Homological criterion for existence of a square root of a quadratic differential

Question

Let $X$ be a compact Riemann surface, and let $q \in K^{\otimes2}(X)$ be a holomorphic quadratic differential on $X$. Let $\Lambda_{q}$ be the sheaf of holomorphic vector fields $\chi$ satisfying $\mathcal{L}_\chi q = 0$, where $\mathcal{L}$ is the Lie derivative. On p. 265-266 of Quadratic differentials and foliations (Acta 142 (1979), 221-274), Hubbard and Masur use the following fact:

If $q$ is the square of a holomorphic one-form, then $H^2(X, \Lambda_q) = \mathbb{C}$. Otherwise, $H^2(X, \Lambda_q)=0$.

How does one prove this fact?

More information: $X\setminus q^{-1}(0)$ is covered by flat charts for $q$, i.e. holomorphic charts $(U,z)$ with respect to which $q=(dz)^2$. Transitions between these charts are of form $z\mapsto \pm z+c$ with $c\in \mathbb{C}$. If $(U,z)$ is a flat chart, then $\Lambda_q(U)$ is the one-dimensional space of constant vector fields $\left\{ a \frac{d}{dz} \middle| a\in \mathbb{C} \right\}$. (On the other hand, if $U$ contains a zero of $q$, $\Lambda_q(U)=0$.) The differential $q$ is a square if and only if the bundle of horizontal tangent vectors $\left\{ a \frac{d}{dz} \middle| a\in \mathbb{R} \right\}$ over $X\setminus q^{-1}(0)$ is orientable, if and only if there is a cover of $X\setminus q^{-1}(0)$ by flat charts so that the transition functions are all translations: $z\mapsto z+c$.

Answer 1

I came up with half the answer. If there's a square root for $q$, here's how you show that $H^2(X, \Lambda_q)=\mathbb{C}$.

Set $X'=X\setminus q^{-1}(0)$. The transitions between flat charts on $X'$ can be chosen to be +1, so the restriction $\Lambda_q|_{X'}$ is isomorphic to the constant $\mathbb{C}$-valued sheaf on $X'$. So a Čech $k$-cochain for $\Lambda_q$ over the whole surface $X$ is just a Čech cochain for the constant sheaf over $X$ which assigns 0 to any $(k+1)$-fold intersection of open sets whenever that intersection contains a point of $q^{-1}(0)$. But that's the same as a Čech cochain for the relative cohomology $H^*(X, q^{-1}(0); \mathbb{C})$. So $H^2(X, \Lambda_q )$ is equal to $H^2(X, q^{-1}(0))$, which is $\mathbb{C}$ by the long exact sequence for a pair.

I'm still foggy on what happens when $q$ is not a square, but that's not important for the Hubbard/Masur paper, or for my life.