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I'm wondering if $\sqrt[3]{z^2}$ is a Riemann surface including point $(0, 0)$ or not.

Treat it as the zero set of $F(z, w) = 0$, where $F(z, w) = z^2-w^3$. Then $F_z = 2z$ and $F_w = 3w^2$ and both are zero in point $(0, 0)$. From this view point it seems it doesn't define a Riemann surface since I cannot apply the implicit function theorem. Hence the Riemann surface it defined doesn't include the point $(0, 0)$.

But if I follow another viewpoint, treating it as a complete analytic function, it seems that it's a Riemann surface and can be extended to include the point $(0,0)$ as a branch point. And it seems similar with the riemann surface defined by $\sqrt[3]{z}$, which has the point $(0, 0)$ as a branched point.

Is that I missed something? Does the riemann surface including point $(0, 0)$ or not?

Thanks!

onRiv
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    The origin is a singular point of the subset of $\Bbb C^2$ defined by the equation $0=z^2-w^3$. – Lubin Jun 12 '22 at 20:30
  • @Lubin Thanks. I occured this example when I was trying to give some example to the Hurwitz formula. Maybe I got it now. Though the singularity of it is quite hard to imagine or visualize – onRiv Jun 13 '22 at 04:13
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    Indeed, @onriv , I think I read that a well-chosen intersection of this singularity (2 dimensions, sitting in 4) with a (3-dimensional) real hyperplane, gives a 1-dimensional intersection in that 3-space. And it’s a trefoil knot! – Lubin Jun 13 '22 at 17:14

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