Before i state the actual problem, here's a premise. In the case of a Spin 1 massive particle it's possible to demonstrate that $$\sum_{\lambda=0,\pm1}\epsilon_{\lambda}^{* \ \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{q^\mu q^\nu}{q^2}$$ for a massless particle it will be $$\sum_{\lambda=\pm1}\epsilon_{\lambda}^{* \ \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{q^\mu n^\nu+ q^\nu n^\mu}{q \cdot n}-\frac{q^\mu q^\nu}{(q\cdot n)^2}$$

where $n^\mu=(1,0,0,0)$ and $$n\cdot \epsilon=0\\

q\cdot \epsilon=0\\

q\cdot n=q^{0}$$

Ok, now to my understeanding in QED due to the gauge invariance of the theory under $U(1)$ follows the Ward identity: $$q_\mu \mathcal{M^\mu}=0$$ which implies that for all practical purposes we can drop all the terms except for $-g_{\mu\nu}$

My problem lies in the calculation of a QCD process (all particles are assumed massless ) $g \ (gluon)\rightarrow q \ \bar{q}$ needed to compute the splitting function $P_{qg}$ (the probability that a gluon converts into a quark wich carries a fraction of the impulse of the gluon) which is paramtrized in the following way $$K_A (gluon)=(p^0,0,p) \\ K_B(q)=(zp+\frac{p_{\perp}^2}{2 \ zp},p_{\perp},zp) \\

K_C(\bar{q})=((1-z)p+\frac{p_{\perp}^2}{2 \ (1-z)p},-p_{\perp},(1-z)p)$$ such that $$K_A=K_B+K_C$$ provided that $$p^0=p+\frac{p_{\perp}^2}{2 \ z(1-z)p}$$ which gives the gluon a small virtuality. Up to $O(p_{\perp}^4)$ the following identities are true: $$\tag 1 K_A\cdot K_B=\frac{K^2_A}{2}\\

K_A\cdot K_C=\frac{K^2_A}{2}\\

K_B\cdot K_C=\frac{K^2_A}{2}\\

K^2_A=\frac{p_{\perp}^2}{z(1-z)}$$now the authors of the article state that is important to consider: $$\sum_{\lambda=\pm1}\epsilon_{\lambda}^{* \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n}-\frac{K_A^\mu K_A^\nu}{(K_A\cdot n)^2}$$ because the middle term $$\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n}$$ when plugged in the trace (which comes from the sum over polarizations of $\mathcal{ M}(g\rightarrow q \ \bar{q})$) $$ tr({K}\!\!\!/_C \gamma^\mu {K}\!\!\!/_B \gamma^\nu)$$ gives a non zero contribution.

I have two problems:

the first is a conceptual one, why in a QCD calculation i have to consider all the polarization sum terms unlike in QED? Is it due to the fact that QCD is non- abelian? If so, where it comes from mathematically speaking?

The second problem is a practical one: the product $$\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n} \cdot tr({K}\!\!\!/_C \gamma^\mu {K}\!\!\!/_B \gamma^\nu)=-8 p_{\perp}^2+O(p_{\perp}^4)$$ but if i actually do the calculation it yelds me zero since $$tr(\gamma^\alpha \gamma^\mu \gamma^\beta \gamma^\nu)=4(g_{\alpha\mu}g_{\beta\nu}-g_{\alpha\beta}g_{\mu\nu}+g_{\alpha\nu}g_{\beta\mu})$$ then we have form the product : $$\frac{1}{K_A \cdot n}[tr(K\!\!\!/_C K\!\!\!/_A K\!\!\!/_B n\!\!\!/)]+tr(K\!\!\!/_C n\!\!\!/ K\!\!\!/_B K\!\!\!/_A)]$$ which should become $$\frac{8}{K_A \cdot n}[(K_C \cdot K_A)(K_B \cdot n)-(K_C \cdot K_B)(K_A \cdot n)+(K_B \cdot K_A)(K_C \cdot n)]$$ but from eq. $(1)$ we know that becomes: $$\frac{8}{K_A \cdot n }\cdot \left( \frac{K_A^2}{2} \right)[(K_B+K_C-K_A)\cdot n]$$ which should be zero for the conservation of energy! what i am doing wrong? thank you for any help.