one 1-form on the Riemann surface of an algebraic function

Question

Given the many-valued algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$, we can get a Riemann surface $S$ that is topologically equivalent to a torus. I am wondering whether the $1$-form $ \displaystyle\frac{dz}{w}$ is a holomorphic form on the torus $S$.

On the surface we know that

$$ \frac{dz}{w}=\frac{dz}{\sqrt{(z-1)(z-2)(z-3)}},$$

I think that $z=1,2,3$ are poles of the above 1-form, so it is a meromorphic 1-form on the $z-$plane. Is it a holomorphic 1-form on the torus $S$ determined by the algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$?

I just know a little about Riemann surface theory. I will appreciate any suggestions and comments. Thanks.

Answer 1

What Riemann surface did you obtain? $C:w^2=(z-1)(z-2)(z-3)$ or $C-(1,0),(2,0),(3,0)$ or the projective closure $E\simeq C\cup (\infty,\infty)$?

Anyway $dz/w$ is holomorphic on them all: near $(1,0)$ we have $z=f(w)w^2$ with $f$ analytic at $0$. At $(\infty,\infty)$ we have $z=(z/w)^{-2} g(z/w),w=(z/w)^{-3} h(z/w)$ with $g,h$ analytic at $0$.

Eventually $dz/w$ is a nowhere vanishing holomorphic 1-form and $p\to \int_{p_0}^p dz/w$ is an isomorphism $E \to \Bbb{C/(uZ+vZ)}$ where $uZ+vZ$ is the lattice obtained by integrating over the closed-loops.

Answer 2

Given the algebraic function $w^2=(z-1)(z-2)(z-3)$, differentiate each side, we get $$2wdw=[(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)]dz .$$ Therefore we have $$\frac{dz}{w}=\frac{2dw}{(z-2)(z-3)+(z-1)(z-3)+(z-1)(z-2)}.$$ It is easy to see that the RHS is holomorphic locally at (1,0), (2,0),(3,0) respectively.