# Math Help - calculation of cohomology group for a 0-dim manifold

1. ## calculation of cohomology group for a 0-dim manifold

Could someone explain to me this example? I know all the basis and definitions but I am new to calculation of cohomology group

here is the example

http://i51.tinypic.com/wb7609.jpg

Why $Z^{k}= \begin{cases} 0 for \mathbb{R} , k=0\\ 0 , k>0\end{cases}$?? why do we get a real numbers and then we get just 0?

and then boundary group $B^{k}(M)=0$ I guess that it is because a point does not have a boundary, is it correct and finally how did they get the cohomology group? how did they obtain again R and 0?

Thank you

2. Originally Posted by rayman
Could someone explain to me this example? I know all the basis and definitions but I am new to calculation of cohomology group

here is the example

http://i51.tinypic.com/wb7609.jpg

Why $Z^{k}= \begin{cases} 0 for \mathbb{R} , k=0\\ 0 , k>0\end{cases}$?? why do we get a real numbers and then we get just 0?

and then boundary group $B^{k}(M)=0$ I guess that it is because a point does not have a boundary, is it correct and finally how did they get the cohomology group? how did they obtain again R and 0?

Thank you
What are your $Z^k$, I can't seem to find their definition on their. Are they your boundary maps? Or, perhaps the kernels of your boundary maps?

3. $Z^{k}$ is just a cyclic group or in the language of cohomology $ker_{d+1}$
boundrary is the other one

4. Originally Posted by rayman
$Z^{k}$ is just a cyclic group or in the language of cohomology $ker_{d+1}$
boundrary is the other one
So, $Z^k$ is the free abelian group with rank $k$?

5. yes

6. Originally Posted by rayman
yes
But then your question doesn't make sense? How could $\mathbb{Z}^0\cong\mathbb{R}$?

7. well that is an example taken from the text book and it seems it can be, I will investigate this and if I come to some good conclusions I will write again

8. Are you sure it's not the 0-th cocycle group?

9. yes, you are right this is 0-th cocycle group