Can one make sense of de Rham cohomology for the complement of a (dense) irrational flow on the torus?

Question

Recent work has led me to consider whether one could define consider the complement of a dense irrational flow on the torus $P_\alpha \subset T^2$ as some kind of generalized smooth space, and consider the complex of smooth forms on it in a naive way and hope that it yields some sort of reasonable cohomology. Knowing very little about notions of generalized smooth spaces, or whether they have a robust notion of de Rham complexes, and having failed to find any reference to this example in a brief search of the literature, I was hoping some expert could help me out.

Answer 1

Any foliation gives rise to two Lie groupoids (monodromy and foliation, see for example the book of Moerdijk–Mrčun), and a Lie groupoid gives rise to a simplicial presheaf on the site of smooth manifolds. The latter have a very well-behaved notion of de Rham cohomology associated to them.

For an expository account of this theory, see Chern–Weil forms and abstract homotopy theory by Freed and Hopkins.

Another good reference is Differential Cohomology by Amabel, Debray, and Haine.

Answer 2

The torus $T^2$ is the quotient of ${\mathbf R}^2$ by ${\mathbf Z}^2$. I denote by $z = [x,y]$ the class of $(x,y) \in {\mathbf R}^2$. I am used to denote $\Delta_\alpha \subset T^2$ what you denote by $P_\alpha$. That said:

The irrational torus $T_\alpha$ is by definition the quotient space $T^2/\Delta_\alpha$. Now the complement of a dense irrational flow $T^2 - \Delta_\alpha$ is the union of all the orbits in $T^2$ of $\Delta_\alpha$, except the orbit $[0]$ of $0 = [0,0]$. That is, $T_\alpha - [0]$. According to the Exercise 19 in Diffeology $T_\alpha - [0]$ is discrete, that is, the sum of its elements. Thus, $T^2 - \Delta_\alpha$ is (diffeologically) the sum of all the orbits of the subgroup $\Delta_\alpha$ which are all diffeomorphic to $\Delta_\alpha$, that is, diffeomorphic to $\mathbf R$. Thus, denoting by $z + \Delta_\alpha$ the orbit of $z$ by $\Delta_\alpha$, we get $$ T^2 - \Delta_\alpha = \coprod_{[z] \in T_\alpha - [0]} z + \Delta_\alpha, $$ where $[z] \in T_\alpha$ is the class modulo $\Delta_\alpha$ of $z \in T^2$. Then, for all $z \in T^2 -\Delta_\alpha$ $$ \Omega^k(z + \Delta_\alpha) \simeq \Omega^k({\mathbf R}). $$ Thus a differential form on $T^2 - \Delta_\alpha$ is non zero only for $0$-forms and $1$-forms, and is the sum, indexed by $T_\alpha - [0]$, of any choice of $0$-forms or $1$-forms for each component. In other words: $$ \Omega^k(T^2 - \Delta_\alpha) = \Omega^k \bigg( \coprod_{[z] \in T_\alpha - [0]} z + \Delta_\alpha\bigg) \simeq \prod_{[z] \in T_\alpha - [0]} \Omega^k({\mathbf R}), $$