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I am stuck on one step that occurs without explanation in several Algebraic geometry books.

Starting from the exact sequence

$$0\rightarrow \Omega_{\mathbb{P}^n}\rightarrow \mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus n+1}\rightarrow \mathcal{O}_{\mathbb{P}^n}\rightarrow 0$$

it is concluded that $$\omega_{\mathbb{P}^n}=\wedge^n \Omega_{\mathbb{P}^n}\cong \mathcal{O}_{\mathbb{P}^n}(-n-1)$$

How does this follow and in particular how does $\Omega_{\mathbb{P}^1}\cong \mathcal{O}_{\mathbb{P}^1}(-2)$ follow ?

Thanks in advance.

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  • $\begingroup$ It is not too hard to work this out using homogeneous coordinates. $\endgroup$
    – S. Carnahan
    Jul 20, 2014 at 23:38
  • $\begingroup$ @S.Carnahan That is true, I was thinking there was some more conceptual proof. $\endgroup$ Jul 20, 2014 at 23:41
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    $\begingroup$ Explicitly: $dz_1 \wedge \cdots \wedge dz_n$ is a meromorphic form, with simple poles along the $n+1$ coordinate hyperplanes. (A good trick on any toric var!) $\endgroup$
    – user47305
    Jul 21, 2014 at 2:01

2 Answers 2

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det of the middle term of a short exact sequence is the tensor product of the dets of the left and right terms (det = top wedge).

The canonical bundle is det of \Omega, det of O is O.

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  • $\begingroup$ So in the case of $n=1$ I would take the second exterior product of the middle term ? I was thinking I should take the $n$ th exterior product of all terms. $\endgroup$ Jul 20, 2014 at 23:44
  • $\begingroup$ @ReneSchipperus: Hartshorne says in Example 8.20.1 of his book Algebraic geometry that "we take the highest exterior powers of the exact sequence (8.13)". This is the same as what the response says above. $\endgroup$
    – GH from MO
    Jul 20, 2014 at 23:49
  • $\begingroup$ @GHfromMO I see. I was taking all to the same power. Is there a reference for this fact ? $\endgroup$ Jul 20, 2014 at 23:54
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    $\begingroup$ @ReneSchipperus: I am no expert, but I think this should follow locally by linear algebra. I found a semi-reference in Hartshorne's book: part (d) of Exercise 5.16 on Page 128. $\endgroup$
    – GH from MO
    Jul 20, 2014 at 23:57
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    $\begingroup$ sorry for writing the answer in a hurry, I was bored and playing around with my phone while waiting for a friend in my car (I should have also added that det of any line bundle is the line bundle itself, but you already figure that out). A good summary of these facts for sheaves can be found in Goertz-Wedhorn (a marvellous book -- we are ALL waiting for Volume II!!!) section (7.20) and the relevant exercise is 7.30. $\endgroup$ Jul 21, 2014 at 2:20
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One could see this in the following way. We have $$\omega_{\mathbb{P}^n} = \mathcal{O}_{\mathbb{P}^n}(c_1)$$ where $c_1 = c_1(\omega_{\mathbb{P}^n}) = c_{1}(\bigwedge^n\Omega_{\mathbb{P}^n}) = c_1(\Omega_{\mathbb{P}^n})$ is the first Chern class. Now, by the Euler's exact sequence $$0\mapsto\Omega_{\mathbb{P}^n}\rightarrow\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus (n+1)}\rightarrow \mathcal{O}_{\mathbb{P}^n}\mapsto 0$$ we get $$c_1(\omega_{\mathbb{P}^n}) = c_1(\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus (n+1)})-c_1(\mathcal{O}_{\mathbb{P}^n}) = -n-1.$$ Therefore $$\omega_{\mathbb{P}^n} = \mathcal{O}_{\mathbb{P}^n}(-n-1).$$

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    $\begingroup$ Can you just use additivity of degree in exact sequences? Or do you need to use Chern classes.? If deg $\bigwedge^n \Omega = \text{deg } \Omega$ then degrees would work, but I am unsure how the degree interacts with exterior powers. $\endgroup$
    – user100272
    Jun 23, 2017 at 14:15

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