1
$\begingroup$

Consider three quadratics in $\mathbb{C}P^4$: $$ x_0^2+4x_1^2+\frac{x^2_2}{4}=0,~ x_1x_4+x_2x_3=0, ~ x_0^2+x_3^2+x_4^2=0. $$

If there intersection was non-singular, then the intersection should be a curve of genus $5$, see this note.

However, the intersection in our problem has $4$ singular points $$ [0,\frac{1}{2},\pm 2i,0,0] \text{ and } [0,0,0, 1, \pm i]. $$

So it seems to me that (the normalization of) the intersection curve should have genus $$ g=5-4=1. $$

I want to know if my guess is correct and if it is part of some general result.

Improve this question
$\endgroup$
6
  • $\begingroup$ The curve is not indecomposable, so what do you mean by its genus? $\endgroup$
    – Sasha
    Commented Jan 8, 2021 at 20:22
  • $\begingroup$ @Sasha Yes the only possibilities are $x_1=\pm x_3$ and $x_2=\mp x_4$ so there are two components. I made some changes on the problem. $\endgroup$
    – Zhaoting Wei
    Commented Jan 8, 2021 at 20:43
  • 1
    $\begingroup$ Not two, but four! $\endgroup$
    – Sasha
    Commented Jan 8, 2021 at 20:44
  • 1
    $\begingroup$ This is a valid deduction if the curve is irreducible and the singularities are all nodes. Can you check this? $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 15:39
  • 1
    $\begingroup$ I believe Hartshorne contains (1) how to compute the arithmetic genus of a complete intersection, (2) resolving a node singularity reduces the arithmetic genus by 1, (3) for a smooth irreducible curve the arithmetic genus matches the geometric genus. $\endgroup$
    – Will Sawin
    Commented Jan 12, 2021 at 15:29

0

Reset to default

You must log in to answer this question.