Formal pseudodifferential residue ($\mathrm{res}P=p_{-1}(x)$)

The formal residue is the coefficient of ${\partial }^{-1}$ after coefficients have been moved to the left by the formal pseudodifferential composition rule. It is not an analytic residue at a point. For scalar coefficients, the residue of a commutator is a total derivative; integrating it with vanishing boundary terms produces the cyclic Adler trace.

Integrate the formal pseudodifferential residue over a period, or over the real line when the integral converges and the relevant boundary terms vanish. More algebraically, apply a linear functional on the coefficient algebra that annihilates derivatives. For $A=a{\partial }^p$, $B=b{\partial }^q$ and $r=p+q+1\ge 0$, the two residues are $\left(\genfrac{}{}{0ex}{}{p}{r}\right)a{b}^{(r)}$ and $\left(\genfrac{}{}{0ex}{}{q}{r}\right)b{a}^{(r)}$. Since $\left(\genfrac{}{}{0ex}{}{q}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{p}{r})$, integration by parts proves $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$. For $r<0$ both residues vanish. Bounded orders ensure only finitely many monomial pairs contribute to the residue of a product.

For a fixed formal pseudodifferential operator $P={\sum }_j{p}_j{\partial }^j$ and a monic differential operator $L={\partial }^m+{\sum }_{l=0}^{m-1}{u}_l{\partial }^l$, cyclicity of the Adler trace gives $l_P(L)=\mathrm{Tr}(PL)$ with $c=\int p_{-m-1}dx$ and $a_l=\mathrm{res}({\partial }^{l}P)={\sum }_{k=0}^l\left(\genfrac{}{}{0ex}{}{l}{k}\right)p_{k-l-1}^{(k)}$. The function is affine in the variable coefficients; its variational derivative with respect to $u_l$ is $a_l$.