## 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.

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 $H_k(K)=\begin{cases}\mathbb{Z}&k=0\\ \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}&k=1\\ 0&\text{otherwise} \end{cases}$