# 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?

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

$$\sum_i (n_i-2)=-1<0,$$

(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.