4
$\begingroup$

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?

$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

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

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.