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

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

• Let me know if I need any modification to my question. It seems like I put so many text in a single thread. Do I need to split the question? Any suggestion will be appreciated. Thanks Commented Mar 23 at 16:01
• I haven't read all of it but this is nowhere near the scale of a "too long" question imo Commented Mar 23 at 19:43
• @N00BMaster It seems we read the same reference at the moment. Have you looked the sections of this line bundle ? math.stackexchange.com/questions/4887317/… Commented Mar 25 at 17:03

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 $$\ell=(z_0:\dots:z_n)\in U_i\cap U_j$$.

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.

• Thank you very much for your answer. Now I got the clear picture, @Threeaggies. Could you confirm if my lifting for sections $\sigma_i$ and $\sigma_j$ is correct? Commented Mar 24 at 5:21
• I edited the post, and hopefully it helps. It takes a while to get used to such calculations. But keep in mind that you need to define concrete numbers from a line $\ell$ with homogenous coordinates, so usually quotients of coordinates are required. Also the local sections need to patch up by the transition functions $\psi_{ij}$ to define something global. Then some sections are holomorphic and some are meromorphic, depending on if you divide by functions with zeros. Commented Mar 24 at 6:16
• Thank you for the detailed response. Yes, these calculations can be quite challenging, especially when you're working through them yourself. I'm considering delving into characteristic classes (specifically Chern classes) for $\mathcal{O}(-1)$ and $\mathcal{O}(1)$ to compare the differences between these bundles. Could you recommend a textbook or resource that provides detailed calculations for the projective case? Thanks again. @Threeaggies Commented Mar 24 at 6:39
• I learned such things from Griffiths & Harris, but only after a long time they appear natural to me. Still, their book is more concrete than some others, and section 1.1 there can be useful. Commented Mar 24 at 7:03
• Sorry for late accepting. I was going through my exam. When I try to generalize this for $\mathcal O(-k)$ or $\mathcal O(k)$ I stuck on. Like, $$\psi_j^{-1}: U_j\times {\mathbb C}^k\to \pi^{-1}(U_j);\ (\ell, w_1,\cdots,w_k)\to (\ell, z),$$ for me it hard to visualize $\pi^{-1}(U_j)$ and get the transition function $\psi_{ij}$ for this case. it will be a great help if you give me some insight on how to understand the generalization? @Threeaggies Commented Apr 24 at 9:45