# Cartan's Identity

1. Dec 21, 2006

### SeReNiTy

Hi can someone help me prove this identity? I'm having trouble understand the role of the interior product or more precisely how to calculate with it.

My professor uses the "cut" notation _| but i don't see this in any text books. Can someone give me some hints on how to prove the identity?

2. Dec 22, 2006

### coalquay404

There's nothing difficult about proving Cartan's formula, although I admit that it can become a nightmare keeping track of the indices when you deal with a general $p$-form. For those who aren't aware of what Cartan's formula states, it says that the Lie derivative of a $p$-form $\alpha$ along the flow generated by a vector field $X$ is equal to the anticommutator of the interior product and the exterior derivative, i.e.,

$$\pounds_X\alpha = \{d,\iota_X\}\alpha = d(\iota_X\alpha) + \iota_X(d\alpha)$$

There's really no mystery here: simply plug in the definitions of the exterior derivative, interior product, and Lie derivative, and show that the left-hand and right-hand sides are equal.

In case you're really having problems understanding the interior product, note that it's simply a map which takes a vector and a $p$-form as arguments and spits out a $(p-1)$-form as an output. That is, the interior product is a map

$$\iota:T_xM\times\Lambda^p_x(M)\to\Lambda^{(p-1)}(M),\,\,\,\iota:(X,\alpha)\mapsto\iota_X\alpha.$$

It is defined so that

$$\iota_X\alpha(X_{(2)},\ldots,X_{(p)}) \equiv \alpha(X,X_{(2)},\ldots,X_{(p)}).[/itex] If this doesn't help, could you be a bit more specific about precisely what it is you're having difficulty with? Last edited: Dec 22, 2006 3. Dec 22, 2006 ### SeReNiTy My problem was i was trying to define the lie derivative of vector fields in the direction of vector fields ie, [X,Y] = LxY to equal the anticommutator. 4. Dec 23, 2006 ### coalquay404 You can define the Lie derivative of one vector field along the flow generated by another to be equivalent to the Lie bracket. That is, if $X,Y\in\mathfrak{X}(\mathcal{M})$ then [tex] \pounds_XY = [X,Y].$$

However, Cartan's identity doesn't deal with the behaviour of vectors at all - it deals with the behaviour of $p$-forms. Look, suppose that you have some $p$-form $\alpha$ and $X\in\mathfrak{X}(\mathcal{M})$. Then Cartan's identity states that

$$\pounds_X\alpha = \{d,\iota_X\}\alpha.$$

In order to prove this identity you need to have explicit knowledge of three things: the Lie derivative, the interior product, and the exterior derivative. Suppose that we have a metric on our manifold with an associated covariant derivative operator $\nabla$. Then recall that, in component notation, the Lie derivative of $\alpha$ along the flow generated by $X$ is

$$(\pounds_X\alpha)_{i_1i_2\ldots i_p} = X^m\nabla_m \alpha_{i_1i_2\ldots i_p} +\sum_{\mu=1}^p\alpha_{i_1\ldots i_{\mu-1}mi_{\mu+1}\ldots i_p}\nabla_{i_\mu}X^m$$

(This is a standard identity for the Lie derivative - you should memorise not only this but also the more general expressions for the Lie derivative of a tensor of type $(r,s)$ along a vector field, and a weight-$w$ tensor density of type $(r,s)$ along a vector field.) The above equation gives you the left-hand side of Cartan's identity. Now focus on the right-hand side. Recall that the interior product has components

$$(\iota_X\alpha)_{i_2\ldots i_p} = \frac{1}{p}X^{i_1}\alpha_{i_1i_2\ldots i_p}.$$

If you take the exterior derivative of this you should then have an expression for the components of $d(\iota_X\alpha)$. Now reverse the process so that you obtain an expression for the components of $\iota_X(d\alpha)$. Add them together to show that

$$(d(\iota_X\alpha))_{i_1\ldots i_p} + (\iota_X(d\alpha))_{i_1\ldots i_p} = (\pounds_X\alpha)_{i_1\ldots i_p}.$$

You've then proved that Cartan's identity holds for a general $p$-form.