Let $f\left(x,y\right)$ be an elementary function of $x$ and $y$.


$$4x^{3}-ax-b-y^{2}$$ $$a^{x}-y^{2}+x^{3}y-1$$ $$x^{3}-3xy+y^{3}$$ $$\frac{\sin^{2}x}{x^{2}+\sin^{2}y}-\frac{1}{3}$$

Let $C$ denote the zero-locus of $f\left(x,y\right)$ over $\mathbb{C}$ (affine and/or projective—whatever makes sense for the given $f\left(x,y\right)$).

I know that any holomorphic or meromorphic differential 1-form defined on $C$ can be written as $g\left(x,y\right)dx$ for some expression $g\left(x,y\right)$.

My question is, given $f\left(x,y\right)$, $C$, and $g\left(x,y\right)$ as described above, what computation do I perform to check if $g\left(x,y\right)dx$ defines a holomorphic or meromorphic differential 1-form on $C$?

Heads up: the terms "chart" and "transition map" give me a headache (they're too abstract), so just speak of "formulas"/"expressions" and what I need to do with them.


  • $\begingroup$ Given meromorphic functions $f,g,h$ on unit disk look at $f(z)dz,f(z)dg(z)=f(z)g'(z)dz$,$\frac{f(z)dg(z)}{dh(z)}=f(z)g'(z)/h'(z)$,$\int_0^zf(s)dg(s)=u(z)+\sum_jc_j\log(z-a_j)$ with $u$ meromorphic, $a_j$ the simple poles of $f(z)g'(z)$. If $u$ is holomorphic and there are no $a_j$ then $f(z)dg(z)$ is a holomorphic one-form. You can replace $z$ by any holomorphic chart, the integral of one-forms is chart independent. On a curve it is a matter of finding some charts to make it locally biholomorphic to the unit disk and get a definition of "meromorphic". Harmonic (forms) are real part of those $\endgroup$ – reuns Mar 24 at 22:08
  • $\begingroup$ How does one find the charts? $\endgroup$ – MCS Mar 24 at 22:56
  • $\begingroup$ Let $F(x,y) = y^2-x^3-1$ and $C = \{ (x,y) \in \Bbb{C}^2, F(x,y)=0\}$ then $(\partial_xF,\partial_yF) = (-3x^2,2y)$ thus for $(x_0,y_0) \in C, y_0 \ne 0$ and $(x,y) \in C, |x-x_0|^2+|y-y_0|^2 = r^2$ small the Taylor expansion of $F(x,y)=0$ gives $(x-x_0)(-3x_0^2)+(y-y_0) 2y_0 = O(r^2)$ so $(x,y) \mapsto y-y_0$ is a local chart from $C$ to $ \Bbb{C}$. Equivalently look at the points where $x \mapsto (x,\sqrt{x^3+1})$ is holomorphic. From there you are left with a few isolated singular points. $\endgroup$ – reuns Mar 24 at 23:15

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