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Im reading a book where they study the X function of the Riemann surface $y^3=(x-x_1)*(x-x_2)$ ($x, y \in \mathbb{S}^2$) and $dX$.

They say that $dX$ has a zero of degree 2 in $x_1$ and $x_2$ and "clearly no other zero". However they don't give any explanation.

My reasoning: a theorem says that X is meromorphic of degree 3 (locally in $x_1$, we have $y'^3=x'$ and a meromorphic function is surjective and takes every value n times). So $dX$ has a zero of degree 2 on $x_1$ and $x_2$.

Now if I look at $\frac{\partial P}{\partial x}$ (with $P=y^3-(x-x_1)*(x-x_2)$) I see that it is not null except where $x = \frac{x_1+x_2}{2} = \alpha$ so we can write $x=f(y)$ where $x \neq \alpha$. This gives $y^3=(f(y)-x_1)*(f(y)-x_2)$ and $3y^2 = f'(y)*Q(y)$. Hence $dX = 0$ if $f'(y) = 0$ which implies $y=0$ and $x = x_1$ or $x = x_2$, where $x \neq \alpha$.

But what about the case $x = \alpha$? Is there something clear that I am missing?

Thomas
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1 Answers1

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At the points with $x = \alpha$, $\frac{\partial P}{\partial x} = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $\mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.

Without abuse of notation, if we write $\pi_x, \pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $\pi_x : U \to V$ is a chart, then $(\pi_x)_* (d\pi_x|_U) = d(\operatorname{id}_V)$.

Bart Michels
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