Suppose I have a closed subvariety $V$ of $\mathbb P^n \times \mathbb A^m$ given by explicit equations. Is it possible in practice to compute the coherent cohomology of $V$ with coefficients in a line bundle pulled back from $\mathbb P^n$ using the Cech complex for the open affine cover of $V$ by intersections of $V$ with the products of the standard affine open subsets of $\mathbb P^n$ with $\mathbb A^m$ (similarly to how cohomology of the projective space with coefficients in a line bundle is computed)? If not, how to compute this cohomology?
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3$\begingroup$ Yes you can but you would need an "acyclic" cover of $V$, i.e., a cover of $V$ by affine (Stein) subvarieties of $V$ such that their intersections are also affine (Stein). This reduces the computation to a simplicial cohomology computation which could be challenging . Even when $m=0$ and $\dim V=1$ this could be tricky. $\endgroup$– Liviu NicolaescuCommented Aug 6 at 10:06
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2$\begingroup$ The cohomology of a line bundle over say curve is sensitive to both the complex structure on the curve and the complex structure on the bundle. Look for example at a curve in $\mathbb{P}^2$. The affine varieties you mention look like lines with points removed, in fact the number of points removed is exactly the degree of the curve. $\endgroup$– Liviu NicolaescuCommented Aug 6 at 12:36
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1$\begingroup$ @Yellow Pig I once asked a question about the correctness of an algorithm that would supply you with a solution to your problem here on MO. It is mathoverflow.net/questions/301923/… I had tested the algorithm successfully on some examples from Mumfords "Lectures on curves on an algebraic surface" and some other test examples I made up and it should be correct for a field $A$. The question was closed and received no answers and I had no other opportunity to get it corroborated, maybe you want to have a look. $\endgroup$– Jürgen BöhmCommented Aug 8 at 20:09
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1$\begingroup$ @Yellow Pig Yes, for $\mathbb{A}^m$ you have to take $A=k[y_1,\ldots,y_m]$. What do you think about the correctness of my argument in the article? I nowhere found the idea to take $H^n(X-)$ as the basic right-exact functor and $H^{n-i}(X,-)$ as its $i$-th satellite functor but it appeared sound to me. It would be great to have a second (and third..) opinion on this! (That $H^n(X,-)$ is right exact follows by the way from the fact that $\mathbb{P}^n_A$ can be covered by $n+1$-affines and Cech-cohomology). $\endgroup$– Jürgen BöhmCommented Aug 9 at 21:48
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1$\begingroup$ @Yellow Pig I just tested the algorithm with an older Macaulay2 and also the newest version in the web interface. Unfortunately it could not compute correct results for even very simple examples where I had thought it could. I think the abstract algorithm as stated in the link is still correct, but it might be impossible to implement it in Macaulay2 - at least not in the way I did in the example code there. The example code should work for A a field, say A=QQ, but a polynomial ring, e.g. A = QQ[a], seems to cause errors the root of which I could not pinpoint. $\endgroup$– Jürgen BöhmCommented Aug 12 at 23:09
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