MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{E}$ be a quasi-coherent sheaf on a scheme $S$. Consider the projective scheme $p : \mathbb{P}(\mathcal{E}) \to S$ and the canonical epimorphism $p^*(\mathcal{E}) \to \mathcal{O}_{\mathbb{P}}(1)$. This corresponds to an epimorphism $p^*(\mathcal{E})(-1) \to \mathcal{O}_{\mathbb{P}}$. It is well-known that its kernel is isomorphic to $\Omega^1_{\mathbb{P}/S}$ if $S$ is affine and $\mathcal{E}$ is free of finite rank (Hartshorne, Theorem II.8.13): This is the famous Euler sequence. This also follows when $S$ is arbitrary and $\mathcal{E}$ is locally free of finite rank. Even for that I don't know a reference in the literature, except for Ravi Vakil's notes, Class 39.

Actually, I have proven that it holds without any assumptions on $\mathcal{E}$, i.e. we always have an exact sequence $0 \to \Omega^1_{\mathbb{P}/S} \to p^*(\mathcal{E})(-1) \to \mathcal{O}_{\mathbb{P}} \to 0$ (generalized Euler sequence). The proof takes two pages, but basically it is a direct coordinate-free generalization of the proof of the special case already mentioned. I don't want to put so much spam in my thesis, and would like to cite this result, which probably has been proven in the 60s. I could not find it in EGA. So my question is: Is this already written down somewhere?

Edit: This Euler sequence can be now found in my thesis as Theorem 4.5.13. I still wonder if there is another reference.

share|cite|improve this question
I'm not sure though, but I don't think such an exact sequence can hold if $\mathcal E$ is not coherent (or at least if $\mathcal E$ is free of infinite rank). – Mike Lowrey Dec 16 '12 at 19:37
@Mike: I am pretty sure that the usual proof works for $\mathcal{O}_S^{\oplus I}$, where $I$ is any set. Locally, one uses that $\Omega^1_{R[\{x_i\}_{i \in I}]/R}$ is free on $\{d(x_i) : i \in I\}$. It follows that the sequence is exact when $\mathcal{E}$ is locally free (without any restriction on the rank). – Martin Brandenburg Dec 16 '12 at 20:33
@Martin. Thank you. – Mike Lowrey Dec 17 '12 at 9:35
I don't want to put so much spam in my thesis, and would like to cite this result, which probably has been proven in the 60s Giving appropriate credit is nice and important of course. Then again, a thesis seems to me to be a good place to write up things that are supposedly well-known to the experts but for which a good reference is lacking. – Olivier Dec 18 '12 at 8:55
I completely agree with Olivier. – diverietti Dec 18 '12 at 9:01

Your Answer


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

Browse other questions tagged or ask your own question.