# Transformation of field strength tensor in non-abelian gauge theory

The field strength tensor is defined as

$$F_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu +g f^{abc} A_\mu^b A_\nu^c$$

where $$f^{abc}$$ are the antisymmetric structure constants and $$A_\mu^a$$ the gauge fields which transform as follows:

$$A_\mu^a\rightarrow A^a_\mu+\frac{1}{g}\partial_\mu \alpha^a-f^{abc}\alpha^b A_\mu^c$$

where $$\alpha^a$$ is infinitesimal and parameterizes the gauge transformation. For example a field transforms as $$\psi\rightarrow U\psi$$, where $$U=\exp i\alpha^a T^a\approx 1 +i\alpha^a T^a$$, where $$T^a$$ are the generators.

I want to calculate the transformation of $$F^a_{\mu\nu}$$ by plugging in the transformation of $$A_\mu^a$$:

$$F_{\mu\nu}^a\rightarrow \partial_\mu (A^a_\nu+\frac{1}{g}\partial_\nu \alpha^a-f^{abc}\alpha^b A_\nu^c)-\partial_\nu (A^a_\mu+\frac{1}{g}\partial_\mu \alpha^a-f^{abc}\alpha^b A_\mu^c) +g f^{abc} (A^b_\mu+\frac{1}{g}\partial_\mu \alpha^b-f^{bde}\alpha^d A_\mu^e) (A^c_\nu+\frac{1}{g}\partial_\nu \alpha^c-f^{chi}\alpha^h A_\nu^i)\\ = F_{\mu\nu}^a-f^{abc}\alpha^b(\partial_\mu A_\nu^c-\partial_\nu A_\mu^c)-f^{chi}\alpha^h gf^{abc}(A_\mu^b A_\nu^i-A_\mu^i A_\nu^b)$$

The last term cannot be correct, since I know that the correct answer is:

$$F_{\mu\nu}^a\rightarrow F_{\mu\nu}^a-f^{abc}\alpha^b(\partial_\mu A_\nu^c-\partial_\nu A_\mu^c+g f^{cde}A_\mu^dA_\nu^e)$$

Can you spot my mistake?

I haven't checked the algebra, but often in these kinds of calculations you need to use the Jacobi identity, $$$$𝑓^{𝑎𝑑𝑒}𝑓^{𝑏𝑐𝑑}+𝑓^{𝑏𝑑𝑒}𝑓^{𝑐𝑎𝑑}+𝑓^{𝑐𝑑𝑒}𝑓^{𝑎𝑏𝑑}=0.$$$$ It would conceptually make sense if you end up needing to use it here, since the Jacobi identity is needed for $$𝑓^{𝑎𝑏𝑐}$$ to be the structure constants for a legitimate Lie algebra, which is necessary for $$F^{a}_{\mu\nu}$$ to transform properly.