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I'm reading about Euler exact sequence in Ravi Vakil's notes, and I need help to check a few things. Given a scheme $X$, and a locally free sheaf $\mathcal{E}$ of rank $n+1$ on $X$, let us start from here https://mathoverflow.net/questions/116526/generalized-euler-sequence-on-a-projective-scheme

and say that we have an exact sequence $$ 0\to \Omega_{\mathbb{P}\mathcal{E}/X}\to \pi^*(\mathcal{E}^\vee)(-1)\to\mathcal{O}_{\mathbb{P}\mathcal{E}}\to 0\quad (I) $$ or dually $$ 0\to \mathcal{O}_{\mathbb{P}\mathcal{E}}\to \pi^*\mathcal{E}\otimes O_{\mathbb{P}\mathcal{E}}(1)\to \mathcal{T}_{\mathbb{P}\mathcal{E}/X}\to 0\quad (II) $$ where $\mathbb{P}{\mathcal{E}}=\mathrm{Proj}({\mathrm{Sym}(\mathcal{E^\vee})})$ with $\pi:\mathbb{P}\mathcal{E}\to X$. In the notes (chapter 13), he writes $$ 0\to \Omega_{\mathbb{P}\mathcal{E}/X}\to \mathcal{E}^\vee\otimes O(-1)\to\mathcal{O}_{X}\to 0\quad (III) $$ I think $(III)$ is the pushforward of $(I)$, but I am not sure why. I can see that $\pi_*\mathcal{O}_{\mathbb{P}{\mathcal{E}}}$ has no higher direct images, so (II) becomes $$ 0\to \mathcal{O}_{X}\to \pi_*\pi^*\mathcal{E}\otimes O_{X}(1)\to \pi_*\mathcal{T}_{\mathbb{P}\mathcal{E}/X}\to 0\quad (IV) $$ or $$ 0\to \mathcal{O}_{X}\to \mathcal{E}\otimes \mathcal{E}^\vee\to \pi_*\mathcal{T}_{\mathbb{P}\mathcal{E}/X}\to 0\quad (V) $$ by projection formula. But I do not see why $\pi_*\pi^*\mathcal{E}=\mathcal{E}$. My question is:

Why is $\pi_*\pi^*\mathcal{E}=\mathcal{E}$ true? if possible, please say something general. say you have a map $f:X\to Y$ so that $f_*O_X=O_Y$ and a sheaf $\mathcal{E}$ on $X$. what conditions on $\mathcal{E}$ do you need for $f_*f^*\mathcal{E}=\mathcal{E}$ to hold?

[Edit: Ok, my question really did not make sense. By $O(-1)$ he means $\pi_* \mathcal{O}_{\mathbb{P}\mathcal{E}}(-1)$, and not $O_X(-1)$ even when $O_X(-1)$ makes sense. so we do not need to have $\pi_*\pi^*\mathcal{E}=\mathcal{E}$].

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adrido
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  • In (III), what is the kernel? your kernel is the same as in (I), but $\Omega_{\mathbb P(\mathscr E)/X}$ is not a sheaf on $X$. – Brenin Oct 23 '14 at 20:23
  • I'm using Vakil's notation, and I think he means $\pi_*\Omega_{\mathbb{P}\mathcal{E}/X}$. – adrido Oct 23 '14 at 20:28

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Projection formula says that $$\pi_\ast(\pi^\ast\mathcal{E}\otimes O_{\mathbb{P}\mathcal{E}}(1))=\mathcal E\otimes \pi_\ast\mathcal O_{\mathbb P\mathcal E}(1).$$ So the pushforward of (II) is $$0\to \mathcal O_X\to \mathcal E\otimes \pi_\ast\mathcal O_{\mathbb P\mathcal E}(1)\to \pi_\ast T_{\mathbb P\mathcal E/X}\to 0.$$ Note that all this has nothing to do with the condition $\pi_\ast\pi^\ast\mathcal E=\mathcal E$ (which may be false): we just used projection formula at the very beginning. Still, your question is interesting and I am not the right person to answer it, unfortunately. Something has already been said here, though.

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Brenin
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