sections of an invertible sheaf, and their support

Question

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.

Answer 1

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

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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 $\mathcal O(D - E)$.

Answer 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.