# Rank of a cohomology group, Betti numbers.

How is the rank of a cohomology group computed and what does it convey? I am trying to understand the concept behind betti numbers in a simplicial homology.

Edited with details:

Given a set of nodes/vertices/points X, let $C_{l}(X)$ denote the subsets of $X$ with cardinality $|C_{l}(X)|=l+1$. $\partial_{l}$ and $\delta_{l}$ are bounded linear maps with $\partial_{l}$:$C_{l-1}(X)\rightarrow C_{l}(X)$ satisfying $$\partial_{l-1}\circ\partial_{l}=0$$ and are also known as the boundary$(\partial_{l})$and co-boundary operators($\delta_{l}$). Here, $\delta_{l}=\partial_{l}^{*}$ is the adjoint operator $\delta_{l}$:$C_{l}(X)\rightarrow C_{l+1}(X)$ with $$\delta_{l}\circ\delta_{l-1}=0$$. The following is the definition of a co-boundary operator $$(\delta_{l}\, f)(x_{0},x_{1},...,x_{l+1})=\sum_{i=0}^{l+1}(-1)^{i}f(x_{0},x_{1},...\hat{x_{i}}..,x_{l+1})$$ and $\hat{x_{i}}$ indicates that it is omitted from the summation. Example to, Verify this: $$f(s_{i},s_{j},s_{k},s_{l})= f(s_{j},s_{k},s_{l})-f(s_{i},s_{k},s_{l})+f(s_{i},s_{j},s_{l})-f(s_{i},s_{j},s_{k})$$ $$= f(s_{k},s_{l})-f(s_{j},s_{l})+f(s_{j},s_{k})-f(s_{k},s_{l})+f(s_{i},s_{l})-f(s_{i},s_{k})+f(s_{j},s_{l})-f(s_{i},s_{l})$$ $$+f(s_{i},s_{j})-f(s_{j},s_{k})+f(s_{i},s_{k})-f(s_{i},s_{j})=0$$

I want a matrix representation of the coboundary and boundary operators. How is that representation done? I am not aware of the computational aspects of building a co-boundary operator in a matrix form from a simplicial complex. I believe if we have a matrix representation for one operator, the other would be its transpose.

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Cohomology groups are just abelian groups, so their rank is computed like the rank of any other abelian group... Your question is not very specific, and is more or less of the form «tell me about X». Can you be more specific about what you want to know? – Mariano Suárez-Alvarez Feb 9 '12 at 20:55
@Mariano- I have edited the question now. – user23600 Feb 9 '12 at 21:43

You are correct. Given a chain complex, and a specified basis for each chain group, consider the dual basis. Then the matrix of the coboundary operator is the transpose of the boundnary operator matrix. Also the rank of $H^i(X)$ is equal to the dimension of $H^i(X;\mathbb Q)$ as a rational vector space, and this is isomorphic to $H_i(X;\mathbb Q)$. So in fact the ranks of the cohomology groups are equal to the ranks of the corresponding homology groups, even though the torsion will shift around.