the expression under the trace, Theorem \[CMzeta\] may be used. We deduce that on the one hand, the sums $$\begin{gathered}
{\mathrm{Res}}_{z=0} \dfrac{1}{z} \sum_{k\geq 1} {\mathrm{Tr}}\left(a \left(U_g \binom{-z/4}{k} b^{(k)}\Delta^{-z/4 -k}  U_{g^{-1}} \right)\right) {\vspace{2mm}}{\mathrm{Res}}_{z=0} \dfrac{1}{z} \sum_{k\geq 1} {\mathrm{Tr}}\left(a \left(U_{g^{-1}} \binom{-z/4}{k} U_g^{(k)}\Delta^{-z/4-k}   \right)\right)\end{gathered}$$ are finite, since the zeta function is holomorphic on a half-plane ${\mathrm{Re}}(z) \gg 0$. On the other hand, as the poles of the zeta function are simple, the terms carrying a power of $z^{2}$ vanish under the residue, and we are respectively left with $$\begin{gathered}
-{\mathrm{Res}}_{z=0} \sum_{k\geq 1} {\mathrm{Tr}}\left(a \left(U_g \dfrac{(-1)^{k-1}}{4k}   b^{(k)}\Delta^{-z/4-k}  U_{g^{-1}} \right)\right) {\vspace{2mm}}-{\mathrm{Res}}_{z=0} \sum_{k\geq 1} {\mathrm{Tr}}\left(a \left(U_{g^{-1}} \dfrac{(-1)^{k-1}}{4k}  U_g^{(k)} \Delta^{-z/4-k}   \right)\right)\end{gathered}$$ We then recognize the commutator with the logarithm of $\Delta^{1/4}$ (cf. [@Rod2013]), and we finally obtain $$\begin{aligned}
\phi(a U_g, b U_{g^{-1}}) &= {\mathop{\ooalign{$\displaystyle{\int}$\cr$-$}}}a \left(U_g [\log \Delta^{1/4},b] U_{g^{-1}} - U_g b U_{g^{-1}} [\log \Delta^{1/4}, U_g] U_{g^{-1}} \right) {\vspace{2mm}}&= {\mathop{\ooalign{$\displaystyle{\int}$\cr$-$}}}a U_g [\log \Delta^{1/4}, b U_{g^{-1}}]\end{aligned}$$ This ends the proof of the proposition. $\hfill{\square}$

The pseudodifferential extension (\[extension\]) is closely related to another extension. Indeed the quotient of $\Psi^0_{H,c}(M)$ by its two-sided ideal $\Psi_{H,c}^{-1}(M)$ of operators of order $\leq -1$ is $G$-equivariantly isomorphic to the commutative algebra of leading symbols ${C^{\infty}_c}(S^*_HM)$. The natural inclusion of smoothing operators in $\Psi_{H,c}^{-1}(M)$ and 