The Construction of a Basis of Holomorphic Differential 1-forms for a given Planar Curve

$\def\CC{\mathbb{C}}$I'll come back later and leave an answer to (1), which is classical and straightforward. I have nothing to say about (3). I thought the most interesting question was (2), but I am not sure that I understood it correctly.

Here is how I understand question (2).

Let $X_0$ be the plane curve $F(x,y)=0$; let $X$ be the smooth projective completion of $X_0$. We will try to construct a uniformization $z \mapsto (\phi(z), \phi'(z))$ from $\CC \to X$. We may try to define $\phi$ either by the ODE $F(\phi, \phi')=0$, or by the condition that $\phi^{-1}(x) = \int_{x_0}^{x} \tfrac{dx}{F_y(x, y(x))}$. Here $F_y$ is the partial derivative with respect to $y$ and $y(x)$ is a branch of the algebraic function given by $F(x,y(x))$. What goes wrong?

We know something must go wrong because, if we had a nonconstant holomorphic map $\CC \to X$, it would lift to the universal cover of $X$. But, if $X$ has genus $\geq 2$, the universal cover of $X$ is isomorphic to the open disc $\mathbb{D}=\{ z : |z|<1 \}$. So we would have a nonconstant map $\CC \to \mathbb{D}$ and thus a bounded entire function, contrary to Louiville's theorem.

Let's note that things are fine locally. Let $U$ be a simply connected open subset of $X_0$. Then $\tfrac{dx}{y(x)}$ is a well defined $1$-form on $U$ with an antiderivative $\psi : U \to \CC$. The derivative of $\psi$ is nonvanishing so the inverse function theorem does provide $\psi$ with a local inverse $\phi$, and this $\phi$ does solve the ODE $F(\phi, \phi')=0$. However, this is a local discussion. When we try to continue $\phi$ to all of $\CC$, we encounter two problems.

First problem: $\psi$ will have ramification. It is clearer to think about $\tfrac{dx}{y(x)}$ as a $1$-form on $X_0$, not on $\CC$. On the curve $X_0$, we have $F_x(x,y) dx + F_y(x,y) dy$, so $\tfrac{dx}{F_y} = - \tfrac{dy}{F_x}$. The condition that $F$ is smooth says that at least one of these two denominators is always nonzero, so we can define a $1$-form $\omega$ by whichever side of this equation doesn't divide by zero. The $1$-form $\omega$ is nonvanishing on $X_0$ and seems to usually extend to a holomorphic $1$-form on $X$. (I don't know a precise theorem ruling out the possibility of a pole at one of the points of $X \setminus X_0$.)

However, if $X$ has genus $g$, then a $1$-form on $X$ must have $2g-2$ zeroes (with multiplicity), and $2g-2>0$ for $g \geq 2$. So $\omega$ must vanish at some of the points of $X \setminus X_0$. For example, consider the curve $y^2 = x^5+ \cdots$, where the right hand side is a degree $5$ polynomial with distinct roots. There is one additional point $p$ in $X \setminus X_0$. Writing $z$ for a local coordinate near $p$, we have $y = z^{-5} + \cdots$ and $x = z^{-2} + \cdots$. We have $\omega = \tfrac{dx}{2 y} = \tfrac{dx}{2 \sqrt{x^5+\cdots}}$. At the point $p$, the numerator $dx$ looks like $z^{-3} dz$, and the denominator looks like $z^{-5}$, so $\omega$ has a zero of order $2$.

If $\omega$ vanishes to order $k$ near a point $p$, then $\psi = \int \omega$ will vanish to order $k+1$, so the inverse function $\phi$ will have branching like a $(k+1)^{\mathrm{th}}$ root. So $\phi$ can't be defined globally on $\CC$; one needs to make some branch cuts to make a region on which it makes sense.

Second problem: Changing the path of integration. We defined $\psi(x) = \int_{x_0}^x \omega$. If we take this integral along two paths that differ in $H_1(X, \mathbb{Z})$, then the integral will change by a constant known as a period. In the genus $1$ case, this is where the double periodicity of $\phi$ comes from. On a genus $g$ curve, $H_1(X, \mathbb{Z})$ has rank $2g$, so there are $2g$ periods, which generate a dense group of translations. We can analytically continue $\phi$ around $\CC$ and come back on a branch which differs from our original branch by any element of this dense translation group.

In the Abel-Jacobi construction One replaces the single one-form $\omega$ with $g$ one-forms $(\omega_1, \omega_2, \ldots, \omega_g)$. So $( \int \omega_1, \cdots, \int \omega_g)$ gives a map $\Psi$ from a cover of $X$ to $\CC^g$. This map $\Psi$ does not have the problems above: The $\omega_j$ have no common zero, so $\Psi$ is locally injective, and the periods form a discrete subgroup of $\CC^g$. But the target is higher dimensional than the source, so there is no hope of inverting $\Psi$.

Example (sorry that I don't have time to make pictures!) let's take our curve $X_0$ given by $y^2 = x^5 + \cdots = \prod_{i=0}^4 (x-x_i)$. Let $p_i$ be the preimage of $x_i$ in $X_0$. We also write $p_{\infty}$ for the point of $X \setminus X_0$ and $x_{\infty}$ for the point $\infty$ of the Riemann sphere. The $x$-coordinate is a degree $2$ map from $X$ to the Riemann sphere, branched over the $x_i$.

Draw rays from $x_1$, $x_2$, $x_3$ and $x_4$ to $x_{\infty}$; let the angles between adjacent rays be $\theta_1$, $\theta_2$, $\theta_3$, $\theta_4$ where $\sum \theta_j = 2 \pi$. The preimages of these rays are four circles, each one passing through $p_{\infty}$ and through one of $p_1$, $p_2$, $p_3$, $p_4$. The circle meet each other at angles $\theta_j/2$, since $X \to \mathbb{CP}^1$ is $2$-fold ramified at $p_{\infty}$. Removing these four circles from $X$ leaves behind an open octagon $U$, and we recover the standard description of a genus two surface as an octagon with sides glued together.

The region $U$ is simply connected, and $\omega = \tfrac{dx}{2 y}$ is a well defined $1$-form on it. On $U$, the $1$-form $\omega$ has an antiderivative $\psi$. The region $\psi(U)$ is a planar octagon. As we discussed before, $\omega$ vanishes to order $2$ at $p_{\infty}$, so $\psi$ is $3$-fold ramified there, and the angles $3 \theta_j/2$, each twice for $1 \leq j \leq 4$. Note that this adds up to $6 \pi$, as the angles of an octagon should.

Opposite sides are parallel and of the same length; namely, the length of one such side is $2 \int_{p_j}^{p_{\infty}} \omega = \int_{x_j}^{x_{\infty}} \tfrac{dx}{\sqrt{x^5+\cdots}}$. The midpoint of this edge corresponds to $p_j$. If I didn't screw up, the displacement between parallel sides are $2 \int_{p_0}^{p_j} \omega = \int_{x_0}^{x_j} \tfrac{dx}{\sqrt{x^5+\cdots}}$. These integrals are the four periods, which generate a dense subgroup of $\CC$. As we try to analytically continue $\phi$ around $\CC$, we find branches of $\phi$ which are defined on translates of this octagon by those periods; if we encircle a vertex of the octagon, then $\phi$ branches like a cube root.

References: For very concrete discussions of the geometry of integrals like $\int \tfrac{dx}{\sqrt{x^5+\cdots}}$ I recommend Zeev Nehari's Conformal Mapping. For modern applications of these ideas in research, I recommend papers on "billiards" by people like Laura Demarco, Curt McMullen or Alex Wright. You might start with these slides as a concrete example of describing Riemann surfaces by gluing polygons just as I have here.

I'm getting back to the question of describing holomorphic $1$-forms on a plane curve.

Affine curves: Let $X$ be a smooth curve in $\mathbb{A}^2$, given by the equation $F(x,y)$. Then $F_x dx + F_y dy=0$. Since $F=0$ is smooth, the functions $F_x$ and $F_y$ have no common zero on $X$ and we can define a $1$-form on $X$ by $\omega = \tfrac{dx}{F_y} = - \tfrac{dy}{F_x}$. Moreover, $\omega$ is nowhere vanishing, so every $1$-form on $X$ is of the form $h \omega$ for some holomorphic function $h$ on $X$. For example, if $X$ is a hyperelliptic curve $y^2 = x^{2g+1} + a_{2g} x^{2g} + \cdots + a_1 x$, then $\omega = \tfrac{dx}{2y}$ and integrating $\omega$ along paths in $X$ corresponds concretely to computing integrals like $\int \tfrac{dx}{\sqrt{x^{2g+1} + a_{2g} x^{2g} + \cdots + a_1 x+a_0}}$.

Incidently, there is a more conceptual way to describe $\omega$: It is the residue of the $2$-form $\tfrac{dx \wedge dy}{F(x,y)}$ along $X$. Since every automorphism of $\mathbb{A}^2$ multiplies $dx \wedge dy$ by a scalar, and every automorphism taking $X$ to itself (setwise) multiplies $F$ by a scalar, this shows that $\omega$ is well defined up to scalar multiple independent of the choice of coordinates on $\mathbb{A}^2$.

Projective curves Let $\overline{X}$ be the smooth projective completion of $X$. Then we can ask whether or not $h \omega$ extends to a global holomorphic $1$-form on $\overline{X}$. The vector space of holomorphic $1$-forms on $\overline{X}$ always has dimension $g$. Describing which $h \omega$ extend to $\overline{X}$ in general involves describing how to compute the smooth projective completion of $X$, which is a little complicated, so I'll just give the most important special cases.

Hyperelliptic curves Let $X$ be given by $y^2 = x^{2g+1} + a_{2g} x^{2g} + \cdots + a_1 x+a_0$ for some square free polynomial of degree $2g+1$. Then a basis for the $1$-forms on $\overline{X}$ is $x^j \omega$ for $0 \leq j < g$. So the periods of this curve are concretely integrals of the form $\int \tfrac{x^j dx}{\sqrt{x^{2g+1} + a_{2g} x^{2g} + \cdots + a_1 x+a_0}}$ for $j$ in this range.

Smooth projective planar curves Let $f(x,y,z)$ be a smooth degree $d$ hypersurface, so $F(x,y) = f(x,y,1)$. Then a basis of $1$-forms on $\overline{X}$ is $x^i y^j \omega$ for $i+j \leq d-3$.

Curves transverse to a toric compatification The following includes both of the previous cases. Let $F(x,y) = \sum_{(i,j) \in A} F_{ij} x^i y^j$ for some finite set $A$ of exponents. Let $\Delta$ be the convex hull of $A$, this is a convex polytope. I will impose a condition for each edge of $\Delta$. Let $(p,q)$, $(p,q)+(u,v)$, $(p,q)+2 (u,v)$, ..., $(p,q) + k (u,v)$ be the lattice points on an edge. Suppose that the single variable polynomial $\sum F_{(p+ku, q+kv)} z^k$ has no nonzero multiple roots. Then the closure of $X$ in the toric variety associated to $\Delta$ is smooth. In that case, a basis for the $1$-forms is $x^{i-1} y^{j-1} \omega$, where $(i,j)$ runs over the lattice points in the interior of $\Delta$.