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.

$\begingroup$ Well you can always define it to be the sheaf cohomology of locally constant $\mathbb{R}$valued functions. I'm not sure if you want just a cohomology theory, or if you want actual forms though... $\endgroup$– Josh LackmanCommented Aug 1, 2022 at 4:15

1$\begingroup$ Most likely you need to look into the noncommutative geometry approach. A detailed analysis of the irrational torus flow case is, I believe, in the ConnesMoscovici paper. $\endgroup$– მამუკა ჯიბლაძეCommented Aug 1, 2022 at 5:19

3$\begingroup$ There are also various approaches of cohomology of foliations, which might give you what you are after $\endgroup$– Thomas RotCommented Aug 1, 2022 at 6:25

1$\begingroup$ the approach via foliations and diffeological spaces seems most promising. do you know of a reference where the basic cohomology of the nonclosed foliation of a torus (with leaves translates of an irrational flow) is computed? $\endgroup$– xirCommented Aug 1, 2022 at 19:07
2 Answers
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 wellbehaved 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.

3$\begingroup$ This is true, how does this relate to the complement of the irrational flow on the torus though? $\endgroup$ Commented Aug 5, 2022 at 3:42

2$\begingroup$ @JoshLackman: One possible formalization of the de Rham cohomology of the complement is the relative de Rham cohomology of the inclusion of the irrational flow in the torus. Such de Rham cohomology can be defined as described in this answer. $\endgroup$ Commented Aug 5, 2022 at 4:03
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: $$ \Omega^k(T^2  \Delta_\alpha) = \Omega^k \bigg( \coprod_{[z] \in T_\alpha  [0]} z + \Delta_\alpha\bigg) \simeq \prod_{[z] \in T_\alpha  [0]} \Omega^k({\mathbf R}), $$

$\begingroup$ Great answer! So we can say this complement, equipped with the subset diffeology, is (diffeologically) diffeomorphic to a sum space, which itself is a diffeological manifold. Am I right? $\endgroup$– ARACommented Jun 30, 2023 at 6:15

1$\begingroup$ @ARA Yes you're right. I omitted to specify that we consider the subset diffeology because it was quite obvious considering your question. Since you're here download the Chinese reprint of the book, I put the link. $\endgroup$ Commented Jun 30, 2023 at 6:39

$\begingroup$ Another thing, it seems that one can continue to compute the de Rham cohomology of this complement. If I am not mistaken, that would be the product of the de Rham cohomology of $\mathbf{R}$, indexed by $T_{\alpha}[0]$. Am I correct? $\endgroup$– ARACommented Jun 30, 2023 at 6:44

1$\begingroup$ @ARA Yes, that's correct: every differential closed from on the sum is a set of closed forms, one on each leaf, that is, equivalent to a product of closed form on $\mathbf R$ indexed by $T_\alpha  [0]$. $\endgroup$ Commented Jun 30, 2023 at 6:51

1$\begingroup$ Thanks, Patrick! this once again shows the capabilities of diffeology in such situations. $\endgroup$– ARACommented Jun 30, 2023 at 7:21