# Lie derivative problem/error

1. Sep 16, 2008

### jostpuur

I'm trying to use an example to make sense out of the equation

$$\mathcal{L}_X = d\circ i_X + i_X\circ d.$$

Some simple equations:

$$\omega = \omega^1 dx_1 + \omega^2 dx_2$$

$$i_X\omega = X_1\omega^1 + X_2\omega^2$$

$$(d\omega)^{11} = (d\omega)^{22} = 0,\quad (d\omega)^{12} = \frac{1}{2}(\partial_1\omega^2 - \partial_2\omega^1) = - (d\omega)^{21}$$

$$\mu = \mu^{12} dx_1\wedge dx_2 + \mu^{21} dx_2\wedge dx_1$$

$$(i_X\mu)^1 = X_1 \mu^{11} + X_2 \mu^{21} = X_2 \mu^{21},\quad\quad (i_X\mu)^2 = X_1 \mu^{12}$$

$$\eta = \eta^0$$

$$(d\eta)^1 = \partial_1 \eta^0,\quad\quad (d\eta)^2 = \partial_2 \eta^0$$

Applying them:

$$((d\circ i_X)\omega)^1 = \partial_1 (i_X\omega) = \partial_1(X_1\omega^1 + X_2\omega^2)$$

$$((d\circ i_X)\omega)^2 = \partial_2 (i_X\omega) = \partial_2(X_1\omega^1 + X_2\omega^2)$$

$$((i_X\circ d)\omega)^1 = X_2 (d\omega)^{21} = -\frac{X_2}{2}(\partial_1\omega^2 - \partial_2\omega^1)$$

$$((i_X\circ d)\omega)^2 = X_1 (d\omega)^{12} = \frac{X_1}{2}(\partial_1\omega^2 - \partial_2\omega^1)$$

$$((d\circ i_X \;+\; i_X\circ d)\omega)^1 = (\partial_1 X_1)\omega^1 \;+\; (\partial_1 X_2)\omega^2 \;+\; X_1\partial_1\omega^1 \;+\; \frac{X_2}{2}\partial_1\omega^2 \;+\; \frac{X_2}{2}\partial_2\omega^1$$

$$((d\circ i_X \;+\; i_X\circ d)\omega)^2 = (\partial_2 X_1)\omega^1 \;+\; (\partial_2 X_2)\omega^2 \;+\; \frac{X_1}{2}\partial_2 \omega^1 \;+\; X_2\partial_2\omega^2 \;+\; \frac{X_1}{2}\partial_1 \omega^2$$

There exists a following formula for the Lie derivative:

$$(\mathcal{L}_X\omega)^{i_1,\ldots ,i_k} = X\cdot \omega^{i_1,\ldots ,i_k} \;+\; \sum_{\alpha = 1}^k (\partial_{i_{\alpha}} X_j) \omega^{i_1,\ldots ,i_{\alpha - 1},j, i_{\alpha + 1}, \ldots , i_k}$$

In this example it becomes

$$(\mathcal{L}_X\omega)^1 = X_1\partial_1 \omega^1 \;+\; X_2\partial_2\omega^1 \;+\; (\partial_1 X_1) \omega^1 \;+\; (\partial_1 X_2) \omega^2$$

$$(\mathcal{L}_X\omega)^2 = X_1\partial_1 \omega^2 \;+\; X_2\partial_2 \omega^2 \;+\; (\partial_2 X_1)\omega^1 \;+\; (\partial_2 X_2) \omega^2$$

But this starts to look like

$$\mathcal{L}_X \omega \neq (d\circ i_X + i_X\circ d)\omega$$

Where is this going wrong?

2. Sep 17, 2008

### haushofer

For your particular example of $\omega$ being a one-form and X the vector field, I get

$$[di_{X}+i_{X}d ]\omega = d[\omega(X)] + d\omega (X) \\ = d(\omega_{\mu}X^{\mu}) + 2[\partial_{[\mu}\omega_{\nu]}dx^{\mu}\wedge dx^{\nu}] (X)$$

This becomes

$$[\omega_{\mu}\partial_{\nu}X^{\mu} + X^{\mu}\partial_{\nu}\omega_{\mu}]dx^{\nu} + X^{\mu}[\partial_{\mu}\omega_{\nu} - \partial_{\nu}\omega_{\mu}]dx^{\nu}$$

We see that two terms cancel and we obtain

$$[\omega_{\mu}\partial_{\nu}X^{\mu} + X^{\mu}\partial_{\mu}\omega_{\nu}]dx^{\nu}$$

This looks like the Lie-derivative:

$$\mathcal{L}_{X}\omega = [X^{\nu}\partial_{\nu}\omega_{\mu} + \partial_{\mu}X^{\nu}\omega_{\nu}]dx^{\mu}$$

If this doesn't clarify your problem I can look at all those explicit index-stuff you wrote down, but hopefully this helps :P

3. Sep 20, 2008

### jostpuur

The mistake was here. That was the same thing as

$$i_X\mu = i_X\big(\mu^{ij} dx_i\wedge dx_j\big) = X_i \mu^{ij} dx_j$$

when it should have been

$$i_X\mu = \frac{1}{2}i_X\big((\mu^{ij} - \mu^{ji})dx_i\otimes dx_j\big) = \frac{1}{2} X_i(\mu^{ij} - \mu^{ji})dx_j$$

I found that when following your calculation, and looking where it started going differently.