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.

  • $\begingroup$ 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 $\endgroup$
    – Mitch
    Commented Apr 4, 2011 at 14:33
  • $\begingroup$ draw a picture, orient everything, and use the definition of the boundary map if you're having trouble $\endgroup$
    – yoyo
    Commented Apr 4, 2011 at 14:38
  • $\begingroup$ 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. $\endgroup$ Commented Apr 4, 2011 at 16:04

2 Answers 2


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 $d_3=0$.

  • $\begingroup$ 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 $\endgroup$
    – Juan S
    Commented Apr 4, 2011 at 23:58
  • $\begingroup$ Why is $d_2$ the sum of the four edges with signs corresponding to whether the arrow is flipped or not when (say) moving in clockwise direction? (and why is this true in general for the $d_i$? Hatcher goes through great lengths to show how the $d_i$ can be computed. How does this fit in to his description of the $d_i$?) $\endgroup$
    – Anon
    Commented Feb 13, 2023 at 18:32
  • 1
    $\begingroup$ @anon in cellular homology, the sign is induced by the degree of the map. For the case of an edge on the boundary of a disk, this degree is $\pm 1$ depending on whether the induced orientation matches. $\endgroup$ Commented Feb 13, 2023 at 20:43

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .