Homology of $n$-dimensional Torus

Question

This webpage describes the homology groups $H_k(T^n)$ of the $n$-dimensional torus as being $\mathbb{Z}^{{n}\choose{k}}$. There is a very helpful example in Hatcher, example 2.39, where he works through the case of $n=3$. It isn't obvious to me, however, what parts of his argument should be generalized for the computation in higher dimension, or are there some parts of his argument that cannot be generalized?

Answer 1

So the cleanest way will be to use the Kunneth formula, but if you do not want to use that you can work it out directly (and this even gives intuition to why the Kunneth formula might hold).

The obvious way to proceed is by using cellular homology. In order to do this, we really must understand the product CW structure. In thinking about how to answer this question, I was hoping it was somewhat obvious what the attaching maps should be (basically you want them to be higher order versions of $aba^{-1}b^{-1}$). Now it might be obvious for someone more geometrically minded than me, but I do not feel comfortable concluding something like that. The next thought is to use induction.

Here is the inductive hypothesis: The differential in the cellular complex of $T^n$ with the CW structure given by $T^n = S^1 \times T^{n-1}$ is $0$.

The base case holds (either $n=0,1,2$ all are obvious).

For the inductive step (if $n>1$):

The $k$-cells of $T^n$ are the products of the $k$-cells of $T^{n-1}$ with the $0$-cell of $S^1$ along with with the products of the $(k-1)$-cells of $T^{n-1}$ with the $1$-cell of $S^1$. The differential clearly is 0 on the first type from how the attaching map on the products are defined and our inductive hypothesis. For the second type, the attaching map of $\partial (e^k \times I)$ touches only the cells the attaching map for $\partial e^k$ touches multiplied by the $1$-cell of $S^1$. Whats more, the coefficient on these is defined to be the degree of a particular map we get after quotienting away most of the complex. It is easy to check (this I am comfortable to say is geometrically clear) that this map is the suspension of the associated map for $T^{n-1}$.

By the inductive hypothesis, the map we are suspending has degree 0. This means it is homotopically trivial. The suspension of such a map is also homotopically trivial, and so is degree 0. This completes the inductive step.

Now basic combinatorics tells us that the number of $k$-cells in $T^{n}$ is $n \choose k$ from which we deduce that $H_k(T^n)=\mathbb{Z}^{n \choose k}$.

Answer 2

We can use Mayer-Vietoris to obtain the result via elementary “purely topological” means. I now present a “purely algebraic” method (not really; under the hood it uses some covering space theory but requires nothing harder such as the excision theorem).

As per Weibel chapter $6$: since $T_n:=(S^1)^n$ is a $K(\Bbb Z^n,1)$ space (that’s the first homotopy group and its universal cover $\Bbb R^n$ is contractible) the topological singular homology $H_k(T_n;\Bbb Z)$ is isomorphic to the group homology $H_k(\Bbb Z^n;\Bbb Z)$. If $n=0$, we obviously have $\Bbb Z^{\binom{n}{k}}$ as the answer. Now proceed by induction.

In the below, the right hand instances of $\Bbb Z$ will denote the trivial $G$-module for $G=\Bbb Z^m$, some $m$. Unwinding definitions could make the notation a bit confusing; $\Bbb Z\Bbb Z$, anyone?

The Lyndon-Hochschild-Serre spectral sequence applied to $G=\Bbb Z^{n+1}$, $H=\Bbb Z^n$ shows us (using the inductive hypothesis): $$E^2_{p,q}\cong H_p\left(\Bbb Z;\Bbb Z^{\binom{n}{q}}\right)\implies H_{p+q}(\Bbb Z^{n+1};\Bbb Z)$$But only for $p=0,1$ are the terms nontrivial and we are on the second page, so the sequence collapses. For $p=0,1$ the corresponding entry is easily ($\mathsf{Tor}$, hence group homology, is additive!) $\Bbb Z^{\binom{n}{q}}$, the only calculation we need being $H_{0,1}(\Bbb Z;\Bbb Z)\cong\Bbb Z$.

Since these terms are free, projective, the exact sequences split so we get, with a minimum of fuss: $$H_k(T_{n+1})\cong\Bbb Z^{\binom{n}{k}}\oplus\Bbb Z^{\binom{n}{k-1}}\cong\Bbb Z^{\binom{n+1}{k}}$$So the induction is complete.