Recent content by i0j1k2l3

your eq.6 is wrong, eq.5 comes from following simple trick: $$ f^{abc} \partial_\mu A^a_\nu A^{b\nu}A^{c\mu} = f^{abc} \partial_\nu A^a_\mu A^{b\mu}A^{c\nu} = f^{acb} \partial_\nu A^a_\mu A^{c\mu}A^{b\nu} = -f^{abc} \partial_\nu A^a_\mu A^{c\mu}A^{b\nu} $$

for those terms with no g, you can use ∂A∂B = ∂(A∂B) - A∂∂B，then you can find the second term εμναβ∂∂ should be zero. for terms with g, you need ∂(ABC) = ∂ABC+A∂BC+AB∂C, and combining them with εμναβ fabc, you will see the right hand three terms are equivalent with each other, that's why you...