# Hatcher 2.1.5

Hatcher, Algebraic Topology, Chapter 2, Section 1

5. Compute the simplicial homology groups of the Klein bottle using the $\Delta$-complex structure described at the beginning of the section.

Proof. Recall that the $\Delta$-complex structure referred to here is as follows:

Under this structure, we have that $C_2(K)=\langle U,L\rangle\cong\mathbb{Z}^2$, $C_1(K)=\langle a,b-c,c\rangle\cong\mathbb{Z}^3$,1 and $C_0(K)=\langle v\rangle\cong\mathbb{Z}$. This gives rise to the following chain complex where $X=K$:

The maps $\partial_i$ behave as follows: $\partial_2:U\mapsto a+b-c$ and $\partial_2:L\mapsto a-b+c$. In particular, then, $\partial_2:n_1U+n_2L\mapsto a(n_1+n_2)+b(n_1-n_2)+c(n_2-n_1)$, which implies that $\text{ker}(\partial_2)=\left\{n_1,n_2\in\mathbb{Z}\,:\,n_1=n_2\text{ and }n_1=-n_2\right\}$, i.e. $\text{ker}(\partial_2)=\left\{n_1,n_2\in\mathbb{Z}\,:\,n_1=n_2=0\right\}=0$. Moreover, the (reduced) image of $\partial_2$ can be computed by way of the following matrix computation:

$\text{Im}(\partial_2)=\left(\begin{array}{c}\partial_2(U)\\[0.5em]\partial_2(L)\end{array}\right)=\left(\begin{array}{ccc}1 & 1 & -1 \\[0.5em]1 & -1 & 1\end{array}\right)\underset{R_2=R_2+R_1}{\longrightarrow}\left(\begin{array}{ccc}1 & 1 & -1 \\[0.5em]2 & 0 & 0\end{array}\right)$.

Hence, $\text{Im}(\partial_2)=\langle a+b-c,2a\rangle$. Fortunately, the other maps are easier to analyze: $\partial_1:a,b,c\mapsto v-v=0$ so that $\text{Im}(\partial_1)=0$ and $\text{ker}(\partial_1)=\langle a,b-c,c\rangle$, and $\partial_0\equiv0$ implies that $\text{Im}(\partial_0)=0$, $\text{ker}(\partial_0)=C_0(K)$. Now then:

$H_2(K)=\text{ker}(\partial_2)/\text{Im}(\partial_3)=0/0\cong 0$,

$H_1(K)=\text{ker}(\partial_1)/\text{Im}(\partial_2)=\langle a,b-c,c\rangle/\langle a+b-c,2a\rangle$,

and

$H_0(K)=\text{ker}(\partial_0)/\text{Im}(\partial_1)=C_0(K)/0\cong\mathbb{Z}$.

Of course, the above representation of $H_1(K)$ is hardly satisfying, so to remedy that, we turn to a little algebra.

In the quotient, $a+b-c=0=2a$. Note that $2a=0$ implies that the homology group will have torsion in the form of a $\mathbb{Z}/2\mathbb{Z}$ summand. Otherwise, $a+b-c=0$ precisely when $b-c=-a$. Therefore, the “numerator group” $\langle a,b-c,c\rangle$ is equivalent to the group $\langle a,c\rangle$, while the “denominator” group is precisely $\langle 0,2a\rangle=\langle 2a\rangle$. Hence, $H_1(K)$ has the form

$H_1(K)=\langle a,c\rangle/\langle 2a\rangle \cong \langle c\rangle\oplus \langle a\rangle/\langle 2a\rangle \cong \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.

$\square$

1. The choice of $b-c$ for the second component is allowable due to the fact that the matrix with rows $(1,0,0), (0,1,-1), (0,0,1)$ row reduces to the identity matrix, i.e. the basis $a,b-c,c$ is equivalent to the basis $a,b,c$. Moreover, considering the computations that follow, this basis makes the final computation more clear. Even clearer would have been the (also equivalent) basis $a,a+b-c,c$, a fact I didn’t realize until later.