Homogeneous polynomials on complex projective line

129 Views Asked by At

I am working with some (holomorphic) sections of line bundles over the complex projective line $\mathbb{CP}^1$, these can be seen to be taken as homogeneous polynomials in two complex variables $$ p \in H^0(\mathbb{CP}^1,\mathcal{O}(d)) \simeq S^d(\mathbb{C}^2)^*. $$ I want to write them in some canonical form by changing my homogenoeus coordinates, ideally writing $$ p' = g^*p = a x_0^k x_1^{d-k} $$ where $g$ is a change of coordinates (I think an element of $PGL(2,\mathbb{C})$). Two questions:

Question 1: is it possible to do this? Is there a constructive manner? I have been playing around with some low degree polynomials but I get stuck at elements like $$p=x_0^3 - x_1^3 = (x_0 - x_1 \xi_1)(x_0 - x_1 \xi_2)(x_0 - x_1 \xi_3),$$ where $\xi_i$ are the 3 cube roots of 1.

Question 2: if such a thing is always possible... how many polynomials can I write in my preferred canonical form with a single change of coordinates? I think this has to do with the degree of transitivity of the action of $PGL(2,\mathbb{C})$ on $\mathbb{CP}^1$.

1

There are 1 best solutions below

2
On BEST ANSWER

This is not possible. The section $$x_0^k x_1^{d-k}$$ has a $k$-th root at $0$ and a ($d-k$)-th root at $\infty$. But as you noted, your polynomial $$p = x_0^3 - x_1^3$$ has three distinct roots $1, \xi$ and $ \xi^2$. If $$g: \mathbb{CP}^1 \to \mathbb{CP}^1$$ is any isomorphism, the polynomial $g^* p$ will also have three distinct roots, so it can never be of the form $x_0^k x_1^{d-k}$.