# Integration along fibers is independent of the lift

According with Wikipedia https://en.wikipedia.org/wiki/Integration_along_fibers

Let $$\pi: E \to B$$ be a fiber bundle over a manifold with compact oriented fibers. If $$\alpha$$ is a $$k$$-form on $$E$$, then for tangent vectors $$w_i$$'s at $$b$$, let $$(\pi_* \alpha)_b(w_1, \dots, w_{k-m}) = \int_{\pi^{-1}(b)} \beta$$ where $$\beta$$ is the induced top-form on the fiber $$\pi^{-1}(b)$$; i.e., an $$m$$-form given by: with $$\widetilde{w_i}$$ lifts of $$w_i$$ to $$E$$, $$\beta(v_1, \dots, v_m) = \alpha(v_1, \dots, v_m, \widetilde{w_1}, \dots, \widetilde{w_{k-m}}).$$ (To see $$b \mapsto (\pi_* \alpha)_b$$ is smooth, work it out in coordinates; cf. an example below.) Then $$\pi_*$$ is a linear map $$\Omega^k(E) \to \Omega^{k-m}(B)$$.

Recall that a lift of $$w \in T_{b}B$$ is any $$\tilde{w} \in T_{a}E$$ such that $$\pi(a)=b$$ and $$d\pi_{a}{\tilde{w}}=w$$.

My questions is:

Why the integration along fibers is independent of the lifts $$\widetilde{w_{i}}$$ of $$w_{i}$$?

I would like to mention the following post as well: push forward of differential form/ integration over fiber asked by Ivan.

My question is in some way the same as Ivan's question. Why the the integration along fibers is independent of the decomposition (which in fact isn't unique).

Take a point $$a\in \pi^{-1} b$$. Take vectors $$\tilde w_i$$ and $$w_i'$$ on $$T_a E$$ such that $$d_a \pi(\tilde w_i)=d_a \pi( w_i') = v_i$$. Notice $$\tilde w_i - w_i' \in \ker d_a \pi= T_a \pi^{-1}b$$.

Since the dimension of $$\pi^{-1}b$$ is $$m$$, the vectors $$v_1,\ldots,v_m, \tilde w_1 - w_1'$$ must be linearly dependent and, therefore, $$\alpha(v_1,\ldots,v_m,\tilde w_1 - w_1',\tilde w_2,\ldots,\tilde w_{k-m})=0.$$

By the above identity we have $$\alpha(v_1,\ldots,v_m,\tilde w_1,\tilde w_2,\ldots,\tilde w_{k-m}) = \alpha(v_1,\ldots,v_m,w_1',\tilde w_2,\ldots,\tilde w_{k-m}).$$

Analogously, we may exchange $$\tilde w_2$$ by $$w_2'$$, $$\tilde w_3$$ by $$w_3'$$, etc.

Therefore, $$\alpha(v_1,\ldots,v_m,\tilde w_1,\tilde w_2,\ldots,\tilde w_{k-m}) = \alpha(v_1,\ldots,v_m,w_1', w_2',\ldots,w_{k-m}').$$

• Could someone explain why is $ker d_a \pi = T_a \pi^{-1} (b)$ , please ?
– Mira
Commented Jan 27, 2021 at 11:23
• This is well known. see lee, 2nd edition intro to smooth manifolds prop 5.38 (page 117) for example. Commented Feb 11, 2021 at 19:03
• @Simplyorange, thanks a lot !
– Mira
Commented Jan 13, 2022 at 13:30

Note that $$v_1,\dots,v_m$$ (except in trivial cases) will be a basis for the tangent space to the fiber at a point $$a\in\pi^{-1}(b)$$. By multilinearity and skew-symmetry, changing one of the remaining vectors by an element of that tangent space will have no effect on the value $$\alpha(v_1,\dots,v_m,\tilde v_{m+1},\dots,\tilde v_{k-m})$$. Think about it in terms of determinants: $$\det\left[\begin{array}{c|c} A&B \\ \hline 0&C \end{array}\right] = \det\left[\begin{array}{c|c} A&0 \\ \hline 0&C \end{array}\right].$$