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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 $H_2(T^3)$.

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Have you seen the Kunneth formula? – Matt Jan 20 '13 at 0:53
Do you already know the answer for $T^2$? – Jason DeVito Jan 20 '13 at 2:12
I think this is better if you know Kunneth formula... – Bombyx mori Jan 20 '13 at 2:21
Yeah, Jason DeVito – user58930 Jan 20 '13 at 4:09
Mayer-Vietoris and induction can show the dimension of $H^k(T^n)$ is at most $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}$. Wedging together the various 1-forms from the $S^1$ components shows it is at least $\binom{n}{k}$. So you have the dimension is exactly $\binom{n}{k}$, and you know what the generators look like. – user641 Jan 20 '13 at 7:50

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 $H^{1}(X)=\mathbb{R}^{3}$, $H^{2}(X)=\mathbb{R}^{3}$.

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I think you may want to explain exactly how you're cutting the $T^3$ to get the pieces. Also, it looks like the pieces are not homeomorphic to a torus, but rather homotopy equivalent to a torus (which is good enough for the rest of your argument to apply.) – Jason DeVito Jan 20 '13 at 2:26
Thanks for pointing out the mistake. – Bombyx mori Jan 20 '13 at 3:45
Why is the intersection not the disjoint union of two tori: $T^2 \times \{pt_1\} \sqcup T^2 \times\{pt_2\}$? – Eric O. Korman Jan 20 '13 at 5:26
i didn't how to use partition of unity? can you detail, please? But thanks you for answer. – user58930 Jan 20 '13 at 14:14
Why $H^1(T^2 \times I)\rightarrow H^2(X)$ has a one dimensional image? – user58930 Jan 20 '13 at 14:57

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