# Lie derivative of tensor field with respect to Lie bracket

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1. Sep 14, 2015

### "Don't panic!"

I'm trying to show that the lie derivative of a tensor field $t$ along a lie bracket $[X,Y]$ is given by $$\mathcal{L}_{[X,Y]}t=\mathcal{L}_{X}\mathcal{L}_{Y}t-\mathcal{L}_{Y}\mathcal{L}_{X}t$$

but I'm not having much luck so far. I've tried expanding $t$ on a coordinate basis, such that $t=t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}$ and then using the properties $$\mathcal{L}_{X}t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}=X\left[t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right]$$ and $$\mathcal{L}_{X}\left(\partial_{\mu_{i}}\otimes dx^{\nu_{j}}\right)=\left(\mathcal{L}_{X}\partial_{\mu_{i}}\right)\otimes dx^{\nu_{j}}+\partial_{\mu_{i}}\otimes \left(\mathcal{L}_{X}dx^{\nu_{j}}\right)$$ In doing so, I end up with $$\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)t=\mathcal{L}_{X}\mathcal{L}_{Y}t-\mathcal{L}_{Y}\mathcal{L}_{X}t\\=[X,Y]\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right)\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\left(\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\partial_{\mu_{1}}\right)\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+\cdots +t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\partial_{\mu_{1}}\otimes\cdots\otimes\left(\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\partial_{\mu_{m}}\right)\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes\left(\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)dx^{\nu_{1}}\right)\otimes dx^{\nu_{m}}+\cdots +t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes\left(\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)dx^{\nu_{m}}\right)\\ =[X,Y]\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right)\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)\\ =\mathcal{L}_{[X,Y]}\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right)\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)$$

Now, if $$t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)=t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\mathcal{L}_{[X,Y]}\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)$$ then I arrive at the required result as $$\mathcal{L}_{[X,Y]}\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right)\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\left(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X}\right)\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)\\=\mathcal{L}_{[X,Y]}\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\right)\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}+t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\mathcal{L}_{[X,Y]}\left(\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)\\=\mathcal{L}_{[X,Y]}\left(t^{\mu_{1}\cdots\mu_{n}}_{\;\;\;\;\nu_{1}\cdots\nu_{m}}\partial_{\mu_{1}}\otimes\cdots\otimes\partial_{\mu_{m}}\otimes dx^{\nu_{1}}\otimes dx^{\nu_{m}}\right)\\=\mathcal{L}_{[X,Y]}t$$

but if not, then I'm stumped (at the moment) as to what to do next?!

Also, if what I've done is correct it still seems a little sloppy - is there a nicer way to show it?

Any help would be much appreciated.

Last edited: Sep 14, 2015
2. Sep 14, 2015

### Ben Niehoff

Maybe you should focus on simpler cases, rather than trying to do a whole arbitrary tensor at once. You ought to be able to show

$$\mathcal{L}_{[X,Y]} f = (\mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X) f$$
where $f$ is a function. Then show

$$\mathcal{L}_{[X,Y]} Z = (\mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X) Z$$
where $Z$ is a vector field. Then show

$$\mathcal{L}_{[X,Y]} \alpha = (\mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X) \alpha$$
where $\alpha$ is a 1-form. You can choose

$$Z = \partial_\mu, \qquad \alpha = d x^\mu$$
if you like.

Once you can show those, then use

$$\mathcal{L}_X (T \otimes S) = (\mathcal{L}_X T) \otimes S + T \otimes (\mathcal{L}_X S)$$
to prove the general case by induction.

3. Sep 14, 2015

### "Don't panic!"

OK, good idea. Here it goes...

1. $\mathcal{L}_{[X,Y]}f=\mathcal{L}_{X}\mathcal{L}_{Y}f-\mathcal{L}_{Y}\mathcal{L}_{X}f$.

To start, note that $\mathcal{L}_{X}f=X[f]$. Given this, it follows that $$\mathcal{L}_{X}\mathcal{L}_{Y}f-\mathcal{L}_{Y}\mathcal{L}_{X}f=\mathcal{L}_{X}Y[f]-\mathcal{L}_{Y}X[f]=X[Y[f]]-Y[X[f]]=[X,Y][f]=\mathcal{L}_{[X,Y]}f$$

2. $\mathcal{L}_{[X,Y]}Z=\mathcal{L}_{X}\mathcal{L}_{Y}Z-\mathcal{L}_{Y}\mathcal{L}_{X}Z$.

Note that $\mathcal{L}_{X}Y=[X,Y]$. Then, we have $$\mathcal{L}_{X}\mathcal{L}_{Y}Z-\mathcal{L}_{Y}\mathcal{L}_{X}Z = \mathcal{L}_{X}[Y,Z]-\mathcal{L}_{Y}[X,Z] =[X,[Y,Z]]-[Y,[X,Z]]=XYZ-ZYX-YXZ+ZXY=[[X,Y],Z]=\mathcal{L}_{[X,Y]}Z$$

I have to admit, I'm not exactly sure how to proceed with the one-form case?!

Last edited: Sep 14, 2015
4. Sep 15, 2015

### Ben Niehoff

For that, you might need Cartan's formula:

$\mathcal{L}_X \alpha = d (\iota_X \alpha) + \iota_X (d \alpha)$

5. Sep 15, 2015

### "Don't panic!"

Thanks for the tip.

So, starting with the interior product $\iota_{X} :\Omega^{r}(M)\rightarrow\Omega^{r-1}(M)$ of some $r$-form $\omega\in\Omega^{r}(M)$ with respect to a vector field $X\in\mathscr{X}(M)$ (where $\mathscr{X}$ is the set of all vector fields on $M$) $$\iota_{X}\omega(X_{1},\ldots,X_{r-1})\equiv\omega(X,X_{1},\ldots,X_{r-1})$$ For $X=X^{\mu}\partial_{\mu}$ and $\omega=\frac{1}{r!}\omega_{\mu_{1}\cdots\mu_{r}}dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{r}}$, we have $$\iota_{X}\omega=\frac{1}{(r-1)!}X^{\nu}\omega_{\nu\mu_{2}\cdots\mu_{r}}dx^{\mu_{2}}\wedge\cdots\wedge dx^{\mu_{r}}$$ From this it follows that $$\iota_{[X,Y]}\omega\left(X_{2},\ldots,X_{r}\right)=\frac{1}{(r-1)!}[X,Y]^{\nu}\omega_{\nu\mu_{2}\cdots\mu_{r}}X_{2}^{\mu_{2}}\cdots X_{r}^{\mu_{r}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ =\frac{1}{(r-1)!}\left(X^{\lambda}\partial_{\lambda}Y^{\nu}-Y^{\lambda}\partial_{\lambda}X^{\nu}\right)\omega_{\nu\mu_{2}\cdots\mu_{r}}X_{2}^{\mu_{2}}\cdots X_{r}^{\mu_{r}}\qquad\qquad\qquad\\=\frac{1}{(r-1)!}\left(X^{\lambda}\partial_{\lambda}Y^{\nu}\right)\omega_{\nu\mu_{2}\cdots\mu_{r}}X_{2}^{\mu_{2}}\cdots X_{r}^{\mu_{r}}-\frac{1}{(r-1)!}\left(Y^{\lambda}\partial_{\lambda}X^{\nu}\right)\omega_{\nu\mu_{2}\cdots\mu_{r}}X_{2}^{\mu_{2}}\cdots X_{r}^{\mu_{r}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\$$
Now, I know this should be equal to $X\left(\iota_{Y}\omega\right)-Y\left(\iota_{X}\omega\right)$, but I'm unsure how to proceed to this result from where I'm up to at the moment?

Last edited: Sep 15, 2015