Then the delta function in the first term  of Eq.~\eqref{A14} implies that $\Omega=(s_2-s_3)\epsilon_\mathbf{k}$. Since we chose $\Omega>0$,   the delta function is non-zero only  if $s_2=+1$ and $s_3=-1$. Therefore, $\Omega=2\epsilon_\mathbf{k}$ or $\epsilon_\mathbf{k}=\Omega/2$. However, the limits of integration over $\omega$ in Eq.~\eqref{A11} along with the condition $\omega=\xi_\bk^{s_3}$ imply that $-\Omega<\xi_k^{s_3}=-\epsilon_\mathbf{k}-\EF$. Thus, we have $-\Omega<-\Omega/2-\EF$ or $\Omega>2\EF=\od$.
The second term in Eq.~\eqref{A14} is analyzed in a similar way and with the same result.

Now we show that the first  term in Eq.~\eqref{A12}, proportional to $\Im\tilde{\Sigma}_{s_5}(\bk+\bq,\bq,\omega+\Omega)$, does contribute to the real part of the conductivity for $\Omega<\od$.
The contribution of this term to $\Im W_{\text{SE}_1,\text{R}}$ (denoted by superscript 1) is given by
\be
\Im W^{(1)}_{\text{SE}_1,\text{R}}(\bk,\bq,
\Omega)=-\frac{1}{\pi}\Re\left[g_{s_1,\text{R}}(\mathbf{k},\Omega+\xi^{s_3}_\mathbf{k})g_{s_2,\text{R}}(\mathbf{k},\Omega+\xi^{s_3}_\mathbf{k})\right]\int^0_{-\Omega}\dee\omega
\Im\tilde{\Sigma}_{s_5,\text{R}}(\mathbf{k+q},\mathbf{q},\omega+\Omega)\Im g_{s_3,\text{R}}(\mathbf{k},\omega).\nn\\
\label{ImWSE1}
\ee
Note that, formally speaking, the real part of the product of two Green's function in the equation above contains a highly singular term: $\left[\delta\left(\Omega+\xi_\bk^{s_3}-\xi^{s_1}_\bk\right)\right]^2$. In fact, it can be shown that such a term only renormalizes the Drude weight but does not contribute to the regular part of the conductivity. Postponing the proof till  Sec.~\ref{sec:delta2}, we now proceed discarding this term.
From Eq.~\eqref{eq:TildeSigmaZeroTForm},
\be
\Im\tilde{\Sigma}_{s_5,\text{R}}(\mathbf{k+q},\mathbf{q},\omega+\Omega)=
-\frac{\pi}{4}\sum\limits_{s_4,s_6=\pm 1}&\int_\mathbf{p}\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})
\times\delta(\omega+\Omega+\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_5})\nn\\
\ee
Because $\Omega+\omega>0$, the delta function in the equation above implies that $\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_5}<0$. Comparing this condition with the ones imposed by the theta functions, we see that only the first set of theta functions is non-zero
Integrating over $\omega$ with the help of Eq.~\eqref{imgs3}, we obtain
\bea
\Im W^{(1)}_{\text{SE}_1,\text{R}}(\bk,\bq,
\Omega)
-\frac{\pi}{4}\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}&&\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})
\ag{\theta(\Omega+\xi^{s_3}_\mathbf{k})}
\theta(-\xi_\mathbf{k}^{s_3})\theta(\xi^{s_{6}}_\mathbf{p})\theta(-\xi^{s_4}_\mathbf{p+q})\theta(\xi_\mathbf{k+q}^{s_5})
\nonumber\\
\eea
Substituting the last result into
$\Im\Pi^{\text{SE}_1}_\text{R}$, obtained by analytic continuation of Eq.~\eqref{A7},
\be
\Im\Pi^{\text{SE}_1}_\text{R}&&(\Omega)=
\sum\limits_{\mathcal{S}_{\text{SE}_1}}\int_\mathbf{k,q}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s_1}\hat{M}_{\mathbf{k+q}}^{s_5}\hat{M}_{\mathbf{k}}^{s_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k}}^{s_3}\right)V^2_\text{st}(\mathbf{q})\times \frac{1}{\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\frac{1}{\Omega-\xi^{s_2}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\times\nonumber\\
\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})
\theta(\Omega+\xi^{s_3}_\mathbf{k})
\theta(-\xi_\mathbf{k}^{s_3})\theta(\xi^{s_{6}}_\mathbf{p})\theta(-\xi^{s_4}_\mathbf{p+q})\theta(\xi_\mathbf{k+q}^{s_5})\delta(\Omega+\xi_\mathbf{k}^{s_3}+\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_5}).\IEEEeqnarraynumspace
\ee
Relabeling $\mathbf{p}\rightarrow\mathbf{p-q}$ and then $\mathbf{p}\rightarrow\mathbf{-p}$, and using that $\epsilon_\bk=\epsilon_{-\bk}$, we find
\bea
\Im\Pi^{\text{SE}_1}_\text{R}(\Omega)
=-\frac{\pi^2}{32} \sum\limits_{\mathcal{S}}&\int_{\mathbf{k,p,q},\nu}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s_1}\hat{M}_{\mathbf{k+q}}^{s_5}\hat{M}_{\mathbf{k}}^{s_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k}}^{s_3}\right)\tr(\hat{M}_\mathbf{-p-q}^{s_6}\hat{M}_\mathbf{-p}^{s_4})V^2_\text{st}(\mathbf{q})\times \frac{1}{\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\frac{1}{\Omega-\xi^{s_2}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\nonumber\\
\theta(\Omega+\xi^{s_3}_\mathbf{k})
\theta(-\xi_\mathbf{k}^{s_3})\theta(\xi^{s_{6}}_\mathbf{p+q})\theta(-\xi^{s_4}_\mathbf{p})\theta(\xi_\mathbf{k+q}^{s_5})\delta(\Omega+\nu+\xi_\mathbf{k}^{s_3}-\xi_\mathbf{k+q}^{s_5})\delta(\nu-\xi^{s_4}_\mathbf{p}+\xi^{s_{6}}_\mathbf{p+q}),\label{A18}
\eea
where we have re-introduced
\be
\Im\Pi^{\text{SE}_1}(\Omega)=-\frac{\pi^2}{32}\sum\limits_{\mathcal{S}}&\int_{\mathbf{k,p,q},\nu}V^2_\text{st}(\mathbf{q})\times\mathcal{T}^{\text{SE}_1}_{\mathcal{S}}\mathcal{G}^{\text{SE}_1}_{\mathcal{S}}\nonumber\\
\theta(\Omega+\xi^{s_3}_\mathbf{k})
\theta(-\xi_\mathbf{k}^{s_3})\theta(\xi^{s_{6}}_\mathbf{p+q})\theta(-\xi^{s_4}_\mathbf{p})\theta(\xi_\mathbf{k+q}^{s_5})\delta(\Omega+\nu+\xi_\mathbf{k}^{s_3}-\xi_\mathbf{k+q}^{s_5})\delta(\nu-\xi^{s_4}_\mathbf{p}+\xi^{s_{6}}_\mathbf{p+q}),
\label{ImSE1}
\ee
\bse
\bea
\mathcal{T}^{\text{SE}_1}_{\mathcal{S}}=\frac{1}{d}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s_1}\hat{M}_{\mathbf{k+q}}^{s_5}\hat{M}_{\mathbf{k}}^{s_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k}}^{s_3}\right)\tr\left(\hat{M}_\mathbf{-p-q}^{s_6}\hat{M}_\mathbf{-p}^{s_4}\right)\\
\mathcal{G}^{\text{SE}_1}_{\mathcal{S}}=\frac{1}{\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\frac{1}{\Omega-\xi^{s_2}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}.
\eea
\ese
The last equation gives the SE$_1$ part of Eq.~\eqref{eq:firsteqofresultlist} in the main text.

\subsubsection{SE$_2$ diagram \label{SE2}}

On the Matsubara axis, the second self-energy diagram ($\text{SE}_2$) in Fig.~\ref{fig:AllDiagrams} is related  to the $\text{SE}_1$ one via
\be
\Pi^{\text{SE}_2}(i\Omega)=
\Pi^{\text{SE}_1}(-i\Omega),\IEEEeqnarraynumspace
\ee
which, upon analytic continuation, implies that
\bea
\Im\Pi^{\text{SE}_2}_\text{R}(\Omega)=\Im\Pi^{\text{SE}_1}_\text{A}(-\Omega)=-\Im\Pi^{\text{SE}_1}_\text{R}(-\Omega),\label{A23}
\eea
where $\Pi^{\text{SE}_1}_\text{A}(\ve)$ is the advanced correlation function. Note that although the total correlation function is a real function of time and, therefore, the imaginary part of its Fourier transform  is an odd function of frequency, the partial contributions to the total correlation function from different diagrams do not have definite parity. Therefore, Eq.~\eqref{A23} cannot be simplified further. In the previous section, we considered $\Im\Pi^{\text{SE}_1}_\text{R}(\Omega)$ for $\Omega>0$; all we have to do now is the extend this analysis for $\Omega<0$. Omitting the steps, which are almost identical to those in the previous section,
we present only the final result:
\be
\Im\Pi^{\text{SE}_2}(\Omega)=
\sum\limits_{\mathcal{S}}&\int_{\mathbf{k,p,q},\nu}V^2_\text{st}(\mathbf{q})\times\mathcal{T}^{\text{SE}_2}_{\mathcal{S}}\mathcal{G}^{\text{SE}_2}_{\mathcal{S}}\nonumber\\
\theta(\Omega+\xi^{s_3}_\mathbf{k})
\theta(-\xi_\mathbf{k}^{s_3})\theta(\xi^{s_{6}}_\mathbf{p+q})\theta(-\xi^{s_4}_\mathbf{p})\theta(\xi_\mathbf{k+q}^{s_5})\delta(\Omega+\nu+\xi_\mathbf{k}^{s_3}-\xi_\mathbf{k+q}^{s_5})\delta(\nu-\xi^{s_4}_\mathbf{p}+\xi^{s_{6}}_\mathbf{p+q}),
\label{ImSE2}
\ee
\bse
\be
\mathcal{T}^{\text{SE}_2}_{\mathcal{S}}=\tr\left(\hat{\bv}
\hat{M}_\mathbf{-k-q}^{s_1}\hat{M}_{\mathbf{-k}}^{s_3}\hat{M}_{\mathbf{-k-q}}^{s_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{-k-q}}^{s_5}\right)\tr\left(\hat{M}_\mathbf{p}^{s_4}\hat{M}_\mathbf{p+q}^{s_6}\right)\\
\mathcal{G}^{\text{SE}_2}_{\mathcal{S}}=\frac{1}{\Omega-\xi^{s_5}_\mathbf{k+q}+\xi^{s_1}_\mathbf{k+q}}\frac{1}{\Omega-\xi^{s_5}_\mathbf{k+q}+\xi^{s_2}_\mathbf{k+q}}.
\ee
\ese
Thus we have derived the SE$_2$ part of Eq.~\eqref{eq:firsteqofresultlist} in the main text.
\subsubsection{Elimination of singular terms} \label{sec:delta2}
In Sec.~\ref{SE1}, we argued that the singularities of the $\delta^2(x)$ type, occurring in e.g., Eq.~\eqref{ImWSE1}, can be ignored. Here, we prove this statement. Because the singularity occurs already for the case when all helicities are the same, we put
$s_1=s_2=s_3=s_5\equiv s$ and suppress the helicity index. Also, since the momentum-dependence of the self-energy is irrelevant for the present argument, we denote temporarily $\tilde S(i\omega)\equiv\tilde\Sigma_s(\bk+\bq,\bk,i\omega)$.
Adding  Eq.~\eqref{ImWSE1} to the corresponding contribution from diagram SE$_2$, we obtain
\bea
g^2(\mathbf{k},i\omega)\tilde S(i\omega)
\left[g(\mathbf{k},i\omega-i\Omega)+g(\mathbf{k},i\omega+i\Omega)\right],\label{WS}
\eea
where we also made a change  of variables  $i\omega\to i\omega+i\Omega$ in $W_{\text{SE}1}$. Applying the identity
\bea
\eea
with $s=s'$ to Eq.~\eqref{WS} twice, we obtain
\bea
W_{\text{SE}}(i\Omega)=\frac{1}{(i\Omega)^2} T\sum_\omega \left[g(\bk,i\omega+i\Omega)+g(\bk,i\omega-i\Omega)-2g(\bk,i\omega)\right] \tilde S(i\omega).
\eea
Summing over $\omega$ and performing analytic continuation, we get
\bea
\label{A30}\eea
\bea
\mathcal{Z}(\Omega)&=&\int\frac{d\omega}{\pi}n_\fer(\omega)\left\{g_{\text{R}}(\bk,\omega+\Omega)\Im \tilde S_{\text{R}}(\omega)+\Im g_{\text{R}}(\bk,\omega)\tilde S_\text{A}(\omega-\Omega)+g_{\text{A}}(\bk,\omega-\Omega)\Im \tilde S_\text{R}(\omega)+\Im g_{\text{R}}(\bk,\omega) \tilde S_\text{R}(\omega+\Omega)\right.\nn\\
&&\left.-2\Im\left[g_{\text{R}}(\bk,\omega)\tilde S_{\text{R}}(\omega)\right]\right\}.
\eea
Taking the imaginary part of Eq.~\eqref{A30}, we find
\bea
\Im W_{\text{SE,R}}(\Omega)=\frac{\Im\mathcal{Z}(\Omega)}{\Omega^2}+\Im\frac{1}{(\Omega+i0^+)^2}\Re\mathcal{Z}(\Omega).
\eea
It is the second term that contains an essential singularity. We re-write the singular part as
\bea
\Im\frac{1}{(\Omega+i0^+)^2}\Re\mathcal{Z}(\Omega)=-\Im\left(\frac{\partial}{\partial\Omega}\frac{1}{\Omega+i0^+}\right)\Re\mathcal{Z}(\Omega)=\pi\delta'(\Omega)\Re\mathcal{Z}(\Omega)=-\pi\delta(\Omega)\frac{\partial}{\partial\Omega}\Re\mathcal{Z}(\Omega)\Big\vert_{\Omega=0}.\label{A34}
\eea
The imaginary part of the retarded current-current correlator must be an odd function of $\Omega$, and the same is true for $\Im W_{\text{SE,R}}(\Omega)$ and
But then $\Re\mathcal{Z}(\Omega)$ is odd in $\Omega$, and its derivative must vanish at $\Omega\to 0$ at least as $\Omega$, which means that left-hand side of Eq.~\eqref{A34} is proportional to $\Omega\delta(\Omega)$. Recalling also that $\Re\sigma(\Omega)=-\Im\Pi_\text{R}(\Omega)/\Omega$, we see that the singular part of Eq.~\eqref{A34} only renormalizes the weight of the delta function term in the conductivity.
Such a term is rendered finite by taking momentum-relaxing scattering, e.g., scattering by impurities, into account, but is of no interest to our study and can be safely discarded.

\subsection{Vertex diagram\label{appen:VertexDetails}}
Explicitly, the vertex diagram (V) in Fig.~\ref{fig:AllDiagrams} reads
\be
\Pi^\text{V}(i\Omega)&=\frac{1}{16d}\sum\limits_{\mathcal{S}_\text{V}}\int_\mathbf{k,q}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s'_1}\hat{M}_{\mathbf{k+q}}^{s'_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k+q}}^{s'_3}\hat{M}_{\mathbf{k}}^{s'_5}\right)V^2_\text{st}(\mathbf{q})\nonumber\\
&\times T\sum\limits_{\omega}T\sum\limits_{\nu}g_{s'_1}(\mathbf{k},i\omega+i\Omega)g_{s'_2}(\mathbf{k+q},i\omega+i\nu+i\Omega)g_{s'_3}(\mathbf{k+q},i\omega+i\nu)g_{s'_5}(\mathbf{k},i\omega)\pi_0(\mathbf{q},i\nu),
\ee
where $\mathcal{S}_{\text{V}}=\{s'_1,s'_2,s'_3,s'_5\}$.
Applying the identity \eqref{ident}
to the pairs of the Green's functions with the same momenta, using Eq.~\eqref{pi0} for $\pi_0(\bq,i\nu)$,
and summing over $\nu$, we obtain
\be
\Pi^\text{V}(i\Omega)&=\frac{1}{16d}
\sum\limits_{\mathcal{S}_\text{V}}\int_\mathbf{k,q}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s'_1}\hat{M}_{\mathbf{k+q}}^{s'_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k+q}}^{s'_3}\hat{M}_{\mathbf{k}}^{s'_5}\right)V^2_\text{st}(\mathbf{q})\frac{1}{i\Omega-\xi^{s'_1}_\mathbf{k}+\xi^{s'_5}_\mathbf{k}}\frac{1}{i\Omega-\xi^{s'_2}_\mathbf{k+q}+\xi^{s'_3}_\mathbf{k+q}}\nn\\
&\times W_\text{V}(\bk,\bq,i\Omega),
\ee
\be
\right]
\left[\tilde{\Sigma}_{s'_3}(\mathbf{k+q},\mathbf{q},i\omega)-\tilde{\Sigma}_{s'_2}(\mathbf{k+q},\mathbf{q},i\omega+i\Omega)\right],
\ee
and $\tilde{\Sigma}_{s}(\mathbf{k},\mathbf{q},i\omega)$ is defined in Eq.~\eqref{eq:TildeSigmaZeroTForm}.

Upon summation over $\omega$ and analytic continuation, we obtain for the real and imaginary parts of the retarded counterpart of $W_\text{V}$ (as for the self-energy diagram, we also switch to
unprimed helicities
at this point, cf. footnote~\ref{ftnt:PivsRexplanation} in the main text):
\be
\Re W_{\text{V},\text{R}}(\bk,\bq,\Omega)=-\frac{1}{\pi}&\int d\omega n_\text{F}(\omega)\left[\Im\left(g_{s_5,\text{R}}(\mathbf{k},\omega)\tilde{\Sigma}_{s_3,\text{R}}(\mathbf{k+q},\mathbf{q},\omega)\right)+\Im\left(g_{s_1,\text{R}}(\mathbf{k},\omega)\tilde{\Sigma}_{s_2,\text{R}}(\mathbf{k+q},\mathbf{q},\omega)\right)\right.\nonumber\\
&-\left(\Re\tilde{\Sigma}_{s_2,\text{R}}(\mathbf{k+q},\mathbf{q},\omega+\Omega)\Im g_{s_5,\text{R}}(\mathbf{k},\omega)+\Re g_{s_5,\text{A}}(\mathbf{k},\omega-\Omega)\Im\tilde{\Sigma}_{s_2,\text{R}}(\mathbf{k+q},\mathbf{q},\omega)\right.\nonumber\\
&\left.\left.+\Re g_{s_1,\text{R}}(\mathbf{k},\omega+\Omega)\Im\tilde{\Sigma}_{s_3,\text{R}}(\mathbf{k+q},\mathbf{q},\omega)+\Re\tilde{\Sigma}_{s_3,\text{A}}(\mathbf{k+q},\mathbf{q},\omega-\Omega)\Im g_{s_1,\text{R}}(\mathbf{k},\omega)\right)\right],\label{ReWV}
\ee
\be
\Im W_{\text{V},\text{R}}(\bk,\bq,\Omega)=-\frac{1}{\pi}&\int d\omega\left[n_\text{F}(\omega+\Omega)-n_\text{F}(\omega)\right]\left[\Im\tilde{\Sigma}_{s_2,\text{R}}(\mathbf{k+q},\mathbf{q},\omega+\Omega)\Im g_{s_5,\text{R}}(\mathbf{k},\omega)
\right.\nonumber\\
+\Im g_{s_1,\text{R}}(\mathbf{k},\omega+\Omega)\Im\tilde{\Sigma}_{s_3,\text{R}}(\mathbf{k+q},\mathbf{q},\omega) \right].\IEEEeqnarraynumspace\label{eq:WvertexImPart}
\ee
The imaginary part of the current-current correlator is then given by
\be
\Im\Pi^\text{V}_\text{R}
\sum\limits_{\mathcal{S}_\text{V}}
\int_\mathbf{k,q}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s_1}\hat{M}_{\mathbf{k+q}}^{s_2}\cdot\hat{\bv}
\hat{M}_{\mathbf{k+q}}^{s_3}\hat{M}_{\mathbf{k}}^{s_5}\right)V^2_\text{st}(\mathbf{q})\mathcal{B}(\bk,\bq,\Omega),
\ee
\bea\mathcal{B}(\bk,\bq,\Omega)&=&\frac{\Im W_\text{V,R}(\bk,\bq,\Omega)}{(\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_5}_\mathbf{k})(\Omega-\xi^{s_2}_\mathbf{k+q}+\xi^{s_3}_\mathbf{k+q})}-\pi\left(\frac{\delta(\Omega-\xi^{s_2}_\mathbf{k+q}+\xi^{s_3}_\mathbf{k+q})}{\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_5}_\mathbf{k}}+\frac{\delta(\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_5}_\mathbf{k})}{\Omega-\xi^{s_2}_\mathbf{k+q}+\xi^{s_3}_\mathbf{k+q}}\right)
\Re W_\text{V,R}(\bk,\bq,\Omega),
\nn\\
\label{calB}
\eea
and where we omitted the term with an essential singularity, which is similar to the singular term in the self-energy diagram discussed in Sec.~\ref{sec:delta2}.
We will now show that only the first term in
Eq.~\eqref{calB} contributes to the real part of the conductivity
for $0<\Omega<2\EF$.
The reasoning is similar to but more involved than the reasoning in Sec.~\ref{sec:SE}.

Using the constraints imposed by the delta functions in front of $\Re W_\text{V}(\bk,\bq,\Omega)$, we obtain that, for $\Omega>0$,
only the sets
\bea
s_2=+1,s_3=-1\; \text{and}\; s_1=+1,s_5=-1\label{2315}
\eea
are allowed for the first and second terms in the round brackets, respectively.
This implies that
$\epsilon_{\bk+\bq}=\Omega/2$ and $\epsilon_\bk=\Omega/2$
in the first and second terms, respectively.
Next, we will show that $\epsilon_\mathbf{k+q}>\EF$ and $\epsilon_\mathbf{k}>\EF$ in the first and second terms, respectively, which will imply that the corresponding contribution to $\Im\Pi^\text{V}_\text{R}(\Omega)$ is non-zero only if $\Omega>2\EF$. To see the constraints imposed on $\epsilon_{\bk+\bq}$ and $\epsilon_\bk$, we write down an explicit form of $\Re W_\text{V,R}(\bk,\bq,\Omega)$, using Eqs.~\eqref{ReWV},
\eqref{eq:ImReMixtureProperty}, and  \eqref{eq:TildeSigmaForm}.
After some re-arrangement, we get
\be
\Re & W_{\text{V},\text{R}}(\bk,\bq,\Omega)=
\frac 14\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}\left[\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})\right]\left[n_\text{F}(\xi^{s_{6}}_\mathbf{p})-n_\text{F}(\xi^{s_4}_\mathbf{p+q})\right]\nonumber\\
\frac{n_\text{F}(\xi_\mathbf{k}^{s_5})-n_\text{F}(-\xi^{s_4}_\mathbf{p+q}+\xi^{s_{6}}_\mathbf{p}+\xi_\mathbf{k+q}^{s_3})
\right.\nonumber\\
\right],\label{eq:ReWVertexsolved}
\ee
where $
The difference of the Fermi functions in the first line of Eq.~\eqref{eq:ReWVertexsolved} is non-zero only if
either $\xi^{s_{6}}_\mathbf{p}>0$ and $\xi^{s_4}_\mathbf{p+q}<0$, or vice versa. Now we analyze these two options one by one.

For $\xi^{s_{6}}_\mathbf{p}>0$ and $\xi^{s_4}_\mathbf{p+q}<0$, we have $f_s=-\theta(\xi_\mathbf{k+q}^{s})$ and thus
$\xi_{\bk+\bq}^{s_{2,3}}>0$. The combination of these conditions implies that the arguments of the Fermi functions containing three terms are positive, and thus the arguments of the Fermi functions containing a single term must be negative. With that, Eq.~\eqref{eq:ReWVertexsolved} is simplified to
\be
\Re W_{\text{V},\text{R}}(\Omega)=-\frac{1}{4}&\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})\theta(\xi^{s_{6}}_\mathbf{p})\theta(-\xi^{s_4}_\mathbf{p+q})\nonumber\\
\right.\nonumber\\
\left(\frac{\theta(\xi_\mathbf{k+q}^{s_{2}})}{\xi_\mathbf{k}^{s_1}+\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_2}}+\frac{\theta(\xi_\mathbf{k+q}^{s_{3}})}{\Omega-\xi_\mathbf{k}^{s_1}-\xi^{s_4}_\mathbf{p+q}+\xi^{s_{6}}_\mathbf{p}+\xi_\mathbf{k+q}^{s_3}}\right)
\right].\label{eq:vertWVintermediate}
\ee
Now, we focus on the first term in the round brackets in Eq.~\eqref{calB}. Using the constraint following from $\delta(\Omega-\xi^{s_2}_\mathbf{k+q}+\xi^{s_3}_\mathbf{k+q})$,
we arrive at
\be
\Re W_{\text{V},\text{R}}(
\xi^{s_2}_\mathbf{k+q}-\xi^{s_3}_\mathbf{k+q}
\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})\theta(\xi^{s_{6}}_\mathbf{p})\theta(-\xi^{s_4}_\mathbf{p+q})\left[\theta(\xi_\mathbf{k+q}^{s_{3}})-\theta(\xi_\mathbf{k+q}^{s_{2}})\right]\nonumber\\
\frac{\theta(-\xi_\mathbf{k}^{s_5})}{
\xi_\mathbf{k}^{s_5}+\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_3}
\right.
\left.
\frac{\theta(-\xi_\mathbf{k}^{s_{1}})}{
\xi_{\bk+\bq}^{s_2}-\xi_\mathbf{k}^{s_1}-\xi^{s_4}_\mathbf{p+q}+\xi^{s_{6}}_\mathbf{p}
\right].
\ee
The RHS of the last equation is non-zero only if the arguments of the theta functions in the first pair square brackets are of opposite signs. According to Eq.~\eqref{2315}, $s_3=-1$ while $s_2=1$.  Therefore, $\xi^{s_3}_{\bk+\bq}<0$ and $\xi^{s_2}_{\bk+\bq}=\epsilon_{\bk+\bq}-\EF>0$.
$\epsilon_\mathbf{k+q}>\EF$. But because $\epsilon_{\bk+\bq}=\Omega/2$ for the first term in the round brackets of Eq.~\eqref{calB}, we have the condition that $\Omega>2\EF$.
Thus, this term only contributes for frequencies above $\omega_\text{D}$ and is not relevant for our analysis.

The second term in the round brackets of Eq.~\eqref{calB} is simplified using the condition following from
\Re W_{\text{V},\text{R}}(\xi^{s_1}_\bk-\xi_\bk^{s_5})=
\frac{1}{4}
\sum\limits_{s_4,s_6=\pm 1}\int_\mathbf{p}\tr(\hat{M}_\mathbf{p}^{s_6}\hat{M}_\mathbf{p+q}^{s_4})\theta(\xi^{s_{6}}_\mathbf{p})\theta(-\xi^{s_4}_\mathbf{p+q})\left[\theta(-\xi_\mathbf{k}^{s_{5}})-\theta(-\xi_\mathbf{k}^{s_1})\right]\nn\\
\frac{\theta(\xi_\mathbf{k+q}^{s_{3}})}{\xi_\mathbf{k}^{s_5}+\xi^{s_4}_\mathbf{p+q}-\xi^{s_{6}}_\mathbf{p}-\xi_\mathbf{k+q}^{s_3}}
\frac{\theta(\xi_\mathbf{k+q}^{s_{2}})}{\xi_\mathbf{k+q}^{s_2}-\xi_\mathbf{k}^{s_1}-\xi^{s_4}_\mathbf{p+q}+\xi^{s_{6}}_\mathbf{p}}
\right]
Using the constraints on $s_1$ and $s_5$ from Eq.~\eqref{2315} again, we conclude that
$\epsilon_\mathbf{k}>\EF$. But because $\epsilon_\bk=\Omega/2$ for the second term in the round brackets of Eq.~\eqref{calB}, we again have the condition that $\Omega>2\EF$.

The case of $\xi^{s_{6}}_\mathbf{p}<0$ and $\xi^{s_4}_\mathbf{p+q}>0$ is analyzed in the same way and with the same result.
Therefore, the term  proportional to $\Re W_{\text{V},\text{R}}(\Omega)$ in Eq.~\eqref{calB} is absent for $\Omega<2\EF$.

Focusing now on the first term of Eq.~\eqref{calB}, we obtain, after some re-arrangements and re-labelling of momenta and helicities,
the following final expression for the vertex diagram
for $\Omega<2\EF$:
\be
\Im\Pi^{\text{V}}_\text{R}(\Omega)=
\frac{\pi^2}{32}&\sum\limits_{\mathcal{S}}\int_{\mathbf{k,p,q},\nu}V^2_\text{st}(\mathbf{q})\left[\mathcal{T}^{\text{V}_1}_{\mathcal{S}}\mathcal{G}^{\text{V}_1}_{\mathcal{S}}+\mathcal{T}^{\text{V}_2}_{\mathcal{S}}\mathcal{G}^{\text{V}_2}_{\mathcal{S}}\right]\nonumber\\
\ee
where, for the sake of convenience, we  re-introduced $\int_\nu=\int\dee\nu/2\pi$  and
\bse
\be
\mathcal{T}^{\text{V}_1}_{\mathcal{S}}=\frac{1}{d}\tr\left(\hat{\bv}
\hat{M}_\mathbf{k}^{s_1}\hat{M}_{\mathbf{k+q}}^{s_5}\cdot\hat{\bv}
\hat{M}_{\mathbf{k+q}}^{s_2}\hat{M}_{\mathbf{k}}^{s_3}\right)\text{Tr}\left(\hat{M}_{\mathbf{-p-q}}^{s_6}\hat{M}_{\mathbf{-p}}^{s_4}\right)\\
\mathcal{T}^{\text{V}_2}_{\mathcal{S}}=\frac{1}{d}\tr\left(\hat{\bv}
\hat{M}_{\mathbf{-k-q}}^{s_5}\hat{M}_{\mathbf{-k}}^{s_1}\cdot\hat{\bv}
\hat{M}_{\mathbf{-k}}^{s_3}\hat{M}_{\mathbf{-k-q}}^{s_2}\right)\text{Tr}\left(\hat{M}_\mathbf{p}^{s_4}\hat{M}_{\mathbf{p+q}}^{s_6}\right)\\
\mathcal{G}^{\text{V}_1}_{\mathcal{S}}=\mathcal{G}^{\text{V}_2}_{\mathcal{S}}=\frac{1}{\Omega-\xi^{s_1}_\mathbf{k}+\xi^{s_3}_\mathbf{k}}\frac{1}{\Omega-\xi^{s_5}_\mathbf{k+q}+\xi^{s_2}_\mathbf{k+q}}.
\ee
\ese
Thus we have derived Eq.~\eqref{eq:VertexTGresults} of the main text.

\subsection{Sum of the self-energy and vertex diagrams\label{appen:SEplusV}}
Adding up Eqs.~\eqref{ImSE1}, \eqref{ImSE2}, and \eqref{ImV}, we obtain the combined contribution of the self-energy and vertex diagram for
\bea
\Im\Pi^{\text{SE+V}}_\text{R}(\Omega)&=&\Im \Pi^{\text{SE}_1}_\text{R}(\Omega)+\Im \Pi^{\text{SE}_2}_\text{R}(\Omega)+\Im \Pi^\text{V}_\text{R}(\Omega)=
-\frac{\pi^2}{32}\sum\limits_{\mathcal{S}}\int_\mathbf{k,p,q}\int_\nu V^2_\text{st}(\mathbf{q})\times\theta(-\xi^{s_3}_\mathbf{k})\theta(\xi^{s_5}_\mathbf{k+q})\theta(-\xi^{s_4}_\mathbf{p})\theta(\xi^{s_6}_\mathbf{p+q})\nn\\
\delta(\Omega+\nu+\xi^{s_3}_\mathbf{k}-\xi^{s_5}_\mathbf{k+q})\delta(\nu+\xi^{s_6}_\mathbf{p+q}-\xi^{s_4}_\mathbf{p})
\left[\mathcal{T}^{\text{SE}_1}_{\mathcal{S}}\mathcal{G}^{\text{SE}_1}_{\mathcal{S}}+\mathcal{T}^{\text{SE}_2}_{\mathcal{S}}\mathcal{G}^{\text{SE}_2}_{\mathcal{S}}-\mathcal{T}^{\text{V}_1}_{\mathcal{S}}\mathcal{G}^{\text{V}_1}_{\mathcal{S}}-\mathcal{T}^{\text{V}_2}_{\mathcal{S}}\mathcal{G}^{\text{V}_2}_{\mathcal{S}}\right].\label{eq:ImPiSEplusV}
\eea
Similar computations for PAL and CAL diagrams give Eqs.~\eqref{eq:PALTGresults} and~\eqref{eq:CALexactexpressions}, respectively.

\section{
Asymptotic expressions for the conductivity\label{appen:asymp_eval}}
As an example, we derive
an asymptotic expression for the purely electron-electron contribution to the conductivity of a 3D Dirac metal with Hubbard interaction in the limit $\Omega\ll\EF$.
We start with the general result  \eqref{Eq:generalzeroTexpression}, which we reproduce below for the reader's convenience:
\be
\mathcal{R}^{J_u
}_{\mathcal{S}}(\Omega)=K^{J_u}\int_{\mathbf{k,p,q}}\int_\nu &V^2_{\text{st}}(\mathbf{q})
\mathcal{T}^{J_u}_\mathcal{S}\left(\bk,\bp,\bq\right)\mathcal{G}^{J_u}_\mathcal{S}(\bk,\bp,\bq,\Omega)\nn\\
&\times \theta(-\xi^{s_3}_\mathbf{k})\theta(-\xi^{s_4}_\mathbf{p})\theta(\xi^{s_5}_\mathbf{k+q})\theta(\xi^{s_6}_\mathbf{p+q})\delta(\Omega+\nu+\xi^{s_3}_\mathbf{k}-\xi^{s_5}_\mathbf{k+q})\delta(\nu+\xi^{s_6}_\mathbf{p+q}-\xi^{s_4}_\mathbf{p}),
\ee
$\mathcal{T}^{J_u}_\mathcal{S}(\mathbf{k,p,q})$  and $\mathcal{G}^{J_u}_\mathcal{S}(\mathbf{k,p,q})$ are defined in Eqs.~\eqref{eq:firsteqofresultlist}-\eqref{eq:CALexactexpressions}, respectively.
We remind the reader that the purely electron-electron contribution corresponds to all helicities being positive, i.e.,
\bea
\mathcal{S}=\mathcal{S}_+\equiv \{s_1=1,s_2=1,s_3=1,s_4=1,s_5=1,s_6=1\}.\label{S+}
\eea
In a more explicit form,
\be
\mathcal{R}^{J_u}_{\mathcal{S}_+}(\Omega)&=\lambda_3^2K^{J_u}
\int^\infty_0\frac{dkk^{2}}{(2\pi)^3}
\int^\infty_0\frac{dpp^{2}}{(2\pi)^3}
\int^\infty_0\frac{dq q^{2}}{(2\pi)^3}\int^\infty_{-\infty}\frac{d\nu}{2\pi}
\nonumber\\
\int\limits_{0}^{\pi}d\theta_\mathbf{q\hat z}\sin\theta_\mathbf{q\hat z}\int\limits_{0}^{2\pi}d\phi_\mathbf{q,\hat z}
\int\limits_{0}^{\pi}d\theta_\mathbf{kq}\sin\theta_\mathbf{kq}\int\limits_{0}^{2\pi}d\phi_\mathbf{k, q}\int\limits_{0}^{\pi}d\theta_\mathbf{pq}\sin\theta_\mathbf{pq}\int\limits_{0}^{2\pi}d\phi_\mathbf{p,q}
\mathcal{T}^{J_u}_\mathcal{S_+}(\mathbf{k,p,q})\mathcal{G}^{J_u}_\mathcal{S_+}(\mathbf{k,p,q})\nn\\
\label{eq:angularpartseparated}
\ee
$\theta_{{\bf nn}'}$ is the angle between vectors ${\bf n}$ and ${\bf n'}$,
$\phi_{{\bf n},{\bf n}'}$
is the
azimuthal angle of vector ${\bf n}$ in a spherical system with vector ${\bf n}'$ taken as the polar axis.
Now we solve the integrals over $\theta_{\bk\bq}$ and $\theta_{\bp\bq}$, using the delta functions in in the expression above, which yields
\bea
\mathcal{T}^{J_u}_\mathcal{S_+}(\mathbf{k,p,q})\mathcal{G}^{J_u}_\mathcal{S_+}(\mathbf{k,p,q})
\delta(\Omega+\nu+\epsilon_\mathbf{k}-\epsilon_\mathbf{k+q})\delta(\nu+\epsilon_\mathbf{p+q}-\epsilon_\mathbf{p})\nn\\
&&=\frac{(\epsilon_\bk+\nu+\Omega)(\epsilon_\mathbf{p}-\nu)}{\epsilon_\bk\epsilon_\bp\vd^2 q^2}
\left[
\Theta(\mu_\bk,\mu_\bp),
\eea
where $\Theta(\mu_\bk,\mu_\bp)$
encapsulates the constraints imposed by the delta functions in Eq.~\eqref{eq:angularpartseparated}, i.e.,
\be
\qquad\,\,\,
\epsilon_\mathbf{k+q}=\Omega+\nu+\epsilon_\mathbf{k},\epsilon_\mathbf{p+q}=\epsilon_\mathbf{p}-\nu,\label{eq:deltafuncconstraint1}
\ee such that
\be
-1\leq \mu_\bk=\frac{(\Omega+\nu+\epsilon_\bk)^2-\epsilon_\bk^2-\vd^2q^2}{2\epsilon_\bk\vd q}\leq 1,\nn\\
-1\leq \mu_\bp=\frac{(-\nu+\epsilon_\mathbf{p})^2-\epsilon_\bp^2-\vd^2q^2
}{2\epsilon_\bp\vd q}\leq 1.\label{eq:cosineconstraints}
\ee
Note that the constraints~\eqref{eq:deltafuncconstraint1} are the same ones as in Eq.~\eqref{cb} of the main text for the special case of all helicities being positive.
The last two equations are resolved in terms of $q$, which imposes constraints on the range of integration over $q$.
In addition, the first two theta functions in Eq.~\eqref{eq:angularpartseparated} imply that $\epsilon_\bk\leq\EF$ and $\epsilon_\bp\geq\EF$  for $s_3=s_4=1$, whereas the second two theta functions, in conjunction with Eq.~\eqref{eq:deltafuncconstraint1}, imply that $\Omega+\nu+\epsilon_\bk\geq\EF$
and $\epsilon_\bp-\nu\geq\EF$.

Next, we switch
from integration over $k$ and $p$
to integration over dispersions $\epsilon_\bk=\vd k$ and $\epsilon_\bp=\vd p$. Imposing explicitly all the constraints described above, we arrive at
\be
\mathcal{R}^{J_u}_{\mathcal{S_+}}(\Omega)=\int\limits_{\EF-\Omega}^{\EF}d\epsilon_\bk\int\limits_{\max\left\{
0 ,2\EF-\Omega-\epsilon_\bk\right\}}^{\EF}
\right\}/\vd}^{
\min\left\{(\Omega+\nu
\right\}/\vd}dq \mathcal{L}^{J_u}_\mathcal{S_+}(\epsilon_\bk,\epsilon_\bp,q,\nu,\Omega),
\label{eq:3Dangularintegratedexpression}
\ee
\be
\mathcal{L}^{J_u}_\mathcal{S_+}(\epsilon_\bk,\epsilon_\bp,q,\nu,\Omega)=\frac{K^{J_u}\lambda_3^2}{(2\pi)^{10}\vd^8}\bar{\mathcal{Q}}^{J_u}_\mathcal{S_+}(\epsilon_\bk,\epsilon_\bp,q,\nu,\Omega)
\epsilon_\bk \epsilon_\bp(\epsilon_\bk+\Omega+\nu)(\epsilon_\mathbf{p}-\nu)
\ee
\be
\bar{\mathcal{Q}}^{J_u}_\mathcal{S_+}(\epsilon_\bk,\epsilon_\bp,q,\nu,\Omega)
\sin\theta_\mathbf{qz}\int\limits_{0}^{2\pi}d\phi_\mathbf{qx}
\int\limits_{0}^{2\pi}d\phi_\mathbf{kq}\int\limits_{0}^{2\pi}d\phi_\mathbf{pq}
\left[\mathcal{T}^{J_u}_\mathcal{S_+}(\bk,\bp,\bq)\mathcal{G}^{J_u}_\mathcal{S_+}(\bk,\bp,\bq)\right]\Big\vert_{\cos\theta_{\bk\bq}=\mu_\bk,\cos\theta_{\bp\bq}=\mu_\bp}.\label{eq:restofangintegrals3D}
\ee

For the particular case of $J_u=\text{SE}_1$ and all helicities positive, we have $\mathcal{G}^{\text{SE}_1}_\mathcal{S_+}(\bk,\bp,\bq)=1/\Omega^2$, and thus the angular integrals in Eq.~\eqref{eq:restofangintegrals3D} involve only
$\mathcal{T}^{\text{SE}_1}_\mathcal{S_+}(\bk,\bp,\bq)$ in  Eq.~\eqref{eq:firsteqofresultlist}, which is evaluated explicitly as
\bea
\mathcal{T}^{\text{SE}_1}_\mathcal{S_+}(\bk,\bp,\bq)&=&-\frac{8 \vd^3q \sin\theta_{\bk\bq}}{\epsilon_{\bk+\bq}}\frac{\epsilon_\bp+\epsilon_\bpq+ q\cos\theta_{\mathbf{pq}}}{\epsilon_\bpq}
\Big[8 \cos2 \theta_{\bk\bq} \cos\phi_{\mathbf{qx}} \sin\theta_{\bq\mathbf{z}}
\left(
\cos\theta_{\bq\mathbf{z}} \cos\phi_{\mathbf{kq}} \cos\phi_{\mathbf{qx}}-\sin\phi_{\mathbf{kq}} \sin\phi_{\mathbf{qx}}\right)
\nn\\
&&+\frac{1}{8} \sin2 \theta_{\bk\bq}\left(
4+12 \cos2 \theta_{\bq\mathbf{z}}
+4 \cos2 \theta_{\bq\mathbf{z}}\cos2\phi_{\mathbf{kq}}
-4 \cos2 \phi_{\mathbf{kq}}
+12\cos2 \theta_{\bq\mathbf{z}}\cos2\phi_{\mathbf{qx}}
\right.\nn\\
\cos2\phi_{\mathbf{kq}}
+12\cos2 \phi_{\bk\mathbf{q}}\cos2\phi_{\mathbf{qx}}
-12 \cos2 \phi_{\mathbf{qx}}
-16 \cos\theta_{\bq\mathbf{z}} \sin2 \phi_{\mathbf{kq}} \sin2 \phi_{\mathbf{qx}}
\right)\Big].
\nn\\
\label{trace}
\eea
Integrating Eq.~\eqref{trace} over the angles and imposing the constraints from Eq.~\eqref{eq:deltafuncconstraint1}, we obtain
\be
\mathcal{L}^{\text{SE}_1}_{\mathcal{S}_+}(\epsilon_\bk,\epsilon_\bp,q,\nu,\Omega)=-\frac{
\lambda_3^2}{
\vd^6 \Omega^2 }
\left
\IEEEeqnarraynumspace
\label{D9}
\ee
where we used $K^{\text{SE}_1}=-\pi^2/32$ as defined in Eq.~\eqref{KJu} of the main text.

In this particular case, the integrand of Eq.~\eqref{eq:3Dangularintegratedexpression} is a purely polynomial function, and thus it is possible to obtain an exact result for all frequencies $\Omega<2\EF$. However, this is not the case for other diagrams (except for SE$_2$) and for the 2D case. For
consistency with these other cases,
we will only evaluate the integral \eqref{eq:3Dangularintegratedexpression} for $\Omega\ll\EF$. To this end, we note that for $\Omega<\EF$ the limits of the integral for $\epsilon_\bp$ in Eq.~\eqref{eq:3Dangularintegratedexpression} become
\be
\ee
Thus, for $\Omega\ll\EF$ it is convenient to define new dimensionless
\be
\,
\\
\label{eq:xyzdefintionlastline}
\ee
Note that $x,y$ and $z$ are all of order $1$.
Also, the upper and lower limits of the integral over $q$ in Eq.~\eqref{eq:3Dangularintegratedexpression} can be replaced by $2k_F$ and $0$, respectively.
substitute $\epsilon_\mathbf{k},\epsilon_\mathbf{p},\nu$ in terms of $x,y,z$ into Eq.~\eqref{D9}, define $r=q/2k_F$,
and expand the result in $x,y,z$  and $\Omega$ for $\Omega\ll\EF$.
After the expansion, Eq.~\eqref{D9} is reduced to
\bea
\mathcal{L}
\vd^6}(2\EF)^2\Big[
\left(\frac{2\EF}{\Omega}\right)^2
\nonumber\\
-2 y-xy+3y^2-yz+z^2\right)
\nonumber\\
\nn\\
+\left(2 x^2-4 xz-2x+2 z^2+2 z+1\right)
\Big].
\eea
Performing the four-dimensional integral
\be
\mathcal{R}^{\text{SE}_1}_{\mathcal{S}_+}(\Omega)=\Omega^3(2\kf)\int\limits_0^1\dee x\int\limits_0^{x}\dee y\int\limits_0^{y}\dee z\int\limits_0^{
\mathcal{L
\ee
we obtain
\bse
\be
\mathcal{R}^{\text{SE}_1}_{\mathcal{S}_+}(\Omega)=
-\frac{\lambda_3^2}{135\pi^5} \frac{\kf^6}{\vd}
\left
\frac{\Omega}{\EF}
\frac{9}{80}\left(\frac{\Omega}{\EF}\right)^3\right
\label{D15a}
\ee
Similar expansions are
performed for other contributions.
Below we just list the results:
\be
\mathcal{R}^{\text{SE}_2}_{\mathcal{S}_+}(\Omega)=
-\frac{\lambda_3^2}{135\pi^5} \frac{\kf^6}{\vd}
\left[
\frac{\Omega}{\EF}+
\frac{9}{80}\left(\frac{\Omega}{\EF}\right)^3\right],
\\
\mathcal{R}^{\text{V}_1}_{\mathcal{S}_+}(\Omega)+\mathcal{R}^{\text{V}_2}_{\mathcal{S}_+}(\Omega)=
-\frac{\lambda_3^2}{135\pi^5} \frac{\kf^6}{\vd}
\left[-
\frac{10}{7}
\frac{\Omega}{\EF}
\frac{129}{280}\left(\frac{\Omega}{\EF}\right)^3\right],
\\
\mathcal{R}^{\text{PAL}_1}_{\mathcal{S}_+}(\Omega)+\mathcal{R}^{\text{PAL}_2}_{\mathcal{S}_+}(\Omega)+\mathcal{R}^{\text{CAL}_1}_{\mathcal{S}_+}(\Omega)+\mathcal{R}^{\text{CAL}_2}_{\mathcal{S}_+}(\Omega)=
-\frac{\lambda_3^2}{135\pi^5} \frac{\kf^6}{\vd}
\left[-
\frac{4}{7}
\frac{\Omega}{\EF}
\frac{71}{140}
\left(\frac{\Omega}{\EF}\right)^3\right].\IEEEeqnarraynumspace\label{D15d}
\ee
\ese
The conductivity is obtained by multiplying the sum of all $\mathcal{R}^{J_u}_{\mathcal{S}_+}$ by $-1/\Omega$. The leading, linear-in-$\Omega$ terms in Eqs.~\eqref{D15a}-\eqref{D15d}, would then produce an $\Omega$-independent conductivity, which would be the case for a Fermi liquid with Galilean invariance broken completely by, e.g., umklapp scattering. However, in our case Galilean invariance is broken only partially, and the conductivity is suppressed compared to the case with fully-broken Galilean invariance. Indeed, adding up
Eqs.~\eqref{D15a}-\eqref{D15d}, we see that linear-in-$\Omega$ terms cancel out, while the sum the subleading, cubic terms
reproduces Eq.~\eqref{3D_sigma_ee} of the main text.

\section{Combining contributions from all diagrams for intra-band absorption due to electron-electron interaction
\label{appen:CombingLFeediagrams}}
In this Appendix, we demonstrate how the contributions from all diagrams are combined together for the case of electron-electron interaction, when all the helicities are positive: $s_i=+1$ for $i=1\dots 6$.
We introduce the following definitions \bse
\bea
\ket{s_\mathbf{k}}&\equiv&\ket{\mathbf{k},s}\equiv\ket{\psi_{\mathbf{k},s}},\\
\hat{M}_\mathbf{k}^{s}&=&\ket{s_\mathbf{k}}\bra{s_\mathbf{k}},\\
\Phi^{s,s'}_\mathbf{k,
\bk'}&=&\braket{
\mathbf{v}^{s,s'}_{\mathbf{k}}&=&\bra{
\eea
\ese
and list the properties that will be used in this section:
\bse
\bea
\braket{s_\mathbf{k}|s'_\mathbf{k'}}&=&\braket{s_\mathbf{-k}|s'_\mathbf{-k'}},\\
\braket{s_\mathbf{k}|s_\mathbf{k}}&=&1,\\
\braket{s_\mathbf{k}|-s_\mathbf{k}}&=&0,\\
\mathbf{v}^{+,+}_{\mathbf{-k}}&=&-\mathbf{v}^{+,+}_{\mathbf{k}}.\\
\eea
\ese
We remind the reader that an algebraic expression for each diagram is the sum of two terms, labeled as $J_1$ and $J_2$, where $J=$SE, V, etc. denotes the type of a diagram.\footnote{The two terms for the self-energy diagram corresponds to two distinct diagrams, labeled as SE$_1$ and SE$_2$ in Fig.~\ref{fig:AllDiagrams}. For the rest of the diagrams, the two terms appear only in the algebraic expressions rather as distinct diagrams.} We first combine the $J_1$ terms together, and then do the same for the $J_2$ terms.
Note that, according to Eqs.~\eqref{eq:firsteqofresultlist}-\eqref{eq:CALexactexpressions}, $\mathcal{G}^{J_1}_{\mathcal{S}_+}=\mathcal{G}^{J_2}
where $\mathcal{S}_+$ is defined in Eq.~\eqref{S+}.
Therefore, we need to combine only the trace parts.
For $\mathcal{T}^{\text{SE}_1}_{\mathcal{S}_+}$ we obtain
\be
\mathcal{T}^{\text{SE}_1}_{\mathcal{S}_+}
\nonumber\\
\nonumber\\
\nonumber\\
\ee
Performing the same steps for the V$_1$, PAL$_1$ and CAL$_1$ contributions and adding them up, we obtain
\be
\mathcal{T}_1\equiv
\sum_{J_1=\text{SE}_1,\text{V}_1,\text{PAL}_1,\text{CAL}_1}K^{J_1}\mathcal{T}^{J_1}_{\mathcal{S}_+}
\Delta\mathbf{v}
\left|\Phi^{+,+}_\mathbf{k,k+q}\right|^2\left|\Phi^{+,+}_\mathbf{p,p+q}\right|^2,
\label{T1}
\ee
where $K_{J_u}$ is defined in Eq.~\eqref{KJu} of the main text, and
\be
\Delta\mathbf{v}
\mathbf{v}^{+,+}_{\mathbf{k}}+\mathbf{v}^{+,+}_{\mathbf{p+q}}-\mathbf{v}^{+,+}_{\mathbf{k+q}}-\mathbf{v}^{+,+}_{\mathbf{p}}
\ee
In the same way, we re-write the $J_2$ terms and combine them together. The only difference compared to the $J_1$ case is that,  when combining the PAL$_2$ and CAL$_2$ contributions, we
to re-label the momenta as $\mathbf{k}\leftrightarrow\mathbf{p}$ in the integrand of Eq.~\eqref{Eq:generalzeroTexpression}. This can be done for the present case,
when all the helicities are positive, and thus the delta and theta functions in Eq.~\eqref{Eq:generalzeroTexpression} can be reduced back to their original forms  by replacing first $\nu\rightarrow\nu-\Omega$ and then $\nu\rightarrow-\nu$. After these manipulations, we obtain for the sum of the $J_2$  terms
\be
\mathcal{T}_2\equiv\sum_{J=\text{SE}_2,\text{V}_2,\text{PAL}_2,\text{CAL}_2}K^{J_2}\mathcal{T}^{J_2}_{\mathcal{S}_+}=\frac{\pi^2}{32}\mathbf{v}^{+,+}_{\mathbf{k+q}}\cdot\Delta\mathbf{v}\left|\Phi^{+,+}_\mathbf{k,k+q}\right|^2\left|\Phi^{+,+}_\mathbf{p,p+q}\right|^2.\label{T2}
\ee
Adding up Eqs.~\eqref{T1} and \eqref{T2},
we find
\be
\mathcal{T}_{\mathcal{S}_+}\equiv  \mathcal{T}_1+\mathcal{T}_2=
\cdot\left(
\mathbf{v}^{+,+}_{\mathbf{k}}
\ee
Using the same reasoning
as for the PAL$_2$ case above, we relabel $\mathbf{k}\leftrightarrow\mathbf{p}$ in Eq.~\eqref{sumT} and rewrite it as
\be
\mathcal{T}_{\mathcal{S}_+}
\left(
\mathbf{v}^{+,+}_{\mathbf{p}+\bq}-
\mathbf{v}^{+,+}_{\mathbf{p}}\right)
\left|\Phi^{+,+}_\mathbf{p,p+q}\right|^2\left|\Phi^{+,+}_\mathbf{k,k+q}\right|^2,
\label{sumT2}
\ee
where we used that $\Delta\mathbf{v}
\rightarrow-\Delta\mathbf{v}
on $\mathbf{k}\leftrightarrow\mathbf{p}$. Adding Eqs.~\eqref{sumT} and \eqref{sumT2}, we obtain a symmetrized form of $\mathcal{T}_{\mathcal{S}_+}$:
\be
\mathcal{T}_{\mathcal{S}_+}=-\frac{\pi^2}{64}
\left(\Delta\mathbf{v}\right)^2
\left|\Phi^{+,+}_\mathbf{p,p+q}\right|^2\left|\Phi^{+,+}_\mathbf{k,k+q}\right|^2,
\ee
which is reproduced in Eq.~\eqref{sumTmain} of the main text.
\end{document}