Show that [J_a,G_a] = 0, commutation relationships

ma18
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Homework Statement


Using the given equations prove that

upload_2015-10-28_12-22-32.png

Homework Equations

upload_2015-10-28_12-22-47.png
,

upload_2015-10-28_12-23-42.png
,[/B]

upload_2015-10-28_12-21-41.png
+
upload_2015-10-28_12-21-50.png


(it won't render together in Maple for whatever reason)

The Attempt at a Solution



So I started with expanding the Jacobi Identity (the third relevant equation) and through tedious algebra arrived at proving it unnecessarily and then finding that

upload_2015-10-28_12-27-32.png


then by applying the second equation this goes to

upload_2015-10-28_12-27-49.png


I am unsure how to proceed from here however.

The question also gives the hint that the trick to solve it is to take the appropriate particular values of the labels a,b,y in the Jacobi identity but I am unsure how to use this.

Any help would be greatly appreciated.
 
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You're not actually given ##[G_a,G_b]##, so it's a bit of a hassle to use that Jacobi identity. Try to find a way to relate ##[J_a,G_b]## to the expression ## [ J_a, [J_b,J_c]]## and use the Jacobi identity for that.
 
fzero said:
You're not actually given ##[G_a,G_b]##, so it's a bit of a hassle to use that Jacobi identity. Try to find a way to relate ##[J_a,G_b]## to the expression ## [ J_a, [J_b,J_c]]## and use the Jacobi identity for that.

Hmm okay, tbh I'm not sure how to proceed with that
 
ma18 said:
Hmm okay, tbh I'm not sure how to proceed with that

To push you along, consider
$$ {\epsilon_a}^{bc} [J_b,J_c] = i {\epsilon_a}^{bc} {\epsilon^d}_{bc} G_d.$$
Using an identity for the Levi-Civita symbol will give you an expression for ##G_a## in terms of the commutator of ##J##s.
 
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