map in cohomology. We shall now compute $\partial [{\mathrm{Tr}}_{[1]}]$. To do this, we lift ${\mathrm{Tr}}_{[1]}$ on $\Psi_c^{-\infty}(M) \rtimes G$ to a linear map on $\Psi_{H,c}^0(M) \rtimes G$ using a zeta function renormalization $${\mathrm{Tr}}_{[1]}'\left(\sum_{g \in G} P_g U_g \right) = {\mathrm{Pf}}_{z=0} {\mathrm{Tr}}\left(P_1  \cdot \Delta^{-z/4}\right)$$ where $\Delta$ is a sub-elliptic Laplacian, and ${\mathrm{Pf}}_{z=0}$ is the constant term in the Laurent series expansion of the zeta-function at $z=0$. Then, $\partial [{\mathrm{Tr}}_{[1]}]$ is represented in ${\mathrm{HP}}^1(S_H^0(M) \rtimes G)$ by the cyclic 1-cocycle $$\phi(aU_g,bU_h) = {\mathrm{Tr}}_{[1]}'([aU_g,bU_h])$$ for all $a,b\in \Psi^0_{H,c}(M)$ and $g,h\in G$. This expression makes sense as a cocycle over $S_H^0(M) \rtimes G$ because it vanishes whenever $a$ or $b$ belongs to the smooting ideal $\Psi^{-\infty}_{H,c}(M)$. Then, because the trace is localized at units, one finds that $\phi(aU_g,bU_h)=0$ if $gh\neq 1$, and $$\phi(a U_g, b U_{g^{-1}}) = {\mathrm{Tr}}_{[1]}'([a U_g, b U_{g^{-1}}]) = {\mathrm{Pf}}_{z=0} {\mathrm{Tr}}\left([a U_g, b U_{g^{-1}}]  \cdot \Delta^{-z/4}\right)$$ otherwise. The formula can be made a little more explicit if we gather accurately the relevant terms. This is the aim of the following proposition.

The cyclic 1-cocycle $\phi$ given above is given in terms of the Connes-Moscovici residue : $$\label{equivariant Radul cocycle} 
\phi(a U_g, b U_{g^{-1}}) = {\mathop{\ooalign{$\displaystyle{\int}$\cr$-$}}}a U_g [\log \Delta^{1/4}, b U_{g^{-1}}]$$

Firstly, 