Local trivialization of $\mathcal O(-1)$, proposition 2.2.6, complex geometry by Huybrechts

Question

I was reading Complex Geometry by Daniel Huybrechts. On page 68, section 2.2 we have a proposition of holomorphic line bundle over $\mathbb P^n$,

Proposition 2.2.6: The projection $\pi:\mathcal O(-1)\rightarrow\mathbb P^n$ is given by projecting to the first factor. Let $\{U_i\}_{i=0}^n$ be an open covering of $\mathbb P^n$. A canonical trivialization of $\mathcal O(-1)$ over $U_i$ is given by, $$\psi_i:\pi^{-1}(U_i)\rightarrow U_i\times\mathbb C,\quad(\ell,z)\mapsto(\ell,z_i)$$ The transition maps $\psi_{ij}(\ell):\mathbb C\rightarrow\mathbb C$ are given by $w\mapsto \frac{z_i}{z_j}\cdot w$, where $\ell=(z_0:\cdots,z_n)$.

Suppose we have $(\ell,z^*)$ where $\ell$ belongs to $U_i$ and $z^*\in\mathbb C\setminus\{0\}$. In this scenario, I assumed that if we map $(\ell,z^*)$ using $\psi_j^{-1}$, it would look like this: $(\ell,z_0,\cdots,z_{j-1},z^*,z_{j+1},\cdots,z_n)$, inserting $z^*$ at position $j$. However, if this option doesn't hold, what alternatives should we consider? Because we need also to satisfy, $z\in \ell$. Now, $\psi_i(\ell,z_0,\cdots,z_{j-1},z^*,z_{j+1},\cdots,z_n)=(\ell,z_i)$. When we apply $\psi_i$ to $(\ell,z_0,\cdots,z_{j-1},z^*,z_{j+1},\cdots,z_n)$, fixing $\ell$, I noticed that the transition function $z^*\mapsto z_i$.

Question: What I got doesn't match with the book. Where did I make the mistake?

Similarly, I want to tackle the same issue for sections: $\sigma_i$ in $\Gamma(U_i,\mathcal O(-1))$ and $\sigma_j$ in $\Gamma(U_j,\mathcal O(-1))$. Here, we're lifting $\ell$ from $\mathbb P^n$ to $\mathcal O(-1)$. So, we have $$U_j\times\mathbb C\stackrel{\psi_j}{\leftarrow}\pi^{-1}(U_i\cap U_j)\stackrel{\psi_i}{\rightarrow}U_i\times\mathbb C$$ and, \begin{align} \sigma_i&:U_i\rightarrow\pi^{-1}(U_i)\stackrel{\psi_i}{\cong}U_i\times\mathbb C\\ \sigma_j&:U_j\rightarrow\pi^{-1}(U_j)\stackrel{\psi_j}{\cong}U_j\times\mathbb C \end{align} Now, I couldn't get how the section $\sigma_i$ lift $\ell=(z_0:\cdots,z_n)\in U_i\cap U_j$, one possible way maybe \begin{align} \sigma_i(\ell)&=\left((z_0:\cdots,z_n),\frac{z_0}{z_i},\cdots,\frac{z_n}{z_i}\right)\stackrel{\psi_i}{\rightarrow}\boxed{(\ell,1)}\\ \sigma_j(\ell)&=\left((z_0:\cdots,z_n),\frac{z_0}{z_j},\cdots,\frac{z_n}{z_j}\right)\stackrel{\psi_j}{\rightarrow}\boxed{(\ell,1)} \end{align} Because $\mathcal{O}(-1):=\{(\ell,z)\in \mathbb{CP}^n \times \mathbb{C}^{n+1}: z\in \ell\}$, that's why I think the lifting might be $\in\mathbb{CP}^n \times \mathbb{C}^{n+1}$ (Let me correct if my understanding is wrong). Then I assume the transition map might be $\frac{z_j}{z_i}$ because $\sigma_i\mapsto \frac{z_j}{z_i}\cdot\sigma_j$ while fixing $\ell$. I think I do mistake on the trivialization part $\boxed{(\ell,1)}$ (Does the $1$ reflect the "matrix representation" mentioned here?).

Question: I'm unsure if my computation is correct, can anyone verify that? And does the trivialization part play any role to get the transition maps?

My confusion arise since I haven't found any information on section for holomorphic bundles up until section 2.2, though I haven't covered everything since I'm studying the book on my own and skip most of the part which seems unfamiliar to me. It will be a great help if anyone suggest an answer or resource from where I can clear my understanding. TIA

Answer 1

Your $\psi_j^{-1}$ is not quite right. We have $$ \psi_j^{-1}: U_j\times {\mathbb C}\to \pi^{-1}(U_j);\ (\ell, w)\to (\ell, z), $$ where $\ell = (z_0:\dots:z_j:\dots:z_n)$ with $z_j\neq 0$ by $\ell\in U_j$, and $$ z=\frac{w}{z_j}(z_0, \cdots, z_j, \cdots, z_n) = \Big(\frac{z_0}{z_j}w,\cdots, w,\cdots, \frac{z_n}{z_j}w\Big). $$

So $z\in {\mathbb C}^{n+1}$ is the unique vector on the line $\ell$ whose $j$th component is $w$. We do this by the multiplication of a suitable scale. Also note that we changed the homogeneous coordinates using $:$ in $\ell$ to ordinary coordinates using $,$ in $z$.

Then we see that $$ \psi_{ij}=\psi_i\psi_j^{-1}: (\ell, w)\to (\ell, z)\to \Big(\ell, \frac{z_i}{z_j}w\Big), $$ since that is the $i$th component of $z$.

That is why Huybrechts writes $$ \psi_{ij}(\ell)(w)=\frac{z_i}{z_j}w, $$ where $\ell=(z_0:\dots:z_n)\in U_i\cap U_j$.

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Let me add a bit about sections. I don't think the $1$ you chose would work. A local section $$s_i: U_i\to \pi^{-1}(U)\overset{\psi_i}\to U\times {\mathbb C}$$ should satisfy $$s_i(\ell) = \psi_{ij}(\ell) s_j(\ell)$$ to patch up to get a global section. In the ${\mathcal O}(-1)$ case, $\psi_{ij}(\ell)=\frac{z_i}{z_j}$. There are no global holomorphic sections, in this case.

However if you consider ${\mathcal O}(1)\to {\mathbb P}^n$, then $$ \psi_{ij}(\ell)=\frac{z_j}{z_i}. $$ Then $$ s_i(\ell)=\frac{z_0}{z_i}, $$ where $\ell=(z_0:\dots:z_n)$ with $z_i\neq 0$, do define a holomorphic section as they patch. Indeed, $$ s_i(\ell)=\frac{z_0}{z_i} = \frac{z_j}{z_i}\frac{z_0}{z_j} = \psi_{ij}(\ell)s_j(\ell). $$ Instead of $z_0$, using $z_k$ for $0\leq k\leq n$ gives $n+1$ holomorphic sections, which form a basis of $$ \Gamma({\mathbb P}^n, {\mathcal O}(1)) = H^0({\mathbb P}^n, {\mathcal O}(1)) \ \text{ with dimension }\ h^0({\mathbb P}^n, {\mathcal O}(1))=n+1. $$

For the original ${\mathcal O}(-1)$, choosing say $$ s_i(\ell) = \frac{z_i}{z_0} $$ only gives a meromorphic section, since on $U_i$, $z_0$ still has zeros.