This site is being phased out.

How to compute Betti numbers

From Mathematics Is A Science
Jump to navigationJump to search

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.

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:

$C_k(K) =$ span$\{k$-cells in $K\}$.

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:

find all $x \in C_k(K)$ such that ${\partial}_k(x) = 0$.

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:

$B_k(K) = {\rm \hspace{3pt} span}\{ \partial_{k+1}(s_i): s$ is a $(k+1)$-cell in $K \}$.

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).

Cell decomposition of sphere with string.jpg

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

$${\partial}_2: C_2(K) {\rightarrow} C_1(K), {\bf R} {\rightarrow} {\bf R}^2,$$

$${\partial}_2 = [0, 0]^T = \left| \begin{array}{} 0 \\ 0 \end{array} \right| ;$$

$${\partial}_1: C_1(K) {\rightarrow} C_0(K), {\bf R}^2 {\rightarrow} {\bf R}^2,$$

$${\partial}_1 = \left| \begin{array}{rr} -1 & -1 \\ 1 & 1 \end{array} \right| ;$$

$${\partial}_0: C_0(K) {\rightarrow} 0, {\bf R}^2 {\rightarrow} 0,$$

$${\partial}_0 = [0, 0]^T = \left| \begin{array}{} 0 \\ 0 \end{array} \right| .$$

Chain complex:

$$0 {\rightarrow} C_0(K) {\rightarrow} C_0(K) {\rightarrow} C_0(K) {\rightarrow} 0.$$

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

$$\beta_2(K) = \dim Z_2(K) - \dim B_2(K) = 1;$$ $$\beta_1(K) = \dim Z_1(K) - \dim B_1(K) = 1;$$ $$\beta_0(K) = \dim Z_0(K) - \dim B_0(K) = 1.$$