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I have come across a proof of Bott vanishing for $\mathbb{P}^{n}$ using the Euler sequence as follows:

From the Euler sequence on $\mathbb{P}^{n}$

$$ 0 \to \Omega \to \mathcal{O}(-1)^{\oplus n+1 } \to \mathcal{O} \to 0 $$

we take the $p$-th exterior power to obtain the sequence

$$ 0 \to \Omega^{p} \to \bigwedge^{p} \mathcal{O}(-1)^{\oplus n+1} \to \Omega^{p-1} \to 0 $$

which should remain exact according to my reference. Why is this the case?

Then we can also twist to obtain short exact sequences

$$ 0 \to \Omega^{p}(k) \to \bigwedge^{p} \mathcal{O}(-1)^{\oplus n+1}(k) \to \Omega^{p-1}(k) \to 0 $$

And the claim now is that from the long exact sequence in cohomology it follows that

$$H^{q}(\Omega^{p}(k))=0$$

unless

  • $k=0$ and $0\leqslant p=q \leqslant n $.

  • $q=0$ and $k>p$.

  • $q=n$ and $k<p-n$.

So for example the first case would follow indeed from the LES in cohomology if we had that the cohomology of the middle term vanishes. But does it? If so, why? Cohomology and direct sums commute, and so do direct sums and exterior powers (direct sums and exterior powers do not commute strictly speaking, but I am referring to the formula of the exterior product of a direct sum). So I expect that one can show that the cohomology of that sheaf vanishes just by formal manipulations of whats inside. But I couldn't find the actual way to do it.

Any hints?

Pedro
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  • I'm not convinced about the exactness of that sequence. (Not saying it's not true, but not convinced.) – Patrick Da Silva Feb 25 '18 at 11:30
  • Indeed, this shouldn't hold in general, so I guess you really need to use that the sheaves are locally free and hence that the sequences on the stalks split – Pedro Feb 25 '18 at 11:38
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    For the exact sequence $0\rightarrow\Omega^p\rightarrow\bigwedge^p\mathcal{O}(-1)^{\oplus n+1}\rightarrow\Omega^{p-1}\rightarrow 0$, you can look at this thread which deal with general vector bundles : https://math.stackexchange.com/questions/2583840/induced-short-exact-sequence-on-wedge-product/2588845#2588845 – Roland Feb 25 '18 at 15:55
  • Thank you! The link clarifies what is the result being used there. Thanks a lot. Regarding commutativity, yes, I shouldn't have said commute, my bad. I meant "commutes" as in the formula which allows you to express the exterior power of a direct sum as the direct sum of the tensor products of exterior powers summing up to the original power. What I had in mind was using this inductively to prove that it is zero – Pedro Feb 25 '18 at 16:07
  • Right, I didn't read that part of the comment. Thanks a lot! – Pedro Feb 25 '18 at 16:12
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    No wait, there is a mistake in the isomorphism of my previous comment, and I can't edit it anymore, so I deleted it. This is $(\bigwedge^p\mathcal{O}(-1)^{n+1})(k)=\mathcal{O}(k-p)^{\oplus\binom{p}{n+1}}$. – Roland Feb 25 '18 at 16:14
  • Okay, thanks again! Feel free to post it as an answer for me to be able to approve it if you want – Pedro Feb 25 '18 at 16:16

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