TITLE: basic differential forms
QUESTION [10 upvotes]: Given a fiber bundle $f: E\rightarrow M$ with connected fibers we call the image $f^*(\Omega^k(M))\subset \Omega^k(E)$ the subspace of basic forms. Clearly, for any vertical vector field $X$ on $E$ we have that the interior product $i_X(f^*\omega)$ and the Lie derivative $L_X(f^*\omega)$ vanish for all $\omega \in \Omega^k(M)$. Is the converse true? That is, if $\alpha \in \Omega^k(E)$ is a form such that $i_X(\alpha)=0$ and $L_X(\alpha)=0$ for all vertical vector fields $X$ on $E$, is it true that $\alpha$ is a basic form? I believe so, but I am not sure how to prove it. Thanks for your help.

REPLY [4 votes]: If you added a connectedness assumption (which is essential, as I mentioned in my comment to your main question), what you desire follows from 
Prop Let $f:M\to N$ be a submersion such that for every $y\in N$, $f^{-1}(y)$ is a connected submanifold of $M$. Then if $\mu\in\Omega^k(M)$ is a differential form satisfying $i_Y\mu = 0$ and $i_Y d\mu = 0$ for any $Y\in \ker df$, then there is a unique $\nu\in\Omega^k(N)$ such that $\mu = f^*\nu$. 
Sketch of proof It suffices to show that the differential form defined by $\nu(f_*X_1,\ldots ,f_*X_k) = \mu(X_1,\ldots, X_k)$ is well-defined. By the assumption that $i_Y\mu = 0$ if $y\in \ker df$, at a fixed base point $x\in M$ you have that if $f_*X_1 = f_*X'_1$, $\mu(X_1,\ldots, X_k) = \mu(X'_1, X_2,\ldots,X_k)$. Now, it is clear that if $X$ is a vector field on $M$ such that its pushforward is well-defined (in the sense that there exists a vectorfield $Z$ on $N$ such that if $x_1,x_2\in f^{-1}(y)$, $df(X)|_{x_1} = df(X)|_{x_2} = Z|_y$) and if $Y$ is a vector field in $\ker df$, we have that $[X,Y] \in \ker df$. This can be easily checked by acting $[X,Y]$ on the pullback of functions defined on $N$. From this you can conclude that using $i_Yd\mu = 0$, for any vector fields $X_1, \ldots, X_k$ such that the pushforward is well-defined, $\mu(X_1,\ldots, X_k)$ is constant along connected components of $f^{-1}(y)$. q.e.d.

Addendum to address comments: Consider the scalar $\mu(X_1,\ldots, X_k)$ on $M$. Let $Y$ be a vertical vector field, then we have that
$$ Y(\mu(X_1,\ldots,X_k)) = (\mathcal{L}_Y\mu)(X_1,\ldots,X_k) + \mu([Y,X_1],X_2,\ldots,X_k) + \cdots + \mu(X_1,\ldots, X_{k-1}, [Y,X_k]) $$
using Leibniz rule for Lie derivatives. So local constancy of the scalar field needs that $\mathcal{L}_Y\mu \equiv 0$ and that $[Y,X_l]$ is vertical.