Compute sheaf cohomology on P1xP1

[Daniel Platt]

I want to define a monad bundle on P1xP1 and compute its sheaf cohomology groups. As a toy example, say I want to compute the cohomology groups of the kernel of {{x,y,u,v}}, where (x:y) and (u:v) are coordinates on the two P1 factors respectively.

I found two ways to define P1xP1. However, in both cases the last line attempting to compute the cohomology gives the error "expected degree length 1".

Way 1:

needsPackage "SegreClasses"
R = makeProductRing(QQ,{1,1})

F = sheaf ker matrix {{a,b,c,d}}

HH^0(F)

Way 2:

needsPackage "TateOnProducts"
(S,E) = productOfProjectiveSpaces{1,1}

F = sheaf ker matrix {{x_(0,0),x_(1,0),x_(0,1),x_(1,1)}}

HH^0(F)

Alternatively, I could view P1xP1 as a smooth quadric in P3 under the Segre embedding. But I don't know how to define the sheaf in terms of the coordinates on P3. How can I compute the cohomology groups I'm after?

[Mahrud Sayrafi]

Hi Daniel,

Try using cohomologyMatrix instead of HH.

It would be good if HH automatically used that algorithm, but that's not currently the case.

Mahrud

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[Daniel Platt]

Neat, thanks for the answer.

So the right thing to do would then be the following? I.e. not convert the module ker matrix {{x_(0,0),x_(1,0),x_(0,1),x_(1,1)}} to a sheaf.

needsPackage "TateOnProducts"
(S,E) = productOfProjectiveSpaces{1,1}

F = ker matrix {{x_(0,0),x_(1,0),x_(0,1),x_(1,1)}}
low = {-2,-2};high={2,2};
cohomologyMatrix(F,low,high)

Does this then compute the sheaf cohomology of F viewed as a sheaf? I don't know much algebra, so I wonder if that could be some different cohomology.