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How to compute Betti numbers
One can acquire the Betti numbers from the homology groups (and cohomology) by taking their dimensions/ranks. However, sometimes we need only to count the topological features. The we should compute the Betti numbers directly.
In this article, we re-write How to compute homology with simplification that this approach allows us.
We summarize the procedure for computing the Betti numbers of a cell complex, by hand.
Contents
Procedure
Step 1: present the cell complex
List the cells of the complex $K$. Try to find the most economical representation - start with the highest dimension. Then merge the cells whenever possible and use cell collapse as well. Indicate the orientations. List the boundaries of the cells.
Step 2: find the chain groups
The $k$-chain group $C_k(K)$ is given as a vector space with basis consisting of the cells of the complex:
Its dimension is then obvious.
Step 3: find the matrices of the boundary operators
The $k$-boundary operator
$${\partial}_k: C_k(K) {\rightarrow} C_{k-1}(K)$$
is a linear operator. As such it is determined by its values on the elements of the basis of the domains space $C_k(K)$, the $k$-cells, as they are expressed as linear combinations of the elements of the basis of the target space $C_{k-1}(K)$, the $(k-1)$-cells. The coefficients of these linear combinations form the columns of the matrix.
Step 4: find the dimensions of the cycle groups
The $k$-cycle group is the kernel of the $k$-boundary operator:
$$Z_k(K) = \ker {\partial}_k.$$
As such it is the solution of the matrix equation:
It is a subspace of the chain group $C_k(K)$. One finds it by finding the rank of the matrix of ${\partial}_k$:
$$\dim Z_k(K) = \dim C_k(K) - {\rm \hspace{3pt} rank \hspace{3pt}} {\partial}_k. $$
Shortcuts:
- if ${\partial}_k$ is $0$, its kernel is the whole domain space and:
$$\dim Z_k(K) = \dim C_k(K).$$
- if ${\partial}_k$ is one-to-one, its kernel is $0$:
$$\dim Z_k(K) = 0.$$
Step 5: find the dimensions of the boundary groups
The $k$-boundary group is the image of the $(k+1)$-boundary operator:
$$B_k(K) = {\partial}_{k+1}.$$
As such it is spanned by the images of the basis of the $(k+1)$-cycle group, $(k+1)$-cells:
It is a subspace of the cycle group $Z_k(K)$. One finds it by finding the rank of the matrix:
$$\dim B_k(K) = {\rm \hspace{3pt} rank \hspace{3pt}} {\partial}_{k+1}. $$
Shortcuts:
- if ${\partial}_{k+1}$ is onto, its image is the whole target space and:
$$\dim B_k(K) = \dim C_k(K).$$
- if ${\partial}_{k+1}$ is $0$, its image is $0$:
$$\dim B_k(K) = 0.$$
Step 6: find the Betti numbers
The homology group is the quotient vector space of cycles over boundaries:
$$H_k(K) = Z_k(K) / B_k(K).$$
It is a vector space of dimension
$$\beta_k(K) = \dim Z_k(K) - \dim B_k(K).$$
Shortcuts: $$\dim Z_k(K) = \dim B_k(K) \rightarrow \beta_k(K) = 0,$$
$$\dim B_k(K) = 0 \rightarrow \beta_k(K) = \dim Z_k(K).$$
Example
Step 1: present the cell complex
One may choose to triangulate the sphere and then add the string (first image). A more efficient way is to cut the sphere into two hemispheres (second image). Turns out, one can remove one of the edges and merge the hemispheres. One $2$-cells is enough (third image).
Cells and their boundaries: $$\dim 2: {\tau}, {\partial}_2{\tau} = a - a = 0;$$ $$\dim 1: a, c, {\partial}_1a = B - A, {\partial}_1c = B - A;$$ $$\dim 0: A, B, {\partial}_0A = 0, {\partial}_0B = 0.$$
Step 2: find the chain groups
$$C_2(K) = {\rm \hspace{3pt} span \hspace{3pt}} \{\tau \} = {\bf R},$$ $$C_1(K) = {\rm \hspace{3pt} span \hspace{3pt}}\{a, c \} = {\bf R}^2,$$ $$C_0(K) = {\rm \hspace{3pt} span \hspace{3pt}} \{A, B \} = {\bf R}^2.$$
Step 3: find the matrices of the boundary operators
$${\partial}_2: C_2(K) {\rightarrow} C_1(K), {\bf R} {\rightarrow} {\bf R}^2,$$
$${\partial}_2 = [0, 0]^T = \left| \begin{array}{} 0 \\ 0 \end{array} \right| ;$$
$${\partial}_1: C_1(K) {\rightarrow} C_0(K), {\bf R}^2 {\rightarrow} {\bf R}^2,$$
$${\partial}_1 = \left| \begin{array}{rr} -1 & -1 \\ 1 & 1 \end{array} \right| ;$$
$${\partial}_0: C_0(K) {\rightarrow} 0, {\bf R}^2 {\rightarrow} 0,$$
$${\partial}_0 = [0, 0]^T = \left| \begin{array}{} 0 \\ 0 \end{array} \right| .$$
$$0 {\rightarrow} C_0(K) {\rightarrow} C_0(K) {\rightarrow} C_0(K) {\rightarrow} 0.$$
Step 4: find the dimensions of the cycle groups
$${\partial}_2 = 0 \rightarrow \dim Z_2(K) = 1.$$
$${\partial}_1a = {\partial}_1c = B - A \rightarrow {\rm \hspace{3pt} rank \hspace{3pt}} {\partial}_1 = 1 \rightarrow \dim Z_1(K) = 1.$$
$${\partial}_0 = 0 \rightarrow \dim Z_0(K) = 2.$$
Step 5: find the dimensions of the boundary groups
$$\dim B_2(K) = 0,$$ $$\dim B_1(K) = 0,$$ $$\dim B_0(K) = 1.$$
Step 6: find the Betti numbers
$$\beta_2(K) = \dim Z_2(K) - \dim B_2(K) = 1;$$ $$\beta_1(K) = \dim Z_1(K) - \dim B_1(K) = 1;$$ $$\beta_0(K) = \dim Z_0(K) - \dim B_0(K) = 1.$$