+ L_\star (\phi \star \varphi))1 + \Sigma + F_A
&i(\Gamma_\mu + \Omega\Gamma_{\mu+4})\Gamma_9 L_\star (D^\mu \phi)\\
i(\Gamma_\mu + \Omega \Gamma_{\mu+4} )\Gamma_9 L_\star (\overline{D^\mu \phi})&
(H^2 + L_\star (\phi \star \phi))1 + \Sigma + F_B
\end{array}\right)$$ with $$\begin{aligned}
D_\mu \phi &=& \partial_\mu\phi - iA_\mu \star \phi + i\phi\star B_\mu\\ 
F_A &=& \{\mathcal{D}_4 , (\Gamma_\mu + \Omega \Gamma_{\mu+4} )L_\star (A^\mu )\} + (\Gamma_\mu + \Omega\Gamma_{\mu+4} )(\Gamma_\nu + \Omega \Gamma_{\nu+4} )L_\star (A^\mu \star A^\nu)\nonumber \\
&=& \{ L_\star(A^\mu ), i\partial_\mu + \Omega^2 M_\bullet(x_\mu)\} + (1 +\Omega^2)L_\star (A_\mu \star A^\mu )  \nonumber \\
&+& i \left(\frac{1}{4} [\Gamma_\mu, \Gamma_\nu]+ \frac{1}{4} \Omega^2 [\Gamma_{\mu+4} , \Gamma_{\nu+4} ] + \Omega\Gamma_\mu \Gamma_{\nu+4} \right) L_\star (F^{\mu\nu}_A ) ,\end{aligned}$$ $(M_\bullet(\tilde{x}_\mu)\psi)(x) = \tilde{x}_\mu\psi(x)$ is ordinary pointwise multiplication and $F_B$ is obtained just replacing $A$ with $B$. We can recognize in previous expression the field strength $F^A_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -i(A_\mu\star A_\nu - A_\nu\star A_\mu )$

Spectral action
---------------

Recalling the spectral action principle, the bosonic action can be defined exclusively by the spectrum of the Dirac operator. The general form for such bosonic action is: $$S(\mathcal{D}_A) = \textrm{Tr}\chi(\mathcal{D}^2_A) \label{Action}$$ Where $\chi$ is a regularization function $\chi: R_+\to R_+$ for which trace exists.\
The trace in is defined on $\mathcal{B}(L^2(\mathbb{R}^4))$ by 