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We have that the line bundle associated to the sheaf $\mathcal{O}_{\mathbb{P}_{\mathbb{C}}^1}(1)$ is given by: $$ \mathbb{P}_{\mathbb{C}}^2\setminus\{(0:0:1)\} \rightarrow \mathbb{P}_{\mathbb{C}}^1, (x_0:x_1:x_2)\mapsto (x_0:x_1). $$ This corresponds the projection from $(0:0:1)$ to the projective line $\mathbb{P}_{\mathbb{C}}^1 \simeq V(x_2)\subset \mathbb{P}_{\mathbb{C}}^2$.

I was wondering If we can find such an explicit description for the line bundle associated to $\mathcal{O}_{\mathbb{P}^1_{\mathbb{C}}}(2)$. In other words, I would like to find a variety $L$ and a map $\pi:L\rightarrow \mathbb{P}_{\mathbb{C}}^1$ such that $(L,\pi)$ is a line bundle on $\mathbb{P}_{\mathbb{C}}^1$ and its sheaf of sections is $\mathcal{O}_{\mathbb{P}_{\mathbb{C}}^1}(2)$. Does the map $\pi$ have a geometric interpretation like before?

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    $\begingroup$ Over $\Bbb R$ there is a very nice geometric interpretation, $\mathcal O(-1)$ is the Moebius band and $\mathcal O(k)$ is a $\Bbb R$ bundle twisted $k$ time, i.e the cylinder if $k$ is even and the Moebius band if $k$ is odd. $\endgroup$
    – Nicolas Hemelsoet
    Oct 6, 2017 at 11:47
  • $\begingroup$ @NicolasHemelsoet Thank you! That is the kind of thing that I am looking for, but I am working with complex numbers. I will add it to the question. $\endgroup$
    – BenicioLima
    Oct 6, 2017 at 13:22
  • $\begingroup$ Yes I know but I am not sure there is an interpretation for $O(k)$ other than $k=1,k=-1$. You probably also know it but for $k=-1$ you get the blow-up of $\Bbb C^n$ at a point. My personal intuition is that $\mathcal O(2) = \mathcal O(1) \otimes \mathcal O(1)$ and tensoring a line bundle is kind of twisting the total space (this is very vague of course). $\endgroup$
    – Nicolas Hemelsoet
    Oct 6, 2017 at 13:28
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    $\begingroup$ Try looking in Daniel Huybrechts' Complex Geometry. He describes $\mathcal O(1)$ as the space of linear forms on the total space of $\mathcal O(-1)$, and its tensor powers as the spaces of quadratic, cubic, etc. forms on the same total space. $\endgroup$
    – Tabes Bridges
    Oct 6, 2017 at 17:32
  • $\begingroup$ @TabesBridges : This is nice ! You should probably put it as an answer. $\endgroup$
    – Nicolas Hemelsoet
    Oct 6, 2017 at 17:40

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At Nicolas' recommendation, I will open up Huybrechts and expand a bit (this is on p. 91). I will use $\mathcal O$ to denote the total space of the trivial line bundle, and the usual variations to denote the associated tensor bundles. We have the inclusion $$ \mathcal O(-1) \subset \mathcal O^{\oplus n+1}, $$

where every fiber of the trivial bundle is the $\mathbb C^{n+1}$ of which we get $\mathbb P^n$ as a quotient, and the fiber of $\mathcal O(-1)$ over $[\ell] \in \mathbb P^n$ is $\ell \subset \mathbb C^{n+1}$. Since any inclusion of bundles induces an inclusion of tensor powers, we have $$ \mathcal O(-k) \subset \mathcal O^{\oplus k(n+1)}. $$

Now any $F \in \mathbb C[z_1,...,z_{n+1}]_k$ induces a holomorphic map $\tilde{F}: O^{\oplus k(n+1)} \to \mathbb C$ which is linear on all fibers of $\mathcal O^{\oplus k(n+1)} \to \mathbb P^n$ (hence the same thing as an $\mathcal O$-linear map to $\mathcal O$). Restricting $\tilde{F}$ to $\mathcal O(-k)$ thus produces a holomorphic section of $\mathcal O(k)$ by definition: recall that $\mathcal O(k) = \mathcal Hom(\mathcal O(-k),\mathcal O)$. There are a bunch of details I'm leaving out (in particular you have to use power series to prove that the homogeneous holomorphic functions you obtain are in fact algebraic), but this hopefully conveys some of the geometry you were looking for.

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  • $\begingroup$ Dear Tabes Bridges, first of all I wanted to thank you for the answer. Now, maybe I am wrong, but it looks like this construction helps to prove that $H^0(\mathbb{P}^1_{\mathbb{C}},\mathcal{O}_{\mathbb{P}^1_{\mathbb{C}}}(2))\simeq \mathbb{C}[x_0,x_1]_2$. Nevertheless, I do not get to see how it could help to find a total space for the line bundle we are interested in. $\endgroup$
    – BenicioLima
    Oct 9, 2017 at 8:40
  • $\begingroup$ So yes, this gives you an explicit picture of the total space of all the negative line bundles, but not the positive ones. To get what you want, at least in the case of bundles on $\mathbb P^1$, first note that your description of $\mathcal O(1)$ is equivalent to first blowing up the center of the projection, then deleting the exceptional divisor. In this case, we can think of the total space as $\mathbb P(\mathcal O(1) \oplus \mathcal O)$ with the section at infinity corresponding to the quotient $\mathcal O(1) \oplus \mathcal O \to \mathcal O \to 0$ deleted... $\endgroup$
    – Tabes Bridges
    Oct 9, 2017 at 18:30
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    $\begingroup$ ...so more generally, take the Hirzebruch surface $\mathbb F_n$ (aka $\mathbb P(\mathcal O(n) \oplus \mathcal O)$) and delete the section at infinity, which corresponds to projection onto the second summand. $\endgroup$
    – Tabes Bridges
    Oct 9, 2017 at 18:32

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