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what is the homology group for The quotient space of $S^1 \times S^1$ obtained by identifying points in the circle $S1 \times\{x_0\} $ that differ by $\frac{2 \pi}{m}$ rotation and identifying points in the circle $\{x_0\} \times S^1 $ that differ by $\frac{2 \pi}{n}$ rotation.

actually it is exercise 2.2.9.d from hatcher and I am really curious about its homology groups.

it will be awesome if you can show me an imagination of this space.and I really don't know what tools should I use to calculate homology groups,cellular one,simplicial one,mayer-vietoris...,please help me,it will be great if you give me guidance or hint,thank you very much.

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  • $\begingroup$ For a start, draw the torus represented by a square in which opposite edge points are identified. We can assume that $x_0$ is just 'the' corner of this square representation and the two principal circles are just the 'two' edges. Now make an $m\times n$ grid, at least on the edges. $\endgroup$
    – Berci
    May 27, 2014 at 12:45
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    $\begingroup$ the reduced homology of the quotient space is the same as the relative homology $H_\bullet(S^1\times S^1,A)$, where $A$ is the set of the identified points. Now use the long exact sequence for the pair $A\subset S^1\times S^1$. $\endgroup$
    – user8268
    May 27, 2014 at 13:09
  • $\begingroup$ thank you very much,I got it. $\endgroup$
    – kpax
    May 27, 2014 at 13:42
  • $\begingroup$ @user8268 Wait, how is $A$ a subspace? In particular, it seems like you're saying that our space is $X/A$ for some contractible subspace $A$? $\endgroup$ Feb 26, 2018 at 4:49
  • $\begingroup$ @user8268 Would you happen to know how to calculate how many points we are identifying? The homology groups depend on the size of $A$. $\endgroup$
    – user5826
    Apr 22, 2019 at 20:55

1 Answer 1

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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:

Hatcher 2.2.9

(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.

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  • $\begingroup$ Can you explain why the attaching map is $a^nb^ma^{−n}b^{−m}$ ? $\endgroup$
    – 6666
    Apr 17, 2016 at 21:49
  • $\begingroup$ Consider the attaching map of the torus, which is $aba^{-1}b^{-1}$. Because of the identification described above, you now have to attach the 2-cell along $a^nb^ma^{-n}b^{-m}$. $\endgroup$ Apr 21, 2016 at 7:40
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    $\begingroup$ Sorry, but I think, after identification, it's just the little square, thus should it be just $a^{1/n}b^{1/m}a^{-1/n}b^{-1/m}$? $\endgroup$
    – 6666
    Apr 21, 2016 at 7:44
  • $\begingroup$ Why is $\partial_2 = 0$? $\endgroup$
    – user5826
    Apr 22, 2019 at 4:53
  • $\begingroup$ Shouldn't there be $mn$ $2$-cells? $\endgroup$
    – user5826
    Apr 22, 2019 at 19:40

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