# Vector fields as sections of O(2) on P^1(C)

by Reginald Anderson   Last Updated September 11, 2019 18:20 PM - source

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: $$x_0^2\longmapsto v_0; \quad x_0x_1\longmapsto x_{1/0}v_0; \quad x_1^2\longmapsto x_{1/0}^2v_0.$$