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Consider a once punctured solid torus $(\mathbb R^2 \times S^1) /\{pt\}$. It is not difficult to see that it is homotopy equivalent to the bouquet of spheres $S^2\vee S^1$. So this guy has a non-trivial second de Rham cohomology.

How can one construct a 2-form representing a cohomology generator?

user79456
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I am assuming the topology and the differentiable structure is induced from the infinite cylinder. Let $x$ be the point which has been removed. Then find a local chart around of $\mathbb{R}\times S^1$ around $x$ say $\phi : U \rightarrow D$ where $U$ is a sufficiently small neighbourhood of $x$ in $\mathbb{R}\times S^1$and $D$ is the unit disk in $\mathbb{R}^2$. Then finding a 2-form on the manifold is equivalent to finding a 2-form on $D\setminus{0}$ since we can always take this 2-form to a 2-form on the manifold defined on $U\setminus{z}$ and then multiply this by a cutoff function having support inside a compact subset of $U\setminus{z}$ and define $0$ everywhere else. Now there is always a non-closed (and obviously exact) 2-form on $D\setminus{0}$ defined as $(1/z) dz\wedge d\overline{z}$ written in complex coordinates. Once can just write it in real coordinates after some calculation. This 2-form has a non-trivial cohomology class and hence is the generator for the second de Rham cohomology group.

random123
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  • Could you please clarify the construction? As far as I understand, you just construct a 2-form on a punctured unit disk in $\mathbb R^2$. But how do you use it to get a form on $\mathbb R^2\times S^1$ – user79456 Sep 21 '14 at 11:42
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    Once can then pull it back to $U\setminus {z}$. Then you have a 2-form on $\mathbb{R} \times S^1$ after multiplying it by a cuttoff which has support within $U$. – random123 Sep 21 '14 at 11:53
  • But you have defined $U$ as an open set in $\mathbb R\times S^1$. – user79456 Sep 21 '14 at 12:24
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    I am not sure I understand what you mean by that. Could you elaborate? – random123 Sep 21 '14 at 12:25
  • I was asking about a form on $\mathbb R^2\times S^1$, not on $\mathbb R\times S^1$. – user79456 Sep 21 '14 at 12:27
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    Ok. I am sorry. my mistake. I will delete the answer but let me ask can't you use a similar method. – random123 Sep 21 '14 at 12:30
  • I think I can. But in the case of $\mathbb R\times S^1$ something clearly goes wrong. $\mathbb R\times S^1$ is homotopy equivalent to $S^1$, so it has trivial second cohomology. – user79456 Sep 21 '14 at 12:34
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    Yes but what i am doing is constructing the 2-form on punctured $\mathbb{R}\times S^1$. – random123 Sep 21 '14 at 12:36
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    I think that if you take the standard volume form on a 3-sphere and consider it as a 2-form on $D^3\setminus {0}$ and make appropriate changes in the obviously wrong answer you might be through. – random123 Sep 21 '14 at 12:46
  • Oh! Your arguments are ok, but I think, one can go even simpler. Take a standard volume form on $S^2$, written as $\omega = \epsilon_{ijk}\frac{x^i dx^j\wedge dx^k}{8\pi |x|^3}$ and make a coordinate change $x^1 = \exp{(i\phi)}-1$. The think you get is obviously exact and non-closed form (as it is not closed in any ball containing the puncture). Thanks for the stimulating discussion and for the time spent :) – user79456 Sep 21 '14 at 13:44