# Differential forms and behaviour respect product

Let $$M$$ a smooth $$n$$-manifold and $$E = M \times \mathbf{R}^m$$ the product manifold. Let $$\pi: E \to M$$ the projection onto the first factor. Let $$k \leq n$$. It's known that $$\{dx_{i_1} \wedge \cdots \wedge dx_{i_k} : i_1 < \cdots < i_k\}$$ is a local frame of $$\Omega^k(M)$$. Let $$\omega \in \Omega^k(M \times \mathbf{R}^n)$$. Locally $$\omega = f(x_1,\dots,x_n,y_1,\dots,y_m)dx_{i_1} \wedge \cdots \wedge dx_{i_r} \wedge dy_{j_1} \wedge \cdots \wedge dy_{j_s}$$ where $$r+s = k$$ and $$y$$ is coordinates in $$\mathbf{R}^m$$.

Therefore, can we deduce that $$\Omega^{k}(M \times \mathbf{R}^m) = \sum_{p =1}^{k} \Omega^{p}(M) \wedge \Omega^{k-p}(\mathbf{R}^m)$$ ?

I'm very confused because it's true that $$\Lambda(V \times W) = \Lambda V \otimes \Lambda W$$ and here appears the wedge product. In particular, I want to prove that every $$k$$-form in $$M \times \mathbf{R}^m$$ is a linear combination of $$\pi^* (\phi) f(x,y) dy_{j_1} \wedge \cdots \wedge dy_{j_s}$$ where $$\phi \in \Omega^{*}(M)$$. Notice that the last formula is global I don't need charts

• Did you mean $r+s = k$? Also, did you mean $\Omega^k(M\times \mathbb{R}^m) = \sum\Omega^p(M)\wedge\Omega^{k-p}(\mathbb{R}^m)$ ? Commented Oct 31, 2020 at 13:45
• yes sorry, I'm going to fix it Commented Oct 31, 2020 at 13:54
• What do you mean by $\Omega^p(M)\wedge\Omega^{k-p}(\mathbb{R}^m)$? Do you have a particular definition in mind for the wedge product of different vector spaces? Commented Oct 31, 2020 at 14:05
• I mean that locally are wedge products Commented Oct 31, 2020 at 14:11
• Yes; such a construction would clearly act fiberwise, but how? Given two vector spaces $U$ and $V$, how would you define $U\wedge V$? Commented Oct 31, 2020 at 14:14

Your last equation gives you the answer: while you indeed have the wedge products of the two coframe, the "function" part may mix the coordinates. In fact you write $$f(x,y)$$ and not every such $$f$$ can be written as a product of a function on $$M$$ with a function on $$R^n$$, for example $$\sin(xy)$$. If you prefer, $$C^\infty (M\times N) \neq C^\infty(M) C^\infty(N)$$.
• But, It's true that every differential form in $M \times \mathbf{R}^m$ is $\pi^*\phi f(x,t)\wedge dt$ or simply $\pi^*(\phi)$ (independent of the $dt$ factor). But how is it possible? $\pi^* : \Omega^k(M ) \to \Omega^k(M \times R^m)$ is not sobreyective... locally is easy true prove, but, globally? I'm referring to Differential Forms in Algebraic Topology of W. Bott, when he construct integration along the fiber. Commented Oct 31, 2020 at 14:21
• I am only saying that what you want to deduce is not true. The fact that any differential form on $M\times R^n$ can be written as $\pi^* (\phi) f(x,y) dy_{j_1} \wedge \cdots \wedge dy_{j_s}$ seems not surprising to me but I don't have a proof right now. I don't see any contradiction with $\pi^*$ not being surjective, you have $f(x,t)$ which is not in $\pi^*\Omega(M)$ and also not in (the pullback of) $\Omega(\mathbb{R}^n)$.
What is true is what you stated. If $$M$$ and $$N$$ are manifolds, then $$\Lambda^k\left(M\times N\right) = \bigoplus_{r+s=k} \Lambda^r(M)\otimes\Lambda^s(N)$$. The projections $$\pi_M : M\times N \to M$$ and $$\pi_N : M\times N \to N$$ give a morphism $$\bigoplus_{r+s=k}\Omega^r(M)\otimes \Omega^s(N) \to \Omega^{k}(M\times N)$$. There is no reason that it will be surjective.
For example, if $$k=0$$, you have a morphism \begin{align} \Omega^0(M)\otimes \Omega^0(N) &\longrightarrow \Omega^0(M\times N) \\ (f,g) &\longmapsto (p,q)\mapsto f(p)g(q) \end{align}but it is clearly not surjetive in many examples. The fact is that in the fibers, everything works fine, but when you take sections, that are global notions, you cannot act like you were working pointwisely.
• I'm referring to W.Bott in his book Differential Forms in Algebraic Topology. He say that in $M \times \mathbb{R}$ there are two types of forms, linear combination of forms $\pi^* \phi \cdot f(x,t)$ and $\pi^*\phi \wedge f(x,t)dt$ (page 38) where $\phi$ lives in $\Omega^*(M)$. After he extend this construction to $M \times \mathbb{R}^m$. Commented Oct 31, 2020 at 14:24