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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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It is not too hard to work this out using homogeneous coordinates. –  S. Carnahan Jul 20 at 23:38
    
@S.Carnahan That is true, I was thinking there was some more conceptual proof. –  Rene Schipperus Jul 20 at 23:41
    
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!) –  Mark Jul 21 at 2:01

2 Answers 2

up vote 10 down vote accepted

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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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. –  Rene Schipperus Jul 20 at 23:44
    
@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. –  GH from MO Jul 20 at 23:49
    
@GHfromMO I see. I was taking all to the same power. Is there a reference for this fact ? –  Rene Schipperus Jul 20 at 23:54
    
@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. –  GH from MO Jul 20 at 23:57
    
@GHfromMO OK thanks ill have a look at that now. –  Rene Schipperus Jul 21 at 0:00

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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