L}, \hat{x}^{\nu L}] = i \theta^{\mu \nu} =
-[\hat{x}^{\mu R}, \hat{x}^{\nu R}].$$ Hence in view of (\[LRcommute\]), $$\hat{x}^{\mu c} = \frac{1}{2} \left( \hat{x}^{\mu L} + \hat{x}^{\mu R}
\right)$$ generates a representation of the commutative algebra ${\cal
  A}_0({\mathbb R}^N)$: $${[}\hat{x}^{\mu c}, \hat{x}^{\nu c}] = 0.$$

We regard elements of the gauge group ${\cal G}$ as maps from ${\cal
A}_0({\mathbb R}^N)$ to the Lie group $G$. $$g:~\hat{x^c}\longrightarrow g(\hat{x^c})~\in G$$ In cases of interest, $G$ is a compact connected Lie group . For convenience, we also think of $G$ concretely in terms of the defining finite-dimensional faithful representation by linear operators.

Fields which transform non-trivially under ${\cal G}$ are modules over ${\cal A}_\theta ({\mathbb R}^N)$. If a $d$-dimensional representation of $G$ is involved, they can be elements of ${\cal A}_\theta ({\mathbb
R}^N) \otimes {\mathbb C}^d$. Compatibility of gauge transformations (on these modules) with the $\ast$-product requires us to twist the coproduct on the gauge group too. The new coproduct is $$\Delta_\theta (g(\hat{x}^c)) = F_\theta^{-1} [g(\hat{x}^c) \otimes
  g(\hat{x}^c)] F_\theta \;.
\label{twistedgauge}$$

This deformation of the coproduct for gauge transformations is neccessary if we want to be able to construct gauge scalars, and other composite operators (see [@bpqv1] for details).

Finally, we need to understand 