Renormalization group invariant objects of a quantum field theory

Consider an arbitrary QFT with $g_b$ as the bare coupling constant. After dimensional regularization, is $g_b \mu^\epsilon$ a renormalization group invariant object of the theory? In other words, is the following relation correct?

$$\frac{d (g_b \mu^\epsilon)}{d \log{\mu^2}}=0$$

Please note that the number of space-time dimensions is $d=4-2\epsilon$ and $g_b \mu^\epsilon$ is a dimensional object.

$$\begin{split} \frac{d(g_b\mu^{\epsilon})}{d\log\mu^2}&=\frac{\mu}{2}\frac{d(g_b\mu^{\epsilon})}{d\mu}\\ &=\frac{\mu}{2}\left[\mu^{\epsilon}\frac{dg_b}{d\mu}+g_b\frac{d\mu^{\epsilon}}{d\mu}\right] \end{split}$$ By definition, the bare coupling does not depend on the renormalization scale $\mu$. Hence $$\frac{d(g_b\mu^{\epsilon})}{d\log\mu^2}=\frac{\epsilon g_b}{2}\mu^{\epsilon},$$ which vanishes as $\epsilon\rightarrow 0$.
Edit: Notice that the authors define $$a_B=Z_{as} a_s,$$ where $a_B$ is the bare coupling and $a_s$ is the renormalized coupling, with $$a_s\equiv\frac{g(\mu^2)}{16\pi^2}.$$ In order to keep the coupling $g$ dimensionless in dimensional regularization, we must introduce the dimensionfull quantity $\mu$, so that in $d=4-2\epsilon$ dimensions we have $$g\rightarrow \mu^{\epsilon}g,$$ or $$a_s\rightarrow \mu^{2\epsilon}a_s.$$ Hence $$a_B\mu^{2\epsilon}=Z_{as}a_s\mu^{2\epsilon}.$$ Ordinarily (in my experience) we conclude that the bare coupling itself is invariant under the renormalization group flow because we have already included the scale $\mu$ in its definition, i.e. $$a_B=Z_{as}a_s\mu^{2\epsilon}.$$ However, based on the author's convention, we must include the scale $\mu$ on both sides of this equation. Now that we have ensured that the dimensions will be preserved, we can say that $$\frac{d(a_B\mu^{2\epsilon})}{d\log\mu^2}=0.$$
• Thanks for the reply; however, the relation I am looking for does not merely stand in the limit $\epsilon \rightarrow 0$. Please take a look eq. 5 in arxiv.org/abs/hep-ph/9701390. The result of solving the equation is the $d$-dimensional $\beta$-function. – moha Jun 8 '16 at 15:42
• Thanks for clarifying this point; I got confused because they emphasized that $a_B \mu^{2\epsilon}$ is dimensional and it does not run; $a_B$ itself did not run as well!! Another question: couldn't we from the beginning assume that both $a_s$ and $a_B$ are dimensionless and had the relation $a_B=Z_{a_s} a_s$? – moha Jun 8 '16 at 17:59
• That is what the authors assume: $a_s$ and $a_B$ are dimensionless in $d=4$ dimensions. When we go to $d=4-2\epsilon$ dimensions, the dimensions of the fields and hence the coupling constants must change. Rather than have a coupling with non-integer dimension, we introduce the scale $\mu$. We don't need the scale $\mu$. It is nonphysical and will never appear in any physical quantity. Including $\mu$ is nice because it prevents us from having logarithms of dimensionfull quantities in dim reg. It also doubles as a fictitious scale which we can use to compute the running of the coupling. – Evan Rule Jun 8 '16 at 18:30