{\rm ``Twist \; element"}.
\label{twistelt}$$ Then $$m_\theta (\alpha \otimes \beta) = m_0 [F_\theta \alpha \otimes
\beta] \label{starmult1}$$ where $m_0$ is the point-wise multiplication map, also defined by (\[starmult\]).

The usual action of the Lorentz group ${\cal L}$ is not compatible with $\ast$-multiplication: transforming $\alpha$ and $\beta$ separately by an arbitrary group element $\Lambda \in {\cal L}$ and then $\ast$-multiplying them is not the same as transforming their $\ast$-product. In other words, the usual coproduct $\Delta_0(\Lambda)
= \Lambda \otimes \Lambda$ on the group algebra ${\mathbb C}\cal{L}$ of $\cal{L}$ is not compatible with $m_\theta$. But a new coproduct $\Delta_\theta$ obtained using the twist is compatible, where $$\Delta_\theta(\Lambda)\,=\,F_\theta^{-1}\,\Delta_0 (\Lambda) F_\theta.
\label{twistedcoprod}$$ Here $\partial_\mu \otimes \theta^{\mu \nu}\partial_\nu$ in $F_\theta$ is to be replaced by $~-~ P_\mu \otimes \theta^{\mu \nu} P_\nu$ where $P_\mu$ are translation generators: we are dealing with ${\cal{P}}_\theta \otimes {\cal{P}}_\theta$ where ${\cal{P}}_\theta$ is a Hopf algebra associated with the Poincaré group algebra ${\mathbb C}{\cal{P}}$ with the coproduct (\[twistedcoprod\]).

This twisted coproduct does not preserve standard (anti-)symmetrization, because it does not commute with the usual flip operator $\tau_0$ defined by $\tau_0:(\alpha \otimes \beta) ~=~
(\beta\otimes\alpha)$:

$$\Delta_\theta(\Lambda) \tau_0 \neq \tau_0 \Delta_\theta (\Lambda).$$

On the other hand, it does preserve twisted (anti-)symmetrization, defined using a 