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I would like to understand how to compute the de Rham cohomology directly, that is, without using Meyer Vietoris theorem. All the examples I saw for the direct computation were very simple.

If we consider $S^1 \times \mathbb{R}$ or $S^1 \times S^1$, then I know how to approach it using the Meyer Vietoris theorem, there also many references (e.g. the answer here). Any starting points as to how to compute these directly?

Edit: for the case of $\mathbb{R} \times S^1$, I believe that this manifold is diffeomorphic to the punctured plane, which has a direct computation here. This leaves the torus.

Johana T
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  • What does "compute" mean? Do you intend to guess the answer and then prove "directly" that it is correct? – Ted Shifrin Sep 25 '24 at 15:54
  • Yes - but it's not about the answers but more of how to directly derive the results, with differentials etc rather than using the Theorem mentioned. – Johana T Sep 25 '24 at 16:13
  • Why don't you use duality and de rham theorem. – Ricci Ten Sep 25 '24 at 16:40
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    In general, you cannot hope to succeed without tools. You can do $S^1$ directly (without Mayer-Vietoris). Presumably you've done this. But have you done $S^n$ for $n>1$ directly? I can do $S^1\times S^1$ "directly" only by knowing the answer and then proving it is correct. – Ted Shifrin Sep 25 '24 at 17:10

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