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We know that the complex line bundles over $\Bbb{CP}^1$ are classified by the integers. Each is isomorphic to one of the bundles $\mathcal{O}(n)$ for $n\in\Bbb Z$, where $\mathcal{O}(-1)$ is the tautological line bundle.

Question: In which class does the following line bundle belong?

Let $U_i\subseteq\Bbb{CP}^1$ be the open set of elements $[z_1,z_2]\in\Bbb{CP}^1$ such that $z_i\neq 0$. Let $L$ be the line bundle on $\Bbb{CP}^1$ defined by the transition function $$\psi_{12}:U_1\cap U_2\to \mathrm{GL}(1,\Bbb C),\quad \psi_{12}([z_1,z_2])=\frac{z_1/z_2}{|z_1/z_2|}.$$

We have that $\mathcal{O}(n)$ is the vector bundle with transition function $$\psi_{12}:U_1\cap U_2\to \mathrm{GL}(1,\Bbb C),\quad \psi_{12}([z_1,z_2])=(z_1/z_2)^n,$$ so our line bundle $L$ is not directly one of those. It should be isomorphic to one of those but I cannot find any explicit isomorphism.

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  • $\begingroup$ Unless I'm mistaken, the map $\psi_{12}$ is not holomorphic, so the corresponding line bundle is not a holomorphic one. $\endgroup$ – Michael Albanese Dec 29 '15 at 10:13
  • $\begingroup$ @MichaelAlbanese That's right, but it is a fact that every complex line bundle is isomorphic (in the real sense) to one of the holomorphic line bundles $\mathcal{O}(n)$. $\endgroup$ – Simon Parker Dec 29 '15 at 10:14
  • $\begingroup$ Sorry, whenever I see the notation $\mathcal{O}(n)$, I think of holomorphic line bundles. $\endgroup$ – Michael Albanese Dec 29 '15 at 10:17
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A complex line bundle on $\mathbb{CP}^1$ is completely determined by the transition map $\psi_{12} : U_1\cap U_2 \to GL(1, \mathbb{C})$, but two different transition maps could lead to isomorphic bundles. When does this occur? Precisely when the transition maps are homotopic through transition maps.

The result mentioned above implies the that there is a natural bijection between the set of isomorphism classes of complex line bundles on $\mathbb{CP}^1$ and the set of homotopy clases of maps $U_1\cap U_2 \to GL(1, \mathbb{C})$. As $U_1\cap U_2$ and $GL(1, \mathbb{C}) = \mathbb{C}\setminus\{0\}$ both deformation retract onto $S^1$, the set of homotopy classes of maps $U_1\cap U_2 \to GL(1, \mathbb{C})$ is the same as the set of homotopy classes of maps $S^1 \to S^1$ which can be identified with $\pi_1(S^1) \cong \mathbb{Z}$. So for each integer $n$, there is a corresponding complex line bundle, namely $\mathcal{O}(n)$.

So to determine which line bundle $L$ is isomorphic to, we just need to check which of the transition maps you mentioned $\psi_{12}$ is homotopic to. The map $H : (U_1\cap U_2)\times[0, 1] \to GL(1, \mathbb{C})$ given by

$$H([z_1, z_2], t) = \frac{z_1/z_2}{(1-t)|z_1/z_2| + t}$$

provides a homotopy (through transition maps) between $\psi_{12}$ and the map $[z_1, z_2] \mapsto z_1/z_2$. Therefore $L$ is isomorphic to $\mathcal{O}(1)$.

More generally, there is a natural bijection between the set of homotopy classes of maps $S^{k-1} \to GL_n(\mathbb{C})$ and the set of isomorphism classes of rank $n$ complex vector bundles on $S^k$ (see Proposition $1.11$ of Hatcher's Vector Bundles & K-Theory).

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  • $\begingroup$ Nice ! Can we use this to construct explicit local trivializations $\pi^{-1}(U_i)\to U_i\times\Bbb C$ of the tautological line bundle $\mathcal{O}(-1)$ whose corresponding transition map is $\frac{z_2/z_1}{|z_2/z_1|}$? $\endgroup$ – Simon Parker Dec 29 '15 at 11:44
  • $\begingroup$ You can do this, but I don't have time to explain it now. Hopefully I can do it tomorrow. $\endgroup$ – Michael Albanese Dec 29 '15 at 11:47
  • $\begingroup$ Thank you very much. I hope you will have time; I am looking forward to it! Any hint would be greatly appreciated. $\endgroup$ – Simon Parker Dec 29 '15 at 11:49
  • $\begingroup$ Hello. Have you got the chance to think of this problem again? I tried many things to construct a trivialization that has this transition map, but nothing works. Can we actually have an explicit formula, or we just know it exists without having one? $\endgroup$ – Simon Parker Dec 31 '15 at 12:23
  • $\begingroup$ Sorry, I'll try to do it tomorrow. $\endgroup$ – Michael Albanese Dec 31 '15 at 15:09

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