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Let $\mathbb{C} \mathbb{P}^1 $ the complex projective space and let consider the tautological bundle over $\mathbb{C} \mathbb{P}^1 $:

In the excerpt below (whole document: here ) the map belonging to tautological bundle of $\mathbb{C} \mathbb{P}^1 $ is defined as the canonical map $s: \mathbb{C}^2 \backslash \{0\} \to \mathbb{C} \mathbb{P}^1 $ which assoziates every $(x_0:x_1)$ in canonical way the fiber $s^{-1}(x_0:x_1) = \mathbb{C}*(x_0,x_1)$, therefore the corresponding one dimensional vector space.

My question is how does this constuction yields $\mathcal{O}(-1)$ as line bundle, therefore a twisted sheaf with empty global sections (because of $-1$)?

Source: enter image description here

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  • $\begingroup$ Dear KarlPeter, what is exactly your question ? $\endgroup$
    – Nicolas Hemelsoet
    Nov 3, 2017 at 21:53
  • $\begingroup$ That's not clear to me how the author (re)constructs $\mathcal{O}(-1)$ from the given morphism $s$ associated to tautological bundle. $\endgroup$
    – user267839
    Nov 3, 2017 at 22:00
  • $\begingroup$ Or more precisely: How does $\mathcal{O}(-1)$ correspond to the map $\mathbb{A}^2 \backslash \{0 \} \to \mathbb{P}^1, (a,b) \to(a:b)$? $\endgroup$
    – user267839
    Nov 3, 2017 at 22:10
  • $\begingroup$ Dear KarlPeter thanks for your edit and the precisions, I wrote an answer I hope it is useful. $\endgroup$
    – Nicolas Hemelsoet
    Nov 3, 2017 at 22:11

1 Answer 1

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First the map $s$ should from $\Bbb C^2 \backslash \{(0,0) \}$ to $\Bbb CP^1$. And this is not exactly this map which gives $\mathcal O(-1)$ as the fiber gives a punctured line for all $[x_0:x_1] \in \Bbb CP^1$.

Your reference didn't wrote the map precisely, so let's do it : the correct construction is to take the projection $X \to \Bbb CP^1$ where $X = \{(z,l) \in \Bbb C^2 \times \Bbb CP^1 : z \in l \}$. Here, the fiber is canonically a line as you said. This shows that $\mathcal O(-1)$ is a line bundle. For see that indeed it has no sections, the simplest thing is to write down the condition with the cocyles.

Remark : it is also interesting to see that the projection $X \to \Bbb C^2$ is the blow-up of $\Bbb C^2$ at the origin.

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  • $\begingroup$ Thank you for the answer. One conclusion isn't clear to me: How do you deduce from the fact that the fibers are lines, that $\mathcal{O}(-1)$ is a line bundle? I don't see the relation between $X$ and $\mathcal{O}(-1)$. $\endgroup$
    – user267839
    Nov 3, 2017 at 22:25
  • $\begingroup$ @KarlPeter : this is a bit more precise that this, the fiber over $[x_0:x_1]$ is canonically identified with $\{z \in \Bbb C^2 : z \in [x_0:x_1]\}$, so the line over $[x_0:x_1]$ (considered as point in $\Bbb CP^1$) is exactly $[x_0:x_1]$ (considered as a line in $\Bbb C^2$) ! This is the definition of $\mathcal O(-1)$. Now, you're right that it's not finished yet, for verifying that it is a line bundle I should check that my bundle is locally trivial. On $U = \Bbb C \subset \Bbb CP^1$ you can write a bundle isomorphism $\mathcal O(-1)_{|U} \to U \times \Bbb C$ : (to be continued) $\endgroup$
    – Nicolas Hemelsoet
    Nov 3, 2017 at 22:30
  • $\begingroup$ We take $U : \{x_1 = 1\}$ and the isomorphism $\mathcal O(-1)_{|U} \to U \times \Bbb C$ is given by $([x_0:1], \lambda(x_0,1)) \mapsto (x_0, \lambda)$. $\endgroup$
    – Nicolas Hemelsoet
    Nov 3, 2017 at 22:32
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    $\begingroup$ Yes, it becomes slowly clearer. One puzzle piece is missed. It refers to the equivalence (?) of the definitions of $ \mathcal {O}(1) $. So on the one hand $ \mathcal {O}(1):= \mathcal {O}(-1)^V$ where $ \mathcal {O}(-1)^V$ as you wrote above and the other one as introduced in Liu’s AG. Here I opened a new thread treating this question separately: math.stackexchange.com/questions/2504257/… $\endgroup$
    – user267839
    Nov 4, 2017 at 14:03
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    $\begingroup$ @KarlPeter : good ! If no one will answer to you I can write you something this afternoon or this evening. $\endgroup$
    – Nicolas Hemelsoet
    Nov 4, 2017 at 14:05

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