Cohomology of tensor product of pullback in $mathbb P^1timesmathbb P^1$.
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Let $X=mathbb P^1timesmathbb P^1$, and let $pi_1$ and $pi_2$ be the projection maps. For each $a,binmathbb Z$, we have a sheaf of $mathcal O_X$-modules $mathscr F_{a,b} = pi_1^*mathcal O(a)otimes pi_2^*mathcal O(b)$, and I want to compute its cohomology using the open affine with four open sets obtained from the usual cover ${U_0,U_1}$ of $mathbb P^1$, or otherwise.
Unless I am missing something, on a product $U=U_itimes U_j$, $mathscr F_{a,b}(U)$ consists of bihomogeneous quotients $f(x)g(y)/x_i^ry_j^s$ such that
$deg f = r+a$ and $deg g = s+b$. On sets of the form $U_{ij}times U_k$ and the remaining others there is an analogous description.
Using the above, I was trying to compute $H^*(X,mathscr F_{a,b})$, but quickly run into cumbersome computations. I did get that $d^2$ is zero since all triple and cuadruple intersections are the same, so $d^2=0$ since there are four triple intersections and everything cancels, hence $H^3$ is just $O(a)(U_{01})otimes O(b)(U_{01})$ unless I am missing something.
I can also describe the kernels of $d^0$ and $d^1$, so probably after some
long computations arrive at an answer. Does anyone have a hint on how to move on? Perhaps a more clever approach? Perhaps the Segre embedding could help out here?
Note. This is the last exercise in Chapter 8 of these notes.
projective-space sheaf-cohomology
add a comment |
up vote
1
down vote
favorite
Let $X=mathbb P^1timesmathbb P^1$, and let $pi_1$ and $pi_2$ be the projection maps. For each $a,binmathbb Z$, we have a sheaf of $mathcal O_X$-modules $mathscr F_{a,b} = pi_1^*mathcal O(a)otimes pi_2^*mathcal O(b)$, and I want to compute its cohomology using the open affine with four open sets obtained from the usual cover ${U_0,U_1}$ of $mathbb P^1$, or otherwise.
Unless I am missing something, on a product $U=U_itimes U_j$, $mathscr F_{a,b}(U)$ consists of bihomogeneous quotients $f(x)g(y)/x_i^ry_j^s$ such that
$deg f = r+a$ and $deg g = s+b$. On sets of the form $U_{ij}times U_k$ and the remaining others there is an analogous description.
Using the above, I was trying to compute $H^*(X,mathscr F_{a,b})$, but quickly run into cumbersome computations. I did get that $d^2$ is zero since all triple and cuadruple intersections are the same, so $d^2=0$ since there are four triple intersections and everything cancels, hence $H^3$ is just $O(a)(U_{01})otimes O(b)(U_{01})$ unless I am missing something.
I can also describe the kernels of $d^0$ and $d^1$, so probably after some
long computations arrive at an answer. Does anyone have a hint on how to move on? Perhaps a more clever approach? Perhaps the Segre embedding could help out here?
Note. This is the last exercise in Chapter 8 of these notes.
projective-space sheaf-cohomology
Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X=mathbb P^1timesmathbb P^1$, and let $pi_1$ and $pi_2$ be the projection maps. For each $a,binmathbb Z$, we have a sheaf of $mathcal O_X$-modules $mathscr F_{a,b} = pi_1^*mathcal O(a)otimes pi_2^*mathcal O(b)$, and I want to compute its cohomology using the open affine with four open sets obtained from the usual cover ${U_0,U_1}$ of $mathbb P^1$, or otherwise.
Unless I am missing something, on a product $U=U_itimes U_j$, $mathscr F_{a,b}(U)$ consists of bihomogeneous quotients $f(x)g(y)/x_i^ry_j^s$ such that
$deg f = r+a$ and $deg g = s+b$. On sets of the form $U_{ij}times U_k$ and the remaining others there is an analogous description.
Using the above, I was trying to compute $H^*(X,mathscr F_{a,b})$, but quickly run into cumbersome computations. I did get that $d^2$ is zero since all triple and cuadruple intersections are the same, so $d^2=0$ since there are four triple intersections and everything cancels, hence $H^3$ is just $O(a)(U_{01})otimes O(b)(U_{01})$ unless I am missing something.
I can also describe the kernels of $d^0$ and $d^1$, so probably after some
long computations arrive at an answer. Does anyone have a hint on how to move on? Perhaps a more clever approach? Perhaps the Segre embedding could help out here?
Note. This is the last exercise in Chapter 8 of these notes.
projective-space sheaf-cohomology
Let $X=mathbb P^1timesmathbb P^1$, and let $pi_1$ and $pi_2$ be the projection maps. For each $a,binmathbb Z$, we have a sheaf of $mathcal O_X$-modules $mathscr F_{a,b} = pi_1^*mathcal O(a)otimes pi_2^*mathcal O(b)$, and I want to compute its cohomology using the open affine with four open sets obtained from the usual cover ${U_0,U_1}$ of $mathbb P^1$, or otherwise.
Unless I am missing something, on a product $U=U_itimes U_j$, $mathscr F_{a,b}(U)$ consists of bihomogeneous quotients $f(x)g(y)/x_i^ry_j^s$ such that
$deg f = r+a$ and $deg g = s+b$. On sets of the form $U_{ij}times U_k$ and the remaining others there is an analogous description.
Using the above, I was trying to compute $H^*(X,mathscr F_{a,b})$, but quickly run into cumbersome computations. I did get that $d^2$ is zero since all triple and cuadruple intersections are the same, so $d^2=0$ since there are four triple intersections and everything cancels, hence $H^3$ is just $O(a)(U_{01})otimes O(b)(U_{01})$ unless I am missing something.
I can also describe the kernels of $d^0$ and $d^1$, so probably after some
long computations arrive at an answer. Does anyone have a hint on how to move on? Perhaps a more clever approach? Perhaps the Segre embedding could help out here?
Note. This is the last exercise in Chapter 8 of these notes.
projective-space sheaf-cohomology
projective-space sheaf-cohomology
asked 2 days ago
Pedro Tamaroff♦
95.5k10149295
95.5k10149295
Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday
add a comment |
Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday
Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday
add a comment |
1 Answer
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The following two references deal with this: Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves, $S4$, by Kempf, and A Kunneth formula for coherent algebraic sheaves, by Sampson and Washnitzer.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The following two references deal with this: Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves, $S4$, by Kempf, and A Kunneth formula for coherent algebraic sheaves, by Sampson and Washnitzer.
add a comment |
up vote
0
down vote
The following two references deal with this: Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves, $S4$, by Kempf, and A Kunneth formula for coherent algebraic sheaves, by Sampson and Washnitzer.
add a comment |
up vote
0
down vote
up vote
0
down vote
The following two references deal with this: Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves, $S4$, by Kempf, and A Kunneth formula for coherent algebraic sheaves, by Sampson and Washnitzer.
The following two references deal with this: Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves, $S4$, by Kempf, and A Kunneth formula for coherent algebraic sheaves, by Sampson and Washnitzer.
answered yesterday
Pedro Tamaroff♦
95.5k10149295
95.5k10149295
add a comment |
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Why don't you want to use Kunneth formula ?
– Nicolas Hemelsoet
2 days ago
@NicolasHemelsoet Mostly because I was not aware there was one. :)
– Pedro Tamaroff♦
yesterday
(N.B.: I was somewhat suspicious that the Cech complexes of the respective covers are quasi-isomorphic, but when computing $H^*(X,mathscr F)$ something led be to believe this guess was not quite right, at least not in full generality. I'll try to write down a proof and post it as an answer.)
– Pedro Tamaroff♦
yesterday
It's true in full generality : you only need $X,Y$ separated and $F,G$ quasi coherent sheaves on $X$, resp. $Y$.
– Nicolas Hemelsoet
yesterday