Given that I have had no success on the mathematics stackexchange (see here), I've decided to try my luck here.
I am attempting to solve the following exercise (original formulation here), which to my eyes consists of essentially two steps:
- Construct a smooth 1-form on $\mathbb{R}$ with $\int |\psi|^2 dt < \infty$ for which there exists no function $f$ on $\mathbb{R}$ such that $\int |f|^2 dt < \infty$ and $df = \psi$.
- Can this be arranged such that the corresponding $L^2$-cohomology class projects to the trivial class in the reduced cohomology group $\overline{H}^{1,2}(\mathbb{R}))$
- Consider the 1-form $ e^{-t^2} \, \mathrm{d}t$ which satisfies $\int_\mathbb{R} |e^{-t^2}|^2 \, \mathrm{d}t < \infty$ and suppose there is $f$ such that $\mathrm{d}{f} = e^{-t^2}$. By construction there exists $a \in \mathbb{R}$ with $f(x) = \int_a^x e^{-t^2} \, \mathrm{d}t$ (up to a constant) and since $e^{-t^2} > 0$, $f$ must be strictly monotonuously increasing. By virtue of the Gaussian being $L^2(\mathbb{R})$ we have $\lim_{x \rightarrow \pm \infty}f(x) \in \mathbb{R}$, where by monotonicity $f$ is non-zero for at least one of the infinities, which we use to choose an $M \in \mathbb{R}$ such that $|f(x)|^2 \geq M^2$ for all but finitely many $x \in \mathbb{R}$. We conclude that $\int_\mathbb{R} |f|^2 \, \mathrm{d}t \rightarrow \infty$ and thus $f \notin L^{2}(\mathbb{R})$.
- Let $\psi = e^{-t^2} \, \mathrm{d}t$ be as before, $\mathrm{d}\psi = \mathrm{d}\left( \psi \right) = (\frac{\mathrm{d}}{\mathrm{d}t}e^{-t^2}) \, \mathrm{d}t \land \, \mathrm{d}t = 0 \implies \mathrm{d}\psi \in \mathrm{ker}(\mathrm{d}|_{\Omega^{1,2}(\mathbb{R})})$ (although I think this is a useless "proof" given that $\mathrm{d}$ of a 1-form in a 1-manifold will always be trivial). By i) we know that $\psi$ isn't exact and therefore cannot lie in $\mathrm{im}(\mathrm{d}|_{\Omega^{0,2}(\mathbb{R})})$ and isn't trivial in $H^{1,2}(\mathbb{R})$. Remains to show that there is no sequence $ L^2(\mathbb{R})\ni\mathrm{d}f_n \overset{n \rightarrow \infty}{\longrightarrow} \psi$.
This is where I am stuck. One one hand I think that perhaps I should prove that a such a sequence can never converge to $\psi$ without contradicting $\mathrm{d}f_n \in L^2(\mathbb{R})$, but in my course, we have following statement:
Poincaré duality (Goldstein Troyanov 1998): Let $\omega \in L^{k,p}(M)$ be a weakly closed form; then $\omega$ represents a non-trivial class in $\overline{H}^{k,p}(M)$ if and only if for $q>1$ such that $1/p + 1/q = 1$ there exists a weakly closed form $\nu \in L^{n-k,q}(M)$ such that $\int \omega \land \nu \neq 0$.
Applied to this case, this would mean that there exists a $0$-form, square integrable and weakly closed $\nu$ such that $\int \psi \land \nu \neq 0$, which I understand as showing that $\int \nu(t) \psi(t) \, \mathrm{d}t \neq 0$. This gives me the feeling that there is a big misunderstanding on my part, because I don't see why it would be hard to find such a 0-form, even in the case where it is exact (for example, taking $-2t e^{-t^2} \mathrm{d}t$, for many $f \in L^2$ functions, the integral $\int -2t e^{-t^2}f(t) \mathrm{d}t \neq 0$, despite $-2t e^{-t^2} \mathrm{d}t$ being exact).
Can anyone give a hint or point at the source of my misunderstanding?
P.S. This is sort of a refinement of this post
