real Grassmannian
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1987xk47
Apr 7, 2011, 9:47:12 PM4/7/11
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Hi all,
I'm thinking about the following problem.
For a smooth 2m-dim'l real vector bundle E over M, we can define Gr(2, E) to be a
fiber bundle over M, with fiber the real Grassmannian Gr(2, E_p), p \in M.
Does anyone know whether the map from H^*(M) to H^*(Gr(2,E)) (De Rham cohlgy)
is injective or not? Or does anyone know some reference about the De Rham cohlgy of
real Grassmannian?
Best
Ke
I'm thinking about the following problem.
For a smooth 2m-dim'l real vector bundle E over M, we can define Gr(2, E) to be a
fiber bundle over M, with fiber the real Grassmannian Gr(2, E_p), p \in M.
Does anyone know whether the map from H^*(M) to H^*(Gr(2,E)) (De Rham cohlgy)
is injective or not? Or does anyone know some reference about the De Rham cohlgy of
real Grassmannian?
Best
Ke
Yi Zhu
Apr 7, 2011, 11:41:49 PM4/7/11
to cocktai...@googlegroups.com
Hi,
Please forgive if I make any mistakes in the email since I leave this field far ago.
Is your question just a fact from Leray spectral sequence?
Try Bott-Tu's Leray spectral sequence, the E_2 part of H^p(E) you have a final filtration piece of H^p(M), lives forever... thus it proves the injectivity.
The cohomology of the Real Grass is the same as complex case (the cellular decomposition not the cohomology group). it is called Schubert calculus. Try Griffiths-Harris on Schubert calculus.
Actually you can ask if a kind of Leray-Hirsch theorem holds for Grassmanian bundles, this is correct for projective space bundle comes from a vector bundle. And I do not know the answer since the Grassmanian cohomology algebra is not simply generated by the first chern classes.
Best,
Yi
Please forgive if I make any mistakes in the email since I leave this field far ago.
Is your question just a fact from Leray spectral sequence?
Try Bott-Tu's Leray spectral sequence, the E_2 part of H^p(E) you have a final filtration piece of H^p(M), lives forever... thus it proves the injectivity.
The cohomology of the Real Grass is the same as complex case (the cellular decomposition not the cohomology group). it is called Schubert calculus. Try Griffiths-Harris on Schubert calculus.
Actually you can ask if a kind of Leray-Hirsch theorem holds for Grassmanian bundles, this is correct for projective space bundle comes from a vector bundle. And I do not know the answer since the Grassmanian cohomology algebra is not simply generated by the first chern classes.
Best,
Yi
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gufan...@gmail.com
Apr 7, 2011, 11:43:27 PM4/7/11
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Hi,
I would vote yes for the 1st question. I would even say H^*(Gr(2,E)) is a free module over H^*(M).
For the 2nd, I know Milne Stasheff talks about it.
Best,
Gufang
I would vote yes for the 1st question. I would even say H^*(Gr(2,E)) is a free module over H^*(M).
For the 2nd, I know Milne Stasheff talks about it.
Best,
Gufang
Pan Xuanyu
Apr 8, 2011, 1:12:59 PM4/8/11
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| Hi, I think it is true( at least rank of E is big), the reason is the cohomology with coeff=Q of real Grass is poly generate by the pontragin class of the universal bundle on it, and using Leray theorem. Please look at milnor's Charateristic Class for the cohomology with Q of real Grass. Best Xuanyu --- 11年4月8日，周五, gufan...@gmail.com <gufan...@gmail.com> 写道：
|
1987xk47
Apr 8, 2011, 1:53:18 PM4/8/11
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Hi Xuanyu,
Thank you! I get it.
Is the reference you mentioned a paper or a book?
Ke
Thank you! I get it.
Is the reference you mentioned a paper or a book?
Ke
在2011-04-09，"Pan Xuanyu" <zh...@yahoo.com.cn> 写道：
-----原始邮件-----
发件人:"Pan Xuanyu" <zh...@yahoo.com.cn>
发送时间:2011年4月9日 星期六
收件人:cocktai...@googlegroups.com
主题:Re: real Grassmannian
Hi, I think it is true( at least rank of E is big), the reason is the cohomology with coeff=Q of real Grass is poly generate by the pontragin class of the universal bundle on it, and using Leray theorem. Please look at milnor's Charateristic Class for the cohomology with Q of real Grass.
Best
Xuanyu
---11年4月8日，周五,gufan...@gmail.com<gufan...@gmail.com>写道：
发件人:gufan...@gmail.com<gufan...@gmail.com>
主题: Re: real Grassmannian
收件人:cocktai...@googlegroups.com
日期: 2011年4月8日,周五,上午11:43
I would vote yes for the 1st question. I would even say H^*(Gr(2,E)) is a free module over H^*(M).
For the 2nd, I know Milne Stasheff talks about it.
Best,
Gufang
On , 1987xk47 <1987...@163.com> wrote:
> Hi all,
>
> I'm thinking about the following problem.
>
> For a smooth 2m-dim'l real vector bundle E over M, we can define Gr(2, E) to be a
> fiber bundle over M, with fiber the real Grassmannian Gr(2, E_p), p \in M.
>
> Does anyone know whether the map from H^*(M) to H^*(Gr(2,E)) (De Rham cohlgy)
> is injective or not? Or does anyone know some reference about the De Rham cohlgy of
> real Grassmannian?
>
> Best
> Ke
>
>
>
>
>
>
>
>
>
>
> --
>
> You received this message because you are subscribed to the Google Groups "Prof. Shiu-chun Wong's Cocktail Seminar" group.
>
> To post to this group, send email tococktai...@googlegroups.com.
>
>
>
> For more options, visit this group at http://groups.google.com/group/cocktailseminar?hl=en.
> --
You received this message because you are subscribed to the Google Groups "Prof. Shiu-chun Wong's Cocktail Seminar" group.
To post to this group, send email tococktai...@googlegroups.com.
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Pan Xuanyu
Apr 8, 2011, 9:51:13 PM4/8/11
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| I mean milnor's book Characteristic Classes ( It is an exercise of Chapter 14 or 15, I do not remember). Best Xuanyu --- 11年4月9日，周六, 1987xk47 <1987...@163.com> 写道：
|
|
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