
#1
Jan112, 07:09 PM

P: 5

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

P: 5

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



#4
Jan112, 07:51 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

A Conceptual Question on de Rham cohomology. 



#5
Jan212, 10:53 PM

Sci Advisor
HW Helper
P: 9,422

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




#6
Jan1212, 03:05 PM

Sci Advisor
P: 1,168

How do you define ncocycles and ncoboundaries?




#7
Jan1212, 06:30 PM

Sci Advisor
HW Helper
P: 9,422

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

Sci Advisor
P: 1,168

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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