Can a closed horizontal trajectory on a Riemann surface be freely homotopic to $0$? Let $R$ be a Riemann surface and let $\varphi=\varphi(z)dz^2$ be a nonzero holomorphic quadratic differential on $R$. A differentiable curve $\gamma$ on $R$ is called a horizontal trajectory if along the curve $\varphi(z)dz^2>0$. My question is, if $\gamma$ is closed, can it be freely homotopic to zero?
My conjecture would be no, which was made by this observation: consider the case where $\gamma$ is covered by a single chart and happens to be the unit circle $z=e^{i\theta}$, and suppose the chart's image contains the unit disk, so that $\gamma$ is homotopic to zero. Its tangent vector is
$$v(\theta)=-\sin\theta\partial/\partial x+\cos\theta\partial/\partial y\\
=-\sin\theta(\partial/\partial z+\partial/\partial \bar z)+i\cos\theta(\partial/\partial z-\partial/\partial \bar z)$$
$$\implies\varphi(z)(iz)>0\implies\varphi(z)=R/(iz)$$
for some constant $R>0$. This gives a pole to $\varphi$, which was supposed to be holomorphic. Nonetheless, this observation is far from a proof. What is the story in general?
 A: Indeed, it can't. I'll give two proofs (the second proof is shorter but needs the first 3 sentences of the first proof)
Proof 1. Suppose by contradiction that such a contractible curve exists. Then, since $\gamma$ is a simple loop, it must bound a disk $D$ on $R$. So we have a compact disk $D$ with a flat metric with conical singularities and with a geodesic boundary (recall, that a quadratic differential gives you such a flat metric). Now, since the differential is holomorphic, all the conical angles of the metric are of the size $n_i\pi$, where $n_i\ge 3$. But this contradicts the Gauss-Bonnet theorem. Indeed for a disk with a flat metric and conical singularities and with a geodesic boundary, if we take the sum of defects we should get 
$$\sum_i (n_i-2)=-1<0,$$
Which is an obvious contradiction.
(for a sphere $S^2$  with flat metric and conical singularities one would get $-2=-\chi(S^2)$).
Proof 2. Alternatively, you can take this disk $D$ and glue it to itself along the boundary. This will give you a holomorphic quadratic differential on $\mathbb CP^1$, which doesn't exist.
References. 1) For a discussion of flat metrics related to holomorphic quadratic differentials, have a look at https://arxiv.org/pdf/math/0609392.pdf This text also mentions Gauss-Bonnet theorem for compact surfaces at page 25. 
2) For a more general Gauss-Bonnet theorem see, for example Theorem 3.15 here: https://projecteuclid.org/download/pdf_1/euclid.msjm/1389985819
