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The tangent bundle of $\mathbb{P}^1(\mathbb{C})$ is given by $\mathcal{O}_{\mathbb{P}^1}(2)$. A section of $\mathcal{O}_{\mathbb{P}^1}(2)$ is, say, $x_0^2$, where $[ x_0 : x_1]$ form homogeneous coordinates on $\mathbb{P}^1$. My question is: which vector field is it? Can one draw the vector field, say, by viewing $\mathbb{P}^1$ as $S^2$, or, perhaps, by looking in one open chart?

An answer to my question would either supply a vector field that I can actually draw on $\mathbb{C}$, and would generalize to $\mathbb{P}^1(\mathbb{R})$ so that we can actually draw a vector field on the real line corresponding to (the homogeneous monomial of total degree 2) $x_0^2$, or show that this is not possible (maybe a vector field is not well-defined from homogeneous coordinates). Thanks

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This can be computed on affine coordinates in a way similar to computing global sections of $\mathcal O_{\mathbb P^1}(n)$. Let $x_{1/0}=x_1/x_0$ and $x_{0/1}=x_0/x_1$ be the coordinates on the standard affine open sets $U_i=\{x_i\neq 0\}$. A vector field on $\mathbb A^1_{x_{1/0}}$ takes the form $v=f(x_{1/0})\frac{\partial}{\partial x_{1/0}}$. When we write $v$ in terms of $x_{0/1}=x_{1/0}^{-1}$, the chain rule says: $$ v= f(x_{0/1}^{-1})\frac{\partial}{\partial (x_{0/1}^{-1})} = f(x_{0/1}^{-1})\frac{1}{\partial (x_{0/1}^{-1})/\partial x_{0/1}} \cdot \frac{\partial}{\partial x_{0/1}} =-x_{0/1}^2 f(x_{0/1}^{-1}) \frac{\partial}{\partial x_{0/1}}. $$ So in order for $v$ to not have a pole at $[0:1]$, $f(x_{0/1}^{-1})$ can have a pole of order at most $2$, i.e. $f(x_{1/0})$ is a polynomial of degree at most $2$. Therefore, the global sections of the tangent bundle have this basis: $$ v_0=\frac{\partial}{\partial x_{1/0}} = -x_{0/1}^2 \frac{\partial}{\partial x_{0/1}};x_{1/0}v_0=x_{1/0}\frac{\partial}{\partial x_{1/0}} = -x_{0/1} \frac{\partial}{\partial x_{0/1}};x_{1/0}^2v_0=x_{1/0}^2\frac{\partial}{\partial x_{1/0}} = - \frac{\partial}{\partial x_{0/1}}. $$ And the tangent bundle is isomorphic to $\mathcal O_{\mathbb P^1}(2)$ by a map that takes: $$ x_0^2\longmapsto v_0; \quad x_0x_1\longmapsto x_{1/0}v_0; \quad x_1^2\longmapsto x_{1/0}^2v_0. $$

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  • $\begingroup$ This is beautiful, thank you. $\endgroup$ – Reginald Anderson Sep 12 at 18:33

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