garrett

Gold Member

- 412

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Hey Joe,

Wow, MG11 looks cool -- have fun. No hurry on the SU(2) metric, get to it when you feel like it.

The derivation of the version of Killing's equation using the frame or vielbein instead of the metric is pretty straightforward. Start with the expression for the Lie derivative of the metric components, then plug in the expression for the metric in terms of the frame components.

Here's what may be causing confusion though: there is no pretty (index free) way, using the notation I've described, to deal with the metric. So I choose to think of Killing's equation in terms of the Lie derivative of the vielbein as fundamental. The two expressions are equivalent, so it's a matter of taste.

(And, oops, I had an index wrong in a previous expression for Killing's equation. It should be:

[tex]

{{\cal L}_\vec{\xi}} \, \vec{e_C} = B_C{}^D \vec{e_D}

[/tex]

with

[tex]

B_C{}^D = - B^D{}_C

[/tex]

)

If you can't prove the equivalence of the two expressions for Killing's equation to your satisfaction, let me know and I'll go through it.

Wow, MG11 looks cool -- have fun. No hurry on the SU(2) metric, get to it when you feel like it.

The derivation of the version of Killing's equation using the frame or vielbein instead of the metric is pretty straightforward. Start with the expression for the Lie derivative of the metric components, then plug in the expression for the metric in terms of the frame components.

Here's what may be causing confusion though: there is no pretty (index free) way, using the notation I've described, to deal with the metric. So I choose to think of Killing's equation in terms of the Lie derivative of the vielbein as fundamental. The two expressions are equivalent, so it's a matter of taste.

(And, oops, I had an index wrong in a previous expression for Killing's equation. It should be:

[tex]

{{\cal L}_\vec{\xi}} \, \vec{e_C} = B_C{}^D \vec{e_D}

[/tex]

with

[tex]

B_C{}^D = - B^D{}_C

[/tex]

)

If you can't prove the equivalence of the two expressions for Killing's equation to your satisfaction, let me know and I'll go through it.

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