Bott Tu Exercise 6.14, integration along the fiber

Question

Suppose $\pi:E\to M$ is an oriented $C^\infty$ vector bundle of rank $n$. We denote by $\Omega_{cv}^k(E)$ the set of all differential $k$-forms $\omega$ on $E$, such that for each compact $K\subset M$, $\pi^{-1}(K)\cap \text{supp}(\omega)$ is compact. In particular, the support of the restriction $\omega|_F$ to each fiber is compact. Assume $\{(U_\alpha,\phi_\alpha)\}$ is an oriented trivialization for $E$. On $\pi^{-1}(U_\alpha)$, such a form $\omega$ is uniquely expreesed as a sum of the forms either of type $(\pi^*\phi)f(x,t_1,\dots,t_n)dt_{i_1}\cdots dt_{i_r}$ with $r<n$, or $(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n$. (Here $x_1,\dots,x_n$ are coordinate functions on $U_\alpha$ and $t_1,\dots,t_n$ are fiber coordinates on $\pi^{-1}(U_\alpha)$ given by $\phi_\alpha$. We define a map $\Omega_{cv}^*(E)\to \Omega^*(M)$ by sending the forms of first type to zero, and the forms of second type to $\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt_1 \dots dt_n$.

Exercise 6.14 askes to show that this map is well-defined. Suppose $U_\alpha \cap U_\beta $ is nonempty. Then on $\pi^{-1}(U_\alpha \cap U_\beta)$, a form of second type can be expressed as $$(\pi^*\phi)f(x,t_1,\dots,t_n)dt_1\cdots dt_n=(\pi^* \tau)g(y,u_1,\dots,u_n) du_1\cdots du_n.$$ Then I have to show that $$\phi \int_{\Bbb R^n} f(x,t_1,\dots,t_n)dt=\tau \int_{\Bbb R^n} g(y,u_1,\dots,u_n)du,$$ but I got stuck. Any hints?

Answer 1

Assume that $U_a$ and $U_b$ are relatively compact and $\varphi$, $\psi$ its charts on $M$. By a linearity argument, you can assume that

$$ \phi = h_1 dx_{i_1} \wedge \cdots dx_{i_k} $$

and the same for $\tau$ with coefficient $h_2$. Without loss of generality assume $k = m = \dim M$. Let $T$ and $S$ the coordinates for $t$ and $s$ in $\mathbb{R}^n$. Take the adaptated chats in $E$ $(U_a \times \mathbb{R}^n,\varphi \times T)$ and $(U_b \times \mathbb{R}^n,\psi \times S)$.

We have $\pi^*\phi = h_1 \circ \pi \pi^*(dx) = h_1dx$ in these coordinates. So

$$ \pi^* \phi f(x,t)dt = h_1(x)f(x,t)dx\wedge dt $$ On the other hand, using the change of variables for diff. forms $$ \pi^* \tau gds = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1}) dx \wedge dt = h_1 fdx\wedge dt $$ We deduce that $$h_1f = (h_2 \circ \psi \circ \varphi^{-1})(g \circ \psi \circ \varphi^{-1}) \cdot Jac(\psi \circ \varphi^{-1}) Jac(S \circ T^{-1})$$

The last expression can be integrated into $(U_c = U_a \cap U_b) \times \mathbb{R}^n$ and using Fubini and the change of variable for integrals and the previous definitions $$ \int_{U_c \times \mathbb{R}^n}\phi f(x,t)dx\wedge dt = \int_{U_c \times \mathbb{R}^n}\tau g(y,s)dy \wedge ds. $$ Finally, by a parametrical argument and using $U_a$ and $U_b$ be compact

$$ \phi \int_{\mathbb{R}^n} f(x,t)dt = \tau \int_{\mathbb{R}^n}g(y,s)ds. $$

You must be a little more cautious for $k < n$ because the change of variable for differential forms is a bit different. If I have some typo please fix it.