field contribution} \ +\mathcal{O}(\sqrt{t})\end{aligned}$$ Where in order to simplify the notation we introduce the functions: $$\begin{aligned}
 \mathcal{T}(f)&=&\textrm{Tr}_{L^2(\mathbb{R}^4)}\left(L_\star(f)e^{-tH_4}\right) \nonumber \\
 \mathcal{T}_{\mu_1\cdots\mu_k}(f)&=&\textrm{Tr}_{L^2(\mathbb{R}^4)}\left(L_\star(f)\nabla^{(\Omega)}_{\mu_1}\cdots\nabla^{(\Omega)}_{\mu_k}e^{-tH_4}\right)  \nonumber \\
 \tilde{\mathcal{T}}_{\mu\nu}(f)&=&\textrm{Tr}_{L^2(\mathbb{R}^4)}\left(L_\star(f)\nabla^{(\Omega)}_{\mu_1}\nabla^{(1)}_{\mu_k}e^{-tH_4}\right) \end{aligned}$$

The contributions for the $B$ fields are obtained operating the following substitutions: $$\left\{ A_\mu \to B_\mu, \ F^A_{\mu\nu}\to F^B_{\mu\nu}, \ V_{A,\phi} \to V_{B,\phi}, \ D_\mu\phi \leftrightarrow \overline{D^\mu\phi}\right\}$$ After the computation of the $\mathcal{T},\mathcal{T}_{\mu_1\cdots\mu_k}$ [^3], we have all the ingredients required to compute the leading part of the action replacing all the traces into the . Using the trace property of the star product and the identities $$\begin{aligned}
\tilde{z}_\mu\star f=\frac{1}{2}\{\tilde{z}_\mu,f\} +\frac{1}{2}[\tilde{z}_\mu,f]=\frac{1}{2}\{\tilde{z}_\mu,f\}+i\partial_\mu f \\
\{\tilde{z}_\mu , f\star g \}_\star=\{\tilde{z}_\mu,f \}\star g -if\star \partial_\mu g \end{aligned}$$ we get after some manipulations: $$\begin{aligned}
&&\textrm{Tr}(e^{-t\mathcal{D}^2})=2 \cosh^4(\tilde{\Omega}t) \nonumber \\
&&+\frac{1}{\pi^2(1+\Omega^2)^2} \int d^4z \Bigg\{ -\frac{1}{t}\Bigg(\left(\phi\star\bar{\phi}+ \frac{4\Omega^2}{1+\Omega^2}(\tilde{X}^A_\mu\star\tilde{X}_A^\mu -\tilde{X}^0_{\mu}\star\tilde{X}_0^\mu)\right)\nonumber \\
&&+\left(\bar{\phi}\star\phi+\frac{4\Omega^2}{1+\Omega^2}(\tilde{X}^B_\mu\star\tilde{X}_B^\mu -\tilde{X}^0_{\mu}\star\tilde{X}_0^\mu)\right)\Bigg)\nonumber \\
&&+\frac{1}{2}\left( \phi\star\bar{\phi}+\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^A_\mu\star\tilde{X}_A^\mu\right)^2-\frac{1}{2}\left(\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^\mu_0\star\tilde{X}^0_\mu\right)^2\nonumber \\
&&+\frac{1}{2}\left( \bar{\phi}\star\phi+\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^B_\mu\star\tilde{X}_B^\mu\right)^2-\frac{1}{2}\left(\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^\mu_0\star\tilde{X}^0_\mu\right)^2\nonumber \\
&&+\frac{1}{2}\left(\frac{(1-\Omega^2)^2}{2}-\frac{(1-\Omega^2)^4}{6(1+\Omega^2)^2}\right)(F^A_{\mu\nu}\star F_A^{\mu\nu}+F^B_{\mu\nu}\star F_B^{\mu\nu})\nonumber \\
&&+\frac{1+\Omega^2}{2}\left(D_\mu\phi\star\overline{D^\mu\phi}+\overline{D_\mu\phi}\star D^\mu\phi\right)\Bigg\}(z)+\mathcal{O}(\sqrt{t})\end{aligned}$$ where $$\tilde{X}^A_\mu=\frac{\tilde{z}_\mu}{2}+A_\mu, \ \tilde{X}^B_\mu=\frac{\tilde{z}_\mu}{2}+B_\mu, \ \tilde{X}^0_\mu=\frac{\tilde{z}_\mu}{2}$$ Using the Laurent expansion of $ \coth^4(t^\prime)=t^{\prime-4}+ \frac{4}{3}t^{\prime-2}+\frac{26}{45}+\mathcal{O}(t^{\prime 2})$ and comparing the previous expression to the expansion and putting $\chi_0=\chi(0)$ we are finally able to write the spectral action as: $$\begin{aligned}
&&S(\mathcal{D}_A)= \frac{\theta^4\chi_{-4}}{8\Omega^4}+\frac{2\theta^2\chi_{-2}}{3\Omega^2}+\frac{52\chi_{0}}{45} \nonumber \\
&&+\frac{\chi_0}{\pi^2(1+\Omega^2)^2}+\frac{1}{\pi^2(1+\Omega^2)^2} \int d^4z \Bigg\{\left(\frac{(1-\Omega^2)^2}{2}-\frac{(1-\Omega^2)^4}{6(1+\Omega^2)^2}\right)(F^A_{\mu\nu}\star F_A^{\mu\nu}+F^B_{\mu\nu}\star F_B^{\mu\nu})\nonumber \\
&&+\left(\bar{\phi}\star\phi+ \frac{4\Omega^2}{1+\Omega^2}\tilde{X}^A_\mu\star\tilde{X}_A^\mu-\frac{\chi_{-1}}{\chi_0}\right)^2\nonumber \\
&&+\left(\phi\star\bar{\phi}+\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^B_\mu\star\tilde{X}_B^\mu-\frac{\chi_{-1}}{\chi_0}\right)^2\nonumber \\
&&-2\left(\frac{4\Omega^2}{1+\Omega^2}\tilde{X}^\mu_0\star\tilde{X}^0_\mu-\frac{\chi_{-1}}{\chi_0} \right)^2+ 2(1+\Omega^2)D_\mu\phi\star\overline{D^\mu\phi}\Bigg\}(z)+\mathcal{O}(\chi_1)\label{faction}\end{aligned}$$ We notice that Higgs 