algebraic-topologyhomology-cohomology
Interesting topological spaces to calculate the homology groups.
I am calculating homology groups of several topological spaces to learn and I have already calculated the homology groups of $\mathbb{S}^m$, $\mathbb{R}P^2$, the Klein bottle, $\mathbb{R}^2$ minus a finite number of points and I am going to calculate the homology groups to the moebius band and I was wondering what other interesting topological spaces I can easily calculate the homology groups, thank you very much.
I think this is easiest to see using cellular homology. There is only one $2$-cell $U$, so $C_2 \cong \mathbb{Z}$. The boundary of this $2$-cell is the product of the commutators of the $2g$ $1$-cells: $[a_1, b_1] \cdots [a_g, b_g]$ where $[a_i, b_i] = {a_i}^{-1} {b_i}^{-1} a_i b_i$. For instance, if you traverse the boundary of the octagon you've drawn to represent a genus $2$ surface, you'll see that we traverse each edge twice, once in the positive direction, and once in the negative direction. See here for a reference. However, since the homology groups are abelian groups, these commutators are trivial, i.e., $$ \partial_2(U) = -a_1 - b_1 + a_1 + b_1 + \cdots + -a_g - b_g + a_g + b_g = 0 \, . $$ Since $\partial_2$ maps the generator of $C_2$ to $0$, then $\partial_2 = 0$, hence $\ker \partial_2 = C_2 \cong \mathbb{Z}$.
Although you've got the answer by yourself, I would like to write an answer solving the problem with cellular homology, so that someone who asks the same question can find an answer here. I solved this problem a few month ago in an Algebraic Topology course as an exercise.
Proof:
Let $X = S^1\times S^1/ \sim$ be the space with the identifications:
$$(e^{2\pi i/m}z,x_0)\sim (z,x_0)$$
$$(x_0,e^{2\pi i/n}z)\sim (x_0,z)$$
Like @Berci said, you should imagine this space as a grid of $m$ and $n$ lines, i.e. there are $m$ vertical and $n$ horizontal repititions:
(OK. The picture is not the nicest one, but it's enough to induce an imagination.)
$X$ consists of one 0-cell ($x_0$ is $e_1^0$), two 1-cells ($a$ is $e_1^1$, $b$ is $e_2^1$) and one 2-cell (we all it $e_1^2$).
The attaching map identifies $x\in \partial D_1^2$ with $a^nb^ma^{-n}b^{-m}$.
This implies the cellular chain complex
$$0\to \mathbb{Z}[e_1^2]\overset{\partial_2=0}{\longrightarrow} \mathbb{Z}[a] \oplus\mathbb{Z}[b]\overset{\partial_1=0}{\longrightarrow} \mathbb{Z}[x_0]\to 0.$$
This implies
$$H_p(X) = \begin{cases} \mathbb{Z}\mbox{ for } p=0,2 \\ \mathbb{Z}^2\mbox{ for } p=1 \\ 0\mbox{ for } p>2 \end{cases}.$$
Otherwise you can just see, that the space $X$ is still a Torus (cf. remark above). So it is not surprising, that we've got the homology group of the Torus.
Best Answer
Not sure whether you consider them to be interesting, but these are some spaces whose homology groups I once computed in the past when I was studying algebraic topology:
1) The torus $T^2 = S^1 \times S^1$ or more generally $T^n = S^1 \times \dots \times S^1$
2) The space you get when you take $S^2$ and identify the north and south poles
3) Take $T^2 = S^1 \times S^1$ and quotient out the circle $S^1 \times \lbrace x \rbrace$ for some point $x \in S^1$
4) Take $T^2 = S^1 \times S^1$ and quotient out two different circles $S^1 \times \lbrace x \rbrace$ and $S^1 \times \lbrace y \rbrace$
5) The space $X$ one gets by glueing two solid tori $S^1 \times D^2$ along their boundaries via the identity $S^1 \times S^1 \rightarrow S^1 \times S^1$
6) The dunce hat/cap (Take a solid triangle and identify the sides by the edge word $a^3$ (or $a^2a^{-1}$ - definition varies a bit in the literature)
7) If you managed to compute your examples plus these and you still want more, then I suggest that you just construct some examples and try to compute the homology groups. Sometimes it will work out and otherwise you can still ask questions here.
I will not be able to remember all the homology groups for you to compare though.