# Cellular Homology of the 3-Torus

I'm trying to compute the homology of the 3-torus $T^3=S^1 \times S^1 \times S^1$. Trying to use the typical construction the 2-torus $T^2$ as a starting point, I identified pairs of opposite faces on a cube as shown below: This gives a single 0-cell, three 1-cells, three 2-cells, and a single 3-cell. Still using my $T^2$ computations as a guide, I've gotten that the cellular chain complex for this space is

$0 \to \mathbb{Z} \overset{d_3} \longrightarrow \mathbb{Z}^3 \overset{d_2}\longrightarrow \mathbb{Z}^3 \overset{d_1}\longrightarrow \mathbb{Z} \to 0$

Since there is only one 0-cell, I get that $d_1=0$. I'm having trouble understanding how to compute $d_2$ and $d_3$. Could anyone shed some light on this?

For reference, this is problem IV.10.1 in Bredon's "Topology and Geometry".

• I saw that example, but I'm having trouble understanding specifically why $d_2=0$; he refers to the construction in the $T^2$ calculation but I couldn't find where he did that. – leeabarnett Apr 5 '13 at 3:11
• Use the cellular boundary formula. In the case of $T^2$, consider it as a square with opposite sides identified. Then the attaching map for the $2$-cell is $aba^{-1}b^{-1}$, so its boundary is the the kernel (the degree of $a$ and $a^{-1}$ cancel, etc) – Potato Apr 5 '13 at 3:15
• What exactly are the attaching maps in the $T^3$ case? – leeabarnett Apr 5 '13 at 3:21