Homology of cube with a twist

Solution 1:

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 $d_3=0$.

Solution 2:

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.