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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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What is a 2-torus of revolution in $\mathbb R^3$? –  Rasmus Mar 27 '11 at 16:47
    
I see. You can also think about the torus as $\mathrm{Circle}\times\mathrm{Circle}$ as I do in my answer below. –  Rasmus Mar 27 '11 at 16:59
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up vote 2 down vote accepted

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?

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