1
$\begingroup$

Let $C\subset\mathbb{P}^n$ be a normal curve over an algebraically closed field of characteristic $0$. Assume that $C$ is not contained in any hyperplane. We may assume that $P=[0:\cdots:0:1]$ is on $C$. Consider the projection $\pi:\mathbb{P}^n\dashrightarrow\mathbb{P}^{n-1}$ away from $P$ i.e. $[x_0:\cdots:x_{n-1}:x_n]\mapsto[x_0:\cdots:x_{n-1}]$. Let $C'=\pi(C)$. What can we say about the degree and genus of $C'$?

In Fisher's The higher secant varieties of an elliptic normal curve Lemma 5.2, it was stated for elliptic curves but without proof. Do we have similar conclusions for higher genera?

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ It seems to me that you should have a look to the well-known Riemann-Hurwitz formula e.g. en.wikipedia.org/wiki/Riemann%E2%80%93Hurwitz_formula $\endgroup$
    – Holonomia
    Commented Aug 11, 2021 at 10:59
  • $\begingroup$ I assume that the induced map $C\rightarrow C'$ has degree one. If $P$ is a smooth point, the degree of $C'$ is just $\deg(C)-1$ (in general, $\deg(C)-m(P)$). Of course the geometric genus is the same; for the arithmetic genus we have $g(C')\geq g(C)$, the difference depending on possible trisecants to $C$ passing through $P$. $\endgroup$
    – abx
    Commented Aug 11, 2021 at 12:30
  • $\begingroup$ @abx P is smooth since the OP assumed the curve to be normal, isn't it? $\endgroup$
    – Holonomia
    Commented Aug 11, 2021 at 13:17
  • $\begingroup$ @Holonomia: Probably right. I am not sure if "normal" means smooth or projectively normal. $\endgroup$
    – abx
    Commented Aug 11, 2021 at 14:23
  • $\begingroup$ @abx Yes. The word "normal" means smooth. But why can we say that the degree is $1$? Or is there a hyperplane such that the induced projection has degree $1$? $\endgroup$
    – Li Li
    Commented Aug 13, 2021 at 16:03

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .