# canonical meromorphic section of a line bundle associated with a divisor

I am reading this paper A logarithmic $$\bar\partial$$-equation on a compact Kähler manifold associated to a smooth divisor. The setting is as follows: let $$X$$ be a compact complex manifold and let $$D=\sum_{i=1}^{r} D_i$$ be a simple normal crossing divisor on $$X$$, namely, $$D_i$$, $$1\leq i\leq r$$, are smooth hypersurfaces on $$X$$ and intersect with each other transversely. Therefore, For any $$z \in X$$, which $$k$$ of these $$D_i$$ pass, we may choose local holomorphic coordinates $$\left\{z^1, \cdots, z^n\right\}$$ in a small neighborhood $$U$$ of $$z=(0, \cdots, 0)$$ such that $$D \cap U=\left\{z^1 \cdots z^k=0\right\}$$ is the union of coordinates hyperplanes.

Now, let $$L$$ be a holomorphic line bundle over $$X$$ satisfying $$L^N=\mathcal{O}_X\left(\sum_{i=1}^r a_i D_i\right)$$ for some $$a_i \in \mathbb{Z}, 1 \leq i \leq r$$. Then, in page 6, the author argues as follows:

Let $$\sigma$$ be the canonical meromorphic section of $$\mathcal{O}_X\left(\sum_{i=1}^r a_i D_i\right)$$, and let $$e$$ be a local frame of $$L$$ satisfying $$e^N=\prod_{i=1}^r\left(z^i\right)^{-a_i} \sigma .\,\,\,\,\,\,\,\,\,\,(*)$$

I am confused about $$(*)$$, recall that a canonical meromorphic section $$\sigma$$ (see for example this post canonical meromorphic section)is a global meromorphic function such that $$\text{div}(\sigma)|_U+\sum_{i=1}^{r}a_iD_i|_U\geq 0$$ So, shouldn't $$(*)$$ be $$e^N=\prod_{i=1}^r\left(z^i\right)^{a_i} \sigma?$$

What am I missing or is it just a mistake? Thanks for any possible suggestions!

• No, be careful. $\sigma$ is not a meromorphic function; it is a meromorphic section of the line bundle. Commented Jul 15, 2023 at 19:14
• Hi @TedShifrin, I think the mistake I made is that I confused with the local frame and local coordinate. Here is my thought, please tell me the mistakes I made if there exists any, thanks a lot! Assume that we have a collection of frames/basics $\left\{\widetilde e_i\right\}$ of $L^N$ on $X=\left\{U_i\right\}$, then we should have $\widetilde e_i=\phi_{ij}^{-1}\widetilde e_j$. Thus, following the notations in my question $e^N:=\widetilde e_i=\frac{\sigma}{\prod_{i=1}^r\left(z^i\right)^{a_i}}$, which is $(*)$. Commented Jul 16, 2023 at 8:02

Now assume that we have a collection of frames/basics $$\left\{\tilde{e}_i\right\}$$ of $$L^N$$ on $$X=\left\{U_i\right\}$$, then we should have $$\tilde{e}_i=\phi_{i j}^{-1} \tilde{e}_j$$. Thus, following the notations in my question $$e^N:=\tilde{e}_i=\frac{\sigma}{\prod_{i=1}^r\left(z^i\right)^{a_i}}$$, which is just $$(*)$$.