Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.