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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!

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  • $\begingroup$ No, be careful. $\sigma$ is not a meromorphic function; it is a meromorphic section of the line bundle. $\endgroup$ Commented Jul 15, 2023 at 19:14
  • $\begingroup$ 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 $(*)$. $\endgroup$
    – Invariance
    Commented Jul 16, 2023 at 8:02

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The mistake I made is that I confused with the local frame and local coordinate.

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 $(*)$.

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