Hatcher Exercise 2.1.6

We compute the simplicial homology of the complex described in the text.

First, we fix notation to refer to all the faces of the complex:

t_0 = \set{[v^0_0 v^0_1 v^0_2]}, \dots, t_n = \set{[v^n_0 v^n_1 v^n_2]}

\begin{align} & e_0 = \set{[v^0_0 v^0_2], [v^0_0 v^0_1], [v^0_1, v^0_2], [v^1_0 v^1_2]} \ & e_1 = \set{[v^1_0 v^1_1], [v^1_1 v^1_2], [v^2_0 v^2_2]} \ & \vdots \ & e_i = \set{[v^i_0 v^i_1], [v^i_1 v^i_2], [v^{i+1}_0 v^{i+1}_2]} \ & \vdots \ & e_n = \set{[v^n_0 v^n_1], [v^n_1 v^n_2]} \end{align}

v = \setbuilder{[v^i_j]}{0 \leq j \leq n, 0 \leq i \leq 2}

\begin{align} & \delta2(t_0) = \delta_2([v^0_0 v^0_1 v^0_2]) = [v^0_1 v^0_2] - [v^0_0 v^0_2] + [v^0_0 v^0_1] = (e_0) - (e_0) + (e_0) = e_0 \ & \delta_2(t_i) = \delta_2([v^i_0 v^i_1 v^i_2]) = [v^i_1 v^i_2] - [v^i_0 v^i_2] + [v^i_0 v^i_1] = (e_i) - (e{i-1}) + (ei) = 2e_i - e{i-1} \ & \delta2(t_n) = \delta_2([v^n_0 v^n_1 v^n_2]) = [v^n_1 v^n_2] - [v^n_0 v^n_2] + [v^n_0 v^n_1] = (e_n) - (e{n-1}) + (en) = 2e_n - e{n-1} \end{align}

\delta2 = \left[\begin{array}{c|cccccc} & t_0 & t_1 & t_2 & t_3 & \dots & t_n \ \hline e_0 & 1 & -1 & 0 & 0 & \dots & 0 \ e_1 & 0 & 2 & -1 & 0 & \dots & 0 \ e_2 & 0 & 0 & 2 & -1 & \dots & 0 \ e_3 & 0 & 0 & 0 & 2 & \dots & 0 \ \vdots & \vdots & \vdots & \vdots & \vdots & & \vdots \ e{n-1} & 0 & 0 & 0 & 0 & \dots & -1 \ en & 0 & 0 & 0 & 0 & \dots & 2 \end{array} \right] \xrightarrow{SNF} \left[\begin{array}{c|cccccc} & t_0 & t_1 & t_2 & t_3 & \dots & t_n \ \hline e_0 & 1 & 0 & 0 & 0 & \dots & 0 \ e_1 & 0 & 1 & 0 & 0 & \dots & 0 \ e_2 & 0 & 0 & 1 & 0 & \dots & 0 \ e_3 & 0 & 0 & 0 & 1 & \dots & 0 \ \vdots & \vdots & \vdots & \vdots & \vdots & & \vdots \ e{n-1} & 0 & 0 & 0 & 0 & \dots & 0 \ e_n & 0 & 0 & 0 & 0 & \dots & 2^n \end{array} \right]

$$

Where the matrix on the right is the Smith Normal Form of .

and higher are the zero map.

We have .

To find , we first conclude that , so is a torsion group. The torsion coefficients are given by the non-trivial entries of . We conclude that .

It's clear that is injective, so .