Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 8 / 3

Put every formal pseudodifferential operator in normal order, with all coefficient functions on the left. Its formal pseudodifferential residue is the coefficient of ${\partial }^{-1}$. The Adler trace isHere and below, use periodic coefficients and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration functional must annihilate total derivatives. This convention is needed for the requested trace property; arbitrary elements of $C^{\infty }(\mathbb{R})$ do not supply it automatically.

It suffices to compute with monomials $A=a{\partial }^p$, $B=b{\partial }^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power ${\partial }^{-1}$, so both residues are zero. If $r\ge 0$, the formal pseudodifferential composition rule givesSince $q=r-1-p$, reversal of the factors in the numerator provesAfter $r$ integrations by parts, $\int a{b}^{(r)}dx=(-1{)}^r\int b{a}^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\ge 1$ their residue difference is the total derivativeFor $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward coefficient indices have a bounded sum. Summing the monomial result therefore proves

The boundary convention is substantive. Without it, take $A={\partial }^2$, $B=(\tanh x){\partial }^{-2}$. Both individual residues vanish, but $\mathrm{res}[A,B]=2(\tanh x{)}^{\prime }=2{\mathrm{sech}}^{2}x$ and its integral over $\mathbb{R}$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.

Write $P={\sum }_{j\le \alpha }{p}_j(x){\partial }^j$, setting unused coefficients to zero. Split the Adler trace into its fixed leading part and variable parts:Cyclicity from part (i) givesThus this Hamiltonian trace functional on monic differential operators has the required formTo make each coefficient explicit, expand ${\partial }^l{p}_j$ by the Leibniz rule. A term with $k$ derivatives of $p_j$ contributes to the residue exactly when $l-k+j=-1$. Consequently,These are smooth functions depending on $P$, not on the variable coefficients of $L$. In particular $a_0=p_{-1}$ and $a_1=p_{-2}+p_{-1}^{\prime }$. The coefficient $a_0$ must be included: it appears in the displayed sum even though the printed coefficient list starts at $a_1$.

The monic operators form an affine space, not a vector space. Accordingly $l_P$ is affine in the coordinates $u_l$, with constant $c$, and its differential is linear in variations. Its variational derivatives are $\delta l_P/\delta u_l=a_l$. It is the restriction of a linear trace pairing on the full operator space, which explains the terminology without discarding the constant term.

Use the canonical representatives of functionals of $u$: $P_f=f{\partial }^{-1}$, $P_g=g{\partial }^{-1}$. The formal pseudodifferential composition rule givesand the reversed product has $f,g$ exchanged. ThereforeMultiplying by $L={\partial }^2+u$ on the left, only the leading displayed commutator term can produce a residue. Differentiation of its coefficient produces powers at most ${\partial }^{-2}$, and multiplication by $u$ produces powers at most ${\partial }^{-3}$. Thusand, with the trace convention of part (i),There is no $u$ dependence in this bracket.

Moreover $\mathrm{res}(P_{f}L)=fu$, so $l_{P_f}=\int fudx$ and $\delta l_{P_f}/\delta u=f$. If $K(x,y)=\{u(x),u(y)\}$ is the distributional field bracket, it is characterized bySince $\int {\partial }_x\delta (x-y)g(y)dy=g^{\prime }(x)$, the answer isThe Dirac delta is periodic in the periodic setting, or the ordinary Dirac delta on the real line with admissible test functions. This is the first Hamiltonian structure of the KdV equation, with Poisson operator $-2{\partial }_x$. Its skewness follows by integration by parts; its coefficients are constant, so the functional Jacobi identity has no coefficient-variation terms.

A qualification is needed if the displayed trace formula is interpreted as a bracket on the reduced space for arbitrary operator representatives. Multiplication by a function $h$ gives $l_{m_h}({\partial }^2+u)=0$, yet direct expansion givesWith $h=g=\cos x$ on a period $[0,2\pi ]$, the latter is $\pi$. Hence the formula does not descend through every possible representative of a functional on this reduced affine space. The requested $f{\partial }^{-1},g{\partial }^{-1}$ representatives, used consistently for variational gradients, give the well-defined constant field bracket just computed; a general restriction from the ambient operator algebra needs an appropriate reduction.