# Lie derivative of general differential form

• etotheipi
The above definition is better because we never use the coordinates.In summary, we discussed the Lie derivative on differential forms and its relation to vector fields. We also looked at the Cartan formula and how it can be easily proved using the vector field straightening theorem. The definition of the Lie derivative on forms involves taking the derivative of a sum of forms along a smooth curve. This can be simplified by using the vector field generating the curve.
etotheipi
Homework Statement
Show that Lie derivative of a 1-form ##\omega## satisfies$$(\mathcal{L}_X \omega)_{\mu} = X^{\nu} \partial_{\nu} \omega_{\mu} + \omega_{\nu} \partial_{\mu} X^{\nu}$$Show that the Lie derivative of a (0,2) tensor ##g## is$$(\mathcal{L}_X g)_{\mu \nu} = X^{\rho} \partial_{\rho} g_{\mu \nu} + g_{\mu \rho} \partial_{\nu} X^{\rho} + g_{\rho \nu} \partial_{\mu} X^{\rho}$$If ##\eta## is a p-form and ##i_X \eta## is a (p-1)-form resulting by contracting a vector field ##X## with the first index of ##\eta##, show that$$\mathcal{L}_X \alpha = i_X (d\alpha) + d(i_X \alpha)$$for a differential form ##\alpha##
Relevant Equations
N/A
The first two parts I think were fine, I expressed the tensors in coordinate basis and wrote for the first part\begin{align*} \mathcal{L}_X \omega = \mathcal{L}_X(\omega_{\nu} dx^{\nu} ) &= (\mathcal{L}_X \omega_{\nu}) dx^{\nu} + \omega_{\nu} (\mathcal{L}_X dx^{\nu}) \\ &= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} d(\mathcal{L}_X x^{\nu} ) \\ &= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} d(X^{\sigma} \partial_{\sigma} x^{\nu}) \\ &= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} dX^{\nu} \\ \\ \implies (\mathcal{L}_X \omega )_{\mu} = (\mathcal{L}_X \omega)[\partial_{\mu}] &= X^{\nu} \partial_{\nu} \omega_{\mu} + \omega_{\nu} \partial_{\mu} X^{\nu} \end{align*}and for the second part\begin{align*} \mathcal{L}_X g = \mathcal{L}_X(g_{\eta \xi} dx^{\eta} \otimes dx^{\xi}) &= (\mathcal{L}_X g_{\eta \xi}) dx^{\eta} \otimes dx^{\xi} + g_{\eta \xi} (\mathcal{L}_X dx^{\eta}) \otimes dx^{\xi} + g_{\mu \nu} dx^{\eta} \otimes (\mathcal{L}_X dx^{\xi}) \\ &= X^{\rho} \partial_{\rho} g_{\eta \xi} dx^{\eta} \otimes dx^{\xi} + g_{\eta \xi} dX^{\eta} \otimes dx^{\xi} + g_{\eta \xi} dx^{\eta} \otimes dX^{\xi} \\ \\ \implies (\mathcal{L}_X g)_{\mu \nu} = (\mathcal{L}_X g)[ \partial_{\mu}, \partial_{\nu}] &= X^{\rho} \partial_{\rho} g_{\eta \xi} \delta^{\eta}_{\mu} \delta^{\xi}_{\nu} + g_{\eta \xi} (\partial_{\mu} X^{\eta}) \delta^{\xi}_{\nu} + g_{\eta \xi} \delta^{\eta}_{\mu}(\partial_{\nu} X^{\eta}) \\ &= X^{\rho} \partial_{\rho} g_{\mu \nu} +g_{\rho \nu} \partial_{\mu} X^{\rho} + g_{\mu \rho} \partial_{\nu} X^{\rho}\end{align*}I am a bit stuck on the third part. Using the contraction operator ##\mathscr{C}## I can write\begin{align*} i_X \eta = \mathscr{C}(1,1)[\eta \otimes X] &= \mathscr{C}(1,1)[ (\eta_{\alpha_i \dots \alpha_p} dx^{\alpha_i} \otimes \dots \otimes dx^{\alpha_p}) \otimes X^{\mu} \partial_{\mu}] \\ &= \eta_{\sigma, \alpha_2, \dots, \alpha_p} X^{\sigma} dx^{\alpha_2} \otimes \dots \otimes dx^{\alpha_p} \\ \end{align*}How can I show the result? I guess maybe we can just expand all the terms like in the previous examples, but that's going to take forever and surely there is a nicer way?

I took the equation as a definition of the Lie derivative on differential forms: (eq. 12)
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

Maybe this was a bit cheating. So whatever your definitions are, it's likely only the application of the definitions, or just a variant of your first equation for higher dimensions.

And to be honest to you: every calculation in Graßmann algebras is troublesome, boring. I don't know who first said this:
"[Smart Experimentalist]: 'Yeah, it is reminiscent of what distinguishes the good theorists from the bad ones. The good ones always make an even number of sign errors, and the bad ones always make an odd number.'"-Anthony Zee, Quantum Field Theory in a Nutshell
but whoever it was, he must have experienced calculations on differential forms. In the end it is always the Leibniz rule, the most disguised principle in mathematics: here in disguise of a derivation I guess.

etotheipi
I will try to say the same in more simple terms.

Let
$$\omega=\sum_{0\le i_1<\ldots<i_k\le m}\omega_{i_1\ldots i_k}(x)dx^{i_1}\wedge\ldots\wedge dx^{i_k},\quad x=(x^1,\ldots,x^m)$$
and
$$\frac{d}{dt}g^t(x)=v(g^t(x)),\quad g^0(x)=x.$$
It is clear
$$d(g^t(x)))^i=\frac{\partial (g^t(x))^i}{\partial x^s}dx^s=dx^i+t\frac{\partial v^i(x)}{\partial x^s}dx^s+o(t),\quad t\to 0.$$
By definition
$$L_v\omega=\frac{d}{dt}\Big|_{t=0}\sum_{0\le i_1<\ldots<i_k\le m}\omega_{i_1\ldots i_k}(g^t(x))d(g^t(x))^{i_1}\wedge\ldots\wedge d(g^t(x))^{i_k}.$$
By the way the Cartan formula ##L_v=i_vd+di_v## is extremely easy to prove by using the vector field straightening theorem.

etotheipi

## 1. What is a Lie derivative of a general differential form?

The Lie derivative of a general differential form is a mathematical operation that describes how a differential form changes along a vector field. It is used to measure the rate of change of a differential form with respect to a given direction.

## 2. How is the Lie derivative of a general differential form calculated?

The Lie derivative of a general differential form is calculated by taking the derivative of the form along the vector field, and then subtracting the result from the Lie bracket of the vector field and the form. This can be written as L_V(ω) = d(ω) - [V, ω], where V is the vector field and ω is the differential form.

## 3. What is the significance of the Lie derivative of a general differential form?

The Lie derivative of a general differential form is significant because it allows us to study how a differential form changes as we move along a given direction. This is particularly useful in fields such as differential geometry and physics, where understanding the behavior of differential forms is crucial.

## 4. Can the Lie derivative of a general differential form be extended to higher dimensions?

Yes, the Lie derivative of a general differential form can be extended to higher dimensions. In fact, it is defined for any differential form on a manifold of any dimension. However, the calculation of the Lie derivative becomes more complicated in higher dimensions.

## 5. How is the Lie derivative of a general differential form related to Lie groups?

The Lie derivative of a general differential form is closely related to Lie groups, which are mathematical objects that describe continuous transformations. The Lie derivative can be used to define the action of a Lie group on a differential form, and vice versa. This connection is important in many areas of mathematics and physics, including differential geometry and gauge theory.

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