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Projection to trivial reduced cohomology class in $L^2(\mathbb{R})$

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 ...
Pastudent's user avatar
  • 111
5 votes
1 answer
637 views

Leafwise de Rham cohomology (A true definition of differential forms along leaves)

For a foliated space $(M, \mathcal{F})$, one associate a leafwise de Rham cohomology. This cohomology and trace-class operators on this cohomology and trace interpretations for closed orbits of ...
Ali Taghavi's user avatar
2 votes
1 answer
201 views

Vanishing product of a closed and coclosed form on a Riemannian manifold

For a (compact) Riemannian manifold $(M,g)$, can it happen that for a non-zero form $\text{d}^*\omega$, and a smooth function $f$ such that $\text{d}f \neq 0$, we can have $$ \text{d}f \wedge \text{d}^...
Max Schattman's user avatar
1 vote
1 answer
239 views

Can every De Rham cohomology class be represented by a closed form $\alpha$ with $L_X \alpha=0$

Assume that $M$ is a manifold and $X$ is a vector field on $M$. Is it true to say that every closed form is De Rham-cohomologue to a closed form $\alpha$ with $L_X \alpha =0$?
Ali Taghavi's user avatar