3
$\begingroup$

The line bundle $\mathcal O_{\mathbb CP^1}(-1) = \{(z,\ell)|~z \in \ell \} $ is a submanifold of $\mathbb C^2 \times \mathbb CP^1$ with bundle map the projection. Thus we can the restrict the coordinates of $\mathbb C^2 \times \mathbb CP^1$ to coordinates of $\mathcal O(-1)$.

What about its powers? Are they also such submanifolds? I was thinking that maybe $$\mathcal O_{\mathbb CP^1}(-n) = \{((z_0^n, z_1^n),[z_0,z_1])\}\subset \mathbb C^2 \times \mathbb CP^1.$$

This would give the right transition functions $(\frac {z_1}{z_0})^n$. But I don't think that it is well defined, because the function $[z_0: z_1] \mapsto (z_0^n, z_1^n)$ is not injective (look at the roots of unity).

So how would one define coordinates on $\mathcal O_{\mathbb CP^1}(-n)$? I am especially interested in the case $n=2$.

$\endgroup$
0
$\begingroup$

One way of constructing coordinates on $\mathcal{O}(n)$ is to observe that $\mathcal{O}(-1)$ is a line bundle, so a local section $s$ over $U \subset \mathbb{CP}$ will trivialise the bundle over $U$.

Taking tensor powers of $s$ gives a local section of $\mathcal{O}(-n) = \mathcal{O}(-1)^{\otimes(n)}$; i.e. $s^{\otimes n}$ trivialises $\mathcal{O}(-n)$ over $U$.

This construction will show that in the intersection of coordinate charts on $M$, we get the desired transition functions, as the tensor product of two 1 dimensional matrices is in fact just scalar multiplication.

As for the first part of your question, I believe your description is valid. Indeed, they are precisely what I described in the above.

If you can't see what I mean, consider $z$ as a section over $U$, where $U$ the coordinate chart \begin{align} \phi: U \subset\mathbb{CP} &\to \mathbb{C} \\ [l:1] &\mapsto l \end{align} If $z: U \to \mathcal{O}(-1)$ is a local section $z \in [l:1] \implies z(l)= c(l)(l, 1)$ where $c: \mathbb{C} \to \mathbb{C}$. A tensor power of this section would then be $c(l)^n(l^n, 1)$ as required.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.