&&E_a=-i_{\xi_a}F \,.\end{aligned}$$ Then the exterior derivatives of the electric one-forms yield $$\begin{aligned}
dE_a=i_{\xi_a}dF-{\cal L}_{\xi_a}F=0 \,,\end{aligned}$$ where $F=dA$ is used. Hence there exist locally the potentials $\psi_a$ such that $$\begin{aligned}
d\psi_a=-\frac{1}{\sqrt{3}}i_{\xi_a}F \,. \end{aligned}$$ Hence, the gauge potential can be written as $$\begin{aligned}
 A = \sqrt{3}\psi_adx^a + A_i dx^i \,,\end{aligned}$$ where $\psi_a$ is also independent of the coordinates $x^a$. Next, define the magnetic one-form $B$ by $$\begin{aligned}
&&B=*(\xi_1\wedge \xi_2\wedge F) \,. \end{aligned}$$ Noting that $B$ can be rewritten as $
B=*(\xi_1\wedge \xi_2\wedge F)=-i_{\xi_2}*(\xi_1\wedge F)=i_{\xi_2}i_{\xi_1}*F
$ and using the identity $ 
di_{\xi_2}i_{\xi_1}=i_{\xi_2}i_{\xi_1}d+i_{\xi_1}{\cal L}_{\xi_2}-i_{\xi_2}{\cal L}_{\xi_1}
$, we can write the exterior derivative of $B$ as $$\begin{aligned}
dB=i_{\xi_2}i_{\xi_1}d*F \,. \end{aligned}$$ Then, using the Maxwell equation (\[Maxeq\]), we find that $$\begin{aligned}
dB= -\frac{1}{\sqrt{3}}i_{\xi_2}i_{\xi_1}F\wedge F 
  =  \frac{2}{\sqrt{3}} E_1\wedge E_2 
  = 2\sqrt{3} d\psi_1\wedge d\psi_2=\sqrt{3} d(\psi_1d\psi_2-\psi_2d\psi_1) \,.\end{aligned}$$ This immediately implies that there exists the magnetic potential $\mu$ such that $$\begin{aligned}
d\mu=\frac{1}{\sqrt{3}}B-\epsilon^{ab}\psi_ad\psi_b \,,
\label{eq:mu} \end{aligned}$$ where $\epsilon^{12}=-\epsilon^{21}=1$. We also introduce the twist one-form by $$V_a=*(\xi_1\wedge \xi_2\wedge d\xi_a) \,.$$ Using the Einstein-equation, we can write the exterior derivative of $V_a$ as $$\begin{aligned}
dV_a &=& 2*(\xi_1\wedge \xi_2\wedge R(\xi_a)) \nonumber \\
    &=&-\tau^{-1}i_{\xi_2}i_{\xi_1}*^2(\xi_1\wedge\xi_2\wedge E_a\wedge B) 
\nonumber \\ 
    &=&-E_a\wedge B \nonumber \\
    &=&-3d\psi_a\wedge (d\mu+\epsilon^{bc}\psi_bd\psi_c) \nonumber \\
    &=&-3d[\psi_ad\mu]-d[\psi_a\epsilon^{bc}\psi_bd\psi_c] \,, \end{aligned}$$ where $R(\xi_a)$ in the first line 