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A Conceptual Question on de Rham cohomology.by T_Mart
Tags: de rham 
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#1
Jan112, 07:09 PM

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Hi everybody,
Currently, I am studying cohomology on my own. I have a question: Why H ^{r}_{D}(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. 


#3
Jan112, 07:22 PM

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H^{r}_{D} (M) = Ker(d_{r})/Im(d_{r1}) 


#4
Jan112, 07:51 PM

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A Conceptual Question on de Rham cohomology.



#5
Jan212, 10:53 PM

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it follows from properties of the wedge product, as is being suggested.



#6
Jan1212, 03:05 PM

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How do you define ncocycles and ncoboundaries?



#7
Jan1212, 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. 


#8
Jan1212, 10:36 PM

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Yes, that was the point I was trying to make. Look up the definition of ncocycles and ncoboundaries 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 nmanifold.



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