Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sorry for my poor English.

Let $X$ be a reducible projective variety.

My question is:

  1. How can I compute the dualizing sheaf of $X$ and express it in an explicit way?
  2. Is there a method to get dualizing sheaf of whole reducible variety $X$ from the information of dualizing sheaves of its irreducible components?

Currently I'm not concern the general case, but I want a few accessible concrete examples such as:

  1. reducible hypersurfaces,
  2. union of toric varieties glued at isomorphic orbits(Alexeev calls it stable toric variety).

The reason why I concern is to understand the limit in moduli spaces of stable pairs.

share|improve this question

2 Answers 2

up vote 10 down vote accepted

With regards part 2.

Let's assume that you have two components $X_1$ and $X_2$ (or even unions of components) such that $X_1 \cup X_2 = X$=. Let $I_1$ and $I_2$ denote the ideal sheaves of $X_1$ and $X_2$ in $X$.

Set $Z$ to be the scheme $X_1 \cap X_2$, in other words, the ideal sheaf of $Z$ is $I_1 + I_2$.

It is easy to see you have a short exact sequence $$0 \to I_1 \cap I_2 \to I_1 \oplus I_2 \to (I_1 + I_2) \to 0$$ where the third map sends $(a,b)$ to $a-b$.

The nine-lemma should imply that you have a short exact sequence

$$0 \to O_X \to O_{X_1} \oplus O_{X_2} \to O_Z \to 0$$

If you Hom this sequence into the dualizing complex of $X$, you get a triangle $$\omega_Z^. \to \omega_{X_1}^. \oplus \omega_{X_2}^. \to \omega_{X}^. \to \omega_Z^.[1]$$

You can then take cohomology and, depending on how things intersect (and what you understand about the intersection), possibly answer your question.

If $X_1$ and $X_2$ are hypersurfaces with no common components (which should imply everything in sight is Cohen-Macualay) then these dualizing complexes are all just sheaves (with various shifts), and you just get a short exact sequence $$0 \to \omega_{X_1} \oplus \omega_{X_2} \to \omega_{X} \to \omega_{Z} \to 0$$

Technically speaking, I should also probably push all these sheaves forward onto $X$ via inclusion maps.

share|improve this answer

For hypersurfaces the answer is as follows. Assume that you have components $X_i$ of a reducible hypersurface $X$. Then $\omega_X|_{X_i} \cong \omega_{X_i}(\sum_{i \neq j} X_j ) := \omega_{X_i} \otimes \mathcal{O}(\sum_{i \neq j} X_j )$.

As for the proof: let $X':= \sum_{i \neq j} X_j$. Then we have a few equations:

$\omega_{X_i} \cong \omega_{P^n}(X_i)|_{X_i}$, $\omega_{X'} \cong \omega_{P^n}(X')|_{X'}$

So:

$\omega_{X_i}(X') \cong \omega_{P^n}(X_i+ X')|_{X_i} \cong \left( \omega_{P^n} (X) |_X \right) $ $|_{X_i} \cong {\omega_X} |_{X_i}$

Concerning how you can get generally the canonical sheaf of a reduced scheme from its component, here is something very similar: if your scheme is semi-normal and $S_2$ (which is true for stable pairs), then when you take its normalization $\pi : \tilde{X} \to x$, the support of the points where $\pi$ is not an isomorphism, is a divisor, say $B$. Then you have the formula $\pi^* \omega_X \cong \omega_{\tilde{X}}(B)$. There is a little bit about this in the 4th section of http://arxiv.org/abs/0801.1541 .

share|improve this answer
    
Thank you! I'll check the paper. –  Moon Apr 12 '10 at 1:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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