Assume we are workling on $\mathbb{P}^n$ for some $n\geq 1$ and we have a coherent sheaf $F$ on it.

Then there are two (well known?) spectral sequences $E_r^{p,q}$ with $E_1$-term:

$E_1^{p,q}=H^q(\mathbb{P}^n,F(p))\otimes \Omega^{-p}(-p)$

$E_1^{p,q}=H^q(\mathbb{P}^n,F\otimes \Omega^{-p}(-p))\otimes O_{\mathbb{P}^n}(p)$

both converging to $F$. Here $\Omega^{p}=\wedge^p((T_{\mathbb{P}^n})^{\*})$, see e.g. Okonek/Spindler/Schneider Ch.2 §3.

In special cases these sequences lead to a monad description for $F$, i.e. a complex $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$, which is exact at $A$ and $C$ such that $F$ is the cohomology of this complex.

The main ingredient of the proof of this fact is the existence of a Koszul resolution for the diagonal $\Delta\subset \mathbb{P}^n\times\mathbb{P}^n$.

Now assume with have an additional "structure" sheaf $R$ of noncommutative rings or algebras on $\mathbb{P}^n$, such that $F$ is also an $R$-module.

Is there a generalization or a way to adjust these spectral sequences that also uses the extra structure as an $R$-module?

Maybe there are more general Koszul resolutions which one can use here? Everything in a more noncommuative setting. Maybe there is something like this in the literature?

One case i'm especially interested in is that of maximal orders on the projective plane. That is $R$ is a sheaf of noncommutative algebras, say of rank $4$, which is an Azumaya algebra $\mathcal{A}$ on the complement of a (smooth) divisor $D\subset \mathbb{P}^2$, such that the generic algbera $R_\eta$ is a nontrivial quaternion algebra.

So we have a trace pairing $tr: R\otimes R \rightarrow O_{\mathbb{P}^2}$ which is nondegenerate away from $D$. For every point $p\in D$ the module $R_p$ is a maximal $O_p$-order in the generic stalk $R_\eta$.


The answer depends very strongly on your algebra $R$. For example, a particular case is when $R = O + L$ (where $L$ is a line bundle) and the multiplication is given by a map $L^2 \to O$ (given by a divisor $D$) the category $Coh(P^n,R)$ is equivalent to $Coh(X)$, where $X$ is the double covering of $P^n$ ramified in $D$. And the homological properties of $D^b(Coh(X))$ very strongly depend on $D$.

  • $\begingroup$ Thanks. That's very interesting, but also dissapointing, if this already gets so difficult in such an "easy" example. I added the type of algebras i'm interested in in the main post, but it seems they are too complex to hope that anything i expected could work. The fact that $Coh(P^n,R)$ is equivalent to $Coh(X)$: does this follow from the fact that if $f: X \rightarrow P^n$ is the double cover defined by $L$, then $f_{\*}O_X=R$? $\endgroup$
    – TonyS
    Oct 4 '11 at 15:49
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Sasha
    Oct 4 '11 at 18:09

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