I know that the holomorphic tangent bundle to the complex projective line $T^{1,0}\mathbb{CP}^1$ is isomorphic to $\mathcal{O}(2)$, i.e. the twofold tensor product of the hyperplane bundle of $\mathbb{CP}^1$.
Question: how do I identify under this correspondence the global holomorphic vector fields as sections of this $\mathcal{O}(2)$?
Setup: the isomorphism between these line bundles can be seen by giving $\mathbb{CP}^1$ the homogeneous atlas given by
$$
U_0=\{z_0\neq 0\}\ni (z_0:z_1) \mapsto \frac{z_1}{z_0}=z \in \mathbb{C}
$$
$$
U_1=\{z_1\neq 0\}\ni (z_0:z_1) \mapsto -\frac{z_0}{z_1}=w \in \mathbb{C}
$$
thus the change of coordinates from $z$ to $w$ reads $-\frac{1}{z}$ and the jacobian, which gives the transition functions for $T^{1,0}\mathbb{CP}^1$ is
$$
g_{01}(z) = \frac{1}{z^2} \equiv \left(\frac{z_0}{z_1}\right)^2
$$
This is precisely the transition function of $\mathcal{O}(2)$!
I already know that global holomorphic sections correspond to homogeneous polynomials in two variables
$$H^0(\mathbb{CP}^1,\mathcal{O}(2))\simeq \langle Z_0^2, Z_0Z_1, Z_1^2\rangle$$
As far as I know, the "derivatives" $\frac{\partial}{\partial z},\frac{\partial}{\partial w}$give global vector holomorphic vector fields over $\mathbb{CP}^1$. But what is the third global holomorphic vector field? How can it be linearly independent from these derivative fields?
Best Answer
$\newcommand{\dd}{\partial}$In the $z$-chart, every holomorphic vector field has the form $$ X = (a_{0} + a_{1}z + a_{2}z^{2}) \frac{\dd}{\dd z}. $$ Since $z = \dfrac{1}{w}$ where both coordinates are non-zero, we have $$ \frac{\dd}{\dd w} = \frac{\dd z}{\dd w}\, \frac{\dd}{\dd z} = -\frac{1}{w^{2}}\, \frac{\dd}{\dd z} = -z^{2}\, \frac{\dd}{\dd z} $$ in the overlap, and therefore $$ X = -(a_{0}w^{2} + a_{1}w + a_{2}) \frac{\dd}{\dd w}. $$ In other words, the space of holomorphic vector fields is spanned by $$ \frac{\dd}{\dd z} = -w^{2}\, \frac{\dd}{\dd w},\qquad z\, \frac{\dd}{\dd z} = -w\, \frac{\dd}{\dd w},\qquad z^{2}\, \frac{\dd}{\dd z} = -\frac{\dd}{\dd w}. $$