Lately, I have been reading Nash and Sen’s wonderful Topology and Geometry for Physicists, and it is while reading it that I stumbled upon the idea of making this website. Incidentally, at that moment I had just parsed through their calculation of the de Rham cohomology groups of the $n$-sphere, following which they urge the reader to calculate the cohomology groups of the $n$-torus $T^n = S^1 \times \cdots \times S^1$ ($n$ factors of $S^1$). They give a hint — namely, use the fact that $H^1(S^1) = \mathbb{R}$ and wedge products — and dropped the punchline too:

$$ H^p(T^n, \mathbb{R}) = \mathbb{R}^{\binom{n}{p}}. $$

I thought I might as well take the challenge and make that into the first post for this website. So here’s my meek attempt at proving the preceding equation.

Judging from the hint, one might think that one can somehow generate all elements of $H^p(T^n)$ (the underlying ring of coefficients is understood to be $\mathbb{R}$, so I’ll omit explicit reference to it from now on) by taking wedge products of elements in $H^1(S^1)$. We will show that something like this is indeed true.

First of all, we should verify that the notion of a wedge product makes sense on cohomology groups. This is indeed the case. For any $[\omega], [\eta] \in H^p(M)$, where $M$ is an arbitrary differentiable manifold, we simply define a product, called a cup product $\cup$, by $[\omega]\cup[\eta] = [\omega\wedge\eta]$. Then to show that product is well-defined, one simply has to verify that (1) the wedge product of closed $p$-forms is also closed, and (2) if $\omega’ = \omega + d\alpha$ and $\eta’ = \eta + d\beta$ for some $(p-1)$-forms, so that $[\omega’]=[\omega]$ and $[\eta’]=[\eta]$, then

$$ \begin{align*} [\omega’]\cup [\eta’] = [\omega'\wedge\eta'] &= [\omega\wedge\eta + \omega\wedge d\beta + d\alpha\wedge\eta + d\alpha\wedge d\beta]\\

&= [\omega\wedge\eta + d(\omega\wedge\beta + \alpha\wedge\eta + \alpha\wedge d\beta)] = [\omega\wedge\eta]. \end{align*} $$

Next, it will be helpful to introduce the projections $\pi_i : T^n\to S^1$, where $i$ refers to the $i$th circle $S^1$ in $T^n$. Of course, this induces a pullback $\pi^_i: H^p(S^1) \to H^p(T^n)$. In particular, if we can show that any $[\omega]\in H^p(T^n)$ can be written as $[\omega] = \pi^{i_1}[\gamma]\cup\cdots\cup\pi^*{i_p}[\gamma]$, for some $i_k \in \{1,\ldots,n\}$ and some $\gamma\in H^1(S^1)$, then due to the wedge products involved, there are $\binom{n}{p}$ ways of choosing these $i_k$’s from $\{1,\ldots,n\}$, and this would establish our desired result.

Let us then show why $[\omega]$ can be written as claimed.