Cohomology groups of coherent sheaves for very small and very big twists.

Question

Let $\mathcal{F}$ be nonzero coherent sheaf over the projective space $\mathbb{P}_k^n$. The Serre vanishing Theorem says that $h^i \mathcal{F}(d)=0$ for $i>0$ and $d\gg 0$.

I am wondering if it is always true that $h^0 \mathcal{F}(d) \neq 0$ for $d \gg 0$, and if itn't, under which conditions can that be true.

In the same context, can we say that $h^n \mathcal{F}(d) \neq 0$ for $d \ll 0$? and if not, under which conditions can that be true.

Thank you.

Answer 1

Just to have this down as an answer.

Yes, such a $d$ does exist. Note since $\mathcal{O}(1)$ (the canonical very ample bundle on $\mathbb{P}^n$) is ample, we know that there exists $d\geqslant 0$ such that $\mathcal{F}\otimes\mathcal{O}(d)$ is globally generated. Suppose that $h^0(\mathcal{F}\otimes\mathcal{O}(d))=0$. Then, by the fact that $\mathcal{F}\otimes\mathcal{O}(d)$ is globally generated, we would have that $(\mathcal{F}\otimes\mathcal{O}(d))_p=0$ for all $p\in\mathbb{P}^n$. Thus, $\mathcal{F}\otimes\mathcal{O}(d)=0$, and so

$$\mathcal{F}=(\mathcal{F}\otimes\mathcal{O}(d))\otimes\mathcal{O}(-d)=0$$

contradictory to assumption. So, $\mathcal{F}\otimes\mathcal{O}(d)$ has non-zero global sections.

Now, we also want to show that there is $d\ll 0$ such that $h^n(\mathcal{F}\otimes\mathcal{O}(d))\ne 0$. But, for any $d\leqslant 0$, we have that

$$H^n(\mathbb{P}^n,\mathcal{F}\otimes\mathcal{O}(d))\cong H^0(\mathbb{P}^n,(\mathcal{F}\otimes\mathcal{O}(d))^\vee\otimes\mathcal{O}(-n-1))^\vee$$ by Serre duality (since $\omega_{\mathbb{P}^n}=\mathcal{O}(-n-1)$). Since $\mathcal{O}(d)$ is a line bundle, we have an isomorphism

$$(\mathcal{F}\otimes\mathcal{O}(d))^\vee\cong \mathcal{F}^\vee\otimes\mathcal{O}(-d)$$

and so

$$H^n(\mathbb{P}^n,\mathcal{F}\otimes\mathcal{O}(d))\cong H^0(\mathbb{P}^n,\mathcal{F}^\vee\otimes\mathcal{O}(-d-n-1))^\vee$$

Since $\mathcal{F}^\vee$ is still coherent, the first part shows that for $d\ll 0$, this right hand side will be non-zero, as desired.

Answer 2

There is another nice answer to the first statement, For $d\gg 0$ we have $$p_\mathcal{F}(d)= h^0 \mathcal{F}(d)$$ where $p_\mathcal{F}(d)$ is the Hilbert polynomial.

So the cohomology groups for very large twists should be nonzero because Hilbert polynomial can't have infinitely many roots. and the proof of the second claim is as the answer above.