remark that $$\begin{aligned}
\phi(a U_g, b U_{g^{-1}}) &= {\mathrm{Pf}}_{z=0} {\mathrm{Tr}}\left([a U_g, b U_{g^{-1}}]  \cdot \Delta^{-z/4}\right) {\vspace{2mm}}&= {\mathrm{Res}}_{z=0} \dfrac{1}{z} {\mathrm{Tr}}\left((a U_g b U_{g^{-1}} - b U_{g^{-1}}) a U_g  \cdot \Delta^{-z/4}\right)\end{aligned}$$

Then, work at $z \in {\mathbb{C}}$ with ${\mathrm{Re}}(z) \gg 0$, so that ${\mathrm{Tr}}\left([a U_g, b U_{g^{-1}}]  \cdot \Delta^{-z/4}\right)$ is well-defined. Then, the trace property yields $${\mathrm{Tr}}\left([a U_g, b U_{g^{-1}}]  \cdot \Delta^{-z/4}\right) = {\mathrm{Tr}}\left(a (U_g b U_{g^{-1}}\Delta^{-z/4} -  U_g \Delta^{-z/4} b U_{g^{-1}} )  \right)$$ Then, write $$\begin{aligned}
U_g b U_{g^{-1}}\Delta^{-z/4} - U_g \Delta^{-z/4} b U_{g^{-1}}  &= U_g b U_{g^{-1}}\Delta^{-z/4} - U_g b \Delta^{-z/4} U_{g^{-1}} - U_g [\Delta^{-z/4},b] U_{g^{-1}}  {\vspace{2mm}}&= - U_g [\Delta^{-z/4},b] U_{g^{-1}} + U_g b U_{g^{-1}} [\Delta^{-z/4}, U_g] U_{g^{-1}} \end{aligned}$$

To end the calculations, we need the following lemma, whose proof may be found in [@Hig2006].

\[CM trick\] *(Connes-Moscovici’s trick, [@CM1995; @Hig2006])* For every $z \in {\mathbb{C}}$, we have the following expansion, $$\begin{aligned}
& [\Delta^{-z}, b] \sim \sum_{k\geq 1} \binom{-z}{k} b^{(k)}\Delta^{-z-k} & U_{g^{-1}} [\Delta^{-z}, U_g] \sim \sum_{k\geq 1} \binom{-z}{k} U_{g^{-1}} U_g^{(k)} \Delta^{-z-k}\end{aligned}$$ where we denote $T^{(k)} = {\mathrm{ad}}(\Delta)^k(T)$, ${\mathrm{ad}}(\Delta) = [\Delta, \, . \,]$.

Moreover, note that for every integer $k \geq 1$, $b^{(k)}$ and $U_{g^{-1}} U_g^{(k)}$ are classical Heisenberg pseudodifferential operators whose order stricly decrease as $k$ grows. Hence, evaluating 