Uniqueness of the Kontsevich-Vishik Trace

Let M be a closed manifold. We show that the Kontsevich-Vishik trace, which is defined on the set of all classical pseudodifferential operators on M, whose (complex) order is not an integer greater than or equal to -dim M, is the unique functional which (i) is linear on its domain, (ii) has the trace property and (iii) coincides with the L^2-operator trace on trace class operators. Also the extension to even-even pseudodifferential operators of arbitrary integer order on odd-dimensional manifolds and to even-odd pseudodifferential operators of arbitrary integer order on even-dimensional manifolds is unique.


Introduction
We denote by M a compact n-dimensional manifold without boundary. A classical pseudodifferential operator (ψdo) A on M is said to have order µ ∈ C, if it belongs to the Hörmander class S Re µ 1,0 (M ) and the local symbols a = a(x, ξ) of A have asymptotic expansions a ∼ ∞ j=0 a µ−j , (1.1) where the a µ−j are positively homogeneous of degree µ − j for large ξ. In general, all pseudodifferential operators will be assumed to act on sections of vector bundles over M . We shall write ord A = µ to express that the order of A is µ.
In two remarkable papers, Kontsevich and Vishik in 1994 and 1995 analyzed the properties of determinants of elliptic ψdo's, [5], [6]. One important tool was the construction of a trace-like mapping TR defined on the set of all classical ψdo's whose order is not an element of Z ≥−n , the set of integers greater than or equal to −n.
We shall denote this domain by D. As the sum of two operator of orders µ and µ ′ in D is an element of D only if µ − µ ′ is an integer, D is not a vector space. Thus it does not make sense to speak about linear functionals on D. The map TR : D → C, however, is as linear as it can be expected to be: It is clear that the Kontsevich-Vishik trace cannot be extended to a trace on the algebra of all ψdo's on M : The only trace there (up to multiples) is the Wodzicki residue [11], which is known to vanish on trace class operators. There also is a simple direct way to see this: We know -e.g. from the Atiyah-Singer index theoremthat there exists an elliptic pseudodifferential operator P on M with nonzero index. Using order reducing operators, we may assume the order of P to be zero. Let Q be a parametrix to P modulo smoothing operators. Then If we could extend TR to a trace on all pseudodifferential operators, the right hand side could be rewritten as the trace of the commutator TR[P, Q] and therefore would have to be zero -a contradiction.
It has been observed, however, by Kontsevich-Vishik and Grubb [3] that TR extends to a slightly larger domain. Recall that the symbol a of an integer order operator A is said to be even-even, if the homogeneous components satisfy The Kontsevich-Vishik trace TR(A) for a ψdo A of order µ then can also be defined if µ ∈ Z ≥−n , provided that (EE) n is odd, and the symbol of A is even-even, or (EO) n is even, and the symbol of A is even-odd. For the sake of brevity we shall denote this larger domain (depending on n) by D + .
In both cases, the component a −n in the asymptotic expansion of the symbol of A is odd in ξ for large |ξ|, say for |ξ| ≥ 1: Hence the density for the Wodzicki residue of the operator A vanishes pointwise, i.e.
Here, dσ is the surface measure on the unit sphere S * x M over x in the cotangent bundle. The Wodzicki residue of A is given by integration of res x A over M and therefore also vanishes.
The trace property (1.3) extends to the case where A and B have integer order and AB and BA belong to D + .
It seems, however, that it never has been noticed that the above properties make the Kontsevich-Vishik trace unique. This is what we show in this short note. The proof, which will be given in the next section, relies on ideas in [1].
Theorem. Also the extension of τ to D + is unique. In fact, τ already is unique on the space of all integer order ψdo's which satisfy (EE) or (EO) when µ ≥ −n.

Proof
In order to establish (a), choose a ψdo A of order µ ∈ C \ Z ≥−n on M . We find a covering of M by open neighborhoods and a finite subordinate partition of unity {ϕ j } such that for every pair (j, k), both ϕ j and ϕ k have support in one coordinate neighborhood. We write Each operator ϕ j Aϕ k may be considered a ψdo on R n . As the map τ has the linearity property (1.2), we may confine ourselves to the case where A = op a with a symbol a on R n having an expansion (1.1). Moreover, we can assume that A = ϕAψ whenever ϕ, ψ ∈ C ∞ c (R n ) are equal to one on a sufficiently large set. To simplify further, we write where a µ−j is a symbol on R n , homogeneous in ξ of degree µ − j for |ξ| ≥ 1, and K is so large that r ∈ S −n−1 Since τ (ϕ op(r)ψ) = tr(ϕ op(r)ψ) by (1.4), we will know τ (op a) as soon as we know τ (ϕ op(a µ−j )ψ) for j = 0, . . . , K.
We may assume that µ is not an integer, since the operator trace determines τ on all operators of order µ < −n. Now we let Euler's relation for homogenous functions implies that, for |ξ| ≥ 1, Hence we can write Since a µ−j − b µ−j is regularizing, the first term on the right hand side is determined by Property (1.4). Now we choose additionally χ ∈ C ∞ c (R n ) with χϕ = ϕ and χψ = ψ. The fact that op(∂ ξ k p) = −i [x k , op p ] for an arbitrary symbol p implies that Assuming that τ has Property (1.3), it vanishes on the last term in (2.9) which is a sum of commutators. Hence the proof of (a) is complete.
Next let us show (b). With the same considerations as before we may assume that A = op a is a pseudodifferential operator on R n with a representation as in (2.7), where now µ is an integer ≥ −n and the a µ−j have property (EE) or (EO). We only have to show that τ (ϕ op(a µ−j )ψ) is uniquely determined, j = 0, . . . , µ+n. For µ − j = −n the argument is as before, using the symbols in (2.8) and noting that a µ−j (x, ξ)ξ k is even-even or even-odd whenever this is the case for a µ−j .
So let us consider a −n . The assumption that n is odd and a −n even-even or n is even and a −n even-odd implies that a −n is odd in ξ: Hence, for each fixed x, the integral over the unit sphere S = {|ξ| = 1} vanishes: The Laplace operator ∆ = n k=1 ∂ 2 /∂ξ 2 k in polar coordinates takes the form where r = |ξ| is the radial variable and ∆ S is the Laplace-Beltrami operator on S. Equation (2.10) implies that, for each x, the function a −n (x, ·) is orthogonal to the constants which form the kernel of the symmetric operator ∆ S . Hence there is a unique function q(x, ·) ∈ C ∞ (S), orthogonal to the constants, such that ∆ S q(x, ·) = a −n (x, ·)| S . As ∆ S commutes with the antipodal map η → −η, we have ∆ S (q(x, −·)) = a −n (x, −·)| S = −a −n (x, ·)| S . Hence q(x, ·) + q(x, −·) belongs to the kernel of ∆ S , thus is constant. On the other hand, both q(x, ·) and q(x, −·) are orthogonal to the constants. Therefore q(x, ·) + q(x, −·) is zero, i.e., q(x, ·) is an odd function on S. Now we choose a smooth function ω on R which vanishes for small r and is equal to 1 for r ≥ 1/2. We let b −n = ω(r)r 2−n q = ω(|ξ|) |ξ| 2−n q(x, ξ/|ξ|). This is a smooth function on R n which is homogeneous of degree 2 − n in ξ for |ξ| ≥ 1. As a −n (x, ξ) vanishes for x outside a compact set, so does b −n (x, ξ). In particular, b −n is an element of S 2−n 1,0 (R n × R n ). Moreover, we have for |ξ| ≥ 1 ∆b −n = ∆(r 2−n q(x, ·)) = r −n a −n (x, ·)| S = a −n .