Formal pseudodifferential operator ($P={\sum }_{j\le N}{p}_j(x){\partial }^j$)

A formal pseudodifferential operator is a series in integer powers of $\partial =d/dx$, bounded above but possibly infinite towards negative powers. Its coefficients are smooth functions. Multiplication uses the formal pseudodifferential composition rule. Negative powers are formal inverses, not globally defined integration operators on all smooth functions. Every coefficient in a product is a finite sum, so multiplication is defined without analytic convergence.

Treat $k$ as a formal spectral parameter near infinity. If $P={\sum }_{j\ge 0}{g}_j{\partial }^{\alpha -j}$, the formal pseudodifferential composition rule gives $Pw=k^{\alpha +\beta }{e}^{kx}{\sum }_{s\ge 0}{k}^{-s}{\sum }_{i+j+r=s}{g}_j\left(\genfrac{}{}{0ex}{}{\alpha -j}{r}\right)w_i^{(r)}$. Each coefficient contains finitely many terms. This describes a formal module action, not a claim that the infinite series converges at finite $k$.

For scalar coefficients, leading symbols commute. If $P=a{\partial }^p+\cdots$ and $Q=b{\partial }^q+\cdots$, their product has leading coefficient $ab$, and their commutator has coefficient $pa{b}^{\prime }-qb{a}^{\prime }$ at order $p+q-1$. Thus product order is at most $p+q$, with equality when $ab$ is not identically zero; commutator order is at most $p+q-1$, with equality when its displayed coefficient is not identically zero. Constant-coefficient operators commute. Nonzero smooth functions with disjoint supports show why product orders need not add over $C^{\infty }(\mathbb{R})$.

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 $a_l$.

The generalized Leibniz rule extends composition of differential operators to all integer $n$. Define $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. For nonnegative $n$ the sum stops at $r=n$. For $n=-s<0$, its coefficients are $(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$; in particular ${\partial }^{-1}f=f{\partial }^{-1}-f^{\prime }{\partial }^{-2}+f^{\prime \prime }{\partial }^{-3}-\cdots$. Applying $\partial$ on the left cancels consecutive derivative terms and recovers multiplication by $f$, which explains the inverse expansion.