# What is the canonical meromorphic section $1_D$ of $\mathcal O_X(D)$?

Let $X$ be a compact Riemann surface and let $D$ be a divisor on $X$. With $1_D$ one generally indicates a particular meromorphic section of the invertible sheaf $\mathcal O_X(D)$ such that $$\operatorname{div }(1_D)=D$$

In other words $1_D$ is a collection of compatible meromorphic functions $\{f_\alpha\}$ on an open cover $\{V_\alpha\}$ of $X$ such that $$\operatorname{ord}_x (f_\alpha)=\operatorname{ord}_x(D)$$

I believe that this "collection" of meromorphic functions exists but how to construct the apparently distinguished one $1_D$? To be more precise: what is exactly $1_D$?

Example: On the stacks project I've found this definition for the algebraic case and when $D$ is effective, but honestly it is not clear to me.

• I think that if $D$ is not effective then $1_D$ is a meromorphic section, otherwise it is holomorphic – Dubious Nov 1 '16 at 15:14

One way of seeing this is "by direct definition:"

Define ${\cal L}={\cal O}_X(D)$ as the sub-sheaf of the constant sheaf ${\cal K}_X$ of meromorphic functions such that $${\cal L} (V) = \{ g \in {\cal K}_X \,| \ \text{div}|_V g + D|_V \ge 0 \}.$$

So a meromorphic section is any element of ${\cal K}_X$. We define $$\text{div}_{\cal L}\ g = D + \text{div}\ g,$$ where the div without the subscript is the usual div.

In particular, $\text{div}_{\cal L} (1) = D$.

Note - buyer beware - or feature - $$\text{div}_{\cal L}\ f g = \text{div}\ f + \text{div}_{\cal L}\ g.$$

Alternatively - and equivalently - using compatible functions as you write above:

Assume that $D|_{V_\alpha} = \text{div }|_{V_\alpha} f_a$. Define ${\cal L}={\cal O}_X(D)$ by $${\cal L}|_{V_\alpha} = {1 \over {f_\alpha} } {\cal O}_X|_{V_\alpha} \subset {\cal K} .$$ This is well defined, because the $f_\alpha$ are unique up to multiplication by elements in ${\cal O}_X(V_\alpha)^*.$

If $s\in {\cal L}\otimes {\cal K} (X)$ is a global section (and so a meromorphic section of $\cal L$), where we write $s|_{V_\alpha} = g_\alpha /f_\alpha$ with $g_\alpha \in {\cal K}$ locally, we define $\text{div}_{\cal L} s$ by gluing (not adding!) the local definitions $$\text{div}_{\cal L}|_{V_\alpha}(s|_{V_\alpha}) = \text {div}|_{V_\alpha} g_\alpha.$$

In particular if $1 \in {\cal L}\otimes {\cal K} (X)$ is the section defined by the local equations $$1|_{V_\alpha} = { f_\alpha \over f_\alpha },$$ then again, $\text{div}_{\cal L} (1) = D$.

In the second approach, the expression $\{g_\alpha\}$ of a meromorphic section $s$ of $\cal L$ depends on the choice of $f_\alpha$ - but not its divisor $\text{div}_{\cal L} (s)$.

Explicit example, in the algebraic geometry setting: take $X= \mathbb P^1$, with $U = \text { spec}\ k[t]$, and $V = \text {spec}\ k[1/t]$. Take $D = n\cdot \infty$, with $f_U = 1$, and $f_V = 1/t^n$.

In the first approach, there's not much to say...

In the second approach, $${\cal L}(U) = k[t]\text {, and}\ {\cal L}(V)= 1/(1/t^n)k[1/t] = t^n k[1/t].$$

$$1_U = {1\over 1} \text{, and, }1 = t^n \cdot (1/t)^n,$$ so $\text {div}_{\cal L}|_U 1 = 0$, and $\text {div}_{\cal L}|_V 1 = \text {div}|_V (1/t)^n = n\cdot \infty$.

Edit/Addition: (To define $1_D$ explicitly, and to clear up some sloppiness above, taking especially into account the comment of Georges below.) The sheaf ${\cal L}$ is a subsheaf of the constant sheaf ${\cal K}_X$, so write $i\colon {\cal L} \to {\cal K}$ for the inclusion. In both approaches, I implicitly identified the domain with the codomain of the isomorphism $${\cal L}\otimes_{\cal O} {\cal K} \xrightarrow{i\otimes 1 }{\cal K}\otimes_{\cal O} {\cal K} \xrightarrow{\text{mult}}\cal K.$$ Under the identification, the well-defined, canonical meromorphic section $1_D$ of the original question is expressed locally (in the second approach) as $1/f_\alpha \otimes f_\alpha$ in the domain, and globally as $1$ in the codomain: $${1\over f_\alpha} \otimes f_\alpha \to {1\over f_\alpha} \otimes f_\alpha \to {1\over f_\alpha}\cdot f_\alpha = 1.$$

• I am an idiot - I see I completely misread your entire question!!!! Though I think I might have answered it tangentially. In either approach $1$ is the usual element of $\cal K$. In the second approach, however, we express $1$ as $\{f_\alpha/f_\alpha\}$. I'll try to clear this up later or tomorrow - sorry! Still I hope the above helps. – peter a g Nov 1 '16 at 17:09
• In all rigor you should define $s\vert _{V_\alpha}=\frac 1f_\alpha \otimes g_\alpha \in ({\cal L}\otimes_{\mathcal O_X} {\cal K}_X)(V_\alpha)$ since your definition would yield a section $\frac {g_\alpha}{f_\alpha} \in \mathcal K_X(U_\alpha)$, and you should not identify ${\cal L}\otimes_{\mathcal O_X} {\cal K}_X$ with ${\cal K}_X$ . Nitpicking apart, yours is a great answer : +1, of course. – Georges Elencwajg Nov 1 '16 at 20:46
• @GeorgesElencwajg Thanks for the nitpicking - and I absolutely, totally mean in it... Except that I don't think it's nitpicking, as it is at heart what the OP was actually asking (as I wrote in the previous comment). I wrote "So any meromorphic section is an element of $\cal K$." In the second approach I am identifying ${\cal L}\otimes_{\cal O} {\cal K}$ with (its image in) $\cal K \otimes_{\cal O}{\cal K} = \cal K$. Am I committing a sin = error? In any case, this is how I was understanding the meromorphic section $1|_{V_\alpha} = f_\alpha / f_\alpha$ here - so please nitpick! – peter a g Nov 1 '16 at 21:49
• @GeorgesElencwajg - namely, crossing my t's, in the approaches above, I am defining $\cal L$ as a sub-sheaf of $\cal K$: the 'image in' in the previous comment comes from the tensor of the inclusion. – peter a g Nov 1 '16 at 22:02
• Dear peter, I completely agree with your comments: indeed there is a canonical isomorphisms $\mathcal L\otimes_\mathcal O\mathcal K=\mathcal K \otimes_\mathcal O\mathcal K\to \mathcal K$ and my interpretation living in the domain is sent to your interpretation in the codomain ! I am quite impressed by your vision of the problem at hand, which I have never seen discussed in books before. Is it an original creation of yours ? Anyway I couldn't imagine a clearer answer and I'm sorry I can't upvote it more than once ! – Georges Elencwajg Nov 2 '16 at 7:44