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.
Differential Geometry – Basic Differential Forms Explained
differential-geometrydifferential-topologyfiber-bundles
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Answer: The answer to my question is that $[dA,dB]$ is not necessarily zero. As a counterexample, consider $\mathbb{R}^2$ with the standard inner product and take $A(x,y) = x^2y^2$, $B(x,y) = xy^3$ (I selected these arbitrarily without thinking about them). Then $\nabla A(x,y) = (dA)^\sharp(x,y) = 2xy^2\frac{\partial}{\partial x} + 2x^2y \frac{\partial}{\partial y}$ and $\nabla B(x,y) = (dB)^\sharp(x,y) = y^3 \frac{\partial}{\partial x} + 3xy^2\frac{\partial}{\partial y}$.
Computing the Lie Bracket of $\nabla A$ and $\nabla B$ in coordinates shows that $[dA,dB](x,y) = [\nabla A, \nabla B](x,y) = (-6x^2y^3 - 2y^5)\frac{\partial}{\partial x} + (2xy^4 + 6x^3y^2)\frac{\partial}{\partial y}$ which is not zero at all points.
Resolving my misunderstanding: The following is an explanation of the misunderstanding which prompted my question. Let $A_1,\ldots,A_k$ be functions on a Riemannian $k$-manifold $(M,g)$ such that the collection of vectors $\{\nabla A_i := (dA_i)^\sharp\}$ are linearly independent at the point $x_0 \in M$. This is true if and only if the collection of forms $\{dA_1,\ldots,dA_k\}$ are themselves linearly independent as linear functionals at $x_0 \in M$. Then the derivative of the map $\varphi: M \to \mathbb{R}^k$, $\varphi: x \mapsto (A_1(x),\ldots,A_k(x))$ is full rank at $x_0 \in M$, so the inverse function theorem guarantees that there exists an open set $U \ni x$ such that $\varphi|_U: U \to \varphi(U)$ is a diffeomorphism. Initially, I thought that the pairwise Lie brackets of all of the $\nabla A_i$ must therefore be zero. However, this is only necessarily be true if the integral curves of the $\nabla A_i$ were the coordinate lines determined by the chart $\varphi$. The definition of $\nabla A_i$ relies on the particular metric $g$, and in general $g$ has nothing to do with $\varphi$; the coordinate lines of $\varphi$ which have nothing do with $g$ are thus not the integral curves of the $\nabla A_i$ in general.
$i_X\circ L_X=i_X\circ (i_X\circ d+d\circ i_X)=i_X\circ d\circ i_X$ since $i_X\circ i_X=0$.
$L_X\circ i_X=(i_X\circ d +d\circ i_X)\circ i_X=i_X\circ d\circ i_X$. You deduce that $i_X\circ L_X=L_X\circ i_X$.
Best Answer
First, this can be checked locally, so we may as well assume $E = F\times M$.
Use coordinates $x_i$ on $F$ and $y_j$ on $M$. Then a $k-$form is given by $\alpha =\sum f_{IJ} dx^I\wedge dy^J$. Here, $I = \{i_1,...,i_s\}$ and $dx^I$ means $dx^{i_1}\wedge...\wedge dx^{i_s}$ and we have $|I|+|J|=k$.
The goal is to show that if $i_X(\alpha) = 0$ for all vertical $X$, then $f_{IJ} = 0$ whenever $I\neq \emptyset$.
So fix any $I\neq \emptyset$. Suppose $i_1\in I$. Let $X = \frac{\partial}{\partial x^{i_1}}$, the dual vector do $dx^{i_1}$. Then $0 = i_X(\alpha)$ so $ 0 = i_X(\alpha)(\frac{\partial}{\partial x^{I-i_1}}, \frac{\partial}{\partial x^{J}}) = \pm f_{IJ}$.
Thus the condition guarantees that $f_{IJ} = 0$ whenever $I\neq \emptyset$, so the only terms which appear in $\alpha$ are of the form $f_J dy^J$.
The only problem now is that $f_J$ could depend on the $F$ factor. This is where the Lie bracket term will come in.
The Lie bracket of a form is given by $L_Y(\alpha) = i_Y d\alpha + d i_Y(\alpha)$.
Taking $Y$ vertical, this reduces to $L_Y(\alpha) = i_Y(d\alpha)$ since we've just shown that $i_Y( \alpha) = 0$.
Now, $d\alpha = \sum \frac{\partial{f_J}}{{\partial x^i}} dx^i\wedge dy^J + \sum \frac{\partial{f_J}}{{\partial y^j}} dy^j\wedge dy^J$.
Just as before, by utilizing a dual basis, we can show that since $0 = L_X(\alpha) = di_X(\alpha)$, that $\frac{\partial{f_J}}{\partial x^i} = 0$.
But this implies that each $f_J$ only depends on the $y^j$ coordinates. It's now clear that $\alpha$ is the pull back of something, mainly of $\sum f_J dy^J$, which makes sense because $f$ is really only a function on $M$.