# Problem of the Week #56 - June 24th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $M$ be a manifold. Let $\alpha$ be a $k$-form on $M$ and let $X$ and $Y$ be vector fields on $M$.
(a) Prove that $L_X(\iota_Y\alpha)=\iota_Y(L_X\alpha) + \iota_{(L_X Y)}\alpha$.
(b) Prove that $L_{[X,Y]}\alpha=0$ whenever $L_X\alpha=0$ and $L_Y\alpha=0$.

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Here, $\iota$ is your inclusion map and $[X,Y]$ is your standard Lie bracket.

#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. You can find my solution below.

Proof: (a) We will tackle this guy in pieces and put them all together in the end. We first note that for any $\alpha\in\Omega^k(M)$ and $X_1,\ldots,X_{k-1}\in \Gamma(TM)$,
\begin{aligned} (\mathcal{L}_X\iota_Y\alpha)(X_1,\ldots,X_{k-1}) &= X((\iota_Y\alpha)(X_1,\ldots,X_{k-1})) - \sum_{i=1}^{k-1}(\iota_Y\alpha)( X_1,\ldots,[X,X_i],\ldots,X_{k-1}) \\ &= X(\alpha(Y,X_1,\ldots,X_{k-1}) - \sum_{i=1}^{k-1}\alpha(Y,X_1,\ldots,[X,X_i],\ldots,X_{k-1}).\end{aligned}
On the other hand, we have
\begin{aligned} (\iota_Y\mathcal{L}_X\alpha)(X_1 ,\ldots,X_{k-1}) &= \mathcal{L}_X\alpha(Y,X_1,\ldots,X_{k-1}) \\ &= X(\alpha(Y,X_1,\ldots,X_{k-1})) - \alpha([X,Y],X_1,\ldots,X_{k-1})\\ &\phantom{=} -\sum_{i=1}^{k-1}\alpha (Y,X_1,\ldots, [X,Xi],\ldots, X_{k-1}).\end{aligned}
Therefore,
$(\mathcal{L}_X\iota_Y\alpha -\iota_Y\mathcal{L}_X\alpha)(X_1 ,\ldots,X_{k-1}) = \alpha([X,Y],X_1,\ldots,X_{k-1})= \iota_{[X,Y]}\alpha(X_1,\ldots,X_{k-1})$
Since we have $\mathcal{L}_XY=[X,Y]$ for vector fields, we now see that
$\mathcal{L}_X(\iota_Y\alpha) - \iota_Y(\mathcal{L}_X\alpha) = \iota_{(\mathcal{L}_XY)}\alpha \implies \mathcal{L}_X(\iota_Y\alpha) = \iota_Y(\mathcal{L}_X\alpha) + \iota_{(\mathcal{L}_XY)}\alpha.$

(b) If $\mathcal{L}_X\alpha=0$ and $\mathcal{L}_Y\alpha=0$, then
$\mathcal{L}_{[X,Y]}\alpha = \mathcal{L}_X(\mathcal{L}_Y\alpha) - \mathcal{L}_Y(\mathcal{L}_X\alpha) = \mathcal{L}_X 0-\mathcal{L}_Y 0 = 0.$
This completes the proof.$\hspace{.25in}\blacksquare$

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