as the tensor product of the 8-dimensional spectral triple $(\mathcal{A}_4,\mathcal{H}_4,\mathcal{D}_4,\Gamma_9)$ with the two point Connes-Lott like spectral triple $(\mathbb{C}\otimes \mathbb{C},\mathbb{C}^2, M\sigma_1 )$. The total Dirac operator of the product triple is: $$\mathcal{D}_T = \mathcal{D}_4 \otimes \textbf{I}  + \Gamma_9 \otimes M\sigma_1$$ Or explicitly: $$\mathcal{D}_T= \left(\begin{array}{cc}
    \mathcal{D}_4& M\Gamma_9 \\
    M\Gamma_9& \mathcal{D}_4
\end{array}\right)$$ The algebra becomes $ \mathcal{A}_T=\mathcal{A}_4 \oplus \mathcal{A}_4 $ and acts by diagonal star multiplication on $\mathcal{H}_T= \mathcal{H}_4 \oplus \mathcal{H}_4$. The fluctuated Dirac operator is found using $\mathcal{D}_A = \mathcal{D}_T + \Sigma_i a_i[\mathcal{D}_T,b_i]$ with $a_i,b_i\in\mathcal{A}_T $ of the form $(f,g)$, the computation of the commutator $\mathcal{D}_T$ with $(f,g)$ gives: $$[\mathcal{D}_T,(f, g)] = 
\left(\begin{array}{cc}
    i(\Gamma^\mu + \Omega \Gamma^{\mu+4} )L_\star(\partial_\mu f) & M\Gamma_9 L_\star(f-g) \\
    M\Gamma_9 L_\star(g-f) & i(\Gamma^\mu + \Omega\Gamma^{\mu+4} )L_\star(\partial_\mu g)
\end{array}\right)$$ $L_\star(f)\psi = f \star \psi$ is the left Moyal multiplication. From the commutator we deduce that the form of selfadjoint fluctuated Dirac has to be: $$\mathcal{D}_A =\left(\begin{array}{cc} 
\mathcal{D}_4 + (\Gamma_\mu + \Omega \Gamma_{\mu+4})L_\star(A^\mu )& \Gamma_9 L_\star(\phi) \\
\Gamma_9 L_\star (\bar{\phi}) &\mathcal{D}_4 + (\Gamma_\mu + \Omega \Gamma_{\mu+4} )L_\star (B^\mu ) \end{array}\right)$$ Where $\phi \in \mathcal{A}_4$ is the Higgs complex field and $A_\mu, B_\mu \in \mathcal{A}_4$ are real fields. The spectral action computation needs the square of $\mathcal{D}_A$: $$\mathcal{D}^2_A =\left(\begin{array}{cc}
(H^2_0 