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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 \begin{equation}\partial_{l-1}\circ\partial_{l}=0\end{equation} 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 \begin{equation}\delta_{l}\circ\delta_{l-1}=0\end{equation}. The following is the definition of a co-boundary operator \begin{equation}(\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})\end{equation} 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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    $\begingroup$ 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? $\endgroup$ Feb 9, 2012 at 20:55
  • $\begingroup$ @Mariano- I have edited the question now. $\endgroup$
    – user23600
    Feb 9, 2012 at 21:43

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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.

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  • $\begingroup$ @ Jim,thanks for clarifying. What is the structure of these matrices? Are they like adjacency (0-1) matrices? $\endgroup$
    – user23600
    Feb 9, 2012 at 22:41
  • $\begingroup$ Suppose the simplicial complex consists of five 0-simplices: 1,2,3,4,5; seven 1-simplices: (12), (23), (13), (34), (14), (45), (15); and two 2-simplices (123) and (145). Note here all these simplices are pre-specified with an orientation. Then a map from 1-simplices to 2-simplices in this example can be written as (in matrix form) [1 1 -1 0 0 0 0; 0 0 0 0 1 1 -1] where the two rows represent the two 2-simplices and the seven columns represent the seven 1-simplices? $\endgroup$
    – user23600
    Feb 9, 2012 at 22:43
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    $\begingroup$ Yes. The coboundary operator sums over all simplices that have the given simplex as a codimension one face, with sign determined by orientation compatibility. $\endgroup$ Feb 9, 2012 at 22:48
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    $\begingroup$ See mathoverflow.net/questions/57166/…. Looks like the program Sage has some tools for this. $\endgroup$ Feb 9, 2012 at 22:54
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    $\begingroup$ Woah-just realized that, you are a Professor at Tennessee! $\endgroup$
    – user23600
    Feb 9, 2012 at 23:10

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