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:

Hatcher 2.1.5

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:

Hatcher 2.1.4

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,



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}.


    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.

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