# Genus 5 curves as intersections of 3 quadric hypersufaces of $\mathbb{P}^4$

Let $C$ a smooth projective integral curve of genus $5$. Suppose that it is not hyperelliptic, so that the canonical map $$|\omega_C|:C\to \mathbb{P}^4$$ is a closed embedding. Then $C$ can be represented as an intersection of three smooth quadric hypersufaces of $\mathbb{P}^4$.

We have that $\deg(\omega_C)=8$. So $h^0(C,\omega_C^{\otimes}2)=\deg(\omega_C^{\otimes}2)+1-g=12$ by Riemann-Roch. But $h^0(\mathbb{P}^4, \mathcal{O}_{\mathbb{P}^4}(2))=15$, it follows that there are $3$ linearly independent quadric surfaces containing $C$. Call them $X_1,X_2,X_3$. If $\dim (X_1\cap X_2\cap X_3)=1$ then by degree's consideration it easily follows that $C=X_1\cap X_2\cap X_3$. The $\textbf{Question}$ is: is it always the case? Why is not possible that the dimension of the intersection is bigger?

• How did you get $h^0(\omega_C)=8$? It is supposed to be $g=5$ in your case. $2g-2$ is the degree of $\omega_C$. Commented Apr 6, 2018 at 0:00
• Yes, you are right, of course i meant $\deg(\omega_C)$, thank you for the correction.
– DDT
Commented Apr 6, 2018 at 8:08

By Enriques-Babbage theorem the canonical (non-hyperelliptic) curve is an intersection of quadrics unless it is trigonal or a plane quintic. So, in case of $g = 5$ you need to assume $C$ is not trigonal.
In the trigonal case the intersection of quadrics through $C$ is a cubic scroll.