Assume that $M$ is a principal $\mathbb{S}^1-$bundle, i.e. an $\mathbb{S}^1-$manifold with the property that the action $M\times \mathbb{S}^1\to M$ is free (no fixed points) and the orbits $\mathcal{O}_p:=\{p\cdot\lambda: \lambda\in \mathbb{S}^1\}$ are precisely the fibers of a surjective submersion $h:M\to B$. Let's consider, moreover, the following vector field, $$V_p=\left.\dfrac{d}{dt}\right|_{t=0} p\cdot e^{2\pi i t}$$
I want to prove that $\forall \mu\in \Omega^k(B)$, $\omega:=h^*(\mu)\in \Omega^k(M)$ is an horizontal form, i.e. the interior product $\iota_V(\omega)=0$. Any hint on how to do this?
In other words I have, $$\forall p\in M \quad \exists \lambda\in \mathbb{S}^1\quad s.t.\quad p\cdot \lambda\neq p \quad \text{and} \quad $$$$\mathcal{O}_p:=\{p\cdot \lambda: \lambda\in \mathbb{S}^1\}=h^{-1}(h(p))$$ For example, if $k=1$, then $(\iota_V(h^*(\mu)))_p=(h^*(\mu))_p(V_p)=\mu_{h(p)}((dh)_p(V_p))$.
$i_V\omega(u_1,...,u_{k-1})=i_Vh^*\mu(u_1,...,u_{k-1})=\mu(dh(V),dh(u_1),..,dh(u_{k-1}))=\mu(0,dh(u_1),..,dh(u_{k-1}))=0$