For $D$ a divisor on a smooth variety $X$ over $\mathbb{C}$, we define as usual the subsheaf $\mathcal{O} (D)=\mathcal{O}_X(D)$ of the sheaf of rational functions $\mathcal{K}_X$ as follows:

$$\mathcal{O}_X(D)(U):=\{ f\in\mathcal{K}_X(U)\;|\; (f)\geq - D|_U \}$$

for every open $U\subseteq X$, where $(f)$ denotes the divisor attached to the rational function $f$.

From now on, suppose $D\geq 0$.

Then $\mathcal{O}(-D)$ is a subsheaf of $\mathcal{O}$ whose local sections are those regular functions which vanish along $D$ with multiplicity at least as indicated by the coefficients of $D$.

$\mathcal{O}(D)$ has as local sections those rational functions that are regular outiside of $D$ and are allowed to have poles at most along $D$, of order at most as indicated by the coefficients of $D$.

Let $\{U_\alpha\}$ be an open cover of $X$ such that $D\cap U_\alpha$ is defined by the regular function $\eta_\alpha\in\mathcal{O}(U_{\alpha\beta})$, that is: $D|_{U_{\alpha}}=(\eta_\alpha)$.

The multiplication by $\eta_\alpha$ induces local isomorphisms of sheaves:


showing that $\mathcal{O}(-D)$ is locally free of rank one. So, setting $U_{\alpha\beta}$, we get transition functions $\psi_{\alpha\beta}:=\eta_\alpha\cdot\eta_\beta^{-1}\in\mathcal{O}^{\;*}(U_{\alpha\beta})$ which give a cocycle defining the (corresponding) line bundle $\mathcal{O}(-D)$.

Analogously, local multiplication by the rational function $\eta_\alpha^{-1}$ gives local trivializations for $\mathcal{O}(D)$, and the cocycle given by the reciprocal $\psi_{\alpha\beta}^\vee:=\psi_{\alpha\beta}^{-1}=\eta_\alpha^{-1}\cdot\eta_\beta$ defines the line bundle $\mathcal{O}(D)$, showing that the line bundles $\mathcal{O}(D)$ and $\mathcal{O}(-D)$ are dual to each other.

Note that both line bundles are trivial when restricted to $X\setminus D$ because $\psi_{\alpha\beta}$ becomes a coboundary out of $D$.

Given a line bundle $\mathcal{L}$ with cocycle $\{g_{\alpha\beta}\}$, a global section $s\in \Gamma(X,\mathcal{L})$ is given by a bunch of local regular functions $s_\alpha\in\mathcal{O}(U_{\alpha\beta})$ such that $s_\beta=g_{\alpha\beta}\cdot s_\alpha$ on $U_{\alpha\beta}$.

More generally, if the local functions are allowed to be rational, $s_\alpha\in\mathcal{K}_X(U_\alpha)$, then this gives a global rational section $\sigma\in \Gamma (X,\mathcal{L}\otimes\mathcal{K}_X)$.

So, since clearly $\eta_\beta=\psi^\vee_{\alpha\beta}\cdot\eta_\alpha$ and $\eta_\beta^{-1}=\psi_{\alpha\beta}\cdot\eta_\alpha^{-1}$, we get a global section


such that $(s_D)=D$, and a rational global section



Question 1. By definition of $\mathcal{O}(D)$ as a subsheaf of $\mathcal{K}_X$, we have $\Gamma(X,\mathcal{O}(D))=\mathcal{O}(D)(X)=\{f\in\mathcal{K}_X(X)\;|\;(f)\geq -D\}$. So $s_D\in\Gamma(X,\mathcal{O}(D))$ must be a global rational function on $X$ such that its associated divisor is $\geq -D$. How can this global rational function $s_D$ be described? To be more precise, for example I would be content with a description of $s_D|_{U_\alpha}$ in terms of $\eta_\alpha, \eta_\alpha^{-1}\in\mathcal{K}_X(U_\alpha)$. Note that even if $\{\eta_\alpha\}$ is the expression of $s_D$ in "local trivializations", it's certainly not true that $s_D|_{U_\alpha}=\eta_\alpha$. Likewise, it's not true that $s_D|_{U_\alpha}=\eta_\alpha^{-1}$, as the identity $s_D|_{U_\alpha}=s_D|_{U_\beta}$ on $U_{\alpha\beta}$ must hold untwisted by any cocycle.

By duality, the natural sheaf inclusion $j_D=s_D^\vee:\mathcal{O}_X(-D)\to\mathcal{O}_X$ gets reversed to the sheaf surjection $s_D:\mathcal{O}_X\to\mathcal{O}_X(D)$, $h\mapsto h\cdot s_D$. In general, there is an isomorphism

$$\mathrm{Hom}(\mathcal{O}(-D),\mathcal{O})\cong\mathrm{Hom}(\mathcal{O},\mathcal{O}(D))=\Gamma(X,\mathcal{O}(D)),$$ where we go from right to left via the map on local sections $s\mapsto j_s$, $j_s(f):=\frac{s}{s_D}\cdot f$ ($f$ local section of $\mathcal{O}(D)$).

Question 2. How do we go from left to right in the above isomorphism, $j\mapsto s_j$, in terms of $j,s_D, s_D^\vee, \eta_\alpha$ etc.?

  • $\begingroup$ For Q1 what you can say is that $s_D|_{U_{\alpha}} \cdot \eta_{\alpha} \in \mathcal{O}_X(U_{\alpha}).$ Is that what you want? $\endgroup$ Nov 8, 2012 at 6:49
  • $\begingroup$ and for Q2 I think, $j_s$ is just multiplication with $\eta_{\alpha}^{-1}$ over $U_{\alpha},$ so to get identity, the morphism you want to describe is the multiplication with $\eta_{\alpha}$ over $U_{\alpha}.$ $\endgroup$ Nov 8, 2012 at 7:04
  • $\begingroup$ I realize maybe there can't be an explicit description of $s_D$ in terms of $\eta_\alpha$, because in fact the $\eta_alpha$'s are determined up to a coboundary for $\mathcal{O}_X^{\;*}$. For $X=\mathbb{P}^1$ and $D=\{0\}$, how can we describe $s_D$ in terms of homogeneus coordinates $[x_0:x_1]$ on $\mathbb{P}^1$? $\endgroup$
    – Simplicius
    Nov 9, 2012 at 9:19
  • $\begingroup$ Well, in this case we can take $\eta_0=x_0$ and $\eta_1=1$ (the constant function $1$ on $U_1$). So $\psi_{01}=x_0$ on $U_{01}=\mathbb{C}\setminus\{0\}$. So I guess it's $s_D=1/x_0$, thought of as a global rational function on $\mathbb{P}^1$. As I remarked in the question, $s_D|_{U_\alpha}$ cannot be equal to $\eta_\alpha^{-1}$ for every $\alpha$: indeed $s_D|_{U_1}=1/x_0\neq 1=\eta_1^{-1}$. Is that all right? $\endgroup$
    – Simplicius
    Nov 9, 2012 at 9:30
  • $\begingroup$ Conjecture: $s_D$ is in fact determined only up to global invertible functions $\lambda\in\mathcal{O}_X^{\;*}(X)$, and, in terms of the $\eta_\alpha$'s, $s_D$ is obtained as follows: fix an $\alpha$ and consider $\eta_\alpha$ as a global rational function on $X$, then, for any $\beta$, $s_D|_{U_\beta}=\eta_\alpha^{-1}|_{U_\beta}$. $\endgroup$
    – Simplicius
    Nov 9, 2012 at 9:35

1 Answer 1


Amazingly, the canonical section of $\mathcal{O}_X(D)$ is the constant 1: $$s_D=1 \in\Gamma(X,\mathcal{O}(D))\subset K(X)$$ Indeed we have isomorphisms of sheaves (=trivializations) of $\mathcal O_{U_\alpha}$-modules $$g _\alpha:\mathcal{O}_X(D)|U_\alpha=\frac {1}{\eta_\alpha}\mathcal O_{U_\alpha}\stackrel {\cong}{\to} \mathcal O_{U_\alpha}:t\mapsto t\cdot \eta_\alpha$$ and $g_{\alpha\beta}=g_\alpha\circ g_\beta^{-1}=\frac{\eta_\alpha}{\eta_\beta}$ is the transition cocycle associated to the covering $(U_\alpha)$ for the line bundle $\mathcal{O}(D)$.
[my convention is the inverse of yours because usually trivializations go from the restriction of the bundle to the trivial bundle while you go in the opposite direction]

So the section $1|U_\alpha=s_D|U_\alpha \in \Gamma(U_\alpha,\mathcal{O}(D))$ is sent by $g_\alpha$ to $1\cdot \eta_\alpha=\eta_\alpha\in \Gamma(U_\alpha,\mathcal{O})$ and of course we have $\eta_\alpha=g_{\alpha\beta}\cdot\eta_\beta$ on $U_{\alpha\beta}$ by the definition of $g_{\alpha\beta}$.

To sum up, the canonical section $s_D=1 \in\Gamma(X,\mathcal{O}(D))\subset K(X)$ is represented in the given trivializations $g_\alpha$ of $\mathcal O(D)$ over the $U_\alpha$'s by the family of regular functions $\eta_\alpha\in \Gamma(U_\alpha,\mathcal{O}) $


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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