How to calculate Homology Groups (Klein Bottle)

This post will be a guide on how to calculate Homology Groups, focusing on the example of the Klein Bottle. Homology groups can be quite difficult to grasp (it took me quite a while to understand it). Hope this post will help readers to get the idea of Homology. Our reference book will be Hatcher’s Algebraic Topology (Chapter 2: Homology). I will elaborate further on the Hatcher’s excellent exposition on Homology.

This is also Exercise 5 in Chapter 2, Section 2.1 of Hatcher.

The first step to compute Homology Groups is to construct a \Delta-complex of the Klein Bottle.

klein bottle

One thing to note for \Delta-complexes, is that the vertices cannot be ordered cyclically, as that would violate one of the requirements which is to preserve the order of the vertices.

The key formula for Homology is: \boxed{H_n=\ker\partial_n/\text{Im}\ \partial_{n+1}}.

We have \ker\partial_0=\langle v\rangle, the free group generated by the vertex v, because there is only one vertex!

Next, we have \partial_1(a)=\partial_1(b)=\partial_1(c)=v-v=0. Thus \text{Im}\ \partial_1=0.

Therefore H_0=\ker\partial_0/\text{Im}\ \partial_1=\langle v\rangle /0\cong\mathbb{Z}.

Next, we have \ker\partial_1=\langle a,b,c\rangle. \partial_2U=a+b-c, \partial_2L=c+a-b. To learn more about calculating \partial_2, check out the diagram on page 105 of Hatcher.

We then have \text{Im}\ \partial_2=\langle a+b-c, c+a-b\rangle=\langle a+b-c, 2a\rangle, where we got 2a from adding the two previous generators (a+b-c)+(c+a-b).

Thus H_1=\ker\partial_1/\text{Im}\ \partial_2=\langle a,b,c\rangle/\langle a+b-c, 2a\rangle=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}.

To intuitively understand the above working, we need to use the idea that elements in the quotient are “zero”. Hence a+b-c=0, implies that c=a+b, thus c can be expressed as a linear combination of a, b, thus is not a generator of H_1. 2a=0 implies that a+a=0, which gives us the \mathbb{Z}/2\mathbb{Z} part.

Finally we note that \ker\partial_2=0, and also for n\geq 3, \ker\partial_n=0 since there are no simplices of dimension greater than or equal to 3. Thus, the second homology group onwards are all zero.

In conclusion, we have H_k(K)=\begin{cases}\mathbb{Z}&k=0\\    \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}&k=1\\    0&\text{otherwise}    \end{cases}

Author: mathtuition88

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