# Dirac matrices in dimensional regularization, get correct order epsilon

Let us work in dimension $$D = 4-2\epsilon$$.

In 4-dimension, we can write $$\text{Tr}[A B]$$, where $$A$$ and $$B$$ are string of gamma matrices, as

$$\sum_m \text{Tr}[A~\Gamma^m]\text{Tr}[B~\Gamma^m]$$, where $$\Gamma^m = \{1,\gamma_5,\gamma^\mu,\gamma_5\gamma^\mu,\sigma_{\mu\nu}\}$$ are complete set of gamma matrices spanning the dirac space in 4-dim.

As it is well-known, generalizing this to non-integer $$D$$ dimension causes difficulties since $$\gamma_5$$ (defined as $$\gamma_5= i\gamma^0\gamma^1\gamma^2\gamma^3$$ in 4-dim.) cannot be well-defined.

One often does not need to work with the explicit form of $$\gamma_5$$, but uses the two relations to evaluate the trace:

i)$$~\{\gamma_5,\gamma^\mu\}=0\,,$$

ii)$$~\text{Tr}[\gamma_5\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}]=-4i\epsilon_{\mu\nu\rho\sigma}\,.$$

However, in $$D$$ dimension, the two relations cannot be satisfied simultaneously; and people use different $$\gamma_5$$-schemes to treat $$\gamma_5$$ in $$D$$-dimension.

However, when you evaluate both of the traces $$\text{Tr}[A B]$$ and $$\sum_m \text{Tr}[A~\Gamma^m]\text{Tr}[B~\Gamma^m]$$ in $$D$$-dimension by using different $$\gamma_5$$-schemes they do not necessarily agree.

As an example,

take $$A = (\gamma\cdot p_1)\gamma^\alpha(\gamma\cdot p_2)$$ and $$B =\gamma^\beta(\gamma\cdot p_1)(\gamma\cdot p_2)$$.

Then evaluation of $$\sum_m \text{Tr}[A~\Gamma^m]\text{Tr}[B~\Gamma^m]$$ requires $$\gamma_5$$ scheme choice.

Then $$\text{Tr}[A B] = -4~(D-2)~(2~(p_1\cdot p_2)^2 - p_1^2~ p_2^2)$$

$$\left(\sum_m \text{Tr}[A~\Gamma^m]\text{Tr}[B~\Gamma^m]\right)_{\text{t'Hooft-Veltman}} = -4~(D-2)~\left((D-2)~(p_1\cdot p_2)^2 - (D-3)p_1^2~ p_2^2\right)$$

$$\left(\sum_m \text{Tr}[A~\Gamma^m]\text{Tr}[B~\Gamma^m]\right)_{\text{NDR}} = -4D~(p_1\cdot p_2)^2 + 8p_1^2~ p_2^2$$

where 'NDR' is naive dimensional regularization scheme and 't'Hooft-Veltman' is t'Hooft-Veltman-scheme.

All three results agree when $$D$$ is taken to be 4, but do not agree in $$\epsilon$$ terms. Is there a way to ensure agreement down to $$\epsilon$$ piece?