# Lie derivative of tensor field with respect to Lie bracket

• "Don't panic!"
L}_{[X,Y]}\alpha=(\mathcal{L}_{X}\mathcal{L}_{Y}-\mathcal{L}_{Y}\mathcal{L}_{X})\alpha##.I'm not sure about this one. I tried using the property \mathcal{L}_{X}\alpha(Y)=X[\alpha(Y)]-\alpha([X,Y]) but it didn't seem to get me anywhere. Any suggestions?Perhaps you could try using the properties of the Lie derivative with respect to a vector field, along with the fact that a 1-form can be written as a linear combination of basis 1-forms?
"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.

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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.

Ben Niehoff said:
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

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:
"Don't panic!" said:
I have to admit, I'm not exactly sure how to proceed with the one-form case?!

For that, you might need Cartan's formula:

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

Ben Niehoff said:
For that, you might need Cartan's formula:

L=d(ιXα)+ιX()

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?

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## 1. What is a Lie derivative of a tensor field?

The Lie derivative of a tensor field is a way to measure how a tensor field changes along a given direction. It is a generalization of the concept of directional derivative to non-linear spaces.

## 2. What does the Lie bracket represent in this context?

The Lie bracket represents the commutator or the algebraic difference between two vector fields. It is used to calculate the Lie derivative of a tensor field along a given direction.

## 3. How is the Lie derivative of a tensor field calculated?

The Lie derivative of a tensor field is calculated by taking the Lie bracket of the tensor field with respect to a given direction. This is often represented using the Lie derivative operator, denoted by ℓ.

## 4. What is the significance of the Lie derivative in mathematics?

The Lie derivative is an important tool in differential geometry and differential equations. It allows us to study the evolution of tensor fields and understand the geometric properties of a given space.

## 5. Can the Lie derivative of a tensor field be used to define a Lie algebra?

Yes, the Lie derivative of a tensor field can be used to define a Lie algebra. This is because the Lie bracket, which is used to calculate the Lie derivative, satisfies the properties of a Lie algebra, namely bilinearity, skew-symmetry, and the Jacobi identity.

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