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Does Kunneth formula work for sheaf cohomology with coefficients on the tangent bundle?
I want to calculate $H^1(X, TX)$, where $X=mathbb{C} times mathbb{C}P^2$ and $TX$ is the sheaf of holomorphic sections of the tangent bundle of $X$. I do not know if the Kunneth formula can be used in this situation. Would there be a simple way to compute this?
algebraic-geometry complex-geometry sheaf-cohomology
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
add a comment |
I want to calculate $H^1(X, TX)$, where $X=mathbb{C} times mathbb{C}P^2$ and $TX$ is the sheaf of holomorphic sections of the tangent bundle of $X$. I do not know if the Kunneth formula can be used in this situation. Would there be a simple way to compute this?
algebraic-geometry complex-geometry sheaf-cohomology
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
1
What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16
add a comment |
I want to calculate $H^1(X, TX)$, where $X=mathbb{C} times mathbb{C}P^2$ and $TX$ is the sheaf of holomorphic sections of the tangent bundle of $X$. I do not know if the Kunneth formula can be used in this situation. Would there be a simple way to compute this?
algebraic-geometry complex-geometry sheaf-cohomology
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
I want to calculate $H^1(X, TX)$, where $X=mathbb{C} times mathbb{C}P^2$ and $TX$ is the sheaf of holomorphic sections of the tangent bundle of $X$. I do not know if the Kunneth formula can be used in this situation. Would there be a simple way to compute this?
algebraic-geometry complex-geometry sheaf-cohomology
algebraic-geometry complex-geometry sheaf-cohomology
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
asked Oct 17 '17 at 21:52
Bruno Suzuki
34118
34118
1
What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16
add a comment |
1
What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16
1
1
What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16
add a comment |
1 Answer
1
active
oldest
votes
I don't know if you are still interested, but just in case I'll complete the hint by Mariano. You don't need the Kunneth formula. We have $T(X_1 times X_2) cong pi_1^*TX_1 oplus pi_2^*TX_2$ where $pi_i : X to X_i$ are the projections.
Now $TX_1 = TBbb C$ is the trivial line bundle so does not contribute to $H^1$ and we get $H^1(X,TX) = H^1(X, pi_2^*TBbb P^2) overset{pi_2 text{ is flat }}{=} H^1(Bbb P^2,TBbb P^2) = 0$ where the last equality can be easily verified using the Euler exact sequence.
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
add a comment |
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I don't know if you are still interested, but just in case I'll complete the hint by Mariano. You don't need the Kunneth formula. We have $T(X_1 times X_2) cong pi_1^*TX_1 oplus pi_2^*TX_2$ where $pi_i : X to X_i$ are the projections.
Now $TX_1 = TBbb C$ is the trivial line bundle so does not contribute to $H^1$ and we get $H^1(X,TX) = H^1(X, pi_2^*TBbb P^2) overset{pi_2 text{ is flat }}{=} H^1(Bbb P^2,TBbb P^2) = 0$ where the last equality can be easily verified using the Euler exact sequence.
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
add a comment |
I don't know if you are still interested, but just in case I'll complete the hint by Mariano. You don't need the Kunneth formula. We have $T(X_1 times X_2) cong pi_1^*TX_1 oplus pi_2^*TX_2$ where $pi_i : X to X_i$ are the projections.
Now $TX_1 = TBbb C$ is the trivial line bundle so does not contribute to $H^1$ and we get $H^1(X,TX) = H^1(X, pi_2^*TBbb P^2) overset{pi_2 text{ is flat }}{=} H^1(Bbb P^2,TBbb P^2) = 0$ where the last equality can be easily verified using the Euler exact sequence.
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
add a comment |
I don't know if you are still interested, but just in case I'll complete the hint by Mariano. You don't need the Kunneth formula. We have $T(X_1 times X_2) cong pi_1^*TX_1 oplus pi_2^*TX_2$ where $pi_i : X to X_i$ are the projections.
Now $TX_1 = TBbb C$ is the trivial line bundle so does not contribute to $H^1$ and we get $H^1(X,TX) = H^1(X, pi_2^*TBbb P^2) overset{pi_2 text{ is flat }}{=} H^1(Bbb P^2,TBbb P^2) = 0$ where the last equality can be easily verified using the Euler exact sequence.
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
I don't know if you are still interested, but just in case I'll complete the hint by Mariano. You don't need the Kunneth formula. We have $T(X_1 times X_2) cong pi_1^*TX_1 oplus pi_2^*TX_2$ where $pi_i : X to X_i$ are the projections.
Now $TX_1 = TBbb C$ is the trivial line bundle so does not contribute to $H^1$ and we get $H^1(X,TX) = H^1(X, pi_2^*TBbb P^2) overset{pi_2 text{ is flat }}{=} H^1(Bbb P^2,TBbb P^2) = 0$ where the last equality can be easily verified using the Euler exact sequence.
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
answered Nov 20 '18 at 0:17
Nicolas Hemelsoet
5,7452417
5,7452417
add a comment |
add a comment |
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What cohomology are you talking about? In any case, the tangent bundle of a product is the direct sum of two bundles, each the pullback of the tangent bundle of one of the factors along the corresponding projection. can you use that?
– Mariano Suárez-Álvarez
Oct 17 '17 at 22:01
It is sheaf cohomology. Maybe I can use this fact, I will try.
– Bruno Suzuki
Oct 18 '17 at 17:16