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Let $\pi : M \rightarrow B$ be a fiber bundle. If $\alpha \in A^i(M)$ is a differential form on M of degree $i$, we denote by $\int_{M/B}\alpha \in A^{i-n}(B)$ the integration of $\alpha$ along the fiber.

In the article " Orbites Coadjointes et Cohomologie équivariante" (page 23) the authors say that if $\xi$ is a vector field on $M$ which we can project to a vector field $\pi_*(\xi)$ on the base $B$, then $$\iota(\pi_*(\xi)) \int_{M/B} \alpha = \int_{M/B} \iota (\xi) \alpha.$$

My question is how is the vector field $\pi_* (\xi) $ defined ? and what does it mean that a vector field on $M$ admit a projection on $B$ ?

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Given a smooth map $F:M\to N$ and a vector field $X\in\mathfrak{X}M$, we say a vector field $Y\in\mathfrak{X}N$ is $F$-related to $X$ if $d_pF(X(p))=Y(F(p))$ for all $p\in M$ where $d_pF:T_pM\to T_{f(p)}N$ denotes the differential of $F$ at $p$.

In the case of a fiber bundle $\pi:M\to B$, one can show that, given a vector field $X\in\mathfrak{X}M$, there may or may not exist a $\pi$-related vector field on $B$, but if it does exist, it is unique. This vector field is what the authors refer to by $\pi_*(X)$.

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  • $\begingroup$ hi @Kajelad! Thank you for your answer. I'm just wondering in which cases does the F-related vector field to $X$ exist ? I guess for example that if $F$ is invertible then it exists, am I right ? $\endgroup$
    – Mira
    Commented Jul 29, 2021 at 0:44
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    $\begingroup$ @asma If $F$ is a diffeomorphism (i.e. smoothly invertible) both existence and uniqueness are guaranteed. $\endgroup$
    – Kajelad
    Commented Jul 29, 2021 at 1:22

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