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I'm trying to work out some simple examples demonstrating the correspondence between line bundles and divisors.

The toric variety $X = \mathrm{Bl}\,_0(\mathbb{C}^2/\mathbb{Z}_2)$ has quotient construction $X=[\mathbb{C}^3\setminus\mathbb{V}(x_1,x_2)]/\mathbb{C}^*$ where $\mathbb{C}^*$ acts via $\lambda\cdot(a_0,a_1,a_2)=(\lambda^{-2}a_0,\lambda a_1,\lambda a_2)$. I'm pretty sure the morphism $\pi:X\rightarrow\mathbb{P}^1$ given by $\pi(a_0:a_1:a_2)=(a_1:a_2)$ makes $X$ a line bundle over $\mathbb{P}^1$ (drawing the picture of the morphism between fans seems to make this obvious). Now, line bundles over $\mathbb{P}^1$ correspond to the sheaves $\mathscr{O}_{\mathbb{P}^1}(\alpha)$ for $\alpha\in\mathbb{Z}$. How do I determine which sheaf the line bundle I've constructed corresponds to? Here's what I have so far:

Using the standard open cover $\mathbb{P}^1=U_1\cup U_2$ where $U_i = \{(a_1:a_2)\in\mathbb{P}^1:a_i\neq0\}$, there are isomorphisms $\phi_i:\pi^{-1}(U_i)\rightarrow\mathbb{C}\times U_i$ given by \begin{align*} \phi_1(a_0:a_1:a_2) &= \phi_1\left(a_1^2 a_0:1:\frac{a_2}{a_1}\right)=a_1^2a_0\times\left(1:\frac{a_2}{a_1}\right) \\ \phi_2(a_0:a_1:a_2) &= \phi_2\left(a_2^2 a_0:\frac{a_1}{a_2}:1\right)=a_2^2a_0\times\left(\frac{a_1}{a_2}:1\right) \\ \end{align*} The transition functions are $\displaystyle g_{ij}=\frac{x_i^2}{x_j^2}$. Where should I go from here?

Brian Fitzpatrick
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  • Is what I have written correct? If so I guess I'm having trouble seeing how $\mathscr{O}_{\mathbb{P}^1}(2)$ relates to these transition functions. – Brian Fitzpatrick Dec 07 '13 at 03:31
  • Dear @Brian it always confuses me, whether it's $\mathcal{O}(-2)$ or $\mathcal{O}(2).$ This case, I think, it's $\mathcal{O}(2).$ Unfortunately, I don't have time but you can take a look at this http://math.stackexchange.com/questions/112926/global-sections-of-mathcalo-1-and-mathcalo1-understanding-structu – Ehsan M. Kermani Dec 07 '13 at 03:42
  • The fact is that the projection map $\mathbb{C}^2 \to \mathbb{C}^2/\mathbb{Z}_2$ is a degree two map (away from the origin) (you can see it from the transition functions), so the lifting to their blow ups at the origin, gives you a map $Bl_0(\mathbb{C}^2) \to Bl_0(\mathbb{C}^2/\mathbb{Z}_2).$ But now, $Bl_0(\mathbb{C}^2) \to \mathbb{C}^2$ is $\mathcal{O}(1)$ if I'm not mistaken, so $Bl_0(\mathbb{C}^2/\mathbb{Z}_2) \to \mathbb{C}^2/\mathbb{Z}_2$ is $\mathcal{O}(2).$ – Ehsan M. Kermani Dec 07 '13 at 03:47

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