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I'm studying Riemann surfaces, and I have this question:

The affine curve $S=\{(z_1,z_2)\in\mathbb{C}^2\ :\ z_1^2-z_2^3=0\}$ is a complex manifold?

What I have thought is that if (by concatenation) $S$ is a complex manifold, then it is a a Riemann surface, but this can not happen because then S must be no-singular, but it is not at point $p=(0,0)$ where $\frac{\partial f}{\partial z_1}=0$. But I am not sure if my reasoning is correct.

Is it possible to define another complex structure in S?

Protágoras
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    It is not a complex manifold in the usual sense. But, as a set, is bijective to $\mathbb{C}$ and thus can be given a manifold structure. – Mohan Jun 12 '17 at 00:40
  • I am not directly answering your question, but it might be productive for you to especially see Chow's theorem in the wiki article https://en.wikipedia.org/wiki/Algebraic_geometry_and_analytic_geometry – Tanner Strunk Aug 22 '17 at 04:04

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1) It doesn't make sense to ask whether a set is a complex manifold: a complex manifold is a set endowed with a supplementary structure, namely a holomorphic atlas (or alternatively a suitable sheaf of local rings).
2) However it makes perfectly good sense to ask whether a subset of $\mathbb C^n$ is a complex submanifold. For your $S\subset \mathbb C^2$, the answer is : "no!"
3) A sufficient condition for a subset $X\subset \mathbb C^n$ to be a complex manifold is given by the constant rank theorem.
In your case the function $f(z_1,z_2)= z_1^2-z_2^3$ has rank $1$ everywhere except at $O=(0,0)$ where it has rank $0$, so that constant rank theorem does not apply and cannot prove that $S$ is a submanifold.
But this does not prove that $S$ is not a submanifold, because the constant rank theorem only yields a sufficient condition for being a submanifold.
4) A possible proof that $S$ is not a submanifold is to show that the tangent cone to $S$ at $O$ is the zero vector space: see the answer here.
[Write $f(z)^2=g(z)^3$ with $f(z)=az+bz^2+\cdots,_\: g(z)=\alpha z+\beta z^2+\cdots$ and show that $ a=\alpha=0$]

Georges Elencwajg
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