Note also that in general, the representation of $\Psi_{H,c}(M) \rtimes G$ as linear operators on ${C^{\infty}}(M)$ does not yield pseudodifferential operators.\
Then, one has an *extension* $$\label{extension}
0 \to \Psi_c^{-\infty}(M) \rtimes G \to \Psi^0_{H,c}(M) \rtimes G \to S^0_{H,c}(M) \rtimes G \to 0$$ The usual trace on the algebra of regularizing operators $\Psi_c^{-\infty}(M)$, given by $${\mathrm{Tr}}(K) = \int_M k(x,x) d{\mathrm{vol}}(x)$$ where $k$ stands for the kernel of $K$, extends to a trace ${\mathrm{Tr}}_{[1]}$ on $\Psi_c^{-\infty}(M) \rtimes G$ by *localization at the unit* of $G$ $${\mathrm{Tr}}_{[1]}\left(\sum_{g \in G} K_g U_g \right) = {\mathrm{Tr}}(K_1)$$ That ${\mathrm{Tr}}_{[1]}$ still remains a trace only comes from the invariance of the ordinary operator trace ${\mathrm{Tr}}$ under conjugation by $U_g$.\
In the same way, the Wodzicki residue on $S_{H,c}(M)$ extends to a trace on $S_{H,c}(M) \rtimes G$ by localization at the unit.\
The pseudodifferential extension gives rise to the following commutative diagram involving *algebraic K-theory* and *periodic cyclic homology* ([@Nis1997])

$$\label{Index diagram}
\vcenter{
\xymatrix{
    K_1^{{\mathrm{alg}}}(S^0_{H,c}(M) \rtimes G) \ar[r]^{{\mathrm{Ind}}} \ar[d]^{{\mathrm{ch}}_1} & K_0^{{\mathrm{alg}}}(\Psi_c^{-\infty}(M) \rtimes G) \ar[d]^{{\mathrm{ch}}_0} {\vspace{2mm}}{\mathrm{HP}}_1(S^0_{H,c}(M) \rtimes G) \ar[r]^{\partial} & {\mathrm{HP}}_0(\Psi_c^{-\infty}(M) \rtimes G)}
}$$

The vertical arrows are respectively the odd and even Chern character.\
Denote again $\partial :  {\mathrm{HP}}^0(\Psi_c^{-\infty}(M) \rtimes G) \rightarrow {\mathrm{HP}}^1(S_{H,c}^0(M) \rtimes G)$ the induced excision 