Steenrod squares and higher cup products for differential forms?

Question

I am physicist, so I am sorry if I am not too rigorous in the following. I have two (closely related I guess) questions:

1. Let me consider a triangulated manifold $M$ and its simplicial cohomology. Here the Steenrod square is an operation $Sq^q: H^p (M,Z_2) \to H^{p+q}(M,Z_2)$. My manifold is smooth and I have also a differential structure on it: is there an analogous operation also on the de Rham cohomology? (eventually by considering forms in $H_{dR}^p (M)$ mod 2 for example)
2. Now, perhaps more important for me, the Steenrod squares can be written on a element $\alpha_p \in H^p(M,Z_2)$ by introducing the so called higher cup products, so that $Sq^q \alpha_p = \alpha_p \cup_{p-q} \alpha_p$ (where $\cup_0 = \cup$ is the standard cup product and $\cup_p$ actually makes sense for $\alpha_p \in H^p(M,G)$ for some Abelian group $G$). Is there a generalization of these products also for differential forms? In my understanding these higher cup products basically come from the fact that the standard cup product is not graded commutative at the level of cochains (and its failure to be commutative is measured by $\cup_1$). For the differential forms the wedge product ('which is the cup product for forms') has no this kind of problem, so I cannot see how similar products could arise for the forms.

Let me end with an intuitive physical picture that I have in mind (just to explain my questions). I start with a manifold $M$ with a triangulation. This is a discretization of my physical space, where I work with cochains on simplexes. In my view this is an approximation: I then take the continuum limit to $M$ and I represent the cochains as forms. As long as I am concerned just with the ordinary cup product I somewhat know how to handle it (physically). But higher cup products? I really do not know how to make sense of expressions like $\alpha \cup_1 \alpha$ for example (they somewhat resemble 'discretization errors' in this intuitive picture).

I have tried to look (and I am looking) in the mathematical literature but I have not found anything for now, especially for the part regarding the higher cup products (at least anything for what I could understand).

Answer 1

Steenrod squares are defined as classes in $H^\bullet(M,\mathbb{Z}/2)$ because they are derived from the Bockstein homomorphism, which is induced by the short exact sequence $$\mathbb{Z}\xrightarrow{\cdot 2}\mathbb{Z}\to\mathbb{Z}/2$$ Now, for the induced sequence on cohomology to be interesting in any way, it is crucial that $\mathbb{Z}$ has non-trivial maximal (or principal) ideals, in this case $2\mathbb{Z}$, the image of multiplication by $2$. If you change the coefficients to $\mathbb{R}$, you have no non-trivial maximal ideals. Any ring that is obtained as a quotient of $\mathbb{R}$ by an ideal of $\mathbb{R}$ is either $0$, or $\mathbb{R}$ itself, in which case the map is an isomorphism. There is no interesting "middle ground" like when you take coefficients in a ring that is not a field. So there can be no construction like Steenrod squares for de Rham cohomology. The ground ring is a field which precludes the required properties.

As for your second question: Steenrod squares are related to Massey products, and subsequently to $A_\infty$-algebras and $A_\infty$-categories. As far as I'm aware, the latter are related to quantum error corrections, and things like this, so that sounds like what you're looking for. But trying to explain anything in that realm is well beyond a stackexchange post.