I want to calculate the homology of the $3$-torus via cellular homology. I figured out a CW-decomposition of the $3$-torus: $1$ $0$-cell, $3$ $1$-cell, $3$ $2$-cell, $1$ $3$-cell. So the chain complex looks like:

$0 \to \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to 0$

The first differential is $0$ since there is only one $0$-cell. The second differential is also $0$.

So my question is what is the third differential? I can't figure out the attaching map, so I can calculate the degree.

  • $\begingroup$ Just to be clear, you know how many cells there are in each dimension, but you do not know how those cells are put together to form the CW decomposition? $\endgroup$
    – Lee Mosher
    Aug 22, 2015 at 15:04
  • $\begingroup$ I know how to attach the $2$-cells, but I don't know how to attach the $3$-cell. $\endgroup$
    – Cosmare
    Aug 22, 2015 at 15:07
  • $\begingroup$ Hint: The third differential is also 0. Consider the fundamental cube of the 3-torus and try to show that the compositions $$f_i:S^2\rightarrow X^2\rightarrow X^2/ \left( X^2\setminus U_i \right)\cong S^2$$ have degree zero where the first map is the attaching map of the 3-cell and the second map is the quotient map collapsing everything except the 2-cell $U_i$ to a point. $\endgroup$ Aug 22, 2015 at 15:40
  • $\begingroup$ if I'm allowed to do cheating then this might work... since 3-torus is a compact orientable manifold so its top dimensional homology group is $\mathbb{Z}$ ... and we know that homology group and cw-homology matches...so the third differential has to be 0. $\endgroup$ Aug 22, 2015 at 15:59
  • $\begingroup$ @iwriteonbananas I don't know the first map though, that's kind of the whole problem. $\endgroup$
    – Cosmare
    Aug 22, 2015 at 16:03

2 Answers 2


You need an understanding of the attaching map of the 3-cell in order to compute the differential $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$.

Think of $T^3$ as the quotient of the cube $[-1,+1]^3$ by identifying $(x,y,-1) \sim (x,y,+1)$, $(x,-1,z)\sim(x,+1,z)$, and $(-1,y,z) \sim (+1,y,z)$.

Let $q : [-1,+1]^3 \to T^3$ be the quotient map.

The $0$-skeleton of $T^3$ consists of one $0$-cell, namely the image of the eight corners of the cube. The $1$ skeleton of $T^3$ consists of three $1$-cells, namely the images of the three sets of four parallel sides of the cube. The $2$-skeleton consists of three $2$-cells, namely the images of the three sets of two parallel faces of the cube.

Let's name the 2-cells of $T^3$:

  1. $\sigma_x = q \bigl( \{\pm 1\} \times [-1,+1] \times [-1,+1] \bigr)$
  2. $\sigma_y = q \bigl( [-1,+1] \times \{\pm 1\} \times [-1,+1] \bigr)$
  3. $\sigma_z = q \bigl( [-1,+1] \times [-1,+1] \times \{\pm 1\} \bigr)$

Let's orient these 2-cells. Using the right hand rule, orient each face of the cube by letting your thumb point in the positive direction of the axis normal to that face; let's refer to these as the "positive orientations" of the six faces of the cube. For example, the faces $$\sigma^-_z = [-1,+1] \times [-1,+1] \times - 1 \quad\text{and}\quad\sigma^+_z = [-1,+1] \times [-1,+1] \times + 1$$ are positively oriented by letting your thumb point in the positive $z$-direction. These two positive orientations are compatible with each other under the gluing, therefore they descend to the same orientation on the quotient 2-cell $\sigma_z \subset T^3$.

Let's compute the differential. The unique 3-cell of $T^3$ is parameterized by the entire cube $[-1,+1]^3$. The boundary of this cube is given the "outward orientation", defined by using the right hand rule with your thumb pointing outward from the inside of the cube to the outside. Notice: the outward orientation on $\sigma^+_z$ agrees with the positive orientation on $\sigma^+_z$, whereas the outward orientation on $\sigma^-_z$ disagrees with the positive orientation on $\sigma^-_z$. It follows that the $z$-coefficient of the difreremtial $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ equals $-1+1=0$. The $x$ and $y$ coefficients are zero for similar reasons.

The differential $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ is therefore the zero map.


This answer is beyond the scope or level of the question, but it may be of interest to some to note that there is a notion which reflects the geometry of the $3$-cube and $3$-torus better then the usual chain complex. This uses the crossed complex $\Pi X_*$ defined for a filtered space $X_*$, and in particular for a CW-complex $X$ with its skeletal filtration $X_*$. Here $\Pi X_*$ is defined using the fundamental groupoid $C_1=\pi_1(X_1,X_0)$ and the relative homotopy groups $$C_n= \pi_n(X_n,X_{n-1},x), n \geqslant 2, x \in X_0,$$ and has boundaries $\delta_n: C_n \to C_{n-1}$ and operations of $C_1$ on $C_n$. This idea for the single point case goes back to Blakers (1948) and to Whitehead's paper "Combinatorial Homotopy II".

In the crossed complex $\Pi I^3_*$, let $c^3$ be the top dimensional cell and for $\alpha=-,+$ let $c^\alpha_i=\partial^\alpha_i c^3$ be the two faces in direction $i$ , the first for $\alpha = -$, the last for $\alpha = +$. The formula for the boundary is then as follows: $$\delta_3 c^3= - c^+_3 - ( c^-_2)^{u_{2}c} - c^-_1 + ( c^-_3)^{u_{3}c} + c^+_2 + ( c^-_1)^{u_{1}c} $$ (where $u_i = \partial^+_1 \cdots \hat{\imath} \cdots \partial^+_{3}$).

Notice that if we identify $c^-_i=c^+_i$ we get a non zero formula for the $3$-torus. This formula gets nearer to modelling the geometry of the attaching map of the $3$-cell of the $3$-torus. If we cancel out the operations, and make it abelian, we get $0$, as given by Lee.

This is a specialisation of a formula for dimension $n$ known as the "Cubical Homotopy Addition Lemma", given in the EMS Tract vol 15 partially titled Nonabelian Algebraic Topology.


You must log in to answer this question.

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