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Take the quotient space of the cube $I^3$ obtained by identifying each square face with opposite square via the right handed screw motion consisting of a translation by 1 unit perpendicular to the face, combined with a one-quarter twist of its face about it's center point.

I am trying to calculate the homology of this space.

It is not too hard to see that the CW decomposition of this space has 2 0-cells, 4 1-cells, 3 2-cells and 1 3-cell.

We end up (drawings would help here, but my MS-Paint skills are poor!) with the 2 0-cells ($P$ and $Q$) connected by the 4 1-cells $a,b,c,d$ with $a,c$ from $P$ to $Q$ and $b,d$ from $Q$ to $P$. Thus we have the closed loops $ab,ad,cb,cd$. They also satisfy the relations $abcd=1,dca^{-1}b^{-1}=1,c^{-1}adb^{-1}=1$ via the identification of opposite 2-cells (top/bottom, left/right, up/down). (There is a relationship between the generator loops - the fundamental group is the quaternion group).

From the CW decomposition we get the cellular chain complex

$0 \to \mathbb{Z} \stackrel{d_3}{\to} \mathbb{Z}^3 \stackrel{d_2}{\to} \mathbb{Z}^4 \stackrel{d_1}{\to} \mathbb{Z}^2 \to 0$

I'm struggling to work out the boundary maps. Can it be 'seen' easily from the relations above?

I tried to use the cellular boundary formula. $d_1$ must be a 2 x 4 matrix. The cellular boundary formula gives the relation

$$d_1(e^1_\alpha) = \sum_{\beta=1}^2 d_{\alpha \beta} e^0_\beta$$

Are the entries of the matrix $d_1$ then given by $$\left(\begin{array}{cccc} d_{11} & d_{21} & d_{31} & d_{41} \\ d_{12} & d_{22} & d_{32} & d_{42} \\ \end{array}\right)?$$

I am pretty sure that $d_{\alpha \beta}$ must be $-1$ or $1$ as the attaching map is a homeomorphism (and is not 0), and is dependent on orientation. Therefore I get that $$d_1 = \left(\begin{array}{cccc} 1 & 1 & 1 & 1\\ -1 & -1 & -1 & -1\\ \end{array}\right).$$

Similar logic says that $d_2$ is a 4x3 matrix. Again all entries must be 1 or -1. I'm struggling to see exactly what the boundary map should be here?

Any thoughts on the best approach are appreciated.

Juan S
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  • This may be tellingly elementary , but what is the homology of the cube with no twists at all? What about with a 1/2 twist in only one opposite pair? What about 1/2 twist in all three pairs – Mitch Apr 04 '11 at 14:33
  • draw a picture, orient everything, and use the definition of the boundary map if you're having trouble – yoyo Apr 04 '11 at 14:38
  • I think your $d_1$ should have some $1$s and $-1$s interchanged. Otherwise $d_2d_1\neq 0$, which it needs to be. Basically you have to make sure that $d_1$ of an oriented edge is the vertex at the head minus the vertex at the tail. – Cheerful Parsnip Apr 04 '11 at 16:04

2 Answers2

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Pick orientations for the three $2$-cells which are the square faces. Then $d_2$ picks up the sum of the four edges with a sign determined by whether the edge orientation is induced by the square's orientation. So for example, the way you've set things up one of the squares has $a,b,c,d$ on the boundary with consistent orientations, so $d_2$ of that cell will be $a+b+c+d$ (or a column vector with 4 ones.) Similarly the cell that gives you $dca^{-1}b^{-1}$ has boundary $c+d-a-b$. So, according to your calculations $$d_2=\left(\begin{array}{ccc}1&-1&1\\ 1&-1&-1\\ 1&1&-1\\ 1&1&1\end{array}\right)$$

You can calculate $d_3$ the same way. Pick an orientation on the $3$-cell which is the interior of the cube, and see how the squares sit on its boundary. In fact, each square appears twice with opposite induced orientation, so $d_3=0$.

Cheerful Parsnip
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  • thanks, that helps a lot. And I think you are correct about $d_2$ it should be alternating colums of 1,-1 and -1,1 - which then gives $d_1 d_2 = 0$ as hoped – Juan S Apr 04 '11 at 23:58
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Just to compute a little more out for myself and others for $H_1(X,\mathbb{Z})$ ...

The boundary map $d_2: \mathbb{Z}^4 \rightarrow \mathbb{Z}^3$ has matrix representation

$$d_2=\left(\begin{array}{ccc}1&-1&1\\ 1&-1&-1\\ 1&1&-1\\ 1&1&1\end{array}\right)$$ which over $\mathbb{Z}$ reduces to

$$d_2=\left(\begin{array}{ccc}1&-1&1\\ 0&2&0\\ 0&0&2\\ 0&0&0\end{array}\right)$$, so $Im(d_2)=<a-b+c,2b,2c>$.

And$$d_1=\left(\begin{array}{ccc}1&-1&1&-1\\ -1&1&-1&1\end{array}\right)$$, so $Ker(d_1)=\{(a,b,c, d)| d=a-b+c\} =<a-b+c,b,c>$.

So $H_1(X,\mathbb{Z})=Ker(d_1)/Im(d_2)=<a-b+c,b,c>/<a-b+c,2b,2c>=\mathbb{Z}_2\oplus \mathbb{Z}_2,$ which makes sense since this is the abelianization of the fundamental group which is the quaternion group.

Ashley
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