5
$\begingroup$

I would like to prove the following statement:

Let $C\subseteq \mathbb{P}^{3}$ be an irreducile nonsingular curve of arithmetic genus $g_{a}(C)=24$ and degree $d(C)=14$. Then there exists an irreducible nonsingular cubic surface $X\subseteq \mathbb{P}^{3}$ such that $C$ lies on $X$.

I am able to prove the converse. The following are my unfruitful attempts.

(1) I thaught that this may be proven by using some properties of the Hilbert polynomial.

(2) Another approach might conists in constructing a cubic surface $X$ and a linear system on that surface that contains $C$. More formally, let $C\subseteq \mathbb{P}^{3}$ be an irreducile nonsingular curve of arithmetic genus $g_{a}(C)=24$ and degree $d(C)=14$. Then there exits a cubic surface $X$ such that

$\qquad \qquad \qquad \qquad \qquad \qquad \mathcal{O}_{C}=\mathcal{O}_{X}(C)=\mathcal{O}_{X}(4H)\otimes \mathcal{O}_{X}(2L)$

with $H$ an hyperplane section of $X$, and $L$ a line lying on $X$.

(3) A third approach might consist in using the Picard group. This approach is similar to (2). Let $C\subseteq \mathbb{P}^{3}$ be an irreducile nonsingular curve of arithmetic genus $g_{a}(C)=24$ and degree $d(C)=14$. Let $X$ an irreducible nonsingular cubic surface in $\mathbb{P}^{3}$, and $l,e_{1},\dots,e_{6}\in \mathbb{Z}^{\oplus 7}$ the generators of the Picard group of $X$. Let $D_{X} \sim al-\sum b_{i}e_{i}$ be a divisor on $X$ such that

$ \qquad \qquad \qquad \qquad \qquad \left \{ \begin{array}{l} b_{1}\geq b_{2} \geq \dots \geq b_{6}>0 \\ a\geq b_{1}+b_{2}+b_{5} \\ 3a-\sum b_{i} =14 \\ \frac{1}{2} (a-1)(a-2)- \frac{1}{2} \sum b_{i}(b_{i}-1) = 24 \end{array} \right. $

Then for any curve $C$ there exists a surface $X$ and a divisor $D_{X}$ such that $\mathcal{O}_{C}=\mathcal{O}_{X}(D_{X})$.

Please note that these are just ideas, and they may very well be wrong. Any suggestions on how to approach this problem? Is this a special case of some more generic concept? Also, I am not entirely sure that what I'm trying to prove is true to begin with. If the statement is in fact false, can a simple counter-example be constructed?

$\endgroup$
5
  • $\begingroup$ I do not understand what you mean by $\mathcal O_C=\mathcal O_X(C)$ or, below, $\mathcal O_C=\mathcal O_X(D_X)$. Do you mean to restrict from $X$ to $C$ (if $C\subset X$)? $\endgroup$
    – Brenin
    Commented Nov 22, 2013 at 18:38
  • $\begingroup$ Maybe I wrote it poorely. Yes that's the idea. $\endgroup$
    – Fq00
    Commented Nov 25, 2013 at 22:34
  • $\begingroup$ What is some motivation? How did this question come up? It might help in figuring it out. $\endgroup$ Commented Nov 27, 2013 at 4:04
  • $\begingroup$ This came to my mind in reading Mumford's famuls example of a Scheme with a non-reduced component (see "Deformation Theory", Robin Hartshone, Springer, pp. 91-94). In Mumord's example, a Hilbert scheme is constructed by taking all irreducible curves defined by diviors of a cubic surface $X$ of the form $|4H+2L|$, with $H$ the hyperplane section, and $L$ a line. This question is kind of the converse of that - i.e. can any curve in 3-dimensional projective space be viewed as an element of a divisor of some kind? $\endgroup$
    – Fq00
    Commented Nov 27, 2013 at 11:06
  • $\begingroup$ Let $C\subset \mathbb{P}^3$ be a smooth curve of genus $g=24$ and degree $14$, and let us consider the divisor $D$ on $C$ given by the restriction of a cubic hypersection. We have then $\mathrm{deg}(D)=42$, so by Riemann-Roch $$\ell(D)-\ell(K_C-D)=42+1-g=19$$ Note that the vectorial dimension of cubic hypersurfaces of $\mathbb{P}^3$ is $20$ and that $\mathrm{deg}(K_C-D)=2g-2-42=4$. If $\ell(K_C-D)=0$, you would have shown that $C$ is contained in a cubic hypersurface. But this seems unclear to me. Hence, I would more be tempted to look for a counterexample. $\endgroup$ Commented Jan 7, 2015 at 7:23

1 Answer 1

0
$\begingroup$

It has been over 10 years since this question was asked, but I would like to say that not every curve of degree $14$ and genus $24$ is contained in a cubic.

In Mumford's paper, it is proved that any such curve $C$ bust be contained in a pencil of quartic surfaces. If this pencil has fixed locus, then $C$ is contained in a unique cubic surface, as shown by Mumford.

If this pencil has no fixed locus, then $C$ is linked to a plane conic by a pair of quartics $Q_1$ and $Q_2$ containing it. That is, $Q_1 \cap Q_1 = Y\cup C$, where $Y$ is a plane conic. Then Liaison Theory shows that $H^0(\mathcal{I}_C(3)) = 0$, cf. Chapter III in https://smf.emath.fr/system/files/filepdf/AST_1990__184-185__1_0.pdf.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .