Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $R_\eta$.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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 $D$.

share|improve this answer
    
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 $f_{\*}O_X=R$? –  TonyS Oct 4 '11 at 15:49
    
Yes, precisely. –  Sasha Oct 4 '11 at 18:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.