For the spinor part a convenient basis for the Dirac matrices of $SO(d+4)$ is[^15],

$$\begin{aligned}
\bar{\Gamma}^{i}=\tau^i\otimes\Gamma,~&~ \bar{\Gamma}^a=\mathds{1}\otimes{\Gamma}^a,\end{aligned}$$

where $\left\{\Gamma,\Gamma^a\right\}=0$ and $\left\{\Gamma^a,\Gamma^b\right\}=2\delta^{ab}\mathds{1}$.[^16] The $\Gamma^a$ and $\Gamma$ can be taken to be in the Dirac representation of $SO(d+2)$.

The only terms we will need in the Dirac Lagrangian on $\mathds{R}^{d+1}$ are then (vielbein and spin connection should be understood.),

$$\begin{aligned}
\cal{L}_{\mathds{R}^{d+1}} &= \int_{S^3}\text{tr}\left\{i\bar{\lambda}\left(\bar{\Gamma^i}\partial_i+\bar{\Gamma^a}\partial_a\right)\lambda +g\bar{\lambda}\bar{\Gamma}^i\left[A_i,\lambda\right]\right\}= \nonumber\\ &= \int_{S^3}\text{tr}\left\{i\lambda^{\dag}\left(\tau^{i}\otimes\Gamma^0\Gamma\partial_{i}+\mathds{1}\otimes\Gamma^0\Gamma^a\partial_a\right)\lambda+g\lambda^{\dag}\tau^i\otimes\Gamma^0\Gamma\left[A_i,\lambda\right]\right\}=\nonumber\\
&=\text{tr}\left\{\bar{\psi}^{\alpha}\left(i\Gamma^a\partial_a+\epsilon_{\alpha}(j_{\alpha}+1)\Gamma\right)\psi^{\alpha}-igG^{\alpha\beta\gamma}\bar{\psi}^{\alpha}\Gamma\left[A^{\gamma},\psi^{\beta}\right]\right\},\end{aligned}$$

where in the last line we expanded $\lambda$ in $S^3$ spinor spherical harmonics , $$\begin{aligned}
\lambda = \cal{Y}^{\alpha}(\Omega)\otimes\psi^{\alpha}(\vec{y}).\end{aligned}$$

From the quadratic parts we extract the propagators,

$$\begin{aligned}
\langle a^{\alpha}(\nu,k)a^{\beta}(-\nu,-k)\rangle &\equiv\delta^{\alpha\bar{\beta}}\bar{\Delta}^{\alpha}(k) = \frac{\delta^{\alpha\bar{\beta}}}{k^2+{j_{\alpha}(j_{\alpha}+2)}}, \\
\langle A^{\alpha}(\nu,k)A^{\beta}(-\nu,-k)\rangle &\equiv\delta^{\alpha\bar{\beta}}\Delta^{\alpha}(\nu,k)= \frac{\delta^{\alpha\bar{\beta}}}{\nu^2+k^2+{(j_{\alpha}+1)}}, \\
\langle\cal{A}^{\alpha}_a(\nu,k)\cal{A}^{\beta}_b(-\nu,-k) \rangle &\equiv\delta^{\alpha\bar{\beta}}\hat{\Delta}_{ab}^{\alpha}(\nu,k)= \nonumber\\
&~~~= \frac{\delta^{\alpha\bar{\beta}}}{\nu^2+k^2+{j_{\alpha}(j_{\alpha}+2)}}(\delta_{ab}-\frac{k_ak_b}{k^2+{j_{\alpha}(j_{\alpha}+2)}}),\\
\langle \phi^{\alpha}_{\bar{a}}(\nu,k)\phi^{\beta}_{\bar{b}}(-\nu,-k)\rangle &\equiv\delta^{\alpha\bar{\beta}}\delta_{\bar{a}\bar{b}}\Delta^{\alpha}(\nu,k)= \frac{\delta^{\alpha\bar{\beta}}\delta_{\bar{a}\bar{b}}}{\nu^2+k^2+{(j_{\alpha}+1)}}, \\
\langle \psi^{\alpha}_I(\nu,k)\psi^{\dag\beta J}(\nu,k)\rangle &\equiv\delta^{\alpha\beta}\delta_I^{~J}\Theta^{\alpha}(\nu,k)=\frac{\delta^{\alpha\beta}\delta_I^{~J}}{k_a\Gamma^a+\Gamma\lambda_{\alpha}}.\end{aligned}$$

Now, besides the diagrams in figure 1 we have additional diagrams which are shown in Figure 3.

\[fig:CurvedSpace\] ![Additional self energy diagrams when using dimensional regularization on $S^3$. Dashed (dotted) lines stand for the scalar ($a$) field, wiggly (wavy) line for $A$ ($\cal{A}$) and solid for the fermion.](CurvDepCount.png "fig:"){width="100.00000%"}\

We compute the gauge field 1PI self-energy $\langle A^{\alpha}(0)A^{\beta}(0)\rangle$, with external momenta in the $\mathds{R}^{d+1}$ directions all set to zero.

The expressions for diagrams which were not already evaluated in [@Aharony:2006rf] are,

