A Conceptual Question on de Rham cohomology.

1. Jan 1, 2012

T_Mart

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.

2. Jan 1, 2012

Hurkyl

Staff Emeritus
Well, how is the group defined?

3. Jan 1, 2012

T_Mart

The group is defined as
HrD (M) = Ker(dr)/Im(dr-1)

4. Jan 1, 2012

Hurkyl

Staff Emeritus
And what groups is dr a homomorphism from and to?

5. Jan 2, 2012

mathwonk

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

6. Jan 12, 2012

Bacle2

How do you define n-cocycles and n-coboundaries?

7. Jan 12, 2012

mathwonk

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. Jan 12, 2012

Bacle2

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.