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Cohomology of figure 8

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Cohomology of figure 8

Unlike the examples above, we compute the cohomology of the figure $8$ here without shortcuts by carrying out all required row operations.

TopologicalFigure8.png

This is our cubical complex:

Cubical figure 8.png

These are the bases: $$\begin{array}{|c|cccccccc|} \hline C^0& \varphi_A & \varphi_B & \varphi_C & \varphi_D & \varphi_E & \varphi_F \\ \hline A & 1 & 0 & 0 & 0 & 0 & 0 \\ B & 0 & 1 & 0 & 0 & 0 & 0 \\ C & 0 & 0 & 1 & 0 & 0 & 0 \\ D & 0 & 0 & 0 & 1 & 0 & 0 \\ E & 0 & 0 & 0 & 0 & 1 & 0 \\ F & 0 & 0 & 0 & 0 & 0 & 1\\ \hline \end{array}, \begin{array}{|c|cccccccc|} \hline C^1& \psi_a & \psi_b & \psi_c & \psi_d & \psi_e & \psi_f & \psi_g \\ \hline a & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ b & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ c & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ d & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ e & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ f & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ g & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \hline \end{array}.$$ And this is our cochain complex: $$C^0 \stackrel{\partial^0}{\longrightarrow} C^1 \stackrel{\partial^1}{\longrightarrow} C^2 = 0.$$

Let us compute what $\partial^0$ does to each basis element of $C^0$ and write the result in terms of bases of $C^1$.

$$\begin{array}{} \partial^0 \varphi_A = -\psi_a + \psi_f; & \partial^0 \varphi_D = -\psi_c+\psi_d ;\\ \partial^0 \varphi_B = \psi_a - \psi_b + \psi_6; & \partial^0 \varphi_E = -\psi_d + \psi_e - \psi_g; \\ \partial^0 \varphi_C = \psi_b + \psi_d; & \partial^0 \varphi_F = -\psi_e - \psi_f. \end{array}$$

Thus as a matrix, the coboundary operator is $$\partial^0 = \left[ \begin{array}{r} -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 & 0 \end{array} \right] \begin{array}{l} r_0 \\ r_1 \\ r_2 \\ r_3 \\ r_4 \\ r_5 \\ r_6 \end{array}$$

We now begin applying elementary row operations. $$\begin{array}{l} r_5^*=r_0+r_5 \end{array} \left[ \begin{array}{r} -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 & 0 \end{array} \right] \begin{array}{l} r_0^* = r_0+r_1 \\ r_5^*=r_5+r_1 \\ r_6^* = r_6+r_1 \end{array} \left[ \begin{array}{l} -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & -1 & 0 \end{array} \right] \\ \begin{array}{} r_0^* = r_0-r_6 \\ r_1^* = r_1 - r_6 \\ r_3^* = r_3 - r_6 \\ r_5^* = r_5 - r_6 \end{array} \left[ \begin{array}{r} -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & -1 & 0 \end{array} \right] \begin{array}{} r_2^*=r_2+r_3 \\ r_6 \leftrightarrow r_2 \end{array} \left[ \begin{array}{r} -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 &- 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \\ \begin{array}{} r_5^* = r_5-r_4 \\ r_2^* = r_2+r_4 \\ r_1^* = r_1-r_4 \\ r_0^* = r_0-r_4 \end{array} \left[ \begin{array}{r} -1 & 0 & 0 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \begin{array}{} r_0^* = -r_0 \\ r_1^* = -r_1 \end{array} \left[ \begin{array}{r} 1 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] $$

This implies that $\operatorname{rank}\partial^0=5$, which means $\operatorname{Im}\partial^0 \cong{\bf R}^5$.

Now notice $\ker \partial^1 = C^1 \cong{\bf R}^7$. So, $$H^1 := \ker \partial^1 / \operatorname{Im}\partial^0 \cong{\bf R}^7 / {\bf R}^5 \cong {\bf R}^2 .$$

Exercise. In a similar fashion compute the cohomology of the following complexes:

  • (a)
    CubicalDisk.png
  • (b)
    CubicalMouse.png