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How I can compute explicitly a set of differential forms generating the de Rham cohomology of a 2-torus of revolution in $\mathbb{R}^3$ ? A 2-torus is embedded in $\mathbb{R}^3$ by $\psi(u,v):=\left(x(u,v),y(u,v),z(u,n)\right),$ where $x(u, v) = (R + r \cos{v}) \cos{u},y(u, v) = (R + r \cos{v}) \sin{u} , z(u, v) = r \sin{v}$ and $ u,v\in [0,\pi).$

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    I see. You can also think about the torus as $\mathrm{Circle}\times\mathrm{Circle}$ as I do in my answer below.2011-03-27

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The classes of the angle forms corresponding to each of the two directions in the torus. It is not hard to check they are closed and span cohomology.

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By the Künneth formula $H^*(S^1\times S^1)$ (a version for de Rham cohomology can be found in Bott & Tu, for instance) is naturally isomorphic to the graded tensor product $H^*(S^1)\otimes H^*(S^1)$.

Hence it suffices for your purpose to understand the de Rham cohomology of the circle. Can you describe generators in this case?