$\iota_{[X,Y]}\omega = \mathcal{L}_X\iota_Y\omega - \iota_Y\mathcal{L}_X\omega$ for every $X, Y \in \mathfrak{X}(M)$ and $\omega \in \Omega^k(M)$.

Question

Prove that $\iota_{[X,Y]}\omega = \mathcal{L}_X\iota_Y\omega - \iota_Y\mathcal{L}_X\omega$ for every $X, Y \in \mathfrak{X}(M)$ and $\omega \in \Omega^k(M)$.

I know that

$\mathcal{L}_X\iota_Y\omega - \iota_Y\mathcal{L}_X\omega=(\iota_X d\iota_Y\omega+d\iota_X\iota_Y\omega)-(\iota_Y d\iota_X\omega+\iota_Y\iota_X d\omega)$

For the another side

$i_{[X,Y]}\omega = i_{L_X(Y)-L_Y(X)}\omega =i_{L_{X(Y)}}\omega-i_{L_{Y(X)}}\omega$

So, how much can I do?

Answer 1

My favourite way to show these sorts of equality is to use the Leibniz rule. Let $X$ and $Y$ be two vector fields, $\omega$ be a $k$-form, and $v_2,\ldots,v_k$ be $k-1$ other vector fields. Then, by applying twice Leibniz' rule $\DeclareMathOperator{\lie}{\mathscr{L}}$ \begin{align} (\lie_X(\iota_Y\omega))(v_2,\ldots,v_k) &= \lie_X(\iota_Y\omega(v_2,\ldots,v_k)) - \sum_{j=2}^k \iota_Y\omega(v_1,\ldots,\lie_Xv_j,\ldots,v_k)\\ &= \lie_X(\omega(Y,v_2,\ldots,v_k)) - \sum_{j=2}^k\omega(Y,v_2,\ldots,\lie_Xv_j,\ldots,v_k)\\ &= (\lie_X\omega)(Y,v_1,\ldots,v_k) + \omega(\lie_XY,v_2,\ldots,v_k)\\ & \quad + \sum_{i=2}^k\omega(Y,v_2,\ldots,\lie_Xv_i,\ldots,v_k)\\ & \quad - \sum_{j=2}^k\omega(Y,v_2,\ldots,\lie_Xv_j,\ldots,v_k)\\ &= (\lie_X\omega)(Y_,v_2,\ldots,v_k) + \omega([X,Y],v_2,\ldots,v_k)\\ &= (\iota_Y(\lie_X\omega))(v_2,\ldots,v_k) + (\iota_{[X,Y]}\omega)(v_2,\ldots,v_k). \end{align} Hence, $\lie_X(\iota_Y\omega) = \iota_Y(\lie_X\omega) + \iota_{[X,Y]}\omega$.