I am reading a Teichmueller theory book and trying to understand elliptic curves as examples of Riemann surfaces. Consider the elliptic curve \[X = \{[z : w : y] \in \mathbb C \mathrm P^2 \mid w^2y = z (z - y) (z - \lambda y)\} \] where $\lambda \in \mathbb C$ is a complex number not equal to $0$ or $1$. I am trying to understand how the formula $\omega = \mathrm d z/ w$ produces a holomorphic differential on the Riemann surface $X$, especially at the point $\infty$. My partial work is below.
There is a related question here - Holomorphic differential of elliptic curve over $\mathbb C$ but I don't understand the answer there. It would be great if someone could clarify that answer too.
What I understand so far: $X$ is a Riemann surface. Using the implicit function theorem, we can conclude the following: On the open set $\\{[z : w : y] \in X \mid y = 1\\}$, the function $W = w/y$ is a coordinate except for at most four points where $\frac{\mathrm d}{\mathrm d z} z (z - 1) (z - \lambda) = 0$. Similarly, the function $Z = z/y$ is a coordinate except for at most three points where $Z (Z - 1) (Z - \lambda) = 0$. Near the point $\infty = [0 : 1 : 0]$, the function $u = z/w$ is a coordinate. This covers all the points of $X$, so $X$ is indeed a Riemann surface.
So the formula $\mathrm d z/w = \mathrm d Z/W$ produces a holomorphic differential except at $\infty$ or when $Z \in \\{0, 1, \lambda\\}$. To see that it is holomorphic at $0, 1, \lambda$, we differentiate \[W^2 = Z (Z - 1) (Z - \lambda)\] to get \[2W \mathrm d W = (3Z^2 - 2 (1 + \lambda) Z + \lambda) \mathrm d Z\] \[\Rightarrow \frac{\mathrm d Z}{W} = \frac{2 \mathrm d W}{3Z^2 - 2 (1 + \lambda) Z + \lambda}\] Since $W$ is a coordinate near $Z = 0, 1, \lambda$ and the polynomial $3Z^2 - 2 (1 + \lambda) Z + \lambda$ does not vanish at these three points (since $Z (Z - 1) (Z - \lambda)$ has no repeated roots), the differential is holomorphic there.
But what to do about the point $\infty$? Here one has to compute the differential $\frac{\mathrm d Z}{W}$ in terms of the local coordinate $u$ and show that there is a limit as $u \to 0$. Now we have \[u = \frac{Z}{W} \Rightarrow Z = Wu \] \[\mathrm d Z = W \mathrm d u + u \mathrm d W\] \[\frac{\mathrm dZ}{W} = \mathrm d u + u \frac{\mathrm dW}{W} \] Now $\mathrm d u$ is a perfectly fine holomorphic differential, and so the first term is taken care of. Now need to take care of the second term on the right. This is where I am stuck. How do I express $W$ in terms of $u$ and simplify to get a holomorphic differential $u \frac{\mathrm dW}{W}$?
The infinity point is $[0:1:0]$, i.e. $w\ne 0$. I will write $a$ instead of $\lambda$ against the history for an easy typing.
We start with the $1$-form $\omega$. But note that $\omega$ is defined w.r.t. a very specific affine piece inside the elliptic curve. It is the piece with affine coordinates $(z,w)$, and the $y$-component, $y\ne 0$, is normed to one. In this world lives an ant, which always uses letters like $z,w,y$ for computations, we will also do so in connection with this affine piece.
Now there is also an affine piece, where from the ant point of view of $[z:w:y]$ we are missing the point on the elliptic curve denoted by $\infty$, which is $[0:1:0]$, the $y$-component is missing.
In order to make computations, it is useful to use $[Z:W:Y]$, and the letters $Z,W,Y$ in connection with this world, where $W\ne 0$. Here is living a bee, and in all books in the library it sees only capital letters.
Our passage is the "simple passage" $[z:w:y]=[Z:W:Y]$, and this makes sense in the "common world", which is for the ant $yw\ne 0$, and for the bee $YW\ne 0$. The differential $\omega$ comes in the world of the ant. We need to move it to the bee. For this we use the common chart intersection in the projective atlas. So the restriction of $\omega$ on $yw\ne 0$ (the ant, $y=1$,) and $YW\ne 0$ (the bee, $W=1$,) is (with an equality sign marked with $!$ for the identified charts): $$ \omega = \frac{dz}w =\frac{d(z/y)}{w/y} \overset!= \frac{d(Z/Y)}{W/Y} = \frac{\frac 1{Y^2}(Y\;dZ-Z\; dY)}{\frac 1Y} =\frac 1Y(Y\;dZ-Z\; dY)\ . $$ So on $\omega$ we take the last formula for $\omega$, first on the intersection of charts, then "as is" on the whole bee affine piece $W=1$.
The equation of the curve is here: $$ Y = Z\underbrace{(Z-Y)(Z-aY)}_{=:h=h(Z,Y)}\qquad\text{ around }(Z_0, Y_0) =(0,0)\ . $$ We will use below $Y=Zh$, with $h=h(Z,Y)$, $h(0,0)=0$, in the form where we plug in this into the R.H.S.: $$ Y=Z(Z-Y)(Z-aY)=Z(Z-Zh)(Z-aZh)=Z^3(1-h)(1-ah)\ . $$ (So $Z$ is a uniformizer, order one in $(0,0)$, and $Y$ has degree three, it is $Z^3$ times a local unit.) We then compute: $$ \begin{aligned} \omega &= \frac 1Y(Y\;dZ-Z\; dY) \\ &=dZ-\frac ZY\; dY \\ &=dZ-\frac Z{Z^3(1-h)(1-ah)}\; d(Z^3(1-h)(1-ah)) \\ &=dZ -\frac 1{Z^2(1-h)(1-ah)}\Big(\ (1-h)(1-ah)\; 3Z^2\; dZ +O(Z^3)\; dZ+O(Z^3)\; dY\Big)\ . \end{aligned} $$ And there is no pole in $(0,0)$, the $Z^2$ in the denominator is removed by a $Z^2$ factor in the numerator of the last term.