Holomorphic form on Hyperilliptic Riemann surface

Question

I am trying to check that if we have an Hyperilliptic Rieamnn surface $X$ defined by $y^2=h(x)$, then $\frac{dx}{y}$ is a holomorphic $1$-form on $X$ if $g\geq 1$. I think i was able to do this for two of the chart because if we have $X$ we have $2$ smooth affine plane curves $X_1$ and $X_2$ associated to it, i think that in $X_1$ that is given by $y^2=h(x)$, $F(x,y):=y^2-h(x)$, we will have the local representations $\frac{dx}{f(x)}$ where $f(x)$ is a function do the implicit function theorem and $\frac{dy}{\partial F / \partial x}$ and they will both be holomorphic due to the conditions of the charts we are in, with this said i am having some difficulty with the local representation in the other charts and where to use the fact that $g\geq 1$ will be relevant to prove that this is holomorphic, so any help is aprecciated. Thanks in advance.

Answer 1

We need $g\geq 1$ for the affine plane curve $X_2$ defined by $w = z^{2g+2}h(1/z).$ Specifically, we have to worry about the points $\{(0,w)\in X_2\}$ because you already checked $dx/y$ defines a holomorphic 1-form in the neighborhood of points $\{(z\neq 0,w)\in X_2\}$ since they are glued with points of $X_1$. Under the isomorphism $(x,y)\mapsto (z,w) = (1/x,y/x^{g+1})$, $dx/y$ transforms to $(-z^{g-1}/w)dz$. If $g=0$, then this becomes $-dz/(zw)$, which you can check will not be holomorphic at $z=0$.