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Feb 10, 2023 at 3:49 comment added R. van Dobben de Bruyn Ah right, in this case $\bigwedge^i \mathcal O_{\mathbf P^n} = 0$ for $i \geq 2$, so you just get a bunch of short exact sequences $0 \to \bigwedge^{i-1}\mathcal T_{\mathbf P^n} \to \bigwedge^i \mathcal O_{\mathbf P^n}(1)^{n+1} \to \bigwedge^i \mathcal T_{\mathbf P^n} \to 0$.
Feb 10, 2023 at 2:26 comment added BVquantization thanks, if I well understood, we have $ \bigwedge^2\mathcal{T}_{\mathbb{P}^n} \cong \frac{\bigwedge^2\mathcal{O}(1)^{n+1}}{\mathcal {T}_{\mathbb {P}^n} } $
Feb 10, 2023 at 0:51 comment added R. van Dobben de Bruyn See for instance Hartshorne, Exercise II.5.16 for turning short exact sequences of vector bundles into filtrations computing their symmetric or exterior powers.
Feb 10, 2023 at 0:49 history asked BVquantization CC BY-SA 4.0