Let $\mathcal{F}$ be a nonzero coherent sheaf on the projective space $\mathbb{P}_{k}^m$. I am trying to show that for every integer $d$ there is $j$ for which $h^j\mathcal{F}(d-j) \neq 0$.
My professor said we can use the dual of Tate resolution and self-injective concept to show that.
Let $\mathcal{F}= \widetilde{M}$, and $E$ the exterior algebra over a $m$-dimensional vector space.
Well I know that Tate resolution is just adjoining a minimal free resolution of the kernel of $Hom_k(E, M_{r}) \rightarrow Hom_k(E, M_{r+1})$ to the complex $$Hom_k(E, M_{r}) \rightarrow Hom_k(E, M_{r+1}) \rightarrow Hom_k(E, M_{r+1}) \dots$$
So it has the form $$T(\mathcal{F}):\dots \rightarrow T_1 \rightarrow T_1\rightarrow T_0 \rightarrow Hom(E,M_r) \rightarrow Hom_k(E, M_{r+1}) \rightarrow \dots$$
From properties of Tate resolution we know that the term of $T(\mathcal{F})$ having cohomology degree $d$ is $$\bigoplus_j h^j\mathcal{F}(d-j)\otimes_k E(j-d)$$
So it is enough to show that every term of the Tate resolution can't be zero.
This formula also shows that all terms of Tate resolution is free $E$-modules, and thus projective.
Now $E$ is self-injective which implies that all $E$-projectives are also injective, and the functor $Hom(-,E)$ is exact.
So If we take the dual of the Tate resolution it will be also exact.
My question is how can this give me that no term in the Tate resolution can be zero.
best regards