$$\begin{aligned}
\mathbf{SE1c} &= 4N_sC^{\gamma\alpha\delta}C^{\bar{\gamma}\beta\bar{\delta}}\int \frac{d\nu d^dk}{(2\pi)^{d+1}}\Delta^{\gamma}(\nu,k)\Delta^{\delta}(\nu,k),\label{eq:SE1cInt}\\
\mathbf{SE1d} &= -2N_fG^{\gamma\delta\alpha}G^{\delta\gamma\beta}\int \frac{d\nu d^dk}{(2\pi)^{d+1}}\mathrm{tr}\left(\Theta^{\gamma}(\nu,k)\Theta^{\delta}(\nu,k)\right),\label{eq:SE1dInt}\\
\mathbf{SE1k} &= -2N_sD^{\alpha\beta\gamma}B^{\bar{\gamma}\bar{\delta}\delta}\int \frac{d\nu d^dk}{(2\pi)^{d+1}}\Delta^{\delta}(\nu,k),\label{eq:SE1kInt}\end{aligned}$$

To evaluate these diagrams we set the quantum numbers of the external lines to $\beta=\bar{\alpha}$ and the total angular momenta on the external lines to $j_{\alpha}=j_{\beta}=1$. We can then sum over all quantum numbers (using appendix \[sapp:AMSums\]) which do not involve the propagators. This usually leaves us with simpler integrals due to the selection rules imposed by the 3-j symbols. After solving the integrals the summation over the rest of the quantum numbers will be regulated by $d$.

For example, in $\mathbf{SE1c}$ the selection rules constrains the summation to $j_{\gamma}=j_{\delta}$ and $j_{\gamma}\geq 1$. The angular momentum sum over all quantum numbers but $j_{\gamma}$ gives: $\sum C^{\gamma\alpha\delta}C^{\bar{\gamma}\bar{\alpha}\bar{\delta}}=2R_c^2(j_{\gamma},1,j_{\gamma})=\frac{j_{\gamma}(j_{\gamma}+1)^2(j_{\gamma}+2)}{\pi^2}$ The integral is easily performed and we get,

$$\begin{aligned}
\label{eq:SE1cCurvedDRExm}
\frac{4N_s}{\pi^2}\frac{\Gamma(\frac{3-d}{2})}{(4\pi)^{\frac{1+d}{2}}}\sum_{j_\gamma=1}^{\infty}j_{\gamma}(j_{\gamma}+2)(j_{\gamma}+1)^{d-1}=\frac{4N_s}{\pi^2}\frac{\Gamma(\frac{3-d}{2})}{(4\pi)^{\frac{1+d}{2}}}(\zeta(-1-d)-\zeta(1-d))=\nonumber\\
=-\frac{N_s}{\pi^2}([\frac{1}{\epsilon}+\frac{1}{2}\ln(\pi)-\frac{\gamma}{2}+1]+\gamma+\frac{1}{12})+\cal O(\epsilon).\end{aligned}$$

Other diagrams can be evaluated in a similar way. The results diagram by diagram are given in appendix \[ssapp:1LoopCurvedDR\]. After employing our subtraction scheme (see below (\[eq:Imn\])), the total result for the 1-loop 1PI self-energy of the gauge field is,

$$\begin{aligned}
G_{DR} &=\frac{6-N_s-4N_f}{\pi^2}\gamma+\frac{9+N_s+16N_f}{6\pi^2}-\frac{10}{\pi^2}\zeta(3).\label{eq:GluonCurvedDR}\end{aligned}$$

### Cutoff regularization on $S^3$ {#sssec:CurvedSpaceCutoff}

For the gauge field we obtain the cutoff regularization expressions by taking the $d\rightarrow 0$ limit 