Ring isomorphism in Leray-Serre spectral sequence
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The Leray-Hirsch theorem: let $k$ be a field. Given a fibration $F to E to B $ with $F, B$ path connected and suppose system of local coefficient is zero and the following condition satisfied (a) $H^n(B;k) $ is finitely dimensional for each $n$ and (b) $ i^*:H^*(E;k) to H^*(F;k)$ is onto, here $i$ is the inclusion of fiber. Then $$H^*(E;k) cong H^*(B;k)otimes_k H^*(F;k) $$ as vector spaces.
My question is: What are some extra condition do we need so that above becomes ring isomorphism. In particular for the field $k=Bbb Z/2 Bbb Z$ what extra condition do we need.
Thank you very much for your kind help.
algebraic-topology spectral-sequences
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
$endgroup$
add a comment |
$begingroup$
The Leray-Hirsch theorem: let $k$ be a field. Given a fibration $F to E to B $ with $F, B$ path connected and suppose system of local coefficient is zero and the following condition satisfied (a) $H^n(B;k) $ is finitely dimensional for each $n$ and (b) $ i^*:H^*(E;k) to H^*(F;k)$ is onto, here $i$ is the inclusion of fiber. Then $$H^*(E;k) cong H^*(B;k)otimes_k H^*(F;k) $$ as vector spaces.
My question is: What are some extra condition do we need so that above becomes ring isomorphism. In particular for the field $k=Bbb Z/2 Bbb Z$ what extra condition do we need.
Thank you very much for your kind help.
algebraic-topology spectral-sequences
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
$endgroup$
$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
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– Mike Miller
Jan 20 at 0:56
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@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53
add a comment |
$begingroup$
The Leray-Hirsch theorem: let $k$ be a field. Given a fibration $F to E to B $ with $F, B$ path connected and suppose system of local coefficient is zero and the following condition satisfied (a) $H^n(B;k) $ is finitely dimensional for each $n$ and (b) $ i^*:H^*(E;k) to H^*(F;k)$ is onto, here $i$ is the inclusion of fiber. Then $$H^*(E;k) cong H^*(B;k)otimes_k H^*(F;k) $$ as vector spaces.
My question is: What are some extra condition do we need so that above becomes ring isomorphism. In particular for the field $k=Bbb Z/2 Bbb Z$ what extra condition do we need.
Thank you very much for your kind help.
algebraic-topology spectral-sequences
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
$endgroup$
The Leray-Hirsch theorem: let $k$ be a field. Given a fibration $F to E to B $ with $F, B$ path connected and suppose system of local coefficient is zero and the following condition satisfied (a) $H^n(B;k) $ is finitely dimensional for each $n$ and (b) $ i^*:H^*(E;k) to H^*(F;k)$ is onto, here $i$ is the inclusion of fiber. Then $$H^*(E;k) cong H^*(B;k)otimes_k H^*(F;k) $$ as vector spaces.
My question is: What are some extra condition do we need so that above becomes ring isomorphism. In particular for the field $k=Bbb Z/2 Bbb Z$ what extra condition do we need.
Thank you very much for your kind help.
algebraic-topology spectral-sequences
algebraic-topology spectral-sequences
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
edited Jan 20 at 16:09
Shivani Sengupta
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
asked Jan 19 at 19:31
Shivani SenguptaShivani Sengupta
30619
30619
$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
$endgroup$
– Mike Miller
Jan 20 at 0:56
$begingroup$
@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53
add a comment |
$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
$endgroup$
– Mike Miller
Jan 20 at 0:56
$begingroup$
@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53
$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
$endgroup$
– Mike Miller
Jan 20 at 0:56
$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
$endgroup$
– Mike Miller
Jan 20 at 0:56
$begingroup$
@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53
$begingroup$
@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53
add a comment |
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$begingroup$
The conditions of Proposition 8 are that $H_i(F;k) to H_i(E;k)$ is injective for all $i geq 0$ (or equivalently that $H^i(E;k) to H^i(F;k)$ is surjective) and that each $H^i(F;k)$ and $H^i(B;k)$ are finite-dimensional. The only thing you haven't already assumed is that $H^i(F;k)$ is finite-dimensional. Once you assume that, you get what you want.
$endgroup$
– Mike Miller
Jan 20 at 0:56
$begingroup$
@MikeMiller I got a translation of the above mentioned J P Serre paper. Thanks.
$endgroup$
– Shivani Sengupta
Jan 20 at 15:53