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?