is at each point $(x, p)$ an endomorphism acting on the fibre $E_x$, and (\[wodzicki residue symbol 0\]) becomes : $${\mathop{\ooalign{$\displaystyle{\int}$\cr$-$}}}P = \frac{1}{(2\pi)^n} \int_{S_H^*M} \iota_L \left({\mathrm{tr}}(\sigma_{-(v+2n)}(x,p)) \, \frac{ \omega^n}{n!}\right)$$ where ${\mathrm{tr}}$ denotes the trace of endomorphisms.\

Excision and equivariant residue index formula
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Let $M$ be a foliated manifold. Consider a discrete subgroup $G\subset{\mathrm{Diff}}(M)$ of diffeomorphisms mapping leaves to leaves. By convention we suppose that $G$ acts from the right, so, for any $g\in G$ the induced linear action $U_g$ on the space of functions ${C^{\infty}}(M)$ reads $$(U_g f)(x) = f(x\cdot g)\ ,\quad \forall f\in{C^{\infty}}(M)\ ,\ x\in M$$ Recall that the *algebraic crossed product* $\Psi_{H,c}(M) \rtimes G$ is the universal algebra generated by Heisenberg pseudodifferential operators and group elements, that is, $$\Psi_{H,c}(M) \rtimes G = \left\{ \sum_{g \in G} P_g U_g \, ; \, P_g \in \Psi_{H,c}(M) \right\}$$ where $P_gU_g$ is a short-hand notation for the tensor product $P_g\otimes U_g$, and the sum only contains a finite number of non-zero terms. The multiplication is given by the rule $$P U_g \cdot Q U_h = P (U_g Q U_{g^{-1}}) U_{gh}$$ To this effect, remark that $U_g Q U_{g^{-1}}$ is still a classical Heisenberg pseudodifferential operator, so that the product makes sense. 