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I'm following the book of Okonek, Schneider and Spindler, Vector Bundles on Complex Projective Spaces, and they say that is a useful exercise try to prove Bott's Formula that calculates the cohomology of exterior powers of the cotangent bundle on a projective space, by using induction.

In order to do it, assuming the statement true for for the case p-1 and we have the long exact sequence of cohomologies:

$$0 \to H^{0}(\Omega^{p}_{\mathbb{P^{n}}}) \to C^{n+1}_{p}H^{0}(\mathbb{P^{n}}) \to H^{0}(\Omega^{p-1}_{\mathbb{P^{n}}}) \to H^{1}(\Omega^{p}_{\mathbb{P^{n}}}) \to 0$$ where $C^{n+1}_{p}$ means combination $n+1$, p to p. Here, if $H^{1}(\Omega^{p}_{\mathbb{P^{n}}})$ was zero it will be possible do this because we know the dimension of the other 2 spaces an so we can calculate $\dim H^{0}(\Omega^{p}_{\mathbb{P^{n}}})$. But I don't know how to see it, anyone has a suggestion in order to do it?

Thank you so much in any advance!

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    $\begingroup$ Hint: Try using the Euler sequence $0\to \Omega^1\to O(-1)^{n+1}\to O\to 0$, take exterior powers, and then compute cohomology. (I wonder if the people who voted to close knew this.) $\endgroup$ – Donu Arapura Aug 9 '14 at 13:10
  • $\begingroup$ Thank you for the hint @DonuArapura! But, the long exact sequence that we got doing this, is the sequence above, the problem is that we have two spaces that we know the dimension, and two spaces that we don't, because of this I'm thinking if I have to do the same process to another exact sequence involving the cotangent bundle... The hint on the book says that we could use the sequence $0 \to \cal{O}(-1) \to \cal{O} \to \cal{O}_{\mathbb{P}^{n-1}} \to 0$ but again taking exterior powers we $0 \to \Omega(k-1) \to \Omega(k) \to \Omega|_{\mathbb{P}^{n-1}}(k) \to 0$ $\endgroup$ – User43029 Aug 9 '14 at 13:53
  • $\begingroup$ but $\Omega|_{\mathbb{P}^{n-1}}(k) \to 0 \neq \Omega_{\mathbb{P}^{n-1}}(k)$ so if we try to use induction in n too, we get nothing too, I guess that all the problem is because we don't know $H^{1}(\Omega^{p})$. $\endgroup$ – User43029 Aug 9 '14 at 13:56
  • $\begingroup$ In fact, above when I said taking exterior powers I mean we tensorize the sequence by the locally free sheaf $\Omega(k)$(I can't correct there anymore...) $\endgroup$ – User43029 Aug 9 '14 at 14:36

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