1
$\begingroup$

Let $\mathbb{P}^1= C_d \subset \mathbb{P}^d$ be the rational normal curve of degree $d$ and $S_d \subset \mathbb{P}^{d+1}$ be the projective cone over $C_d$. $S_d$ is a typical example of a log del Pezzo surface.

I want to know about the singularity of general elements $D_d$ of the linear system $|-K_{S_d}|$. I think $D_d$ is singular at the vertex. Does $D_d$ have only a nodal singularity at the vertex?

$\endgroup$

1 Answer 1

2
$\begingroup$

The linear system of $|-K_{S_d}|$ can be identified with a multiple of the linear system of hyperplane sections.

A general member of this will actually miss the vertex, and it is smooth everywhere.

Those elements that contain the vertex are identified with hyperplane sections through the vertex. In turn, these hyperplanes are themselves cones over hyperplanes in $\mathbb P^d$. In other words a general member containing the vertex is isomorphic to $d+2$ lines in general position in $\mathbb P^{d+1}$ going through a fixed point (the vertex).

$\endgroup$
8
  • $\begingroup$ Thank you for the reply. I think that one of the elements would be that $d+2$ rays. Let $\mu:\mathbb{F}_d→S_d$ be the minimal resolution of singularity. I think that there is a push-forward map $\mu_{\ast}:{\rm Cl}(\mathbb{F}_d)→{\rm Cl}(S_d)$. Let $h$ be the negative section on $\mathbb{F}_d$ and $f$ be the fiber class. I think that the image of sections $h+(d+2)f$ are linearly equivalent to the anticanonical divisor of $S_d$ and it's irreducible. Is there a mistake in this argement? $\endgroup$
    – tarosano
    Commented May 9, 2011 at 9:05
  • $\begingroup$ It should actually be $2h+(d+2)f$. Since $h$ is $\mu$-exceptional, $\mu_*h=0$. $\endgroup$ Commented May 9, 2011 at 10:00
  • $\begingroup$ In the case $d=2$ the singularity is of Du-Val type, hence the complete anticanonical linear system of $S_2$ is cut out by the complete linear system of quadrics in $\mathbb{P}^3$. So in this case the general element is a smooth curve of degree $4$ (not passing through the vertex). Of course this is no more true when $d \geq 3$. This remark is just to point out that probably some additional argument involving discrepancies is needed. $\endgroup$ Commented May 9, 2011 at 10:07
  • $\begingroup$ Francesco, you are right. For some reason I interpreted the question as asking about the general member that contains the vertex. $\endgroup$ Commented May 9, 2011 at 16:06
  • $\begingroup$ Dear Sandor, there is still something I do not understand. If the general element of $|-K_{S_d}|$ misses the vertex, then it seems to me that $K_{S_d}$ would be a Cartier divisor, hence $S_d$ would be a Gorenstein variety. But $S_d$ is Gorenstein if and only if $d=2$. So it appears that, after all, all the sections of $|-K_{S_d}|$ must contain the vertex for $d \geq 3$. Am I missing something? $\endgroup$ Commented May 9, 2011 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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