# Elimination of three gluon vertex

I am reading this paper by Dixon, where it is mentioned that one can diagrammatically compute the colour factor of certain Feynman graphs, such as shown in figure 1, where they expand a three gluon vertex as a difference of two diagrams with fermion loops. Now, my question is, is this elimination of three gluon vertex a "trick" to just calculate the colour amplitude, or is it a valid factorization of the complete diagram? More precisely, can I write down:

$$\displaystyle f^{mnp}A^m(x)A^n(x)A^p(x)=\left(f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^n(y)\partial CA^p(z)-f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^p(z)\partial CA^n(y) \right)$$

where, the three colour trace is defined as

$$Tr[\partial CA(x)\partial CA(y)\partial CA(z)] \equiv f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^n(y)\partial CA^p(z)$$

• Why different order? The three gluon vertex should carry three factors of the coupling from $g^2A^2 g\partial A$ – lux Dec 21 '19 at 16:06
• @lux fermion loop diagrams are proportional to $g^3$, and the 3-gluon vertex is proportional to $g$. – Prof. Legolasov Dec 21 '19 at 19:11