Let $X$ be a Riemann surface with $2-$differential $w = f(z) dz^2$ on open set $U$.
If $f(z_0)\neq 0$, $z_0 \in U$, there exists $z_0 \in V \subseteq U$ such that the complex square root $f(z)^{1/2}$ is well-defined on $V$.
Define on $v \in V$, $$q(v) = \int_{z_0}^vf(z)^{1/2}dz.$$ Then $q$ defines a new coordinate. (Hint : consider $dq^2$ on $V$).
So I am not certain what $dq^2$ is. It should be $\omega = 2q \ dq$ ?. Then I think I have to show that $(V, \omega)$ is a chart on $X$ ?.
It is confusing about transition function and charts. So $(V, q)$ is my new set of charts ? So I need to show that transition function is holomorphic ? Then what is have to do with $\omega$ and $dq^2$ ?
Any help please ?
No, they mean the quadratic differential $(dq)^2 = dq\otimes dq$, just as they've been writing $f(z)\,dz^2$. Since $dq = f(z)^{1/2}\,dz$, what is $dq\otimes dq$? (Hint: The answer is $w=dq^2$.) $q$ gives a holomorphic local coordinate because $dq$ is (locally) a holomorphic $1$-form.