4

Suppose I have a section $s$ of an invertible sheaf $L$, vanishing along a divisor $D$. Then there is an isomorphism $(L, s) \simeq (O(D), 1)$. In the next paragraph I'll pick $D=K_X$, but that is just how I stumbled upon my question/confusion, it works for any divisor $D$.

Let $X$ be a smooth variety of dimension $n$, let $\omega$ be an $n$-form, and let $K_X=div(\omega)$. To help me explain my confusion, assume $K_X$ is an effective Weil divisor (although it's not really crucial). The sections of $O(K_X)$ are rational functions $f$ such that $div(f) + K_X \geq 0$, in other words, sections of $O(K_X)$ consist of rational functions that (assuming for convenience $K_X$ is effective) can have \emph{poles} along $K_X$, which is confusing me because $K_X$ is the divisor of zeroes (minus poles) of a rational $n$-form of the canonical bundle $\omega_X \simeq O(K_X)$!

Can someone help sort out my confusion ... I'm having trouble formulating my precise question. How should I think of the sections of $O(K_X)$? As n-forms or as rational functions with poles along $K_X$?

I think what may be happening is that if I think about a section of $O(K_X)$ as rational function $f$, and then if I want to switch to thinking about the divisor of the corresponding section of $\omega_X$, I have to take not $div(f)$ but $div(f) + K_X$. That is basically how the isomorphism $\omega_X \simeq O(K_X)$ goes.

In the definition of terminal singularities, the condition is, for a resolution $f:X \to Y$, we require $f_*O(mK_X -E) = O(mK_Y)$ How should I think of an element of $f_*O(mK_X -E)$, particularly to figure out the following statement, which I found on p. 8 here:

"Assume $Y$ is smooth and let $s \in H^0(Y,mK_Y)$. Then $f^\ast s$ as a section of $mK_X$ will vanish along the exceptional divisor of $f$."

So does that mean $s \in f_*O(mK_X-E)$, or is it $s \in f_*O(mK_X+E)$? Why? I can see reasons for both options even though only one is correct.

ykm
  • 1,621

2 Answers2

4

Regarding your second, more specific question: since $f^*s$ is a section of $ mK_X$ which vanishes along $E$, it lies in $\mathcal O(m K_X - E)$.

In general, if $D$ is a Cartier divisor, then $\mathcal O(D)$ can be thought of as: an invertible sheaf whose meromorphic sections have divisors of zeroes and poles that are linearly equivalent to $D$; or the subsheaf of the sheaf of rational functions consisting of functions such that $div(f) + D \geq 0$. Note that in the first description, we have a particular meromorphic section $s_0$ whose divisor is precisely $D$. We then go from the first description to the second via $s \mapsto s/s_0,$ and from the second to the first via $f \mapsto f s_0$.

Given these definition, you can see that if $E$ is another divisor, then the subsheaf of $\mathcal O(D)$ consisting of sections which furthermore vanish along $E$ is naturally identified with $\mathcal O(D - E)$.

Matt E
  • 127,227
2

One point that may help is that $K_X$ is the divisor of zeroes minus poles of $\omega$. The concrete case to always bear in mind is $\mathbb{P}^n$, where we have an $n$-form $dx_1/x_1\wedge...\wedge dx_n/x_n$. This is nondegerate, i.e. has no zeroes, but has poles along hyperplanes $x_i=0$. So sections of $O(K_X)$ become rational functions vanishing on these hyperplanes, rather than with permitted poles along them.

Kevin Carlson
  • 54,515
  • Yes, I meant zeroes minus poles, I have now edited it to include that. And your example highlights what I find confusing: the sections of $O(K_{\mathbb{P}^n})$ vanish on the hyperplanes, while the n-form has poles on the hyperplanes. I guess that's just how things are. – ykm Jun 15 '14 at 00:40
  • 2
    @yogesh: Dear yogesh, Your comment hints at some confusion that "sections of $\mathcal O(K_{\mathbb P^n})$ vanish on hyperplanes, while the $n$-form has poles on hyperplanes".

    These statements are not in the tension that you seem to imagine they are. Rather, what this expresses is that if we want to multiply an $n$-form on $\mathbb P^n$ by a rational function in such a way as to get an $n$-form without poles, then this rational function would need to vanish along the locus of poles of the $n$-form. Regards,

     – Matt E Jun 16 '14 at 01:35
  • @MattE That is a nice way to think of it. Thank you very much Matt. – ykm Jun 16 '14 at 04:16