A Conceptual Question on de Rham cohomology.


by T_Mart
Tags: de rham
T_Mart
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#1
Jan1-12, 07:09 PM
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Hi everybody,

Currently, I am studying cohomology on my own. I have a question:

Why H rD(M) = 0, when r > n

n is the dimension of manifold M
My book says it is obvious, but to me it is not obvious.

I wish someone could explain this question to me.
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Hurkyl
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#2
Jan1-12, 07:12 PM
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Well, how is the group defined?
T_Mart
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#3
Jan1-12, 07:22 PM
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Quote Quote by Hurkyl View Post
Well, how is the group defined?
The group is defined as
HrD (M) = Ker(dr)/Im(dr-1)

Hurkyl
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Jan1-12, 07:51 PM
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A Conceptual Question on de Rham cohomology.


Quote Quote by T_Mart View Post
The group is defined as
HrD (M) = Ker(dr)/Im(dr-1)
And what groups is dr a homomorphism from and to?
mathwonk
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Jan2-12, 10:53 PM
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it follows from properties of the wedge product, as is being suggested.
Bacle2
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Jan12-12, 03:05 PM
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How do you define n-cocycles and n-coboundaries?
mathwonk
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Jan12-12, 06:30 PM
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there aren't even any ≠0 cochains in dimensions above the dimension of the manifold.

the reason is essentially that an nbyn determinant is always zero if the matrix has rank < n.
Bacle2
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#8
Jan12-12, 10:36 PM
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Yes, that was the point I was trying to make. Look up the definition of n-cocycles and n-coboundaries to see what the cohomology groups are . Or, if you have the right conditions for Poincare Duality, see why you cannot have (n+k)-cycles; k>0, in an n-manifold.


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