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How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes...
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Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles.
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
$endgroup$
add a comment |
$begingroup$
Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles.
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
$endgroup$
5
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
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– Qiaochu Yuan
Feb 28 '12 at 4:03
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This is a great comment.
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– Kerry
Feb 28 '12 at 4:24
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The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27
add a comment |
$begingroup$
Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles.
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
$endgroup$
Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles.
algebraic-geometry schubert-calculus
algebraic-geometry schubert-calculus
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
edited Jan 2 at 0:49
Matt Samuel
37.5k63665
37.5k63665
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
asked Sep 17 '11 at 4:04
LJRLJR
6,57341749
6,57341749
5
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
$endgroup$
– Qiaochu Yuan
Feb 28 '12 at 4:03
$begingroup$
This is a great comment.
$endgroup$
– Kerry
Feb 28 '12 at 4:24
$begingroup$
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27
add a comment |
5
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
$endgroup$
– Qiaochu Yuan
Feb 28 '12 at 4:03
$begingroup$
This is a great comment.
$endgroup$
– Kerry
Feb 28 '12 at 4:24
$begingroup$
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27
5
5
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
$endgroup$
– Qiaochu Yuan
Feb 28 '12 at 4:03
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
$endgroup$
– Qiaochu Yuan
Feb 28 '12 at 4:03
$begingroup$
This is a great comment.
$endgroup$
– Kerry
Feb 28 '12 at 4:24
$begingroup$
This is a great comment.
$endgroup$
– Kerry
Feb 28 '12 at 4:24
$begingroup$
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27
$begingroup$
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27
add a comment |
1 Answer
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I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
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add a comment |
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1 Answer
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$begingroup$
I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
$endgroup$
add a comment |
$begingroup$
I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
$endgroup$
add a comment |
$begingroup$
I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
$endgroup$
I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.
There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
answered Feb 28 '12 at 3:57
KerryKerry
1,366725
1,366725
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5
$begingroup$
If you mean singular cohomology over $mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post: rigtriv.wordpress.com/2009/03/27/…
$endgroup$
– Qiaochu Yuan
Feb 28 '12 at 4:03
$begingroup$
This is a great comment.
$endgroup$
– Kerry
Feb 28 '12 at 4:24
$begingroup$
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm.
$endgroup$
– Jyrki Lahtonen
Feb 28 '12 at 4:27