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Is there a symmetry between the gauge fields $A_{\sigma}$ and $A_{\lambda}$ in the expression: $$f^{abm}f^{bcn}f^{cap}\partial_{\rho}C(x-y)A^m_{\lambda}(y)\partial_{\sigma}C(y-z)A^n_{\sigma}(z)\partial_{\lambda}C(z-x)A^p_{\rho}(x),$$ that allows us to swap them in the expression, so that we end up with the final expression: $$f^{abm}f^{bcn}f^{cap}\partial_{\rho}C(x-y)A^m_{\sigma}(y)\partial_{\sigma}C(y-z)A^n_{\lambda}(z)\partial_{\lambda}C(z-x)A^p_{\rho}(x).$$

Here, the $C(x-y)$ is the scalar propagator defined as $$\Box C(x-y)=-\delta(x-y).$$

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In general the answer is no. But notice that $y$ and $z$ are dummy variables in the sense that they must be integrated over at some point, if that is the case you can swap the names so that the $\lambda$ and $\sigma$ indices agree almost with what you want. This translates the problem into the question, can you switch $m$ and $n$? This will depend on the particular properties of your structure constants. Without any more context that is as far as you could go, just antisymmetry of $f^{abc}$ won't be enough.

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  • $\begingroup$ The x, y and z are integrated over in the expression, and the $f^{abc}$ are the structure constants of an SU(N) gauge group. However I don't see how this follows from dummy index relabeling. Can you elaborate on your answer? $\endgroup$ – Chetan Pandey Dec 17 '19 at 10:41

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