Computing the genus of $y^2=x(x^2-1)$ using 1-forms

Question

I'm trying to compute the genus of the projective curve $C:=V(Y^2Z-X(X^2-Z^2))\subset\Bbb{P}^2_\Bbb{C}$ explicitly using differential forms.

I know beforehand that this is an elliptic curve, so the expected answer is $g=1$.

So I must find a globally defined differential $1$-form $\omega\in\Omega_C$.

In $U_Z:=\{Z\neq 0\}$, we define $x:=\frac{X}{Z}$ and $y:=\frac{Y}{Z}$, so that $y^2=x(x^2-1)$ $(*)$.

I've read in more than one source that the desired form is $\omega:=\frac{dx}{y}$.

Using $(*)$, we have $\frac{dx}{y}=\frac{2dy}{3x^2-1}$ and since $3x^2-1\neq 0$ at the points $(0:0:1),(1:0:1),(-1:0:1)\in C$, we see that $\omega$ has no poles in $U_Z$.

We still have to check that $\omega$ has no pole at infinity $(0:1:0)$.

So we restrict to $U_Y$ and define $u:=\frac{X}{Y}, v:=\frac{Z}{Y}$, so that $v-u^3+uv^2=0$ $(**)$. This way: $$\frac{dx}{y}=v\cdot d\left(\frac{u}{v}\right)=du-\frac{dv}{v}$$

I still can't see how to use $(**)$ to rewrite $\frac{dv}{v}$ so that the pole will vanish.

Am I missing something?

Answer 1

You should have $$\omega = \frac{dx}y = du - \frac{u\,dv}v.$$ Using $v-u^3+uv^2=0$, we have $(1+2uv)dv + (v^2-3u^2)du = 0$. Then \begin{align*} \omega &= du-\frac uv\left(\frac{3u^2-v^2}{1+2uv}\right)du \\ &= \left(\frac{v(1+2uv)-u(3u^2-v^2)}{v(1+2uv)}\right)du \\ &= \left(\frac{2uv^2-2u^3}{v(1+2uv)}\right)du \\ &= \left(\frac{-2v}{v(1+2uv)}\right) du = -2\frac{du}{1+2uv}, \end{align*} which is clearly holomorphic at $(u,v)=(0,0)$.

Answer 2

I think you've made a small mistake in your calculation. I get \begin{align*} \omega := \frac{dx}{y}=v\cdot d\left(\frac{u}{v}\right)=v \frac{v du - u dv}{v^2} = du - \frac{u}{v} dv \, . \end{align*} Differentiating the affine equation $v = u^3 - uv^2$ you found in a neighborhood of $(0:1:0)$, we have \begin{align*} dv = 3u^2du - \left(v^2 du + 2uv dv \right) &\implies (1 + 2uv)dv = (3u^2 - v^2)du\\ &\implies dv = \frac{3u^2 - v^2}{1 + 2uv} du \, . \end{align*}

I think it's a little easier to start with the other representation for $\omega$ in $x,y$ coordinates: \begin{align*} \omega = \frac{2dy}{3x^2 - 1} = \frac{2 d(1/v)}{3(u/v)^2 - 1} = 2 \frac{-\frac{1}{v^2} dv}{3\frac{u^2}{v^2} - 1} = -2 \frac{dv}{3u^2 - v^2} \, . \end{align*} Rewriting $dv$ in terms of $du$ yields \begin{align*} \omega = -2 \frac{1}{3u^2 - v^2} \frac{3u^2 - v^2}{1 + 2uv} du = \frac{-2}{1 + 2uv} du \, . \end{align*} Since the denominator doesn't vanish at $(u,v) = (0,0)$, we see that $\omega$ doesn't have a pole at $\infty$.