One the one hand, any topological surface has a complex structure, so for any natural number $g$, there exists a complex curve with genus $g$. On the other hand, we have a Chow's theorem, which says that any complex analytic variety is algebraic. This means that there exists a non-singular algebraic curve with genus $g$. But algebraic curves have a well-defined degree $d$ which connect to the notion of genus using the formula $g=(d-1)(d-2)/2$. This suggests that $g$ cannot be any natural number.

Obviously this argument has a flaw, but where?


1 Answer 1


The formula you gives concerns plane algebraic curves, but, in general a curve cannot embed in $\bf CP^2$ but in $\bf CP^3$.

  • $\begingroup$ Is there a formula for the genus of a degree $d$ curve in $\mathbb{C}\mathbb{P}^3$ ? $\endgroup$
    – Mihail
    Apr 20, 2021 at 4:47
  • 5
    $\begingroup$ @Mihail one way to produce a curve of any genus in $\Bbb P^3$ is as a divisor on $\Bbb P^1\times\Bbb P^1\subset \Bbb P^3$: a divisor of type $(a,b)$ on this surface has genus $(a-1)(b-1)$ by adjunction. $\endgroup$
    – KReiser
    Apr 20, 2021 at 4:55
  • $\begingroup$ @Mihail, there exists smooth rational (= genus 0) curves of any degree in $CP^3$. $\endgroup$
    – Thomas
    Apr 20, 2021 at 9:51

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