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let $E$ be a vector bundle over a smooth projective variety $X$ and $\pi:\mathbb{P}(E)\rightarrow X$ its projectivization, $T_{\pi}:=ker(\pi_{ * })$ where $\pi_{*}:T_{\mathbb{P}(E)}\rightarrow \pi^{*}T_X$, let $\mathcal{O}_E(-1)\hookrightarrow\pi^{*}E$ be the "tautological" bundle over $\mathbb{P}(E)$. The following it is not a proof but a first reasoning to understand things, if i restrict to a point $x\in X$ i have the usual Euler sequence

$0 \rightarrow \mathcal{O}_{E_x}(-1)\rightarrow ({\pi^{*}}E)_x \rightarrow (T_{\pi})_{x}\otimes \mathcal{O}_{E_x}(-1)\rightarrow 0 $

so my thought is that the generalized euler sequence becomes

$0 \rightarrow \mathcal{O}_{E}(-1)\rightarrow \pi^{*}E\rightarrow T_{\pi}\otimes \mathcal{O}_{E}(-1)\rightarrow 0 $

is this the right path or i'm wrong?

thank you in advance.

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up vote 3 down vote accepted

This sequence is indeed exact. Once you check that the maps are globally defined, exactness can be checked fiberwise (remember that we are dealing with a sequence of vector bundles, not an arbitrary sequence of sheaves), and in this case it follows by the usual Euler sequence for the projective space.

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