# Why does Castelnuovo-Mumford regularity require additional twisting?

Let $F$ be a coherent sheaf on $\mathbb P^n$. For simplicity, let it be locally free. It is called $m$-regular (by Castelnuovo-Mumford) if for all $i > 0$ $$H^i(F(m-i)) = 0.$$ If $F$ is $m$-regular, then the induction on $\dim \mathbb P^n$ and the cohomological exact sequences for $$0 \to F(m-i-1) \to F(m-i) \to F_H(m-i) \to 0,$$ where $H$ is a linear hypersection, give that it is $(m+1)$-regular.

Now let us call $F$ naïvely $m$-regular if for all $i>0$ $$H^i(F(m)) = 0.$$ Then the proof does not work as above, but is there really an example when $F$ is naïvely $m$-regular, but not naïvely $(m+1)$-regular?

• Take $\Omega^1_{\mathbb{P}^2}(-1)$. This is 0-regular, but not 1-regular in your naive sense. – Mohan Nov 10 '16 at 22:38
• @Mohan, could you make your comment an answer, please? – evgeny Dec 19 '16 at 7:01