Hi, 


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.



Which is the correct generalization of Euler sequence to the projectivization of a vector bundle?
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.


