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

Let $X$ be a quasi-projective variety. Suppose that we (perhaps partially, if either enough is known) compactify to $\bar{X}$ with $\bar{X}\setminus X=D$ is a divisor. Say that we know the canonical bundle $K_X$. Then $K_{\bar{X}}=K_X+nD$ for some $n$.

  1. Is $n$ always negative? The examples I'm thinking of are for $X=\mathcal{M}_g,\mathcal{A}_g$
  2. Is there a good method for computing this $n$?

In both cases, I'm particularly interested in finite covers of $\mathcal{M}_g,\mathcal{A}_g$ and other moduli spaces, if that helps to know how the variety is given.

share|improve this question
    
Charles, when you say that $\bar{X} \setminus X = D$ is a divisor, do you mean that $D$ is a prime divisor? –  Karl Schwede Dec 11 '10 at 4:41
    
@Karl, I don't, but perhaps I mean that it's a sum of distinct prime divisors. I'm coming to the conclusion that I've not thought this question through properly. –  Charles Siegel Dec 11 '10 at 12:07
    
My only point here was that, as unknown pointed out below, there's no reason that $K_{\bar{X}} = K_X + nD$ where you use the same $n$. You certainly do have $K_{\bar{X}} = K_X + \sum n_i D_i$. And it's possible that some $n_i$ can be positive while others are negative. –  Karl Schwede Dec 11 '10 at 21:03
    
@Karl, I realize this. I was too focused on some specific examples, and missed a few obvious facts. I think at this point I'll just let this question die and see if I can reformulate it more precisely to what I'm interested in in the future. –  Charles Siegel Dec 12 '10 at 0:13
    
I guess the rest of us understood the question saying something like $K_{\bar X}=K_X + \sum_i n_iD_i$ and asking about negativity of $n_i$. –  Sándor Kovács Dec 12 '10 at 1:38
add comment

3 Answers

up vote 3 down vote accepted

Dear Charles, perhaps I am misunderstanding what you ask, but the answer is no to both questions. Take an arbitrary projective variety $\overline X$, actually for simplicity let $\overline X$ be normal. Then it has a canonical divisor, say $K_{\overline X}$ and choose an actual representative, $K=\sum_i a_i K_i$ where the $a_i$ may be negative or positive. Now let $X=\overline X\setminus ({\rm Supp}\\, K)$. Then $K_X$ is trivial and there is no way you can tell the $a_i$ just from knowing $K_X$. Of course, if you know something else, that's a whole different question.

In any case, this at least shows that your $n$ is not always negative. If you know things like $X$ is covered by rational curves (or does not contain any) that could help, but as the above example shows, you need more information.

Some vaguely related thoughts on the canonical divisor are available at this answer to another MO question.

share|improve this answer
    
I guess that the extra information is that I'm mostly looking at moduli space of curves and abelian varieties with either the toric partial (corank 1) compactification or, if going all the way up, the second Voronoi compactification, and looking at finite covers of these spaces, which I know to be ramified along components of the boundary. –  Charles Siegel Dec 10 '10 at 22:05
add comment

To supplement Sandor's answer, let me note that any curve of positive genus is a counterexample to this claim.

For example, remove any number of points from an elliptic curve, and you will get an example where n=0.

If $g>1$, then the canonical divisor has positive degree, and so when written as a sum of points must have some positive coefficients. Removing one of those points gives a counter-example where $n>0$.

share|improve this answer
add comment

Why should $K_{\bar{X}}=K_X+nD$? Consider the following counterexample.

Let $E$ be an elliptic curve and let $\bar{X}=E\times \mathbb{P}^1$. Then $K_{\bar{X}}=-2E\times 0$. Let $X$ be the open subset $\bar{X}\setminus (e\times \mathbb{P}^1\cup E\times 0)$. Then $K_X$ is trivial as we have deleted $E\times 0$. So if $K_{\bar{X}}=K_X+n(e\times \mathbb{P}^1\cup E\times 0)$, then we would have that $(-n-2)(E\times 0)$ is rationally equivalent to $n(e\times \mathbb{P}^1)$, which is not correct as the picard group of $E\times \mathbb{P}^1$ is isomorphic to $Pic(E)\oplus\mathbb{Z}\cdot p_2^*\mathscr{O}(1)$.

share|improve this answer
    
I guess we all understood the question saying something like $K_{\bar X}=K_X + \sum_i n_iD_i$ and asking about negativity of $n_i$. –  Sándor Kovács Dec 12 '10 at 1:37
add comment

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.