Is there a coordinate-free proof of this Lie derivative identity?

Question

Wikipedia mentions (here and here) that the Lie derivative has the following appealing commutator: $$[\mathcal{L}_X,\iota_Y]=\iota_{[X,Y]}$$ The only way I know to demonstrate this identity relies on coordinate-based manipulations, which are summarized (with a mistake) in this problem. I would like a coordinate-free (purely algebraic, one might say) proof of the same result.

Here's one possible route to a proof, but I can't seem to stick the landing. Let $\phi_t$ be the diffeomorphisms infinitesimally generated by $X$ and recall that $\mathcal{L}_X\omega=\left.\partial_t(\phi_t^*\omega)\right|_{t=0}$. Similarly, $[X,Y]=\mathcal{L}_XY=\left.\partial_t(\phi_{-t}^*Y)\right|_{t=0}$.

Now $\iota_{(\cdot_1)}(\cdot_2)$ is bilinear and smooth, so it commutes with the (Gateaux) time derivative when the equation is interpreted sufficiently weakly. So it suffices to show that $$\left.\partial_t\left([\phi_t^*,\iota_Y]-\iota_{\phi_{-t}^*Y}\right)\right|_{t=0}=0$$ But, applying LHS to a test form, I don't see how to deduce the result.

This answer suggests that it should be obvious from the Leibniz rule, but I don't see why the contraction operator $C$ in that answer should commute with $\mathcal{L}_X$ (as it does, moving from the first to second line).

How can I prove the commutator claim in a coordinate-independent manner?

Answer 1

Here is a coordinate-free proof using the definition of $i_X$ and the properties of $\mathscr{L}$

$\color{teal}{\textbf{Proof}}$
Let $X,Y$ be vector fields and $\omega$ a $k$-form.
To show that the relation $i_{[X,Y]} = [\mathscr{L}_X, i_Y]$ holds, recall that the Lie derivative of $i_Y\omega$, as for any tensor field, is given by \begin{align*} \mathscr{L}_X \big( i_Y\omega(Y_1 ,\dots, Y_{k-1} )\big) & = \mathscr{L}_X(i_Y\omega)(Y_1 ,\dots, Y_{k-1}) \\\\[0pt] & \qquad+ i_Y\omega(\mathscr{L}_X Y_1, Y_2, \dots, Y_{k-1}) \\\\[0pt] & \qquad\qquad+ i_Y\omega(Y_1, \mathscr{L}_X Y_2, Y_3, \dots, Y_{k-1}) \\\\[0pt] & \qquad\qquad\qquad+ \cdots + i_Y\omega(Y_1, \dots, Y_{k-2}, \mathscr{L}_X Y_{k-1}). \end{align*}

By antisymmetry of $i_Y\omega$ and the fact that $\mathscr{L}_Y X = [Y,X]$, we can rewrite the above expression as \begin{align*} \mathscr{L}_X \big( i_Y\omega(Y_1 ,\dots, Y_{k-1} )\big) & = \mathscr{L}_X(i_Y\omega)(Y_1 ,\dots, Y_{k-1}) \\\\[0pt] & \qquad+ \sum_j (-1)^{j+1} i_Y\omega ( [Y,Y_j], Y_1, \dots, \widehat{Y_j}, \dots, Y_{k-1}) , \tag{*} \end{align*}

where $\widehat{Y_j}$ denotes the removal of $Y_j$ (and placement into $[X,Y_j]$).
On the other hand, by definition of the interior product, we have \begin{align*} \mathscr{L}_X \big(i_Y\omega(Y_1 ,\dots, Y_{k-1})\big) & = \mathscr{L}_X \big(\omega(Y, Y_1, \dots, Y_{k-1})\big) \\\\[0pt] & = \mathscr{L}_X \omega(Y, Y_1, \dots, Y_{k-1}) + \omega([X,Y], Y_1, \dots, Y_{k-1}) \\\\[0pt] & \qquad + \sum_j (-1)^{j+1} \omega(Y, [X,Y_j], Y_1, \dots \widehat{Y_j}, \dots, Y_{k-1}) \\\\[0pt] & = i_Y (\mathscr{L}_X\omega)(Y_1, \dots, Y_{k-1}) + i_{[X,Y]} \omega(Y_1, \dots, Y_{k-1}) \\\\[0pt] & \qquad + \sum_j(-1)^{j+1} i_Y\omega ([X,Y_j], Y_1, \dots, \widehat{Y_j}, \dots, Y_{k-1}) . \end{align*}

Equating these two expressions gives \begin{equation*} i_{[X,Y]}\omega = \mathscr{L}_X(i_Y\omega) - i_Y(\mathscr{L}_X\omega) \end{equation*}

as desired. $\color{teal}{\quad\blacksquare}$

If you are exploring $i_{[X,Y]} = [\mathscr{L}_X, i_Y]$, then I bet you are playing around with $\mathscr{L}_X = i_X \circ d + d \circ i_X$ and $d \circ \mathscr{L}_X = \mathscr{L} \circ d$ too. I had trouble finding satisfactory proofs of these, so I will offer you what I have managed to work out. The first comes from a sequence of problems from Spivak's A Comprehensive Introduction to Differential Geometry Vol I. Enjoy! ^_^

$\textbf{Bonus (}\textit{Cartan's Magic Formula}\text{)}$
The above result allows us to formulate a coordinate-free proof of Cartan's Magic Formula, which says that the Lie derivative operator on forms can be computed by the equation \begin{equation*} \mathscr{L}_X = i_X \circ d + d \circ i_X . \end{equation*}

$\color{teal}{\textbf{Proof}}$
We first derive an expression for $d\omega$ as a summation over a single index so that the interior product is more straight forward to investigate. Recall that $d\omega$ is given by \begin{align*} d\omega(X_1, \dots, X_{k+1}) & = \sum_j (-1)^{j+1} X_j \big(\omega(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) \\\\[0pt] & \quad + \sum_{j<l} (-1)^{j+l} \omega([X_j,X_l], X_1, \dots, \widehat{X_j}, \dots, \widehat{X_l}, \dots, X_{k+1}) \tag{**} \\\\[0pt] & = \Sigma_1 + \Sigma_2, \text{say}. \end{align*}

Note that by supplying $\omega$ with $k$ arguments, we get a real number (at every point). Therefore, since $\mathscr{L}_X f = Xf$ for smooth functions $f$, we can apply $(*)$ to the term in $\Sigma_1$, but we must account for the offset indices (due to the removal of $X_j$) by adjusting the sign. Namely, \begin{align*} X_j \big(\omega(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) & = \mathscr{L}_{X_j} \big(\omega(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) \\\\[0pt] & = \mathscr{L}_{X_j}\omega (X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \\\\[0pt] & \quad + \sum_{l=1}^{j-1} (-1)^{l+1} \omega([X_j,X_l], X_1, \dots, \widehat{X_l}, \dots, \widehat{X_j}, \dots, X_{k+1}) \\\\[0pt] & \qquad + \sum_{l=j+1}^{k+1} (-1)^l \omega([X_j,X_l], X_1, \dots, \widehat{X_j}, \dots, \widehat{X_l}, \dots, X_{k+1}) . \end{align*}

Summing over $j$ then gives us \begin{align*} \Sigma_1 & = \sum_j (-1)^{j+1} \mathscr{L}_{X_j}\omega \,(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \\\\[0pt] & \qquad + \sum_{l<j} (-1)^{l+j} \omega \,([X_j,X_l], X_1, \dots, \widehat{X_l}, \dots, \widehat{X_j}, \dots, X_{k+1}) \\\\[0pt] & \qquad\qquad + \sum_{l>j} (-1)^{l+j+1} \omega\,([X_j,X_l], X_1, \dots, \widehat{X_j}, \dots, \widehat{X_l}, \dots, X_{k+1}) \\\\[0pt] & = \sum_j (-1)^{j+1} \mathscr{L}_{X_j}\omega \,(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) - \Sigma_2 - \Sigma_2 . \end{align*}

Thus, \begin{align*} d\omega(X_1, \dots, X_{k+1}) & = \sum_j (-1)^{j+1} \mathscr{L}_{X_j}\omega (X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \\\\[0pt] & \quad- \sum_{j<l} (-1)^{j+l} \omega([X_j,X_l], X_1, \dots, \widehat{X_j}, \dots, \widehat{X_l}, \dots, X_{k+1}) . \end{align*}

Adding this equation to $(**)$ cancels the double sum and leaves us with \begin{align*} d\omega(& X_1, \dots, X_{k+1}) \\\\[0pt] & = \frac{1}{2} \sum_j (-1)^{j+1} \Big[ X_j \big(\omega(X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) + \mathscr{L}_{X_j}\omega (X_1, \dots, \widehat{X_j}, \dots, X_{k+1}) \Big] \\\\[0pt] & = \sum_j a_j , \hspace{0.3cm}\text{say.} \end{align*}

Now, with the aim of computing $i_{X_1}(d\omega)$, we group the terms where $X_1$ is present as an argument, which is the case for $j \geq 2$. At $j = 1$, we have \begin{align*} a_1 & = \frac{1}{2} \Big[ X_1 \big(\omega (X_2, \dots, X_{k+1}) \big) + \mathscr{L}_{X_1}\omega(X_2, \dots, X_{k+1}) \Big] \\\\[0pt] & = \frac{1}{2} \Big[ 2\mathscr{L}_{X_1}\omega(X_2, \dots, X_{k+1}) + \sum_{j \geq 2} (-1)^j \omega([X_1,X_j], X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \Big] \tag*{by (*)} \\\\[0pt] & = \mathscr{L}_{X_1}\omega(X_2, \dots, X_{k+1}) + \frac{1}{2} \sum_{j \geq 2} (-1)^j i_{[X_1,X_j]}\omega(X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) . \end{align*}

The remaining terms are then \begin{align*} \sum_{j\geq 2} a_j & = \frac{1}{2} \sum_{j \geq 2} (-1)^{j+1} \Big[ X_j \big(\omega(X_1, X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) + \mathscr{L}_{X_j}\omega (X_1, X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \Big] \\\\[0pt] & = - \frac{1}{2}\sum_{j \geq 2} (-1)^j \Big[ X_j \big(i_{X_1}\omega(X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \big) + i_{X_1}(\mathscr{L}_{X_j}\omega) (X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \Big] . \end{align*}

So, by definition of the interior product, we have \begin{align*} i_{X_1}(d\omega) & (X_2, \dots X_{k+1}) \\\\[0pt] & = d\omega(X_1, \dots X_{k+1}) \\\\[0pt] & = a_1 + \sum_{j\geq 2} a_j \\\\[0pt] & = \mathscr{L}_{X_1} \omega(X_2, \dots, X_{k+1}) \\\\[0pt] & \quad - \frac{1}{2} \sum_{j \geq 2} (-1)^j \Big[X_j\big(i_{X_1}\omega(X_2, \dots, \widehat{X_j}, \dots, X_{k+1})\big) + \mathscr{L}_{X_j}(i_{X_1}\omega) (X_2, \dots, \widehat{X_j}, \dots, X_{k+1}) \Big] \\\\ & \hspace{6cm}\text{since } i_{[X,Y]} = [\mathscr{L}_X, i_Y] \\\\ & = \mathscr{L}_{X_1}\omega(X_2, \dots, X_{k+1}) - d( i_{X_1}\omega ) (X_2, \dots, X_{k+1}) . \color{teal}{\quad\quad\blacksquare} \end{align*}

From here it is pretty straight forward to see that the $\mathscr{L}$ derivative and exterior derivative commute. Applying the exterior derivative to both sides of Cartan's equation, we have \begin{align*} d \circ \mathscr{L}_X &= d \circ (i_X \circ d) + d \circ (d \circ i_X) \\\\[0pt] &= (d \circ i_X) \circ d + 0 \\\\[0pt] &= (\mathscr{L}_X - i_X \circ d) \circ d \tag*{by Cartan's equation} \\\\[0pt] &= \mathscr{L}_X \circ d - 0 . \end{align*}