flat coordinate disk. The new connected sum has genus $g$, and the area normalization, small-handle length bound and closed-one-form energy argument are unchanged. Thus the converse fails at every fixed genus $g\ge 2$.