How calculate the De Rham cohomology group of $3$-torus: $T^3$?

How do I calculate the De Rham cohomology group of the $3$-torus $T^3$? Here $T^3=S^1 \times S^1 \times S^1 $.

Using the Mayer-Vietoris sequence, I can show that $\dim H_3(T^3)=\dim H_0(T^3)=1$. But I don't know how to find $H_1(T^3)$ and $H_2(T^3)$.

Solution 1:

We cut $T^{3}$ into two parts, each part is homotopic to a torus. One visualize this by considering $T^{3}=T^{2}\times \mathbb{S}^{1}$, and the two parts are $\mathbb{T}^{2}\times I$ respectively, with the $I$ coming out of considering $\mathbb{S}^{1}$ as gluing two intervals together. The intersection of the two parts is again homotopic to the torus. Nowe we have:

$$\rightarrow H^{2}(X)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\oplus H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{3}(X)\rightarrow0$$

We know $H^{2}(\mathbb{T}^{2})=\mathbb{R}^{1}$ via induction. So we have

$$H^{0}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow H^{1}(X)\rightarrow \mathbb{R}^{4}\rightarrow \mathbb{R}^{2}\rightarrow H^{2}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow \mathbb{R}\rightarrow0$$

This gives $H^{2}(X)=\mathbb{R}^{3}$ because last map is an isomorphism and the map $H^{1}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(X)$ has a one dimensional image. Consider a closed one-form $w$ on $\mathbb{T}^{2}$, if we use partition of unity to split it into two parts, then no matter which choice we use we would end up with the same class in $H^{2}(X)$ if one thinks geometrically. This together with the first part gives us $H^{1}(X)=\mathbb{R}^{3}$, $H^{2}(X)=\mathbb{R}^{3}$.