Differential forms on $\mathbb{R}^n \times\mathbb{R}^1$

Question

Let $\omega\in\Omega^{q+1}(\mathbb{R}^n\times\mathbb{R}^1$). I understand that $\omega$ can be uniquely written as a linear combination of $$\alpha(X,T)\text{d}\!X_{i_1}\dots\text{d}\!X_{i_q}\text{d}\!T$$ and $$\beta(X,T)\text{d}\!X_{j_1}\dots\text{d}\!X_{j_{q+1}}$$ with $\alpha,\beta\in C^\infty(\mathbb{R}^n\times\mathbb{R}^1)$. In Bott/Tu it is claimed that $\omega$ is uniquely a linear combination of $$\pi^\ast\phi\cdot f(X,T)\text{d}\!T\,,\qquad\phi\in\Omega^q(\mathbb{R}^n),\quad f\in C^\infty(\mathbb{R}^n\times\mathbb{R}^1)$$ and $$\pi^\ast\phi\cdot f(X,T)\,,\qquad\phi\in\Omega^{q+1}(\mathbb{R}^n),\quad f\in C^\infty(\mathbb{R}^n\times\mathbb{R}^1)\,.$$ As for $\phi=\gamma(X)\text{d}\!X_{i_1}\dots\text{d}\!X_{i_q}$ we have $$\pi^\ast\phi=\gamma(X)\text{d}\!X_{i_1}\dots\text{d}\!X_{i_q}\quad\text{interpreted as a form on }\mathbb{R}^n\,,$$ this seems to imply that every function $\alpha(X,T)$ on $\mathbb{R}^n\times\mathbb{R}^1$ is uniquely a product $\gamma(X)\cdot f(X,T)$ where $\gamma$ only depends on $X$ and $f$ may depend on both $X$ and $T$. What do I misunderstand?

Answer 1

Bott–Tu's claim follows immediately from your observation with a little notational trickery. In the one case, you can write $$ \alpha(x,t)\, dx^{i_1} \wedge \cdots \wedge dx^{i_q} \wedge dt = \pi^\ast(dx^{i_1} \wedge \cdots \wedge dx^{i_q}) \wedge a(x,t)\,dt, $$ where you view $dx^{i_1} \wedge \cdots \wedge dx^{i_q}$ on the right-hand side as a $q$-form on $\mathbb{R}^n$. In the other case, you can write $$ \beta(x,t) \, dx^{i_1} \wedge \cdots \wedge dx^{i_{q+1}} = \pi^\ast(dx^{i_1} \wedge \cdots \wedge dx^{i_{q+1}}) \wedge \beta(x,t), $$ where you view $dx^{i_1} \wedge \cdots \wedge dx^{i_{q+1}}$ on the right-hand side as a $(q+1)$-form on $\mathbb{R}^n$.