Where did the error occur in calculating the Lie derivative using an example?

In summary, the conversation discusses the use of an example to make sense of the equation \mathcal{L}_X = d\circ i_X + i_X\circ d and explores its application in simple equations involving forms and vector fields. It also introduces the Lie derivative and discusses its relationship to the equation, pointing out a mistake in the calculation.
  • #1
jostpuur
2,116
19
I'm trying to use an example to make sense out of the equation

[tex]
\mathcal{L}_X = d\circ i_X + i_X\circ d.
[/tex]

Some simple equations:

[tex]
\omega = \omega^1 dx_1 + \omega^2 dx_2
[/tex]

[tex]
i_X\omega = X_1\omega^1 + X_2\omega^2
[/tex]

[tex]
(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}
[/tex]


[tex]
\mu = \mu^{12} dx_1\wedge dx_2 + \mu^{21} dx_2\wedge dx_1
[/tex]

[tex]
(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}
[/tex]


[tex]
\eta = \eta^0
[/tex]

[tex]
(d\eta)^1 = \partial_1 \eta^0,\quad\quad (d\eta)^2 = \partial_2 \eta^0
[/tex]

Applying them:

[tex]
((d\circ i_X)\omega)^1 = \partial_1 (i_X\omega) = \partial_1(X_1\omega^1 + X_2\omega^2)
[/tex]

[tex]
((d\circ i_X)\omega)^2 = \partial_2 (i_X\omega) = \partial_2(X_1\omega^1 + X_2\omega^2)
[/tex]

[tex]
((i_X\circ d)\omega)^1 = X_2 (d\omega)^{21} = -\frac{X_2}{2}(\partial_1\omega^2 - \partial_2\omega^1)
[/tex]

[tex]
((i_X\circ d)\omega)^2 = X_1 (d\omega)^{12} = \frac{X_1}{2}(\partial_1\omega^2 - \partial_2\omega^1)
[/tex]

[tex]
((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
[/tex]

[tex]
((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
[/tex]

There exists a following formula for the Lie derivative:

[tex]
(\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}
[/tex]

In this example it becomes

[tex]
(\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
[/tex]

[tex]
(\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
[/tex]

But this starts to look like

[tex]
\mathcal{L}_X \omega \neq (d\circ i_X + i_X\circ d)\omega
[/tex]

Where is this going wrong?
 
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  • #2
For your particular example of [itex]\omega[/itex] being a one-form and X the vector field, I get

[tex]

[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)

[/tex]

This becomes

[tex]
[\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}
[/tex]

We see that two terms cancel and we obtain

[tex]

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

[/tex]

This looks like the Lie-derivative:

[tex]

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

[/tex]

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
jostpuur said:
[tex]
(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}
[/tex]

The mistake was here. That was the same thing as

[tex]
i_X\mu = i_X\big(\mu^{ij} dx_i\wedge dx_j\big) = X_i \mu^{ij} dx_j
[/tex]

when it should have been

[tex]
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
[/tex]

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

What is a Lie derivative problem/error?

A Lie derivative problem/error refers to a discrepancy or inconsistency between the actual value of a mathematical object and the value predicted by the Lie derivative. The Lie derivative is a mathematical tool used to describe how a geometric object changes along a given direction.

How is a Lie derivative problem/error detected?

A Lie derivative problem/error can be detected by comparing the predicted value of a geometric object using the Lie derivative to the actual value of the object. If there is a significant difference between the two values, then there may be a problem with the Lie derivative calculation or the underlying data.

What causes a Lie derivative problem/error?

A Lie derivative problem/error can be caused by various factors, such as errors in the input data, incorrect assumptions made in the calculation, or limitations of the mathematical tools used. It can also arise due to the non-linearity of the system being studied.

How can a Lie derivative problem/error be fixed?

The first step in fixing a Lie derivative problem/error is to identify the source of the discrepancy. This may involve double-checking the input data, reviewing the assumptions made in the calculation, or using alternative mathematical tools. Once the source of the error is identified, appropriate adjustments can be made to improve the accuracy of the Lie derivative.

What are the implications of a Lie derivative problem/error in scientific research?

A Lie derivative problem/error can have significant implications in scientific research, as it can lead to incorrect conclusions and hypotheses. It is important for scientists to be aware of the potential for Lie derivative problems/errors and to thoroughly check their calculations to ensure the accuracy of their results.

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