CHAPTER 3 Isotropic and scattering calculi^{[C.IandS.calculi]}

In this chapter many of the general constructions with pseudodifferential operators are carried out in the context of two global calculi on ℝ^{n}. Partly this is done for the obvious reason, that these calculi and results have interesting applications, and partly it is preparatory to the discussion of the geometric algebras of pseudodifferential operators on a compact manifold without boundary and for the scattering algebra on a compact manifold with boundary. Thus, while this chapter is somewhat interstitial, it is designed to clarify the later discussions by separating the generalities of the construction from the particulars of the calculus involved. It should be noted that in this chapter it is generally the polyhomogeneous calculus which is under discussion unless it is explicitly stated to the contrary.

3.1. Isotropic operators^{[S.Sect.isotropic.calculus]}

As noted in the discussion in Chapter 2, there are other sensible choices of the class of amplitudes which can be admitted in the definition of a space of pseudodifferential operators than the basic case of S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) discussed there. One of the smallest such choices is the class which is completely symmetric in the variables x and ξ and consists of the symbols on ℝ^{2n}. Thus, a ∈ S_{∞}^{m}(ℝ_{x,ξ}^{2n}) satisfies the estimates

|D_{x}^{α}D_{ξ}^{β}a(x, ξ)| ≤ C_{α,β}(1 + |x| + |ξ|)^{m−|α|−|β|}^{[1.2.2000.304]} (3.1)

for all multiindices α and β. If m ≤ 0 this is in the space S_{∞}^{m}(ℝ_{x}^{n}; ℝ_{ξ}^{n}); if m > 0 it is not, however,

[1.2.2000.303]

Lemma 3.1. For any p and n

S_{∞}^{m}(ℝ^{p+n}) ⊂ \begin{cases}⋂_{0≥r≥m}(1 + |x|^{2})^{r}S^{m−r}(ℝ_{x}^{p}; ℝ_{ξ}^{n}) & m ≤ 0 \\ (1 + |x|^{2})^{m/2}S^{m}(ℝ_{x}^{p}; ℝ_{ξ}^{n}), & m > 0. \end{cases} ^{[8.2.1998.98]} (3.2)

Proof. This follows from (3.1) and the inequalities

1 + |x| + |ξ| ≤ (1 + |x|)(1 + |ξ|),
1 + |x| + |ξ| ≥ (1 + |x|)^{t}(1 + |ξ|)^{1−t}, 0 ≤ t ≤ 1.

In view of these estimates the following definition makes sense.

[1.2.2000.313]

Definition 3.1. For any m ∈ ℝ we define

64 3. ISOTROPIC AND SCATTERING CALCULI^{[C.IANDS.CALCULI]}

(3.3) Ψ_{iso}^{m}(ℝ^{n}) ⊂ Ψ_{∞−iso}^{m}(ℝ^{n}) ⊂ ⟨x⟩^{m_{+}}Ψ_{∞}^{m}(ℝ^{n})^{[1.2.2000.314]} as the subspaces determined by

A ∈ Ψ_{iso}^{m}(ℝ^{n}) ⟺ σ_{L}(A) ∈ S_{ph}^{m}(ℝ^{2n})
A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) ⟺ σ_{L}(A) ∈ S_{∞}^{m}(ℝ^{2n}).

[1.2.2000.315] (3.4)

As in the discussion of the traditional algebra in Chapter 2 we show the ∗- invariance and composition properties of these spaces of operators by proving an appropriate ‘reduction’ theorem. Note however that there is a small difficulty here. Namely it might be supposed that it is enough to analyse I(a) for a ∈ S_{∞}^{m}(ℝ^{3n}). This however is not the case. Indeed the definition above is in terms of left-reduced symbols. If a ∈ S_{∞}^{m}(ℝ^{2n}) is regarded as a function on ℝ^{3n} which is independent of one of the variables then it is in general not an element of S_{∞}^{m}(ℝ^{3n}). For this reason we consider some more ‘hybrid’ estimates.

Consider a subdivision of ℝ^{3n} into two closed regions:

\begin{matrix}R_{1}(ϵ) = \{(x, y, ξ) ∈ ℝ^{3n}; |x − y| ≤ ϵ(1 + |x|^{2} + |y|^{2} + |ξ|^{2})^{\frac{1}{2}} \} \\ R_{2}(ϵ) = \{(x, y, ξ) ∈ ℝ^{3n}; |x − y| ≥ ϵ(1 + |x|^{2} + |y|^{2} + |ξ|^{2})^{\frac{1}{2}} \}. \end{matrix}^{[1.2.2000.316]} (3.5)

If a ∈ C^{∞}(ℝ^{3n}) consider the estimates

|D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a(x, y, ξ)| ≤ C_{α,β,γ} \begin{cases}⟨(x, y, ξ)⟩^{m−|α|−|β|−|γ|} & in R_{1}(\frac{1}{8}) \\ ⟨(x, y)⟩^{m_{+}}⟨ξ⟩^{m−|γ|} & in R_{2}(\frac{1}{8}). \end{cases} ^{[1.2.2000.317]} (3.6)

The choice ϵ = \frac{1}{8} here is rather arbitrary. However if ϵ is decreased, but kept positive the same estimates continue to hold for the new subdivision, since the estimates in R_{1} are stronger than those in R_{2} (which is increasing at the expense of R_{1} as ϵ decreases). Notice too that these estimates do in fact imply that a ∈ ⟨(x, y)⟩^{m_{+}}S_{∞}^{m}(ℝ^{2n}; ℝ^{n}).

[1.2.2000.318]

Proposition 3.1. If a ∈ C^{∞}(ℝ^{3n}) satisfies the estimates (3.6) then A = I(a) ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) and (2.58) holds for σ_{L}(A).

Proof. We separate a into two pieces. Choose χ ∈ C_{c}^{∞}(ℝ) with 0 ≤ χ ≤ 1, with support in [−\frac{1}{8}, \frac{1}{8}] and with χ ≡ 1 on [−\frac{1}{9}, \frac{1}{9}]. Then consider the cutoff function on ℝ^{3n}

ψ(x, y, ξ) = χ (\frac{|x − y|^{2}}{1 + |x|^{2} + |y|^{2} + |ξ|^{2}} ) .^{[1.2.2000.319]} (3.7)

Clearly, ψ has support in R_{1}(\frac{1}{8}) and ψ ∈ S_{∞}^{0}(ℝ^{3n}). It follows then that a′ = ψa ∈ S_{iso}^{m}(ℝ^{3n}). On the other hand, a′′ = (1 − ψ)a has support in R_{2}(\frac{1}{9}). In this region |x − y|, ⟨(x, y)⟩ and ⟨(x, y, ξ)⟩ are bounded by constant multiples of each other. Thus a′′ satisfies the estimates

(3.8) |D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a′′(x, y, ξ)| ≤ C_{α,β,γ}|x − y|^{m_{+}}⟨ξ⟩^{m−|γ|}

≤ C′_{α,β,γ}⟨(x, y, ξ)⟩^{m_{+}}⟨ξ⟩^{m−|γ|}, supp(a′′) ⊂ R_{2}(\frac{1}{9}).^{[1.2.2000.320]}

First we check that I(a′′) ∈ S(ℝ^{2n}). On R_{2}(\frac{1}{9}) it is certainly the case that

|x − y| ≥ \frac{1}{9}⟨(x, y)⟩ and by integration by parts

|x − y|^{2p}D_{x}^{α}D_{y}^{β}I(a′′) = I(|D_{ξ}|^{2p}D_{x}^{α}D_{y}^{β}a′′).

3.1. ISOTROPIC OPERATORS^{[S.SECT.ISOT ROP IC.CALCULUS]} 65

For all sufficiently large p it follows from (3.8) that this is the product of ⟨(x, y)⟩^{m_{+}} and a bounded continuous function. Thus, I(a′′) ∈ S(ℝ^{2n}) is the kernel of an operator in Ψ_{iso}^{−∞}(ℝ^{n}).

So it remains only to show that A′ = I(a′) ∈ Ψ_{∞−iso}^{m}(ℝ^{n}). Certainly this is an element of ⟨x⟩^{m_{+}}Ψ_{∞}^{m}(ℝ^{n}). The left-reduced symbol of A′ has an asymptotic expansion, as ξ → ∞, given by the usual formula, namely (2.58). Each of the terms in this expansion

a_{L}(A′) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{y}^{α}D_{ξ}^{α}a(x, x, ξ)

is in the space S_{∞}^{m−2|α|}(ℝ^{2n}). Thus we can actually choose an asymptotic sum in the stronger sense that

b′ ∈ S_{∞}^{m}(ℝ^{2n}), b_{N} = b′ − ∑_{|α|<N} \frac{i^{|α|}}{α!} D_{x}^{α}D_{ξ}^{α}a(x, ξ) ∈ S_{∞}^{m−2N}(ℝ^{2n}) ∀ N.

Consider the remainder term in (2.47), given by (2.44) and (2.45). Integrating by parts in ξ to remove the factors of (x − y)^{α} the remainder, R_{N}, can be written as a pseudodifferential operator with amplitude

r_{N}(x, y, ξ) = ∑_{|α|=N} \frac{i^{|α|}}{α!} ∫_{0}^{1} dt(1 − t)^{N}(D_{ξ}^{α}D_{y}^{α}a)((1 − t)x + ty, ξ).

This satisfies the estimates (3.6) with m replaced by m − 2N. Indeed from the symbol estimates on a′ the integrand satisfies the bounds

|D_{x}^{β}D_{y}^{γ}D_{ξ}^{δ}D_{ξ}^{α}D_{y}^{α}a′((1 − t)x + ty, ξ)|

≤ C(1 + |(x + t(x − y)| + |ξ|)^{m−2N−|β|−|γ|−|δ|}.

In R_{1}(\frac{1}{8}), |x − y| ≤ \frac{1}{8}⟨(x, y, ξ)⟩ so |x + t(x − y)| + |ξ| ≥ \frac{1}{2}⟨(x, y, ξ)⟩ and these estimates imply the full symbol estimates there. On R_{2} we immediately get the weaker estimates in (3.7).

Thus, for large N, the remainder term gives an operator in ⟨x⟩^{\frac{m}{2} −N}Ψ_{iso}^{\frac{m}{2} −N}(ℝ^{n}). The difference between A′ and the operator B′ ∈ Ψ_{∞−iso}^{m}(ℝ^{n}), which is R_{N} plus an operator in Ψ_{∞−iso}^{m−2N}(ℝ^{n}) for any N is therefore in Ψ_{iso}^{−∞}(ℝ^{n}). Thus A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}).

This is a perfectly adequate replacement in this context for our previous reduction theorem, so now we can show the basic result.

[8.2.1998.99]

Theorem 3.1. The spaces Ψ_{∞−iso}^{m}(ℝ^{n}) (resp. Ψ_{iso}^{m}(ℝ^{n})) of isotropic (resp. polyhomogeneous isotropic) pseudodifferential operators on ℝ^{n}, defined by (3.4) form an order-filtered ∗-algebra with residual space Ψ_{iso}^{−∞}(ℝ^{n}) = S(ℝ^{2n}) (resp. the same) as spaces of kernels.

Proof. The condition that a continuous linear operator A on S(ℝ^{n}) be an element of Ψ_{∞−iso}^{m}(ℝ^{n}) is that it be an element of (1 + |x|^{2})^{m/2}Ψ_{∞}^{m}(ℝ^{n}) if m ≥ 0 or Ψ_{∞}^{m}(ℝ^{n}) if m < 0 with left-reduced symbol an element of S_{∞}^{m}(ℝ_{x,ξ}^{2n}) :

q_{l} : S_{∞}^{m}(ℝ^{2n}) ⟷ Ψ_{∞−iso}^{m}(ℝ^{n}).^{[8.2.1998.101]} (3.9)

Thus A^{∗} has right-reduced symbol in S_{∞}^{m}(ℝ^{2n}). This satisfies the estimates (3.6) as a function of x, y and ξ. Thus Propostion 3.1 shows that A^{∗} ∈ Ψ_{∞−iso}^{m}(ℝ^{n}), since its left-reduced symbol is in S_{∞}^{m}(ℝ^{2n}). This shows the ∗-invariance. Moreover it also follows that any B ∈ Ψ_{∞−iso}^{m′}(ℝ^{n}) has right-reduced symbol in S_{∞}^{m′}(ℝ^{2n}). Thus if A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) and B ∈ Ψ_{iso}^{m′}(ℝ^{n}) then using this result to right-reduce B we see that the composite opertor has kernel I(a_{L}(x, ξ)b_{R}(y, ξ)) where a_{L} ∈ S_{∞}^{m}(ℝ^{2n}) and b_{R} ∈ S_{∞}^{m′}(ℝ^{2n}). Now it again follows that this product satisifes the estimates (3.6) of order m + m′. Hence, again applying Proposition 3.1, we conclude that A ∘ B ∈ Ψ_{∞−iso}^{m+m′}(ℝ^{n}). This proves the theorem for Ψ_{∞−iso}^{∗}(ℝ^{n}).

The proof for the polyhomogeneous space Ψ_{iso}^{m}(ℝ^{n}) follows immediately, since the symbol expansions all preserve polyhomogeneity.

One further property of the isotropic calculus that distinguishes it strongly from the traditional calculus is that it is invariant under Fourier transformation.

[21.2.1998.107]

Proposition 3.2. If A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) (resp. Ψ_{iso}^{m}(ℝ^{n})) then Â ∈ Ψ_{∞−iso}^{m}(ℝ^{n})

(resp. Ψ_{iso}^{m}(ℝ^{n})) where ˆ Âu = Aû with û being the Fourier transform of u ∈ S(ℝ^{n}).

The proof of this is outlined in Problem 2.20.

3.2. Scattering operators^{[S.Scattering.operators]}

There is another calculus of pseudodifferential operators which is ‘smaller’ than the traditional calculus. It arises by taking amplitudes in (2.2) which treat the base and fibre variables symmetrically, but not ‘simultaneously.’ Thus consider the spaces

(3.10) S_{∞}^{l,m}(ℝ_{z}^{p}, ℝ_{ξ}^{n}) = {a ∈ C^{∞}(ℝ^{p+n});

sup_{ℝ^{p+n}}(1 + |z|)^{−l+|α|}(1 + |ξ|)^{−m+|β|}|D_{z}^{α}D_{ξ}^{β}a(z, ξ)| < ∞, ∀ α, β}.^{[21.2.1998.113]}

Observe that

S_{∞}^{l,m}(ℝ_{z}^{p}; ℝ_{ξ}^{n}) ⊂ (1 + |z|^{2})^{l/2}S_{∞}^{m}(ℝ_{z}^{p}; ℝ_{ξ}^{n}).^{[21.2.1998.115]} (3.11)

We can then define

(3.12) A ∈ Ψ_{∞−sc}^{l,m}(ℝ^{n}) ⟺ A = (1 + |x|^{2})^{l/2}B,

B ∈ Ψ_{∞}^{m}(ℝ^{n}) and σ_{L}(B) ∈ S_{∞}^{0,m}(ℝ_{x}^{n}, ℝ_{ξ}^{n}).^{[21.2.1998.114]}

It follows directly from this definition and the properties of the ‘traditional’ operators that the left symbol map is an isomorphism

σ_{L} : Ψ_{∞−sc}^{l,m}(ℝ^{n}) ⟶ S_{∞}^{l,m}(ℝ_{x}^{n}, ℝ_{ξ}^{n}).^{[1.2.2000.300]} (3.13)

To prove that this is an algebra, we need first the analogue of the asymptotic completeness, Proposition 2.3, for symbols in S_{∞}^{∗,∗}(ℝ^{p}; ℝ^{n}).

[1.2.2000.292]

Lemma 3.2. If a_{j} ∈ S_{∞}^{l−j,m−j}(ℝ^{p}, ℝ^{n}) for j ∈ ℕ_{0} then there exists

a ∈ S^{l,m}(ℝ^{p}, ℝ^{n}) s.t. a − ∑_{j=0}^{N} a_{j} ∈ S_{∞}^{l−N,m−N}(ℝ^{p}, ℝ^{n}) ∀ N ∈ ℕ_{0}.^{[1.2.2000.293]} (3.14)

3.2. SCATTERING OPERATORS^{[S.SCAT T ERING.OP ERAT ORS]}67

Even though there is some potential for confusion we write a ∼ ∑_{j} a_{j} for a symbol

a satisfying (3.14).

Proof. We use the same strategy as in the proof of Proposition 2.3 with the major difference that there are essentially two different symbolic variables. Thus with the same notation as in (2.54) we set

a = ∑_{j} ϕ(ϵ_{j}z)ϕ(ϵ_{j}ξ)a_{j}(z, ξ)^{[1.2.2000.294]} (3.15)

and we proceed to check that if the ϵ_{j} ↓ 0 fast enough as j → ∞ then the series converges in S_{∞}^{l,m}(ℝ^{p}, ℝ^{n}) and the limit satisfies (3.14).

The first of the seminorms, for convergence, is

A_{j} = sup_{z} sup_{ξ}(1 + |z|)^{−l}(1 + |ξ|)^{−m}ϕ(ϵ_{j}z)ϕ(ϵ_{j}ξ)|a_{j}(z, ξ)|.

On the support of this function either |z| ≥ 1/ϵ_{j} or |ξ| ≥ 1/ϵ. Thus

A_{j} ≤ sup_{z} sup_{ξ}(1 + |z|)^{−l+j}(1 + |ξ|)^{−m+j}|a_{j}(z, ξ)|

× sup_{z} sup_{ξ}(1 + |z|)^{−j}(1 + |ξ|)^{−j}ϕ(ϵ_{j}z)ϕ(ϵ_{j}ξ)

≤ ϵ_{j}^{j} sup_{z} sup_{ξ}(1 + |z|)^{−l+j}(1 + |ξ|)^{−m+j}|a_{j}(z, ξ)|

The last term on the right is a seminorm on S_{∞}^{l−j,m−j}(ℝ^{p}, ℝ^{n}) so convergence follows by choosing the ϵ_{j} eventually smaller than a certain sequence of positive numbers. The same argument follows, as in the discussion leading to (2.56), for convergence of the series for the derivatives and also for the stronger convergence leading to (3.14). Since overall this is a countable collection of conditions, all can be arranged by diagonalization and the result follows.

With this result on asymptotic completeness the proof of Theorem 3.1 can be followed closely to yield the analogous result on products. In fact we can also define polyhomogeneous operators. This requires a little work if we try to do it directly. However see (1.97) and Problem 1.17 which encourages us to identify

RC_{p}^{∗} × RC_{n}^{∗} : S_{ph}^{0,0}(ℝ^{p}, ℝ^{n}) ⟷ C^{∞}(𝕊^{p,1} × 𝕊^{n,1}),
S_{ph}^{l,m}(ℝ^{p}, ℝ^{n}) = (1 + |z|^{2})^{l/2}(1 + |ξ|^{2})^{m/2}S_{ph}^{0,0}(ℝ^{p}, ℝ^{n}), l, m ∈ ℝ.

[1.2.2000.297] (3.16)

These definitions are discussed as problems starting at Problem 1.18. Thus we simply define

(3.17) Ψ_{sc}^{l,m}(ℝ^{n}) = {A ∈ Ψ_{∞−sc}^{l,m}; σ_{L}(A) ∈ S_{ph}^{l,m}(ℝ^{n}, ℝ^{n})} .^{[1.2.2000.299]}

[1.2.2000.295]

Theorem 3.2. The spaces Ψ_{∞−sc}^{l,m}(ℝ^{n}) (resp. Ψ_{sc}^{l,m}(ℝ^{n})) of scattering (resp. polyhomogeneous scattering) pseudodifferential operators on ℝ^{n}, form an orderbifiltered ∗-algebra

(3.18) Ψ_{∞−sc}^{l,m}(ℝ^{n}) ∘ Ψ_{∞−sc}^{l′,m′}(ℝ^{n}) ⊂ Ψ_{∞−sc}^{l+l′,m+m′}(ℝ^{n})^{[1.2.2000.296]}

with residual spaces

(3.19)  ⋂_{l,m} Ψ_{∞−sc}^{l,m}(ℝ^{n}) = ⋂_{l,m} Ψ_{sc}^{l,m}(ℝ^{n})Ψ_{iso}^{−∞}(ℝ^{n}) = S(ℝ^{2n}).^{[1.2.2000.355]}

3.3. The residual algebra, Ψ_{iso}^{−∞}(ℝ^{n})^{[S.Residual.isotropic]}

The residual isotropic (and scattering) algebra has two important properties not shared by the residual algebra Ψ_{∞}^{−∞}(ℝ^{n}), of which it is a subalgebra (and in fact in which it is an ideal). The first is that as operators on L^{2}(ℝ^{n}) the residual isotropic operators are compact.

[1.2.2000.309]

Proposition 3.3. Elements of Ψ_{iso}^{−∞}(ℝ^{n}) are characterized amongst continous operators on S(ℝ^{n}) by the fact that they extend by continuity to define continuous linear maps

A : S′(ℝ^{n}) ⟶ S(ℝ^{n}).^{[1.2.2000.356]} (3.20)

In particular the image of a bounded subset of L^{2}(ℝ^{n}) under an element of Ψ_{iso}^{−∞}(ℝ^{n}) is contained in a compact subset.

Proof. The kernels of elements of Ψ_{iso}^{−∞}(ℝ^{n}) are in S(ℝ^{2n}) so the mapping property (3.20) follows.

The norm sup_{|α|≤1} |⟨x⟩^{n+1}D^{α}u(x)| is continuous on S(ℝ^{n}). Thus if S ⊂ L^{2}(ℝ^{n}) is bounded and A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) the continuity of A : L^{2}(ℝ^{n}) ⟶ S(ℝ^{n}) implies that A(S) is bounded with respect to this norm. The theorem of Arzela-Ascoli shows that any sequence in A(S) has a strongly convergent subsequence in ⟨x⟩^{n}C_{∞}^{0}(ℝ^{n}) and such a sequence converges in L^{2}(ℝ^{n}). Thus A(S) has compact closure in L^{2}(ℝ^{n}).

The second important property of the residual algebra is that it is ‘bi-ideal’ or a ‘corner’ in the bounded operators on L^{2}(ℝ^{n}). Note that it is not an ideal.

[1.2.2000.310]

Lemma 3.3. If A_{1}, A_{2} ∈ Ψ_{iso}^{−∞}(ℝ^{n}) and B is a bounded operator on L^{2}(ℝ^{n}) then A_{1}BA_{2} ∈ Ψ_{iso}^{−∞}(ℝ^{n}).

Proof. The kernel of the composite C = A_{1}BA_{2} can be written as a distributional pairing

C(x, y) = ∫_{ℝ^{2n}} B(x′, y′)A_{1}(x, x′)A_{2}(y′, y)dx′dy′ = (B, A_{1}(x, ·)A_{2}(·, y)) ∈ S(ℝ^{2n}).^{[1.2.2000.321]} (3.21)

Thus the result follows from the continuity of the exterior product, S(ℝ^{2n}) × S(ℝ^{2n}) ⟶ S(ℝ^{4n}).

In fact the same conclusion, with essentially the same proof, holds for any continuous linear operator B from S(ℝ^{n}) to S′(ℝ^{n}).

3.4. The ring Id +Ψ_{iso}^{−∞}(ℝ^{n})^{[S.Isotropic.ring]}

Recall that a bounded operator is said to have finite rank if its range is finite dimensional. If we consider a bounded operator B on L^{2}(ℝ^{n}) which is of finite rank then we may choose an orthonormal basis f_{j}, j = 1, …, N of the range BL^{2}(ℝ^{n}). The functionals u ⟼ ⟨Bu, f_{j}⟩ are continuous and so define non-vanishing elements g_{j} ∈ L^{2}(ℝ^{n}). It follows that the Schwartz kernel of B is

B = ∑_{j=1}^{N} f_{j}(x)\overline{g_{j}(y)}.^{[1.2.2000.311]} (3.22)

3.4. THE RING Id +Ψ_{iso}^{−∞}(ℝ^{n})^{[S.ISOT ROP IC.RING]} 69

If B ∈ Ψ_{iso}^{−∞}(ℝ^{n}) then the range must lie in S(ℝ^{n}) and similarly for the range of the adjoint, so the functions f_{j} and g_{j} are also in S(ℝ^{n}). Clearly the finite rank elements of Ψ_{iso}^{−∞}(ℝ^{n}) form an ideal in Ψ_{∞−iso}^{∞}(ℝ^{n}).

[1.2.2000.312]

Proposition 3.4. If A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) then Id +A has, as an operator on L^{2}(ℝ^{n}), finite dimensional null space and closed range which is the orthocomplement of the null space of Id +A^{∗}. There is an element B ∈ Ψ_{iso}^{−∞}(ℝ^{n}) such that

(3.23) (Id +A)(Id +B) = Id −Π_{1}, (Id +B)(Id +A) = Id −Π_{0}^{[1.2.2000.322]}

where Π_{0}, Π_{1} ∈ Ψ_{iso}^{−∞}(ℝ^{n}) are the orthogonal projections onto the null spaces of Id +A and Id +A^{∗} and furthermore, there is an element A′ ∈ Ψ_{iso}^{−∞}(ℝ^{n}) of rank equal to the dimension of the null space such that Id +A + sA′ is an invertible operator on L^{2}(ℝ^{n}) for all s ≠ 0.

Proof. Most of these properties are a direct consequence of the fact that A is compact as an operator on L^{2}(ℝ^{n}); nevertheless we give brief proofs.

We have shown, in Proposition 3.3 that each A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) is compact. It follows that

N_{0} = Nul(Id +A) ⊂ L^{2}(ℝ^{n})^{[1.2.2000.327]} (3.24)

has compact unit ball. Indeed the unit ball, B = {u Nul(Id +A)} satisfies B = A(B), since u = −Au on B. Thus B is closed and precompact. Any Hilbert space with a compact unit ball is finite dimensional, so Nul(Id +A) is finite dimensional.

Now, let R_{1} = Ran(Id +A) be the range of Id +A; we wish to show that this is a closed subspace of L^{2}(ℝ^{n}). Let f_{k} → f be a sequence in R_{1}, converging in L^{2}(ℝ^{n}). For each k there exists a unique u_{k} ∈ L^{2}(ℝ^{n}) with u_{k} ⊥ N_{0} and (Id +A)u_{k} = f_{k}. We wish to show that u_{k} → u. First we show that ∥u_{k}∥ is bounded. If not, then along a subsequent v_{j} = u_{k(j)}, ∥v_{j}∥ → ∞. Set w_{j} = v_{j}/∥v_{j}∥. Using the compactness of A, w_{j} = −Aw_{j} + f_{k(j)}/∥v_{j}∥ must have a convergent subsequence, w_{j} → w. Then (Id +A)w = 0 but w ⊥ N_{0} and ∥w∥ = 1 which are contradictory. Thus the sequence u_{k} is bounded in L^{2}(ℝ^{n}). Then again u_{k} = −Au_{k} + f_{k} has a convergent subsequence with limit u which is a solution of (Id +A)u = f; hence R_{1} is closed. The orthocomplement of the range of a bounded operator is always the null space of its adjoint, so R_{1} has a finite-dimensional complement N_{1} = Nul(Id +A^{∗}). The same argument applies to Id +A^{∗} so gives the orthogonal decompositions

L^{2}(ℝ^{n}) = N_{0} ⊕ R_{0}, N_{0} = Nul(Id +A), R_{0} = Ran(Id +A^{∗})
L^{2}(ℝ^{n}) = N_{1} ⊕ R_{1}, N_{1} = Nul(Id +A^{∗}), R_{1} = Ran(Id +A).

[1.2.2000.328] (3.25)

Thus we have shown that Id +A induces a continuous bijection Ã : R_{0} ⟶ R_{1}. From the closed graph theorem the inverse is a bounded operator B̃ : R_{1} ⟶ R_{0}. In this case continuity also follows from the argument above.^{1} Thus B̃ is the generalized inverse of Id +A in the sense that B = B̃ − Id satisfies (3.23). It only remains to show that B ∈ Ψ_{iso}^{−∞}(ℝ^{n}). This follows from (3.23), the identities in which show that

1We need to show that ∥B̃f∥ is bounded when f ∈ R_{1} and ∥f∥ = 1. This is just the boundedness of u ∈ R_{0} when f = (Id +A)u is bounded in R_{1}.

(3.26) B = −A − AB − Π_{1}, −B = A + BA + Π_{0}

⟹ B = −A + A^{2} + ABA − Π_{1} + AΠ_{0}.^{[1.2.2000.329]}

All terms here are in Ψ_{iso}^{−∞}(ℝ^{n}); for ABA this follows from Proposition 3.3.

It remains to show the existence of the finite rank perturbation A′. This is equivalent to the vanishing of the index, that is

(3.27) Ind(Id +A) = dim Nul(Id +A) − dim Nul(Id +A^{∗}) = 0.^{[1.2.2000.323]}

Indeed, let f_{j} and g_{j}, j = 1, …, N, be respective bases of the two finite dimensional spaces Nul(Id +A) and Nul(Id +A^{∗}). Then

A′ = ∑_{j=1}^{N} g_{j}(x)\overline{f_{j}(y)}^{[1.2.2000.324]} (3.28)

is an isomorphism of N_{0} onto N_{1} which vanishes on R_{0}. Thus Id +A + sA′ is the direct sum of Id +A as an operator from R_{0} to R_{1} and sA′ as an operator from N_{0} to N_{1}, invertible when s ≠ 0.

There is a very simple proof^{2} of the equality (3.27) if we use the trace functional discussed in Section 3.14 below; this however is logically suspect as we use (although not crucially) approximation by finite rank operators in the discussion of the trace and this in turn might appear to use the present result via the discussion of ellipticity and the harmonic oscillator. Even though this is not really the case we give a clearly independent, but less elegant proof.

Consider the one-parameter family of operators Id +tA, A ∈ Ψ_{iso}^{−∞}(ℝ^{n}). We shall see that the index, the difference in dimension between Nul(Id +tA) and Nul(Id +tA^{∗}) is locally constant. To see this it is enough to consider a general A near the point t = 1. Consider the pieces of A with respect to the decompositions L^{2}(ℝ^{n}) = N_{i} ⊕ R_{i}, i = 0, 1, of domain and range. Thus A is the sum of four terms which we write as a 2 × 2 matrix

A = [\begin{matrix}A_{00} & A_{01} \\ A_{10} & A_{11} \end{matrix}] .

Since Id +A has only one term in such a decomposition, Ã in the lower right, the solution of the equation (Id +tA)u = f can be written

(t − 1)A_{00}u_{0} + (t − 1)A_{01}u_{⊥} = f_{1}, (t − 1)A_{10}u_{0} + (A′ + (t − 1)A_{11})u_{⊥} = f_{⊥}^{[1.2.2000.325]} (3.29)

2Namely the trace of a finite rank projection, such as either Π_{0} or Π_{1}, is its rank, hence the dimension of the space onto which it projects. From the identity satisfied by the generalized inverse we see that

Ind(Id +A) = Tr(Π_{0}) − Tr(Π_{1}) = Tr ((Id +B)(Id +A) − (Id +A)(Id +B)) = Tr([B, A]) = 0 from the basic property of the trace.

3.6. FREDHOLM PROPERTY^{[S.F REDHOLM.P ROP ERT Y ]} 71

Since Ã is invertible, for t − 1 small enough the second equation can be solved uniquely for u_{⊥}. Inserted into the first equation this gives

(3.30) G(t)u_{0} = f_{1} + H(t)f_{⊥},

G(t) = (t − 1)A_{00} − (t − 1)^{2}A_{01}(A′ + (t − 1)A_{11})^{−1}A_{10},
H(t) = −(t − 1)A_{01}(A′ + (t − 1)A_{11})^{−1}.^{[1.2.2000.326]}

The null space is therefore isomorphic to the null space of G(t) and a complement to the range is isomorphic to a complement to the range of G(t). Since G(t) is a finite rank operator acting from N_{0} to N_{1} the difference of these dimension is constant in t, namely equal to dim N_{0} − dim N_{1}, near t = 1 where it is defined.

This argument can be applied to tA so the index is actually constant in t ∈ [0, 1] and since it certainly vanishes at t = 0 it vanishes for all t. In fact, as we shall note below, Id +tA is invertible outside a discrete set of t ∈ ℂ.

[1.2.2000.330]

Corollary 3.1. If Id +A, A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) is injective or surjective on L^{2}(ℝ^{n}), in particular if it is invertible as a bounded operator, then it has an inverse in the ring Id +Ψ_{iso}^{−∞}(ℝ^{n}).

[1.2.2000.333]

Corollary 3.2. If A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) then as an operator on S(ℝ^{n}) or S′(ℝ^{n}), Id +A is Fredholm in the sense that its null space is finite dimensional and its range is closed with a finite dimensional complement.

Proof. This follows from the existence of the generalized inverse of the form Id +B, B ∈ Ψ_{iso}^{−∞}(ℝ^{n}).

3.5. Exponential and logarithm

[1.2.2000.350]

Proposition 3.5. The exponential

(3.31) exp(A) = ∑_{j} \frac{1}{j!}A^{j} : Ψ_{iso}^{−∞}(ℝ^{n}) ⟶ Id +Ψ_{iso}^{−∞}(ℝ^{n})^{[1.2.2000.351]}

is a globally defined, entire, function with range containing a neighbourhood of the identity and with inverse on such a neighbourhood given by the analytic function

(3.32) log(Id +A) = ∑_{j} \frac{(−1)^{j}}{j} A^{j}, A ∈ Ψ_{iso}^{−∞}(ℝ^{n}), ∥A∥_{L^{2}} < 1^{[1.2.2000.352]}

3.6. Fredholm property^{[S.F redholm.property]}

An element A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) is said to be elliptic (of order m in the isotropic calculus) if its left-reduced symbol is elliptic in S_{∞}^{m}(ℝ^{2n}).

[21.2.1998.106]

Theorem 3.3. Each elliptic element A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) has a two-sided parametrix B ∈ Ψ_{∞−iso}^{−m}(ℝ^{n}) in the sense that

(3.33) A ∘ B − Id, B ∘ A − Id ∈ Ψ_{iso}^{−∞}(ℝ^{n})^{[1.2.2000.365]}

and it follows that any u ∈ S′(ℝ^{n}) satisfying Au ∈ S(ℝ^{n}) is an element of S(ℝ^{n}); if A ∈ Ψ_{iso}^{m}(ℝ^{n}) is elliptic then its parametrix is in Ψ_{iso}^{−m}(ℝ^{n}).

Proof. This is just the inductive argument used to prove Lemma 2.7. Nevertheless we repeat it here.

The existence of a right inverse for σ_{k}(A) means that the equation σ_{k}(A)c = d always has a solution c ∈ A_{j−k} for given d ∈ A_{j}, namely c = bd. This in turn means that given C_{j} ∈ Ψ_{∞−iso}^{j}(ℝ^{n}) there always exists B_{j} ∈ Ψ_{∞−iso}^{j−k}(ℝ^{n}) such that AB_{j} − D_{j} ∈ Ψ_{∞−iso}^{j−1}(ℝ^{n}). Choosing B_{0} ∈ Ψ_{∞−iso}^{−k}(ℝ^{n}) to have σ_{−k}(B_{0}) = b we can define C_{1} = Id −AB_{0} ∈ Ψ_{∞−iso}^{−1}(ℝ^{n}). Then, proceeding inductively we may assume that B_{j} for j < l have been chosen such that A(B_{0} + ⋯ + B_{l−1}) − Id = −C_{l} ∈ Ψ_{∞−iso}^{−l}(ℝ^{n}). Then using the solvability we may choose B_{l} so that AB_{l} − C_{l} = −C_{l+1} ∈ Ψ_{∞−iso}^{−l−1}(ℝ^{n}) which completes the induction, since A(B_{0} + ⋯ + B_{l}) − Id = AB_{l} − C_{l} = −C_{l+1}. Finally by the asymptotic completeness we may choose B ∼ B_{0} + B_{1} + … which is a right parametrix.

The argument showing the existence of a left parametrix for a left-elliptic operator is completely analogous.

Combining the earlier symbolic discussion and these analytic results we can see that elliptic operators in these calculi are Fredholm.

[21.3.1998.169]

Proposition 3.6. If A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}) (resp. A ∈ Ψ_{∞−sc}^{l,m}(ℝ^{n})) is elliptic then it has a generalized inverse B ∈ Ψ_{∞−iso}^{−m}(ℝ^{n}) (resp. B ∈ Ψ_{∞−sc}^{−l,−m}(ℝ^{n})) satisfying

(3.34) AB − Id = Π_{1}, BA − Id = Π_{0} ∈ Ψ_{iso}^{−∞}(ℝ^{n})^{[21.3.1998.170]}

where Π_{1} and Π_{0} are the finite rank orthogonal (in L^{2}(ℝ^{n})) projections onto the null spaces of A^{∗} and A.

Proof. In the case of an elliptic isotropic operator or order m we know that it has a parametrix B′ ∈ Ψ_{iso}^{−m}(ℝ^{n}) modulo Ψ_{iso}^{−∞}(ℝ^{n}). Thus

AB′ = Id −E_{R}, E_{R} ∈ Ψ_{iso}^{−∞}(ℝ^{n}),
B′A = Id −E_{L}, E_{L} ∈ Ψ_{iso}^{−∞}(ℝ^{n}).

Using Proposition 3.4 it follows that the null space of A is contained in the null space of B′A = Id −E_{L}, hence is finite dimensional. Similarly, the range of A contains the range of AB′ = Id −E_{R} so is closed with a finite codimensional complement. Defining B as the linear map which vanishes on Nul(A^{∗}), and inverts A on Ran(A) with values in Ran(A^{∗}) = Nul(A)^{⊥} gives (3.34). Furthermore these identities show that B ∈ Ψ_{∞−iso}^{−m}(ℝ^{n}) since applying B′ gives

(3.35) B − E_{L}B = B′AB = B′ − B′Π_{1}, B − BE_{R} = BAB′ = B′ − Π_{0}B′ ⟹

B = B′ − B′Π_{1} + E_{L}B′ + E_{L}BE_{R} − E_{L}Π_{0}B′ ∈ Ψ_{∞−iso}^{−m}(ℝ^{n}).^{[1.2.2000.331]}

The proof in the scattering case is essentially the same.

[1.2.2000.332]

Corollary 3.3. If A ∈ Ψ_{iso}^{m}(ℝ^{n}) is elliptic then its generalized inverse lies in Ψ_{iso}^{−m}(ℝ^{n}) and similarly if A ∈ Ψ_{sc}^{l,m}(ℝ^{n}) is elliptic then its generalized inverse lies in Ψ_{sc}^{−l,−m}(ℝ^{n}).

3.7. THE HARMONIC OSCILLATOR^{[S.HARMONIC.OSCILLAT OR]}73

3.7. The harmonic oscillator^{[S.Harmonic.oscillator]}

The harmonic oscillator is the differential operator on ℝ^{n}

H = ∑_{j=1}^{n}(D_{j}^{2} + x_{j}^{2}) = Δ + |x|^{2} .

This is an elliptic element of Ψ_{iso}^{2}(ℝ^{n}). The main immediate interest is in the spectral decomposition of H. The ellipticity of H − λ, λ ∈ ℂ, shows that

(H − λ)u = 0, u ∈ S′(ℝ^{n}) ⟹ u ∈ S(ℝ^{n}) .^{[eq:HO.1]} (3.36)

Since H is (formally) self-adjoint, i.e., H^{∗} = H, there are no non-trivial tempered solutions of (H − λ)u = 0, λ ∉ ℝ. Indeed if (H − λ)u = 0,

0 = ⟨Hu, u⟩ − ⟨u, Hu⟩ = (λ − \overline{λ})⟨u, u⟩ ⟹ u = 0.^{[eq:HO.2]} (3.37)

As we shall see below in more generality, the spectrum of H is a discrete subset of ℝ. In this case we can compute it explicitly.

The direct computation of eigenvalues and eigenfunctions is based on the properties of the creation and annihilation operators

C_{j} = D_{j} + ix_{j}, C_{j}^{∗} = A_{j} = D_{j} − ix_{j}, j = 1, …, n.^{[eq:HO.3]} (3.38)

These satisfy the elementary identities

\begin{matrix}= [A_{j}, A_{k}] = 0, [A_{j}, C_{k}] = 2δ_{jk}, j, k = 1, …, n \\ H = ∑_{j=1}^{n} C_{j}A_{j} + n, [C_{j}, H] = −2C_{j}, [A_{j}, H] = 2A_{j} . \end{matrix}^{[eq:HO.4]} (3.39)

Now, if λ is an eigenvalue, Hu = λu, then

\begin{matrix}H(C_{j}u) = C_{j}(Hu + 2u) = (λ + 2)C_{j}u, \\ H(A_{j}u) = A_{j}(Hu − 2u) = (λ − 2)A_{j}u. \end{matrix}^{[eq:HO.5]} (3.40)

[prop:HO.6]

Proposition 3.7. The eigenvalues of H are

σ(H) = {n, n + 2, n + 4, … }.^{[eq:HO.7]} (3.41)

Proof. We already know that eigenvalues must be real and from the decomposition of H in (3.40) it follows that, for u ∈ S(ℝ^{n}),

⟨Hu, u⟩ = ∑_{j} ∥A_{j}u∥^{2} + n∥u∥^{2} .^{[eq:HO.9]} (3.42)

Thus if λ ∈ σ(H) is an eigenvalue then λ ≥ n.

By direct computation we see that n is an eigenvalue with a 1-dimensional eigenspace. Indeed, from (3.42), Hu = nu iff A_{j}u = 0 for j = 1, …, n. In each variable separately

A_{j}u(x_{j}) = 0 ⇔ u(x_{j}) = c_{1} exp (−\frac{x_{j}^{2}}{2} ) .

Thus the only tempered solutions of A_{j}u = 0, i = 1, …, n are the constant multiples of

u_{0} = exp (−\frac{|x|^{2}}{2} ) ,^{[eq:HO.10]} (3.43)

which is often called the ground state.

Now, if λ is an eigenvalue with eigenfunction u ∈ S(ℝ^{n}) it follows from (3.40) that λ − 2 is an eigenvalue with eigenfunction A_{j}u. Since all the A_{j}u cannot vanish unless u is the ground state, it follows that the eigenvalues are contained in the set in (3.41). We can use the same argument to show that if u is an eigenfunction with eigenvalue λ then C_{j}u is an eigenfunction with eigenvalue λ + 2. Moreover, C_{j}u ≡ 0 would imply u ≡ 0 since C_{j}v = 0 has no non-trivial tempered solutions, the solution in each variable being exp(x_{j}^{2}/2).

Using the creation operators we can parameterize the eigenspaces quite explicitly.

[prop:HO.11]

Proposition 3.8. For each k ∈ ℕ_{0} there is an isomorphism

(3.44) {Polynomials, homogeneous of degree k on ℝ^{n}} ∋ p

⟼ p(C) exp (−\frac{|x|^{2}}{2} ) ∈ E_{k}^{[eq:HO.12]}

where E_{k} is the eigenspace of H with eigenvalue n + 2k.

Proof. Notice that the C_{j}, j = 1, …, n are commuting operators, so p(C) is well-defined. By iteration from (3.40),

HC^{α}u_{0} = C^{α}(H + 2|α|)u_{0} = (n + 2|α|)C^{α}u_{0} .^{[eq:HO.13]} (3.45)

Thus (3.44) is a linear map into the eigenspace as indicated.

To see that (3.44) is an isomorphism consider the action of the annihilation operators. Again from (3.40)

|β| = |α| ⟹ A^{β}C^{α}u_{0} = \begin{cases}0 & β ≠ α \\ 2^{|α|}α!u_{0} & β = α. \end{cases} ^{[eq:HO.14]} (3.46)

This allows us to recover the coefficients of p from p(C)u_{0}, so (3.44) is injective. Conversely if v ∈ E_{k} ⊂ S(ℝ^{n}) is orthogonal to all the C^{α}u_{0} then

⟨A^{α}v, u_{0}⟩ = ⟨v, C^{α}u_{0}⟩ = 0 ∀ |α| = k .^{[eq:HO.15]} (3.47)

From (3.40), the A^{α}v are all eigenfunctions of H with eigenvalue n, so (3.47) implies that A^{α}v = 0 for all |α| = k. Proceeding inductively in k we see that A^{α′}A_{j}v = 0 for all |α′| = k − 1 and A_{j}v ∈ E_{k−1} implies A_{j}v = 0, j = 1, …, n. Since v ∈ E_{k}, k > 0, this implies v = 0 so Proposition 3.8 is proved.

Thus H has eigenspaces as described in (3.44). The same argument shows that for any integer p, positive or negative, the eigenvalues of H^{p} are precisely (n + 2k)^{p} with the same eigenspaces E_{k}. For p < 0, H^{p} is a compact operator on L^{2}(ℝ^{n}); this is obvious for large negative p. For example, if p ≤ −n − 1 then

x_{i}^{β}D_{j}^{α}H ∈ Ψ_{iso}^{0}(ℝ^{n}), |α| ≤ n + 1, |β| ≤ n + 1^{[eq:HO.16]} (3.48)

are all bounded on L^{2}. If S ⊂ L^{2}(ℝ^{n}) is bounded this implies that H^{−n−1}(S) is bounded in ⟨x⟩^{n+1}C_{∞}^{1}(ℝ^{n}), so compact in ⟨x⟩^{n}C_{∞}^{0}(ℝ^{n}) and hence in L^{2}(ℝ^{n}). It is a general fact that for compact self-adjoint operators, such as H^{−n−2}, the eigenfunctions span L^{2}(ℝ^{n}). We give a brief proof of this for the sake of ‘completeness’.

3.7. THE HARMONIC OSCILLATOR^{[S.HARMONIC.OSCILLAT OR]}75

[lem:HO.17]

Lemma 3.4. The eigenfunction of H, u_{α} = π^{− \frac{n}{4}} (2^{|α|}α!)^{−1/2}C^{α}u_{0} form an orthonormal basis of L^{2}(ℝ^{n}).

Proof. Let V ⊂ L^{2}(ℝ^{n}) be the closed subspace consisting of the orthocomplements of all the u_{α}’s. Certainly H^{−n−2} acts on it as a compact self-adjoint operator. Since we have found all the eigenvalues of H, and hence of H^{−n−1}, it has no eigenvalue in V. We wish to conclude that V = {0}. Set

τ = ∥H^{−n−1}∥_{V} = sup{∥H^{−n−1}φ∥; φ ∈ V, ∥φ∥ = 1} .

Then there is a weakly convergent sequence φ_{j} ⇀ φ, ∥φ_{j}∥ = 1, so ∥φ∥ ≤ 1, with ∥H^{−n−1}φ_{j}∥ → τ. The compactness of H^{−n−2} allows a subsequence to be chosen such that H^{−n−1}φ_{j} → ψ in L^{2}(ℝ^{n}). So, by the continuity of H^{−n−1}, H^{−n−1}φ = ψ and ∥H^{−n−1}φ∥ = τ, ∥φ∥ = 1. If φ′ ∈ V, φ′ ⊥ φ, ∥φ′∥ = 1 then

τ^{2} ≥ ∥H^{−n−2} (\frac{φ + tφ′}{\sqrt{1 + t^{2}}} ) ∥^{2} = τ^{2} + 2t⟨H^{−2n−2}φ, φ′⟩ + 0(t^{2})

⟹ ⟨H^{−2n−2}φ, φ′⟩ = 0 ⟹ H^{−2n−2}φ = τ^{2}φ .

This contradicts the fact that H^{−2n−2} has no eigenvalues in V, so V = {0} and the eigenbasis is complete.

Thus, if u ∈ L^{2}(ℝ^{n}) then

u = ∑_{α} c_{α}u_{α}, c_{α} = ⟨u, u_{α}⟩ .^{[eq:HO.18]} (3.49)

[lem:HO.19]

Lemma 3.5. If u ∈ S(ℝ^{n}) the convergence in (3.49) is rapid, i.e., |C_{α}| ≤ C_{N}(1 + |α|)^{−N} for all N and the series converges in S(ℝ^{n}).

Proof. Since u ∈ S(ℝ^{n}) implies H^{N}u ∈ L^{2}(ℝ^{n}) we see that
C_{N} ≥ |⟨H^{N}u, u_{α}⟩| = |⟨u, H^{N}u_{α}⟩| = (n + 2|α|)^{N}|c_{α}| ∀ α .

Furthermore, 2ix_{j} = C_{j} − A_{j} and 2D_{j} = C_{j} + A_{j} so the polynomial derivatives of the u_{α} can be estimated (using the Sobolev embedding theorem) by polynomials in α; this implies that the series converges in S(ℝ^{n}).

[1.2.2000.407]

Corollary 3.4. Finite rank elements are dense in Ψ_{iso}^{−∞}(ℝ^{n}) in the topology of S(ℝ^{2n}).

Proof. Consider the approximation (3.49) to the kernel A of an element of Ψ_{iso}^{−∞}(ℝ^{n}) as an element of S(ℝ^{2n}). In this case the ground state is

U_{0} = exp (−\frac{|x|^{2}}{2} − \frac{|y|^{2}}{2} ) = exp (−\frac{|x|^{2}}{2} ) exp (−\frac{|y|^{2}}{2} )

and so has rank one as an operator. The higher eigenfunctions

C^{α}U_{0} = Q_{α}(x, y)U_{0}

are products of U_{0} and a polynomial, so are also of finite rank.

3.8. L^{2} boundedness and compactness

Recall that Ψ_{∞−iso}^{0}(ℝ^{n}) ⊂ Ψ_{∞}^{0}(ℝ^{n}) so, by Proposition 2.6, these operators are bounded on L^{2}(ℝ^{n}). Using the same argument the bound on the L^{2} norm can be related to the norm of the principal symbol and an N × N matrix.

[1.2.2000.357]

Proposition 3.9. If A ∈ Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}) has principal symbol
a = σ_{L}(A)|_{𝕊^{2n−1}} ∈ C^{∞}(𝕊^{2n−1}; M(N, ℂ)

then

(3.50)  inf_{E∈Ψ_{iso}^{−∞}(ℝ^{n};ℂ^{N})} ∥A + E∥_{B(L^{2}(ℝ^{n};ℂ^{N})} ≤ sup_{p∈𝕊^{2n−1}} ∥a(p)∥.^{[1.2.2000.358]}

Proof. It suffices to prove (3.50) for all single operators A ∈ Ψ_{iso}^{0}(ℝ^{n}). Indeed if j_{v}(z) = zv is the linear map from ℂ to ℂ^{N} defined by v ∈ ℂ^{N} then

∥A∥_{B(L^{2}(ℝ;ℂ^{N}))} = sup_{{v,w∈ℂ^{N};∥v∥=∥w∥=1}} ∥j_{w}^{∗}Aj_{v}∥_{B(L^{2}(ℝ))}.^{[1.2.2000.363]} (3.51)

Since the symbol of j_{w}^{∗}Aj_{v} is just j_{w}^{∗}σ(A)j_{v}, (3.50) follows from the corresponding equality for a single operator:

(3.52)  inf_{E∈Ψ_{iso}^{−∞}(ℝ^{n})} ∥A + E∥_{B(L^{2}(ℝ^{n})} ≤ sup_{p∈𝕊^{2n−1}} |a(p)|, a = σ_{L}(A)|_{𝕊^{2n−1}}.^{[1.2.2000.364]}

The construction of the approximate square-root of C −A^{∗}A in Proposition 2.7 only depends on the existence of a positive smooth square-root for C − |a|^{2}, so can be carried out for any

C > sup_{p∈𝕊^{2n−1}} |a(p)|^{2}.^{[1.2.2000.359]} (3.53)

Thus we conclude that with such a value of C

∥Au∥^{2} ≤ C∥u∥^{2} + ∥⟨Gu, u⟩| ∀ u ∈ L^{2}(ℝ^{n}),

where G ∈ Ψ_{iso}^{−∞}(ℝ^{n}). Since G is an isotropic smoothing operator, for any δ > 0 there is a finite dimensional subspace W ⊂ S(ℝ^{n}) such that

∥⟨Gu, u⟩∥ ≤ δ∥u∥^{2} ∀ u ∈ W^{⊥}.^{[1.2.2000.360]} (3.54)

Thus if we replace A by A(Id −Π_{W}) = A + E where E is a (finite rank) smoothing operator we see that

∥(A + E)u∥^{2} ≤ (C + δ)∥Gu∥^{2} ∀ u ∈ L^{2}(ℝ^{n}) ⟹ ∥(A + E)∥ ≤ (C + δ)^{\frac{1}{2}} .^{[1.2.2000.361]}

This proves half of the desired estimate (3.51), namely

(3.55)  inf_{E∈Ψ_{iso}^{−∞}(ℝ^{n})} ∥A + E∥_{B(L^{2}(ℝ^{n})} ≤ sup_{p∈𝕊^{2n−1}} |a(p)|.^{[1.2.2000.362]}

To prove the opposite inequality, leading to (3.55), it is enough to arrive at a contradiction by supposing to the contrary that there is some A ∈ Ψ_{iso}^{0}(ℝ^{n}) satisfying the strict inequality

∥A∥_{B(L^{2}(ℝ^{n})} < sup_{p∈𝕊^{2n−1}} |a(p)|.

From this it follows that we may choose c > 0 such that c = |a(p)|^{2} for some p ∈ 𝕊^{2n−1} and yet A′ = A^{∗}A−c has a bounded inverse, B. By making an arbitrariy small perturbation of the full symbol of A′ we may assume that it vanishes identically

3.9. SOBOLEV SPACES 77

near p. By (3.55) we may choose G ∈ Ψ_{iso}^{0}(ℝ^{n}) with arbitrariy small L^{2} such that Ã = A′ + B has left symbol rapidly vanishing near p. When the norm of the perturbation is small enough, Ã will still be invertible, with inverse B̃ ∈ B(L^{2}(ℝ^{n}). Now choose an element G ∈ Ψ_{iso}^{0}(ℝ^{n}) with left symbol supported sufficiently near p, so that G ∘ Ã ∈ Ψ_{iso}^{−∞}(ℝ^{n}) but yet the principal symbol of G should not vanish at p. Thus

G = G ∘ Ã ∘ B̃ : L^{2}(ℝ^{n}) ⟶ S(ℝ^{n}), G∗ = G = B̃^{∗} ∘ Ã^{∗} ∘ G^{∗} : S′(ℝ^{n}) ⟶ L^{2}(ℝ^{n}).

It follows that G^{∗}G : S′(ℝ^{n}) ⟶ S(ℝ^{n}) is an isotropic smoothing operator. This is the desired contrdiction, since G, and hence G^{∗}G, may be chosen to have nonvanishing principal symbol at p. Thus we have proved (3.55) and hence the Proposition.

It is then easy to characterize the compact operators amongst the polyhomogeneous isotropic operators as those of negative.

[1.2.2000.366]

Lemma 3.6. If A ∈ Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}) then A is compact as an operator on L^{2}(ℝ^{n}; ℂ^{N}) if and only if it has negative order.

Proof. The necessity of vanishing of the principal symbol for a compact operator follows from Proposition 3.9 and the sufficiency follows from the density of Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}) in Ψ_{iso}^{−1}(ℝ^{n}; ℂ^{N}) in the topology of Ψ_{∞−iso}^{− \frac{1}{2}}(ℝ^{n}; ℂ^{N}) and hence in the topology of bounded operators. Thus such an operator is the norm limit of compact operators so itself is compact.

Also as a consequence of Proposition 3.9 we can see the necessity of the assumption of ellipticity in Proposition 3.6.

[1.2.2000.367]

Corollary 3.5. If A ∈ Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}) then A is Fredholm as an operator on L^{2}(ℝ^{n}; ℂ^{N}) if and only if it is elliptic.

3.9. Sobolev spaces

There is a natural collection of Sobolev spaces associated to the isotropic calculus, and another associated to the scattering calculus.

Note that a differential operator P (x, D_{x}) on ℝ^{n} is an isotropic pseudodifferential operator if and only if its coefficients are polynomials. The fundamental symmetry between coefficients and differentiation suggest that the isotropic Sobolev spaces of non-negative integral order be defined by

H_{iso}^{k}(ℝ^{n}) = {u ∈ L^{2}(ℝ^{n}); x^{α}D_{x}^{β}u ∈ L^{2}(ℝ^{n}) if |α| + |β| ≤ k}, k ∈ ℕ.^{[1.2.2000.337]} (3.56)

The norms

∥u∥_{k,iso}^{2} = ∑_{|α|+|β|≤k} ∫_{ℝ^{n}} |x^{α}D_{x}^{β}u|^{2} dx^{[1.2.2000.339]} (3.57)

turn these into Hilbert spaces. For negative integral orders we identify the isotropic Sobolev spaces with the duals of these spaces

H_{iso}^{k}(ℝ^{n}) = (H_{iso}^{−k}(ℝ^{n}))′ ↪ S′(ℝ^{n}), k ∈ −ℕ.^{[1.2.2000.340]} (3.58)

The (continuous) injection into tempered distributions here arises from the density of the image of the inclusion S(ℝ^{n}) ⟶ H_{iso}^{k}(ℝ^{n}).

[1.2.2000.338]

Lemma 3.7. For any k ∈ ℤ,

(3.59) H_{iso}^{k}(ℝ^{n}) = {u ∈ S′(ℝ^{n}); Au ∈ L^{2}(ℝ^{n}) ∀ A ∈ Ψ_{iso}^{−k}}

= {u ∈ S′(ℝ^{n}); ∃ A ∈ Ψ_{iso}^{−k} elliptic and such that Au ∈ L^{2}(ℝ^{n})}^{[1.2.2000.341]}

and S(ℝ^{n}) ↪ H_{iso}^{k}(ℝ^{n}) is dense for each k ∈ ℤ.

Proof. ^{3} For k ∈ ℕ, the functions x^{α}ξ^{β} for |α| + |β| = k are ‘collectively elliptic’ in the sense that

q_{k}(x, ξ) = ∑_{|α|+|β|=k}(x^{α}ξ^{β})^{2} ≥ c(|x|^{2} + |ξ|^{2})^{k}, c > 0.^{[1.2.2000.344]} (3.60)

Thus Q_{k} = ∑_{|α|+|β|≤k}(D^{β}x^{α}x^{α}D^{β}) ∈ Ψ_{iso}^{2k}(ℝ^{n}), which has principal reduced symbol

q_{k}, has a left parameterix A_{k} ∈ Ψ_{iso}^{−2k}(ℝ^{n}). This gives the identity

(3.61) ∑_{|α|+|β|≤k} R_{α,β}x^{α}D^{β} = A_{k}Q_{k} = Id +E, where

R_{α,β} = A_{k}D^{β}x^{α} ∈ Ψ_{iso}^{−2k+|α|+|β|}(ℝ^{n}), E ∈ Ψ_{iso}^{−∞}(ℝ^{n}).^{[1.2.2000.345]}

Thus if A ∈ Ψ_{iso}^{k}(ℝ^{n})

Au = −AEu + ∑_{|α|+|β|≤k} AR_{α,β}x^{α}D^{β}u.

If u ∈ H_{iso}^{k}(ℝ^{n}) then by definition x^{α}D^{β}u ∈ L^{2}(ℝ^{n}). By the boundedness of operators of order 0 on L^{2}, all terms on the right are in L^{2}(ℝ^{n}) and we have shown the inclusion of H_{iso}^{k}(ℝ^{n}) in the first space space on the right in (3.59). The converse is immediate, so this proves the first equality in (3.59) for k > 0. Certainly the third space in (3.59) contains in the second. The existence of elliptic parametrix B for the ellipic operator A proves the converse since any isotropic pseudodifferential operator of order A′ of order k can be effectively factorized as

A′ = A′(BA + E) = B′A + E′, B′ ∈ Ψ_{∞−iso}^{0}(ℝ^{n}), E′ ∈ Ψ_{iso}^{−∞}(ℝ^{n}).

Thus, Au ∈ L^{2}(ℝ^{n}) implies that A′u ∈ L^{2}(ℝ^{n}).

It also follows from second identification that S(ℝ^{n}) is dense in H_{loc}^{k}(ℝ^{n}). Thus, if Au ∈ L^{2}(ℝ^{n}) and we choose f_{n} ∈ S(ℝ^{n}) with f_{n} → Au in L^{2}(ℝ^{n}) then, with B a parametrix for A, u′_{n} = Bf_{n} → BAu = u + Eu. Thus u_{n} = u′_{n} − Eu ∈ S(ℝ^{n}) → u in L^{2}(ℝ^{n}) and Au_{n} → u in L^{2}(ℝ^{n}) proving the density.

The Riesz representation theorem shows that vS′(ℝ^{n}) is in the dual space, H_{iso}^{−k}(ℝ^{n}), if and only if there exists v′ ∈ H_{iso}^{k}(ℝ^{n}) such that

(3.62) v(u) = ⟨u, v′⟩_{k,iso} = ⟨u, Q_{2k}v′⟩_{L^{2}}, ∀ u ∈ S(ℝ^{n}) ↪ H_{iso}^{k}(ℝ^{n})

with Q_{2k} = ∑_{|α|+|β|≤k} D^{β}x^{2α}D^{β}.^{[1.2.2000.343]}

3This is an essentially microlocal proof.

3.10. THE GROUP G_{iso}^{−∞}(ℝ^{n})^{[SEC:T G]} 79

This shows that Q_{2k} is an isomorphism of H_{iso}^{k}(ℝ^{n}) onto H_{iso}^{−k}(ℝ^{n}) as subspaces of S′(ℝ^{n}). Notice that Q_{2k} ∈ Ψ_{iso}^{2k}(ℝ^{n}) is elliptic, self-adjoint and invertible, since it is strictly positive. This now gives the same identification (3.59) for k < 0.

The case k = 0 follows directly from the L^{2} boundedness of operators of order 0 so the proof is complete.

In view of this identification we define the isotropic Sobolev spaces or any real order the same way

H_{iso}^{s}(ℝ^{n}) = {u ∈ S′(ℝ^{n}); Au ∈ L^{2}(ℝ^{n}) ∀ A ∈ Ψ_{iso}^{−s}} , s ∈ ℝ.^{[1.2.2000.347]} (3.63)

These are Hilbertable spaces, with the Hilbert norm being given by ∥Au∥_{L^{2}(ℝ^{n}} for any A ∈ Ψ_{iso}^{s}(ℝ^{n}) which is elliptic and invertible.

[1.2.2000.346]

Proposition 3.10. Any element A ∈ Ψ_{∞−iso}^{m}(ℝ^{n}), m ∈ ℝ, defines a bounded linear operator

A : H_{iso}^{s}(ℝ^{n}) ⟶ H_{iso}^{s−m}(ℝ^{n}), ∀ s ∈ ℝ.^{[1.2.2000.348]} (3.64)

This operator is Fredholm if and only if A is elliptic. For any s ∈ ℝ, S(ℝ^{n}) ↪ H_{iso}^{s}(ℝ^{n}) is dense and H_{iso}^{−s}(ℝ^{n}) may be identified as the dual of H_{iso}^{s}(ℝ^{n}) with respect to the continuous extension of the L^{2} pairing.

Proof. A straightforward application of the calculus, with the exception of the necessity of ellipticity for an isotropic pseudodifferential operator to be Fredholm. This is discussed in the problems beginning at Problem 3.10.

3.10. The group G_{iso}^{−∞}(ℝ^{n})^{[sec:T G]}

By definition, G_{iso}^{−∞}(ℝ^{n}) is the set (if you want to be concrete you can think of them as operators on L^{2}(ℝ^{n})) of invertible elements of the ring Id +Ψ_{iso}^{−∞}(ℝ^{n}). If we identify this topologically with Ψ_{iso}^{−∞}(ℝ^{n}) then, as follows from Corollary 3.1, G_{iso}^{−∞}(ℝ^{n}) is open. We will think of it as an infinite-dimensional manifold modeled, of course, on the linear space Ψ_{iso}^{−∞}(ℝ^{n}) ≃ S(ℝ^{2n}). Since I have no desire to get too deeply into the general theory of such Fréchet manifolds I will keep the discussion as elementary as possible.

The dual space of S(ℝ^{p}) is S′(ℝ^{p}). If we want to think of S(ℝ^{p}) as a manifold we need to consider smooth functions and forms on it. In the finite-dimensional case, the exterior bundles are the antisymmetric parts of the tensor powers of the dual. Since we are in infinite dimensions the tensor power needs to be completed and the usual choice is the ‘projective’ tensor product. In our case this is something quite simple namely the k-fold completed tensor power of S′(ℝ^{p}) is just S′(ℝ^{kp}). Thus we set

(3.65) Λ^{k}S(ℝ^{p}) = {u ∈ S′(ℝ^{kp}); for any permutation

e, u(x_{e(1)}, … x_{e(h)}) = sgn(e)u(x_{1}, … x_{k})} .^{[T G.1]}

In view of this it is enough for us to consider smooth functions on open sets F ⊂ S(ℝ^{p}) with values in S′(ℝ^{p}) for general p. Thus

v : F ⟶ S′(ℝ^{p}), F ⊂ S(ℝ^{n}) open^{[T G.2]} (3.66) is continuously differentiable on F if there exists a continuous map

v′ : F ⟶ S′(ℝ^{n+p}) and each u ∈ F has a neighbourhood U such that for each N ∃ M with

∥v(u + u′) − v(u) − v′(u; u′)∥_{N} ≤ C∥u′∥_{M}^{2}, ∀ u, u + u′ ∈ U.

Then, as usual we define smoothness as infinite differentiability by iterating this definition. The smoothness of v in this sense certainly implies that if f : X ⟶ S(ℝ^{n}) is smooth then v ∘ F is smooth.

Thus we define the notion of a smooth form on F ⊂ S(ℝ^{n}), an open set, as a smooth map

α : F → Λ^{k}S(ℝ^{p}) ⊂ S′(ℝ^{kp}) .^{[T G.3]} (3.67)

In particular we know what smooth forms are on G_{iso}^{−∞}(ℝ^{n}).

The de Rham differential acts on forms as usual. If v : F → ℂ is a function then its differential at f ∈ F is dv : F ⟶ S′(ℝ^{n}) = Λ^{1}S(ℝ^{n}), just the derivative. As in the finite-dimensional case d extends to forms by enforcing the condition that dv = 0 for constant forms and the identity distribution over exterior products

d(α ∧ β) = (dα) ∧ β + (−1)^{deg α}α ∧ dβ .^{[T G.5]} (3.68)

3.11. Representations^{[S.Representations]}

In §1.9 the compactification of Euclidean space to a ball, or half-sphere, is described. We make the following definition, recalling that ρ ∈ C^{∞}(𝕊^{n,+}) is a boundary defining function.

[28.2.1998.153]

Definition 3.2. The space of of ‘Laurent functions’ on the half-sphere is

(3.69) L(𝕊^{n,+}) = ⋃_{k∈ℕ_{0}} ρ^{−k}C^{∞}(𝕊^{n,+}),

ρ^{−k}C^{∞}(𝕊^{n,+}) = {u ∈ C^{∞}(int(𝕊^{n,+})); ρ^{k}u ∈ C^{∞}(𝕊^{n,+}).^{[28.2.1998.154]}

More generally if m ∈ ℝ we denote by ρ^{m}C^{∞}(𝕊^{n,+}) the space of functions which can be written as products u = ρ^{m}v, with v ∈ C^{∞}(𝕊^{n,+}); again it can be identified with a subspace of the space of C^{∞} functions on the open half-sphere.

[28.2.1998.155]

Proposition 3.11. The compactification map (1.94) extends from (1.96) to give, for each m ∈ ℝ, an identification of ρ^{−m}C^{∞}(𝕊^{n,+}) and S_{cl}^{m}(ℝ^{n}).

Thus, the fact that the Ψ_{iso,cl}^{ℤ}(ℝ^{n}) form an order-filtered ∗-algebra means that ρ^{ℤ}C^{∞}(𝕊^{2n,+}) has a non-commutative product defined on it, with C^{∞}(𝕊^{2n,+}) a subalgebra, using the left symbol isomorphism, followed by compactification.

3.12. Symplectic invariance of the isotropic product^{[S.Symplectic.invariance]}

The composition law for the isotropic calculus, and in particular for it smoothing part, is derived from its identification as a subalgebra of the (weighted) spaces of pseudodifferential operator on ℝ^{n}. There is a much more invariant formulation of the product which puts into evidence more of the invariance properties.

3.12. SYMPLECTIC INVARIANCE OF THE ISOTROPIC PRODUCT^{[S.SY MP LECT IC.INV ARIANCE]} 81

Let W be a real symplectic vector space. Thus, W is a vector space equipped with a real, antisymmetic and non-degenerate bilinear form

(3.70) ω : W × W ⟶ ℝ, ω(w_{1}, w_{2}) + ω(w_{2}, w_{1}) = 0 ∀ w_{1}, w_{2} ∈ W,

ω(w_{1}, w) = 0 ∀ w ∈ W ⟹ w_{1} = 0.^{[23.3.1998.174]}

A Lagrangian subspace of W is a vector space V ⊂ W such that ω vanishes when restricted to V and such that 2 dim V = dim W.

[23.3.1998.175]

Lemma 3.8. Every symplectic vector space has a Lagrangian subspace and for any choice of Lagrangian subspace U_{1} there is a second Lagrangian subspace U_{2} such that W = U_{1} ⊕ U_{2} is a Lagrangian decomposition.

Proof. First we show that there is a Lagrangian subspace. If dim W > 0 then the antisymmetry of ω shows that any 1-dimensional vector subspace is isotropic, that is ω vanishes when restricted to it. Let V be a maximal isotropic subspace, that is an isotropic subspace of maximal dimension amongst isotropic subspaces. Let U be a complement to V in W. Then

ω : V × U ⟶ ℝ^{[23.3.1998.176]} (3.71)

is a non-degenerate pairing. Indeed u ∈ U and ω(v, u) = 0 for all v ∈ V then V + ℝ{u} is also isotropic, so u = 0 by the assumed maximality. Similarly if v ∈ V and ω(v, u) = 0 for all u ∈ U then, recalling that ω vanishes on V, ω(v, w) = 0 for all w ∈ W so v = 0. The pairing (3.71) therefore identifies U with V′, the dual of V. In particular dim w = 2 dim V.

Now, choose any Lagrangian subspace U_{1}. We proceed to show that there is a complementary Lagrangian subspace. Certainly there is a 1-dimensional subspace which does not meet U_{1}. Let V be an isotropic subspace which does not meet U_{1} and is of maximal dimension amongst such subspaces. Suppose that dim V < dim U_{1}. Choose w ∈ W with w ∉ V ⊕ U_{1}. Then V ∋ v ⟶ ω(w, v) is a linear functional on U_{1}. Since U_{1} can be completed to a complement, any such linear functional can be written ω(u_{1}, v) for some u_{1} ∈ U_{1}. It follows that ω(w − u_{1}, v) = 0 for all v ∈ V. Thus V ⊕ ℝ{w − u_{1}} a non-trivial isotropic extension of V, contradicting the assumed maximality. Thus V = U_{2} is a complement of U_{1}.

Given such a Lagrangian decomposition of the symplectic vector space W, let X_{1}, … X_{n} be a basis for the dual of U_{1}, and let Ξ_{1}, …, Ξ_{n} be the dual basis, of U_{1} itself. The pairing (3.71) with U = U_{1} and V = U_{2} identifies U_{2} = U′_{1} so the Ξ_{i} can also be regarded as a basis of the dual of U_{2}. Thus X_{1} … X_{n}, Ξ_{1}, …, Ξ_{n} gives a basis of W′ = U′_{1} ⊕ U′_{2}. The symplectic form can then be written

ω(w_{1}, w_{2}) = ∑_{i=1}^{n}(Ξ_{i}(w_{1})X_{i}(w_{2}) − Ξ_{i}(w_{2})X_{i}(w_{1})).^{[23.3.1998.178]} (3.72)

This is the Darboux form of ω. If the X_{i}, Ξ_{i} are thought of as linear functions x_{i}, ξ_{i} on W now considered as a manifold then these are Darboux coordinates in which(3.72) becomes

ω = ∑_{i=1}^{n} dξ_{i} ∧ dx_{i}.^{[23.3.1998.179]} (3.73)

The symplectic form ω defines a volume form on W, namely the n-fold wedge product ω^{n}. In Darboux coordinates this is just, up to sign, the Lebesgue form dξdx.

[23.3.1998.180]

Proposition 3.12. On any symplectic vector space, W, the bilinear map on S(W),

a#b(w) = (2π)^{−2n} ∫_{W^{2}} e^{iω(w_{1},w_{2})}a(w + w_{1})b(w + w_{2})ω^{n}(w_{1})ω^{2}(w_{2}), dim W = 2n^{[23.3.1998.181]} (3.74)

defines an associative product isomorphic to the composition of Ψ_{iso}^{−∞}(U_{1}) for any Lagrangian decomposition W = U_{1} ⊕ U_{2}.

[23.3.1998.182]

Corollary 3.6. Extended by continuity in the symbol space (3.74) defines a filtered product on S^{∞}(W) which is isomorphic to the isotropic algebra on ℝ^{2n} and is invariant under symplectic linear transformation of W.

Proof. Written in the form (3.74) the symplectic invariance is immediate. That is, if F is a linear transformation of W which preserves the symplectic form, ω(F w_{1}, F w_{2}) = ω(w_{1}, w_{2}) then

F^{∗}(a#b) = (F^{∗}a)#(F^{∗}b) ∀ a, b ∈ S(W).^{[23.3.1998.183]} (3.75)

The same result holds for general symbols once the continuity is established.

Let us start from the Weyl quatization of the isotropic algebra. As usual for computations we may assume that the amplitudes are of order −∞. Thus, A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) may be written

Au(x) = ∫ A(x, y)u(y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(\frac{1}{2}(x + y), ξ)u(y)dydξ.^{[23.3.1998.184]} (3.76)

Both the kernel A(x, y) and the amplitude a(x, ξ) are elements of S(ℝ^{2n}). The relationship (3.76) and its inverse may be written

\begin{matrix}A(s + \frac{t}{2}, s − \frac{t}{2}) = (2π)^{−n} ∫ e^{it·ξ}a(s, ξ)dξ, \\ a(x, ξ) = ∫ e^{−it·ξ}A(x + \frac{t}{2}, x − \frac{t}{2})dt. \end{matrix}^{[23.3.1998.185]} (3.77)

If A has Weyl symbol a and B has Weyl symbol b let c be the Weyl symbol of the composite A ∘ B. Using (3.77) and (3.76)

c(s, ζ) = ∫ e^{−it·ζ}A(s + \frac{t}{2}, z)B(z, s − \frac{t}{2})dt

= (2π)^{−2n} ∫ ∫ ∫ dt dz dξ dηe^{iΦ}a(\frac{s}{2} + \frac{t}{4} + \frac{z}{2}, ξ)a(\frac{z}{2} + \frac{s}{2} − \frac{t}{4}, η)

where Φ = −t · ζ + (s + \frac{t}{2} − z) · ξ + (z − s + \frac{t}{2}) · η.

3.13. COMPLEX ORDER^{[S.COMP LEX.ORDER]} 83

Changing variables of integration to X = \frac{z}{2} + \frac{t}{4} − \frac{s}{2}, Y = \frac{z}{2} − \frac{t}{4} − \frac{s}{2}, Ξ = ξ − ζ and H = η − ζ this becomes

c(s, ζ) = (2π)^{−2n}4^{n} ∫ ∫ ∫ dY dX dΞ dH

e^{2i(X·H−Y ·Ξ)}a(X + s, Ξ + ζ)a(Y + s, H + ζ).

This reduces to (3.74), written out in Darboux coordinates, after the change of variable H′ = 2H, Ξ′ = 2Ξ and ζ′ = 2ζ. Thus the precise isomorphism with the product in Weyl form is given by

A(x, y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a_{ω}(\frac{1}{2}(x + y), 2ξ)u(y)dydξ^{[23.3.1998.186]} (3.78)

so that composition of kernels reduces to (3.74).

Discuss metaplectic group here.

3.13. Complex order^{[S.Complex.order]}

The identification of polyhomogeneous symbols of order zero on ℝ^{2n} with the smooth functions on the radial compactification allows us to define the isotropic operators of a given complex order z ∈ ℂ. Namely, we use the left quantization map to identify

(3.79) Ψ_{iso}^{z}(ℝ^{n}) = ρ^{−z}C^{∞}(𝕊^{2n,1}) ⊂ Ψ_{∞−iso}^{ℜz}(ℝ^{n}).^{[1.2.2000.334]}

Here, ρ ∈ C^{∞}(𝕊^{2n,1}) is a boundary defining function. Any other boundary defining function is of the form aρ with 0 < a ∈ C^{∞}(𝕊^{2n,1}). It follows that the definition is independent of the choice of ρ since a^{z} ∈ C^{∞}(𝕊^{2n,1}) for any z ∈ ℤ.

In fact it is even more useful to consider holomorphic families. Thus if Ω ⊂ ℂ is an open set and h : Ω ⟶ ℂ is holomorphic then we may consider holomorphic families of order h as elements of

(3.80) Ψ_{iso}^{h(z)}(ℝ^{2n}) = {A : Ω ⟶ Ψ_{∞−iso}^{∞}(ℝ^{2n});

Ω ∋ z ⟼ ρ^{h(z)}A(z) ∈ C^{∞}(𝕊^{2n,1}) is holomorphic.}^{[1.2.2000.335]}

Note that a map from Ω ⊂ ℂ into C^{∞}(𝕊^{2n1,}) is said to be holomorphic it is defines an element of C^{∞}(Ω × 𝕊^{2n,1}) which satisies the Cauchy-Riemann equation in the first variable.

[1.2.2000.336]

Proposition 3.13. If h and g are holomorphic functions on an open set Ω ⊂ ℂ and A(z), B(z) are holomorphic familes of isotropic operators of orders h(z) and g(z) then the composite family A(z) ∘ B(z) is holomorphic of order h(z) + g(z).

Proof. It suffices to consider an arbitrary open subset Ω′ ⊂ Ω with compact closure inside Ω. Then h and g have bounded real parts, so A(z), B(z) ∈ Ψ_{∞−iso}^{M}(ℝ^{2n}) for z ∈ Ω′ for some fixed M. It follows that the composite A(z)∘B(z) ∈ Ψ_{∞−iso}^{2M}(ℝ^{2n}). The symbol is given by the usual formula. Furthermore

3.14. Traces on the residual algebra^{[S.T races.residual]}

The algebras we are studying are topological algebras, so it makes sense to consider continuous linear functionals on them. The most important of these is the trace. To remind you what it is we consider first its properties for matrix algebras.

Let M(N; ℂ) denote the algebra of N × N complex matrices. We can simply define

Tr : M(N; ℂ) → ℂ, Tr(A) = ∑_{i=1}^{N} A_{ii}^{[eq:1]} (3.81)

as the sum of the diagonal entries. The fundamental property of this functional is that

Tr([A, B]) = 0 ∀ A, B ∈ M(N; ℂ) .^{[eq:2]} (3.82)

To check this it is only necessary to write down the definition of the composition in the algebra. Thus

(AB)_{ij} = ∑_{k=1}^{N} A_{ik}B_{kj} .

It follows that

Tr(AB) = ∑_{i=1}^{N}(AB)_{ii} = ∑_{i,k=1}^{N} A_{ik}B_{ki}

= ∑_{k=1}^{N} ∑_{i=1}^{N} B_{ki}A_{ik} = ∑_{k=1}^{N}(BA)_{kk} = Tr(BA)

which is just (3.82).

Of course any multiple of Tr has the same property (3.82) but the normalization condition

Tr(Id) = N^{[eq:3]} (3.83)

distinguishes it from its multiples. In fact (3.82) and (3.83) together distinguish Tr ∈ M(N; ℂ)′ as a point in the N^{2} dimensional linear space which is the dual of M(N; ℂ).

[lem:trace]

Lemma 3.9. If F : M(N; ℂ) → ℂ is a linear functional satisfying (3.82) and B ∈ M(N; ℂ) is any matrix such that F (B) ≠ 0 then F (A) = \frac{F (B)}{Tr(B)} Tr(A).

Proof. Consider the basis of M(N; ℂ) given by the elementary matrices E_{jk}, where E_{jk} has jk-th entry 1 and all others zero. Thus

E_{jk}E_{pq} = δ_{kp}E_{jq}.

If j ≠ k it follows that

E_{jj}E_{jk} = E_{jk}, E_{jk}E_{jj} = 0.

Thus

F ([E_{jj}, E_{jk}]) = F (E_{jk}) = 0 if j ≠ k.

3.14. TRACES ON THE RESIDUAL ALGEBRA^{[S.T RACES.RESIDUAL]}85

On the other hand, for any i and j

E_{ji}E_{ij} = E_{jj}, E_{ij}E_{ji} = E_{ii}

so

F (E_{jj}) = F (E_{11}) ∀ j.

Since the E_{jk} are a basis,

F (A) = F ( ∑_{j,k=1}^{N} A_{ij}E_{ij})

= ∑_{j,l=1}^{N} A_{jj}F (E_{ij})

= F (E_{11}) ∑_{j=1}^{N} A_{jj} = F (E_{11}) Tr(A).

This proves the lemma.

For the isotropic smoothing algebra we have a similar result.

[isotropictrace]

Proposition 3.14. If F : Ψ_{iso}^{−∞}(ℝ^{n}) ≃ S(ℝ^{2n}) ⟶ ℂ is a continuous linear functional satisfying

F ([A, B]) = 0 ∀ A, B ∈ Ψ_{iso}^{−∞}(ℝ^{n})^{[eq:4]} (3.84)

then F ([A, B]) = 0 for all A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) and B ∈ Ψ_{iso}^{∞}(ℝ^{n}) and F is a constant multiple of the functional

Tr(A) = ∫_{ℝ^{n}} A(x, x) dx.^{[eq:5]} (3.85)

Proof. Recall that Ψ_{iso}^{−∞}(ℝ^{n}) ⊂ Ψ_{iso}^{∞}(ℝ^{n}) is an ideal so A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) and B ∈ Ψ_{iso}^{∞}(ℝ^{n}) implies that AB, BA ∈ Ψ_{iso}^{−∞}(ℝ^{n}) and it follows that the equality F (AB) = F (BA), or F ([A, B]) = 0, is meaningful. To see that it holds we just use the continuity of F. We know that if B ∈ Ψ_{iso}^{∞}(ℝ^{n}) then there is a sequence B_{n} → B in the topology of Ψ_{iso}^{m}(ℝ^{n}) for some m. Since this implies AB_{n} → AB, B_{n}A → BA in Ψ_{iso}^{−∞}(ℝ^{n}) we see that

F ([A, B]) = lim_{n→∞} F ([A, B_{n}]) = 0 .

We use this identity to prove (3.85). Take B = x_{j} or D_{j}, j = 1, …, n. Thus for any A ∈ Ψ_{iso}^{−∞}(ℝ^{n})

F ([A, x_{j}]) = F ([A, D_{j}]) = 0 .

Now consider F as a distribution acting on the kernel A ∈ S(ℝ^{2n}). Since the kernel of [A, x_{j}] is A(x, y)(y_{j} − x_{j}) and the kernel of (A, D_{j}) is −(D_{y_{j}} + D_{x_{j}})A(x, y) we conclude that, as an element of S′(ℝ^{2n}), F satisfies

(x_{j} − y_{j})F (x, y) = 0, (D_{x_{j}} + D_{y_{j}})F (x, y) = 0.

If we make the linear change of variables to p_{i} = \frac{x_{i}+y_{i}}{2} , q_{i} = x_{i} − y_{i} and set F̃(p, q) = F (x, y) these conditions become

D_{q_{i}} F̃ = 0, p_{i}F̃ = 0, i = 1, …, N.

As we know from Lemmas 1.2 and 1.3, this implies that F̃ = cδ(p) so

F (x, y) = cδ(x − y)

as a distribution. Clearly δ(x − y) gives the functional Tr defined by (3.85), so the proposition is proved.

We still need to justify the use of the same notation, Tr, for these two functionals. However, if L ⊂ S(ℝ^{n}) is any finite dimensional subspace we may choose an orthonal basis φ_{i} ∈ L, i = 1, …, l,

∫_{ℝ^{n}} |φ_{i}(x)|^{2} dx = 0, ∫_{ℝ^{n}} φ_{i}(x)\overline{φ_{j}}(x) dx = 0, i ≠ j .

Then if a_{ij} is an l × l matrix,

A = ∑_{i,j=1}^{ℓ} a_{ij}φ_{i}(x)\overline{φ_{j}(y)} ∈ Ψ_{iso}^{−∞}(ℝ^{n}) .

From (3.85) we see that

Tr(A) = ∑_{ij} a_{ij} Tr(φ_{i}φ̄_{j})

= ∑_{ij} a_{ij} ∫_{ℝ^{n}} φ_{i}(x)\overline{φ_{j}}(x) dx

= ∑_{i=1}^{n} a_{ii} = Tr(a) .

Thus the two notions of trace coincide. In any case this already follows, up to a constant, from the uniqueness in Lemma 3.9.

3.15. Fredholm determinant

For N × N matrices, the linear space of which we denote M(N; ℂ), the determinant is a multiplicative polynomial map

(3.86) det : M(N; ℂ) ⟶ ℂ, det(AB) = det(A) det(B), det(Id) = 1.^{[1.2.2000.404]}

It is not quite determined by these conditions, since det(A)^{k} also satisfies then. The fundamental property of the determinant is that it defines the group of invertible elements

GL(N, ℂ) = {A ∈ M(N; ℂ); det(A) ≠ 0}.^{[1.2.2000.405]} (3.87)

A reminder of a direct definition is given in Problem 3.7.

The Fredholm determinant is an extension of this definition to a function on the ring Id +Ψ_{iso}^{−∞}(ℝ^{n}), or even further to Id +Ψ_{∞−iso}^{−2n−ϵ}(ℝ^{n}) for ϵ > 0. This can be done in several ways using the density of finite rank operators, as shown in Corollary 3.4. We proceed by generalizing the formula relating the determinant to the trace. Thus, for any smooth curve with values in GL(N; ℂ) for any N,

(3.88) \frac{d}{ds} det(A_{s}) = det(A_{s}) tr(A_{s}^{−1} \frac{A_{s}}{ds}).^{[1.2.2000.408]}

In particular if (3.86) is augmented by the normalization condition

\frac{d}{ds} det(Id +sA)|_{s=0} = tr(A) ∀ A ∈ M(N; ℂ)^{[iml.1]} (3.89)

3.15. FREDHOLM DETERMINANT 87

then it is determined.

A branch of the logarithm can be introduced along any curve, smoothly in the parameter, and then (3.88) can be rewritten

(3.90) d log det(A) = tr(A^{−1}dA).^{[1.2.2000.409]}

Here GL(N; ℂ) is regarded as a subset of the linear space M(N; ℂ) and dA is the canonical identification, at the point A, of the tangent space to M(N, ℂ) with M(N, ℂ) itself. This just arises from the fact that M(N, ℂ) is a linear space. Thus dA(\frac{d}{ds}(A + sB)|_{s=0} = B. This allows the expression on the right in (3.90) to be interpreted as a smooth 1-form on the manifold GL(N; ℂ). Note that it is independent of the local choice of logarithm.

To define the Fredholm determinant we shall extend the 1-form

α = Tr(A^{−1}dA)^{[iml.2]} (3.91)

to the group G_{iso}^{−∞}(ℝ^{n}) ↪ Id +Ψ_{iso}^{−∞}(ℝ^{n}). Here dA has essentially the same meaning as before, given that Id is fixed. Thus at any point A = Id +B ∈ Id +Ψ_{iso}^{−∞}(ℝ^{n}) it is the identification of the tangent space with Ψ_{iso}^{−∞}(ℝ^{n}) using the linear structure:

dA(\frac{d}{ds}(Id +B + sE)|_{s=0}) = E, E ∈ Ψ_{iso}^{−∞}(ℝ^{n}).^{[iml.3]}

Since dA takes values in Ψ_{iso}^{−∞}(ℝ^{n}), the trace functional in (3.91) is well defined.

The 1-form α is closed. In the finite-dimensional case this follows from (3.90). For (3.91) we can compute directly. Since d(dA) = 0, essentially by definition, and

dA^{−1} = −A^{−1}dAA^{−1}^{[iml.4]} (3.92)

we see that

dα = − Tr(A^{−1}(dA)A^{−1}(dA)) = 0.^{[iml.5]} (3.93)

Here we have used the trace identity, and the antisymmetry of the implicit wedge product in (3.93), to conlcude that dα = 0. For a more detailed discussion of this point see Problem 3.8.

From the fact that dα = 0 we can be confident that there is, locally near any point of G_{iso}^{−∞}(ℝ^{n}), a function f such that df = α; then we will define the Fredholm determinant by det_{Fr}(A) = exp(f). To define det_{Fr} globally we need to see that this is well defined.

[iml.9]

Lemma 3.10. For any smooth closed curve γ : 𝕊^{1} ⟶ G_{iso}^{−∞}(ℝ^{n}) the integral

∫_{γ} α = ∫_{𝕊^{1}} γ^{∗}α ∈ 2πiℤ.^{[iml.10]} (3.94)

That is, α defines an integral cohomology class, [\frac{α}{2πi}] ∈ H^{1}(G_{iso}^{−∞}(ℝ^{n}); ℤ).

Proof. This is where we use the approximability by finite rank operators. If π_{N} is the orthogonal projection onto the span of the eigenspaces of the smallest N eigenvalues of the harmonic oscillator then we know from Section 3.7 that π_{N}Eπ_{N} → E in Ψ_{iso}^{−∞}(ℝ^{n}) for any element. In fact it follows that for the smooth curve that γ(s) = Id +E(s) and E_{N}(s) = π_{N}E(s)π_{N} converges uniformly with all s derivatives. Thus, for some N_{0} and all N > N_{0}, Id +E_{N}(s) is a smooth curve in G_{iso}^{−∞}(ℝ^{n}) and hence γ_{N}(s) = Id_{N} +E_{N}(s) is a smooth curve in GL(N; ℂ). Clearly

∫_{γ_{N}} α ⟶ ∫_{γ} α as N → ∞,^{[iml.11]} (3.95)

and for finite N it follows from the identity of the trace with the matrix trace (see Section 3.14) that ∫_{N} γ_{N}^{∗}α is the variation of arg log det(γ_{N}) around the curve. This gives (3.94).

Now, once we have (3.94) and the connectedness of G_{iso}^{−∞}(ℝ^{n}) we may define

(3.96) det_{Fr}(A) = exp(∫_{γ} α), γ : [0, 1] ⟶ G_{iso}^{−∞}(ℝ^{n}), γ(0) = Id, γ(1) = A.^{[iml.12]}

Indeed, Lemma 3.10 shows that this is independent of the path chosen from the identity to A. Notice that the connectedness of G_{iso}^{−∞}(ℝ^{n}) follows from the connectedness of the GL(N, ℂ) and the density argument above.

The same arguments and results apply to G_{∞−iso}^{−2n−ϵ}(ℝ^{n}) using the fact that the trace functional extends continuously to Ψ_{∞−iso}^{−2n−ϵ}(ℝ^{n}) for any ϵ > 0.

[iml.13]

Proposition 3.15. The Fredholm determinant, defined by (3.96) on G_{iso}^{−∞}(ℝ^{n}) (or G_{iso}^{−2n−ϵ}(ℝ^{n}) for ϵ > 0) and to be zero on the complement in Id +Ψ_{iso}^{−∞}(ℝ^{n}) (or Id +Ψ_{iso}^{−2n−ϵ}(ℝ^{n})) is an entire function satisfying

(3.97) det_{Fr}(AB) = det_{Fr}(A) det_{Fr}(B), A, B ∈ Id +Ψ_{iso}^{−∞}(ℝ^{n})

(or Id +Ψ_{iso}^{−2n−ϵ}(ℝ^{n})), det_{Fr}(Id) = 1.^{[iml.14]}

Proof. We start with the multiplicative property of det_{Fr} on G_{iso}^{−∞}(ℝ^{n}). Thus is γ_{1}(s) is a smooth curve from Id to A_{1} and γ_{2}(s) is a smooth curve from Id to A_{2} then γ(s) = γ_{1}(s)γ_{2}(s) is a smooth curve from Id to A_{1}A_{2}. Consider the differential on this curve. Since

\frac{d(A_{1}(s)A_{2}(s))}{ds} = \frac{dA_{1}(s)}{ds}A_{2}(s) + A_{1}(s)\frac{dA_{2}(s)}{ds}

the 1-form becomes

γ^{∗}(s)α(s) = Tr(A_{2}(s)^{−1} \frac{dA_{2}(s)}{ds}) + Tr(A_{2}(s)^{−1}A_{1}(s)^{−1} \frac{dA_{2}(s)}{ds}A_{2}(s)).^{[iml.15]} (3.98)

In the second term on the right we can use the trace identity, since Tr(GA) = Tr(AG) if G ∈ Ψ_{iso}^{ℤ}(ℝ^{n}) and A ∈ Ψ_{iso}^{−∞}(ℝ^{n}). Thus (3.98) becomes

γ^{∗}(s)α(s) = γ_{1}^{∗}α + γ_{2}^{∗}α.

Inserting this into the definition of det_{Fr} gives (3.97) when both factors are in G_{iso}^{−∞}(ℝ^{n}). Of course if either factor is not invertible, then so is the product and hence both det_{Fr}(AB) and at least one of det_{Fr}(A) and det_{Fr}(B) vanishes. Thus (3.97) holds in general when det_{Fr} is extended to be zero on the non-invertible elements.

Thus it remains to establish the smoothness. That det_{Fr}(A) is smooth in any real parameters in which A ∈ G_{iso}^{−∞}(ℝ^{n}) depends, or indeed is holomorphic in holomorphic parameters, follows from the definition since α clearly depends smoothly, or holomorphically, on parameters. In fact the same follows if holomorphy is examined as a function of E, A = Id +E, for E ∈ Ψ_{iso}^{−∞}(ℝ^{n}). Thus it is only smoothness across the non-invertibles that is at issue. To prove this we use the multiplicativity just established.

If A = Id +E is not invertible, E ∈ Ψ_{iso}^{−∞}(ℝ^{n}) then it has a generalized inverse Id +E′ as in Proposition 3.6. Since A has index zero, we may actually replace E′ by E′+E′′, where E′′ is an invertible linear map from the orthocomplement of the range of A to its null space. Then Id +E′+E′′ ∈ G_{iso}^{−∞}(ℝ^{n}) and (Id +E′+E′′)A = Id −Π_{0}. To prove the smoothness of det_{Fr} on a neighbourhood of A it is enough to prove the smoothness on a neighbourhood of Id −Π_{0} since Id +E′+E′′ maps a neighbourhood of the first to a neighbourhood of the second and det_{Fr} is multiplicative. Thus consider det_{Fr} on a set Id −Π_{0} + E where E is near 0 in Ψ_{iso}^{−∞}(ℝ^{n}), in particular we may assume that Id +E ∈ G_{iso}^{−∞}(ℝ^{n}). Thus

det_{Fr}(Id +E − Π_{0}) = det(Id +E) det(Id −Π_{0} + (G_{E} − Id)Π_{0})^{[iml.16]}

were G_{E} = (Id +E)^{−1} depends holomorphically on E. Thus it suffices to prove the smoothness of det_{Fr}(Id −Π_{0} + HΠ_{0}) where H ∈ Ψ_{iso}^{−∞}(ℝ^{n})

Consider the deformation H_{s} = Π_{0}HΠ_{0} +s(Id −Π_{0})HΠ_{0}, s ∈ [0, 1]. If Id −Π_{0} + H_{s} is invertible for one value of s it is invertible for all, since its range is always the range of Id −Π_{0} plus the range of Π_{0}HΠ_{0}. It follows that det_{Fr}(Id −Π_{0} + H_{s}) is smooth in s; in fact it is constant. If the family is not invertible this follows immediately and if it is invertible then

\frac{d det_{Fr}(Id −Π_{0} + H_{s})}{ds}

= det_{Fr}(Id −Π_{0} + H_{s}) Tr ((Id −Π_{0} + H_{s})^{−1}(Id −P i_{0})HΠ_{0})) = 0

since the argument of the trace is finite rank and off-diagonal with respect to the decomposition by Π_{0}.

Thus finally it is enough to consider the smoothness of det_{Fr}(Id −Π_{0} +Π_{0}HΠ_{0}) as a function of H ∈ Ψ_{iso}^{−∞}(ℝ^{n}). Since this is just det(Π_{0}HΠ_{0}), interpreted as a finite rank map on the range of Π_{0} the result follows from the finite dimensional case.

[iml.18]

Lemma 3.11. If A ∈ G_{iso}^{ℤ}(ℝ^{n}) and B ∈ G_{iso}^{−∞}(ℝ^{n}) then ABA^{−1} ∈ G_{iso}^{−∞}(ℝ^{n}) and

(3.99) det_{Fr}(ABA^{−1}) = det_{Fr}(B).^{[iml.19]}

Proof. If ABA^{−1} is not invertible then neither is B so both sides of (3.99) vanish. Thus we may assume that B = Id +E is invertible and let Id +E(s) be a smooth curve in G_{iso}^{−∞}(ℝ^{n}) connecting it to the identity. Consider the function

(3.100) det_{Fr}(A(Id +E(s))A^{−1}) = det_{Fr}(Id +AE(s)A^{−1}).^{[iml.20]}

This is certainly smooth and non-vanishing and its logarithm has derivative

Tr(A(Id +E(s))^{−1}A^{−1} \frac{d (A(Id +E(s)A^{−1})}{ds})

= Tr(A(Id +E(s))^{−1} \frac{E(s)}{ds}A^{−1}) = Tr((Id +E(s))^{−1} \frac{E(s)}{ds}).

This is also the derivative of the logarithm of det_{Fr}(Id +E(s)) so the result follows.

3.16. Fredholm alternative^{[S.F redholm.alternative]}

Since we have shown that det_{Fr} : Id +Ψ_{iso}^{−∞}(ℝ^{n}) ⟶ ℂ is an entire function, we see that G_{iso}^{−∞}(ℝ^{n}) is the complement of a (singular) holomorphic hypersurface, namely the surface {Id +E; det_{Fr}(Id +E) = 0}. This has the following consequence, which is sometimes call the ‘Fredholm alternative’ and also part of ‘analytic Fredholm theory’.

[iml.21]

Lemma 3.12. If Ω ⊂ ℂ is an open, connected set and A : Ω ⟶ Ψ_{iso}^{−∞}(ℝ^{n}) is a holomorphic function then either Id +A(z) is invertible of all but a discrete subset of Ω and (Id +A(z)) is meromorphic on Ω with all residues of finite rank, or else it is invertible at no point of Ω.

Proof. Of course the point here is that det_{Fr}(Id +A(z)) is a holomorphic function on Ω. Thus, either det_{Fr}(A(z)) = 0 is a discrete set, D ⊂ Ω or else det_{Fr}(Id +A(z)) ≡ 0 on Ω; this uses the connectedness of Ω. Since this corresponds exactly to the invertibility of Id +A(z) the main part of the lemma is proved. It remains only to show that, in the former case, (Id +A(z))^{−1} is meromorphic. Thus consider a point p ∈ D. Thus the claim is that near p

(3.101) (Id +A(z))^{−1} = Id +E(z) + ∑_{j=1}^{N} z^{−j}E_{j}, E_{j} ∈ Ψ_{iso}^{−∞}(ℝ^{n}) of finite rank^{[iml.22]}

and where E(z) is locally holomorphic with values in Ψ_{iso}^{−∞}(ℝ^{n}).

If N is sufficiently large and Π_{N} is the projection onto the first N eigenspaces of the harmonic oscillator then B(z) = Id +E(z) − Π_{N}E(z)Π_{N} is invertible near p with the inverse being of the form Id +F (z) with F (z) locally holomorphic. Now

(Id +F (z))(Id +E(z)) = Id +(Id +F (z))Π_{N}E(z)Π_{N}

= (Id −Π_{N}) + Π_{N}M(z)Π_{N} + (Id −Π_{N})M′(z)Π_{N}.

It follows that this is invertible if and only if M(z) is invertible as a matrix on the range of Π_{N}. Since it must be invertible near, but not at, p, its inverse is a meromorphic matrix K(z). It follows that the inverse of the product above can be written

(3.102) Id −Π_{N} + Π_{N}K(z)Π_{N} − (Id −Π_{N})M′(z)Π_{N}K(z)Π_{N}.^{[iml.23]}

This is meromorphic and has finite rank residues, so it follows that the same is true of A(z)^{−1}.

This result for the smoothing operators, which really follows from the corresponding result for matrices, gives a similar result for a holomorphic family of elliptic operators.

[iml.24]

Proposition 3.16. Suppose Ω ⊂ ℂ is open and connected, h(z) is holomorphic on Ω and A(z) ∈ Ψ_{iso}^{h(z)}(ℝ^{n}) is a holomorphic and elliptic family. Then if A(z) is invertible at one point it is invertible for all but a discrete set and its inverse is locally of the form B(z) + H(z) where B(z) ∈ Ψ_{iso}^{−h(z)}(ℝ^{n}) is a locally holomorphic family and H(z) is meromorphic with values in Ψ_{iso}^{−∞}(ℝ^{n}) and has all residues of finite rank.

3.18. RESIDUE TRACE^{[S.RESIDUE.T RACE]} 91

Proof. By the results of Section 3.13 we can choose, at least locally near any point, a holomorphic parametrix B(z) for the family in Ψ_{iso}^{−h(z)}(ℝ^{n}). Suppose p ∈ Ω is a point at which A(z) is invertible. Then the parametrix differs from the inverse by a smoothing operator, so modifying B(z) by a constant smoothing operator if follows that B(z)A(z) = Id +E(z) is a holomorphic family in Id +Ψ_{iso}^{−∞}(ℝ^{n}) with E(p) = 0. It follows from the Fredholm alternative that Id +E(z) is invertible with holomorphic inverse near p. Thus, A(z) is invertible for z in an open set around p; the set at which it is invertible is therefore open. Let Z ⊂ Ω be the closed set at which A(z) is not invertible. If p is a boundary point of Z then it follows, using the notation above for a parametrix of A(z) near p, that Id +E(z) is invertible near p and hence p is an isolated point of non-invertibility. Thus all boundary points of Z are isolated so, given that Ω is connected, it is discrete. The local structure of the inverse follows from this discussion.

3.17. Resolvent and spectrum

For elliptic operators with index zero and a point of invertibility.

3.18. Residue trace^{[S.Residue.trace]}

We have shown, in Proposition 3.14, the existence of a unique trace functional on the residual algebra Ψ_{iso}^{−∞}(ℝ^{n}). We now follow ideas originating with Seeley, [11], and developed by Guillemin [5], [6] and Wodzicki [14], [13] to investigate the traces on the full algebra Ψ_{iso}^{ℤ}(ℝ^{n}) of polyhomogeneous operators of integral order. We will prove the existence of a trace but defer until later the proof of its uniqueness.

Observe that for A ∈ Ψ_{iso}^{−∞}(ℝ^{n}) the kernel can be written

A(x, y) = (2π)^{−n} ∫ e^{i(x−y)ξ}a_{L}(x, ξ) dξ

and hence the trace, from (3.85), becomes

Tr(A) = (2π)^{−n} ∫_{ℝ^{2n}} a_{L}(x, ξ) dx dξ ,^{[F eb.17.2000.eq:1]} (3.103)

just the integral of the left-reduced symbol. In fact this is true for any amplitude (of order −∞) representing A :

A = (2π)^{−n} ∫ e^{i(x−y)}a(x, y, ξ) dξ ⟹ Tr(A) = (2π)^{−n} ∫_{ℝ^{2n}} a(x, x, ξ) dx dξ .^{[F eb.17.2000.eq:2]} (3.104)

The integral in (3.103) extends by continuity to a_{L} ∈ S_{∞}^{m}(ℝ^{2n}) provided m < −2n. Thus, as a functional,

(3.105) Tr : Ψ_{∞,iso}^{−2n−ϵ}(ℝ^{n}) → ℂ, for any ϵ > 0.^{[F eb.17.2000.eq:3]}

To extend it further we need somehow to regularize the resultant divergent integral in (3.103) (and to pay the price in terms of properties). One elegant way to do this is to use a holomorphic family as discussed in Section 3.13. Notice that we are passing from the algebra-with-bounds in (3.105) to polyhomogeneous operators.

Lemma 3.13. If A(z) ∈ Ψ_{iso}^{z}(ℝ^{n}) is a holomorphic family then f(z) = Tr(A(z)), defined by (3.104) when ℜ(z) < −2n, extends to a meromorphic function of z with at most simple poles on the divisor

{−2n, −2n + 1, …, −1, 0, 1, … } ⊂ ℂ .

Proof. We know that A(z) ∈ Ψ_{iso}^{z}(ℝ^{n}) is a holomorphic family if and only if its left-reduced symbol is of the form

σ_{L}(A(z)) = (1 + |x|^{2} + |z|^{2})^{z/2}a(z; x, ξ)

where a(z; x, y) is an entire function with values in S_{phg}^{0}(ℝ^{n}). For ℜz < −2n the trace of A(z) is

f(z) = (2π)^{−n} ∫_{ℝ^{2n}}(1 + |x|^{2} + |ξ|^{2})^{z/2}a(z; x, ξ) dx dξ .

Consider the part of this integral on the ball

f_{1}(z) = (2π)^{−n} ∫_{|x|^{2}+|ξ|^{2}≤1}(1 + |x|^{2} + |ξ|^{2})^{z/2}a(z, x, y) dx dξ .

This is clearly an entire function of z, since the integrand is entire and the domain compact.

To analyze the remaining part f_{2}(z) = f(z) − f_{1}(z) let us introduce polar coordinates

r = (|x|^{2} + |ξ|^{2})^{1/2}, θ = \frac{(x, ξ)}{r} ∈ 𝕊^{2n−1} .

The integral, convergent in ℜz < −2n, becomes

f_{2}(z) = (2π)^{−n} ∫_{1}^{∞} ∫_{𝕊^{2n−1}}(1 + r^{2})^{z/2}ã(z; r, θ) dθr^{2n−1} dr.

Let us now pass to the radical compactification of ℝ^{2n} or more prosaically, introduce t = 1/r ∈ [0, 1] as variable of integration, so

f_{2}(z) = (2π)^{−n} ∫_{0}^{1} ∫_{𝕊^{2n−1}} t^{−z}(1 + t^{2})^{z/2}ã(z; \frac{1}{t}, θ) dθt^{−2n} \frac{dt}{t} .

Now the definition of S_{phg}^{0}(ℝ^{2n}) reduces to the statement that

b(z; t, θ) = (1 + t^{2})^{z/2}ã(z; \frac{1}{t}, θ) ∈ C^{∞}(ℂ × [0, 1] × 𝕊^{2n−1})^{[F eb.17.2000.eq:4]} (3.106)

is holomorphic in z.

If we replace b by its Taylor series at t = 0 to high order,

b(z; t, θ) = ∑_{j=0}^{k} \frac{t^{j}}{j!} b_{j}(z; θ) + t^{k+1}b_{(k)}(z; t, θ) ,^{[F eb.17.2000.eq:5]} (3.107)

where b_{(k)}(z; t, θ) has the same regularity (3.106), then f_{2}(z) is decomposed as

f_{2}(z) = (2π)^{−n} ∑_{j=0}^{k} ∫_{0}^{1} ∫_{𝕊^{2n−1}} \frac{t^{−z+j}}{j!} b_{j}(z; θ)t^{−2n} \frac{dt}{t} + f_{2}^{(k)}(z).^{[F eb.17.2000.eq:6]} (3.108)

The presence of this factor t^{k} in the remainder in (3.107) shows that f_{2}^{(k)}(z) is holomorphic in ℜz < −2n + k. On the other hand the individual terms in the sum in (3.108) can be computed (for ℜz < −2n) as

(2π)^{−n} [\frac{t^{−z+j−2n}}{(−z + j − 2n)}]_{0}^{1} ∫_{𝕊^{2n−1}} b_{j}(z, θ)\frac{dθ}{j!}

= (2π)^{−n} \frac{1}{(z − j + 2n)} ∫_{𝕊^{2n−1}} b_{j}(z, θ)\frac{dθ}{j!} .

Each of these terms extends to be meromorphic in the entire complex plane, with a simple pole (at most) at z = −2n + j. This shows that f(z) has a meromorphic continuation as claimed.

By this argument we have actually computed the residues of the analytic continuation of Tr(A(z)) as

lim_{z→−2n+j}(z − j + 2n) Tr(A(z)) = (2π)^{−n} ∫_{𝕊^{2n−1}} a_{j}(θ) dθ^{[1.2.2000.283]} (3.109)

when a_{j}(θ) ∈ C^{∞}(S^{2n−1}) is the function occurring in the asymptotic expansion of the left symbol of A(z):

(3.110) σ_{L}(A(z)) ∼ ∑_{j=0}^{∞}(|x|^{2} + |ξ|^{2})^{z/2−j}ã_{j}(z, θ)

|x|^{2} + |ξ|^{2} → ∞, θ = \frac{(x, ξ)}{(|x|^{2} + |ξ|^{2})^{1/2}} , a_{j}(θ) = ã_{j}(−2n + j, θ).^{[F eb.17.2000.eq:7]}

More generally, if m ∈ ℤ and A(z) ∈ Ψ_{iso}^{m+z}(ℝ^{n}) is a holomorphic family then

Tr(A(z)) is meromorphic with at most simple poles at − 2n − m + ℕ_{0} .^{[F eb.24.2000.eq:1]}

Indeed this just follows by considering the family A(z − m).

We are especially interested in the behavior at z = 0. Since the residue there is an integral of the term of order −2n, we know that

A(z) ∈ Ψ_{iso}^{m+z}(ℝ^{n}) holomorphic with A(0) = 0

⟹ Tr(A(z)) is regular at z = 0 .

[F eb.24.2000.eq:2] (3.111)

This allows us to make the following definition:

Tr_{Res}(A) = lim_{z→0} z Tr(A(z)) if

A(z) ∈ Ψ_{iso}^{m+z}(ℝ^{n}) is holomorphic with A(0) = A .^{[F eb.24.2000.eq:3]}

We know that such a holomorphic family exists, since we showed in Section 3.13 the existence of a holomorphic family F (z) ∈ Ψ_{iso}^{z}(ℝ^{n}) with F (0) = Id; A(z) = AF (z) is therefore an example. Similarly we know that Tr_{Res}(A) is independent of the choice of holomorphic family A(z) because of (3.111) applied to the difference, which vanishes at zero.

Lemma 3.14. The residue functional Tr_{Res}(A), A ∈ Ψ_{iso}^{ℤ}(ℝ^{n}), is a trace:

(3.112) Tr_{Res}([A, B]) = 0 ∀ A, B ∈ Ψ_{iso}^{ℤ}(ℝ^{n})^{[F eb.24.2000.eq:4]}

which vanishes on Ψ_{iso}^{−2n−1}(ℝ^{n}) and is given explicitly by

Tr_{Res}(A) = (2π)^{−n} ∫_{S^{2n−1}} a_{−2n}(θ) dθ^{[F eb.24.2000.eq:5]} (3.113)

where a_{−2n}(θ) is the term of order −2n in the expansion of the left (or right) symbol of a.

Proof. We have already shown that Tr_{Res}(A) is well-defined and (3.113) follows from (3.109) with a_{−2n}(θ) the term of order −2n in the left-reduced symbol of A = A(0). On the other hand, the same argument applies for the right-reduced symbol.

To see (3.112) just note that if A(z) and B(z) are holomorphic families with A(0) = A, and B(0) = B then C(z) = [A(z), B(z)] is a holomorphic family with C(0) = [A, B]. On the other hand, Tr(C(z)) = 0 when ℜz ≫ 0, so the analytic continuation of Tr(C(z)) vanishes identically and (3.112) follows.

As we shall see below, Tr_{Res} is the unique trace (up to a multiple of course) on Ψ_{iso}^{ℤ}(ℝ^{n}).

3.19. Exterior derivation

Let A(z) ∈ Ψ_{iso}^{z}(ℝ^{n}) be a holomorphic family with A(0) = Id . Then

G(z) = A(z) · A(−z) ∈ Ψ_{iso}^{0}(ℝ^{n})

is a holomorphic family of fixed order with G(0) = Id . By analytic Fredholm theory

(3.114) G^{−1}(z) ∈ Ψ_{iso}^{0}(ℝ^{n}) is a meromorphic family with finite rank poles.^{[1.2.2000.284]}

It follows that A^{−1}(z) = A(−z)G^{−1}(z) is a meromorphic family of order −z with at most finite rank poles and regular near 0. Set

Ψ_{iso}^{m}(ℝ^{n}) ∋ B ↦ A(z)BA^{−1}(z) = B(z) .^{[1.2.2000.286]} (3.115)

Thus B(z) is a meromorphic family of order m with B(0) = B. The derivative gives a linear map.

(3.116) Ψ_{iso}^{m}(ℝ^{n}) ∋ B ↦ D_{A}B = \frac{d}{dz} A(z)BA^{−1}(z)|_{z=0} ∈ Ψ_{iso}^{m}(ℝ^{n}) .^{[1.2.2000.285]}

Proposition 3.17. For any holomorphic family of order z, with A(0) = Id, the map (3.116), defined through (3.115), is a derivation and for two choices of A(z) the derivations differ by an inner derivation.

Proof. Since

A(z)B_{1}B_{2}A^{−1}(z) = A(z)B_{1}A^{−1}(z)A(z)B_{2}A^{−1}(z)

it follows that

\frac{d}{dz} A(z)B_{1}B_{2}A^{−1}(z)|_{z=0} = (D_{A}B_{1}) ∘ B_{2} + B_{1} ∘ (D_{A}B_{2}) .^{[1.2.2000.287]}

If A_{1}(z) and A_{2}(z) are two holomorphic families of order z with A_{1}(0) = A_{2}(0) = Id then

A_{2}(z) = A_{1}(z)G(z)

when G(z) ∈ Ψ_{iso}^{∞}(ℝ^{n}) is a meromorphic family, with finite rank poles. Thus

A_{2}(z)BA_{2}^{−1}(z) = A_{1}(z)G(z)BG^{−1}(z)A_{1}^{−1}(z)
= A_{1}(z)BA^{−1}(z) + zA_{1}(z)H(z)A_{1}^{−1}(z) .

Here H(z) = (G(z)BG^{−1}(z) − B)/z is a holomorphic family of degree m with H(0) = G′(0)B − BG′(0). Thus

\frac{d}{dz} A_{2}(z)BA_{2}^{−1}(z)|_{z=0} = \frac{d}{dz} A_{1}(z)BA^{−1}(z)|_{z=0} + [G′(0), B]

which shows that the two derivations differ by an inner derivation, which is to say commutation with an element of Ψ_{iso}^{0}(ℝ^{n}).

3.20. REGULARIZED TRACE^{[S.REGULARIZED.T RACE]} 95

Note that in fact

D_{A} : Ψ_{iso}^{m}(ℝ^{n}) → Ψ_{iso}^{m−1}(ℝ^{n}) ∀ m

since the symbol of A(z)BA^{−1}(z) is equal to the principal symbol of B for all z. For the specific choice of A(z) = H(z) given by

σ_{L}(H(z)) = (1 + |x|^{2} + |ξ|^{2})^{z/2}

we shall set

D_{A}B = D_{H}B .

Observe that \frac{1}{2} log(1 + |x|^{2} + |ξ|^{2}) ∈ S_{∞}^{ϵ}(ℝ^{2n}) ∀ ϵ > 0. Thus log(1 + |x|^{2} + |ξ|^{2}), defined by Weyl quantization, is an element of Ψ_{∞−iso}^{−ϵ}(ℝ^{n}) for all ϵ > 0. By differentiation the symbols satisfy

D_{H}B = [\frac{1}{2} log(1 + |x|^{2} + |D|^{2}), B] + [G, B]

where G ∈ Ψ_{iso}^{−1}(ℝ^{n}). Thus D_{H} is not itself an interior derivation. It is therefore an exterior derivation.

3.20. Regularized trace^{[S.Regularized.trace]}

In Section 3.18 we defined the residue trace of B as the residue at z = 0 of the analytic continuation of Tr(BA(z)), where A(z) is a holomorphic family of order z with A(0) = Id . Next we consider the functional

(3.117) \overline{Tr}_{A}(B) = lim_{z=0}(Tr(BA(z)) − \frac{1}{z} Tr_{Res}(B)) .^{[F eb.24.2000.eq:E]}

In contrast to the residue trace, \overline{Tr}_{A}(z) does depend on the choice of analytic family A(z).

Lemma 3.15. If A_{i}(z), i = 1, 2, are two holomorphic families of order z with A_{i}(0) = Id and G′(0) = \frac{d}{dz} A_{2}(z)A_{1}^{−2}(z)|_{z=0} then

(3.118) \overline{Tr}_{A_{2}}(B) − \overline{Tr}_{A_{1}}(B) = Tr_{Res}(BG′(0)) .^{[F eb.24.2000.eq:F]}

Proof. Writing G(z) = A_{2}(z)A_{1}^{−1}(z), which is a meromorphic family of order 0 with G(0) = Id,

Tr(BA_{2}(z)) = Tr(BG(z)A_{1}(z))

= Tr(BA_{1}(z)) + z Tr(BG′(0)A_{1}(z)) + z^{2} Tr(H(z)A_{1}(z))

where H(z) = \frac{B}{z^{2}} (G(z) − Id −zG′(0)) is then meromorphic with only finite rank poles and is regular near z = 0. Thus the analytic continuation of z^{2} Tr(H(z)A(z)) vanishes at zero from which (3.118) follows.

This regularized trace \overline{Tr}_{A}(B) therefore only depends on the first order, in z, term in A(z) at z = 0. It is important to note that it is not itself a trace.

Lemma 3.16. If B_{1}, B_{2} ∈ Ψ_{iso}^{ℤ}(ℝ^{n}) then

(3.119) \overline{Tr}_{A}([B_{1}, B_{2}]) = Tr_{Res}(B_{2}D_{A}B_{1}) .^{[F eb.24.2000.eq:G]}

Proof. Since \overline{Tr}_{A}([B_{1}, B_{2}]) is the regularized value at 0 of the analytic continuation of the trace of

(3.120) B_{1}B_{2}A(z) − B_{2}B_{1}A(z) = B_{2}[A(z), B_{1}] + [B_{1}, B_{2}A(z)]

= B_{2}([A(z), B]A^{−1}(z))A_{1}(z) + [B_{1}B_{2}A(z)] .^{[F eb.24.2000.eq:H]}

The second term on the right in (3.120) has zero trace before analytic continuation. Thus \overline{Tr}_{A}([B_{1}, B_{2}]) is the regularized value of the analytic continuation of the trace of Q(z)A(z) where

Q(z) = B_{2}[A(z), B_{1}]A^{−1}(z) = zD_{A}B_{1} + z^{2}L(z)

with L(z) meromorphic of fixed order and regular at z = 0. Thus (3.119) follows.

Note that

(3.121) Tr_{Res}(D_{A}B) = 0 ∀ B ∈ Ψ_{iso}^{ℤ}(ℝ^{n})^{[F eb.24.2000.eq:H1]}

and any family A. Indeed the residue trace is the residue of z = 0 of the analytic continuation of Tr(H(z)A(z)) when A(z) is any meromorphic family of fixed order with H(0) = D_{A}B. In particular we can take

H(z) = \frac{1}{z}(A(z)BA^{−1}(z) − B) .

Then H(z)A(z) = \frac{1}{z}[A(z), B] so the trace vanishes before analytic continuation.

3.21. Projections

3.22. Complex powers

3.23. Index and invertibility^{[S.Index.and.invertibility]}

We have already seen that the elliptic elements

E_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⊂ Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}) ↪ B(L^{2}(ℝ^{n}; ℂ^{N}))^{[1.2.2000.369]} (3.122)

define Fredholm operators. The index of such an operator

(3.123) Ind(A) = dim Nul(A) − dim Nul(A^{∗})^{[1.2.2000.370]}

is a measure of its non-invertibility. Set

(3.124) E_{iso,k}^{0}(ℝ^{n}; ℂ^{N}) = {A ∈ E_{iso}^{0}(ℝ^{n}; ℂ^{N}); Ind(A) = k} , k ∈ ℤ.^{[1.2.2000.371]}

[1.2.2000.372]

Proposition 3.18. If A ∈ E_{iso}^{0}(ℝ^{n}; ℂ^{N}) and Ind(A) = 0 then there exists E ∈ Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}) such that A + E is invertible in B(L^{2}(ℝ^{n}; ℂ^{N})) and the inverse then lies in Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}).

Proof. Let B be the generalized inverse of A, assumed to be elliptic. The assumption that Ind(A) = 0 means that Nul(A) and Nul(A^{∗}) have the same dimension. Let e_{1}, ⋯, e_{p} ∈ S(ℝ^{n}; ℂ^{N}) and f_{1}, ⋯, f_{p} ∈ S(ℝ^{n}; ℂ^{N}) be bases of Nul(A) and Nul(A^{∗}). Then consider

E = ∑_{j=1}^{p} f_{j}(x)\overline{e_{j}(y)} ∈ Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}). (3.125)

By construction E is an isomorphism (in fact an arbitrary one) between Nul(A) and Nul(A^{∗}). Thus A+E is continuous, injective and surjective, hence has an inverse in

3.23. INDEX AND INVERTIBILITY^{[S.INDEX.AND.INV ERT IBILIT Y ]} 97

B(L^{2}(ℝ^{n}; ℂ^{N})). Indeed this inverse is B + E^{−1} where E^{−1} is the inverse of E as a map from Nul(A) to Nul(A^{∗}). This shows that A can be perturbed by a smoothing operator to be invertible.

Let

(3.126) G_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⊂ E_{iso,0}^{0}(ℝ^{n}; ℂ^{N}) ⊂ E_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⊂ Ψ0_{iso}(ℝ^{n}; ℂ^{N})^{[1.2.2000.374]}

denote the group of the invertible elements (invertibility being either in B(L^{2}(ℝ; ℂ^{N}) or in Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N})) in the ring of elliptic elements of index 0.

[1.2.2000.373]

Corollary 3.7. The first inclusion in (3.126) is dense in the topology of Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N}).

Proof. This follows from the proof of Proposition 3.18, since A + sE is invertible for all s ≠ 0.

We next derive some simple formulæ for the index of an element of E_{iso}^{0}(ℝ^{n}; ℂ^{N}). First observe that the trace of a finite dimensional projection is its rank, the dimension of its range. Thus

(3.127) Ind(A) = Tr(Π_{Nul(A)}) − Tr(Π_{Nul(A^{∗})})^{[1.2.2000.376]}

where the trace may be reinterpreted as the trace on smoothing operators. The identities, (3.34), satisfied by the generalized inverse of A shows that this can be rewritten

(3.128) Ind(A) = − Tr(BA − Id) + Tr(AB − Id) = Tr([A, B]).^{[1.2.2000.377]}

Here [A, B] = Π_{Nul(A)} − Π_{Nul(A^{∗})} is a smoothing operator, even though both A and B are elliptic of order 0.

[1.2.2000.378]

Lemma 3.17. If A ∈ E_{iso}^{0}(ℝ^{n}; ℂ^{N}) the identity (3.128), which may be rewritten

Ind(A) = Tr([A, B]),^{[1.2.2000.379]} (3.129)

holds for any parametrix B.

Proof. If B′ is a parametrix and B is the generalized inverse then B′ − B = E ∈ Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}). Thus

[A, B′] = [A, B] + [A, E].

Since Tr([A, E] = 0, one of the arguments being a smoothing operator, (3.129) follows in general from the particular case (3.128).

Note that it follows from (3.129) that Ind(A) = Ind(A + E) if E is smoothing. In fact the index is even more stable than this as we shall see, since it is locally constant on E_{iso}^{0}(ℝ^{n}; ℂ^{N}). In any case this shows that

(3.130) Ind : E_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⟶ ℤ, Ind(a) = Ind(A) if a = [A],

E_{iso}^{0}(ℝ^{n}; ℂ^{N}) = E_{iso}^{0}(ℝ^{n}; ℂ^{N})/Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N})
⊂ A_{iso}^{0}(ℝ^{n}; ℂ^{N}) = Ψ_{iso}^{0}(ℝ^{n}; ℂ^{N})/Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N})^{[1.2.2000.381]}

is well-defined.

The argument of the trace functional in (3.129) is a smoothing operator, but we may still rewrite the formula in terms of the regularized trace, with respect to the standard regularizer H(z) with left symbol (1 + |x|^{2} + |ξ|^{2})^{\frac{z}{2}} . The advantage of doing so is that we can then use the trace defect formula (3.119). Thus for any elliptic isotropic operator of order 0

(3.131) Ind(A) = Tr_{Res}(BD_{H}A).^{[1.2.2000.380]}

Here B is a parametrix for A. The residue trace is actually a functional

Tr_{Res} : A_{iso}^{ℤ}(ℝ^{n}; ℂ^{N}) ⟶ ℂ,

so if we write a^{−1} for the inverse of a in the ring E_{iso}^{0}(ℝ^{n}; ℂ^{N}) then

(3.132) Ind(a) = Tr_{Res}(a^{−1}D_{H}a), D_{H} : A_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⟶ A_{iso}^{0}(ℝ^{n}; ℂ^{N})^{[1.2.2000.382]}

being the induced derivation (since D_{H} clearly preserves the ideal Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}).

From this simple formula we can easily deduce two elementary properties of elliptic operators. These actually hold in general for Fredholm operators, although the proofs here are not valid in that generality. Namely

(3.133) Ind : E_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⟶ ℤ is locally constant and^{[1.2.2000.383]}

(3.134) Ind(a_{1}a_{2}) = Ind(a_{1}) + Ind(a_{2}) ∀ a_{1}, a_{2} ∈ E_{iso}^{0}(ℝ^{n}; ℂ^{N}).^{[1.2.2000.384]}

The first of these follows the continuity of the formula (3.132) since under deformation of a in E_{iso}^{0}(ℝ^{n}; ℂ^{N}) the inverse a^{−1} varies continuously, so Ind is continuous and integer-valued, hence locally constant. Similarly the second, logarithmic additivity, property follows from the fact that D_{H} is a derivation, so

D_{H}(a_{1}a_{2}) = (D_{H}a_{1})a_{2} + a_{1}D_{H}a_{2}

and the the trace property of Tr_{Res} which shows that

(3.135) Ind(a_{1}a_{2}) = Tr_{Res}((a_{1}a_{2})^{−1}D_{H}(a_{1}a_{2}) = Tr(a_{2}^{−1}a_{1}^{−1} ((D_{H}a_{1})a_{2} + a_{1}D_{H}a_{2})

= Tr(a_{2}^{−1}a_{1}^{−1}(D_{H}a_{1})a_{2}) + Tr(a_{2}^{−1}D_{H}a_{2}) = Ind(a_{1}) + Ind(a_{2}).^{[1.2.2000.385]}

3.24. Variation 1-form

In the previous section we have seen that the index

Ind : E_{iso}^{0}(ℝ^{n}; ℂ^{N}) ⟶ ℤ^{[1.2.2000.386]} (3.136)

is a multiplicative map which is the obstruction to perturbative invertibility. In the next two sections we will derive a closely related obstruction to the perturbative invertibility of a family of elliptic operators. Thus, suppose

Y ∋ y ⟼ A_{y} ∈ E_{iso,0}^{0}(ℝ^{n}; ℂ^{N})^{[1.2.2000.387]} (3.137)

is a family of elliptic operators depending smoothly on a parameter in the compact manifold Y. We are interested in the families perturbative invertibility question. That is, does there exist a smooth family

(3.138) Y ∋ y ⟼ E_{y} ∈ Ψ_{iso}^{−∞}(ℝ^{n}; ℂ^{N}) such that (A_{y} + E_{y}) ∈ G_{iso}^{0}(ℝ^{n}; ℂ^{N}) ∀ y.^{[1.2.2000.388]}

We have assumed that the operators have index zero since this is necessary (and sufficient) for E_{y} to exist for any one y ∈ Y. Thus the issue is the smoothness (really just the continuity) of the perturbation E_{y}.

We shall start by essentially writing down such a putative obstruction directly and then subsequently we shall investigate its topological origins.

3.24. VARIATION 1-FORM 99

[1.2.2000.389]

Proposition 3.19. If a smooth family (3.137), parameterized by a compact manifold Y, is perturbatively invariant in the sense that there is a smooth family as in (3.138), then the closed 2-form on Y

(3.139) β = \frac{1}{2} Tr_{Res}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} · a_{y}^{−1}D_{H}a_{y}) ∈ C^{∞}(Y; Λ^{2}),

a_{y} = [A_{y}] ∈ E_{iso,0}^{0}(ℝ^{n}; ℂ^{N}),^{[1.2.2000.390]}

is exact.

Proof. Note first that β is indeed a smooth form, since the full symbolic inverse depends smoothly on parameters. Next we show that β is always closed. The 1-forms a_{y}^{−1}d_{y}a_{y}a_{y}^{−1} and da_{y} are exact so differentiating directly gives

dβ = \frac{1}{2} Tr_{Res}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} ∧ d(a_{y}^{−1}D_{H}a_{y}))

= −\frac{1}{2} Tr_{Res}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y}^{−1}D_{H}a_{y}))

+\frac{1}{2} Tr_{Res}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}D_{H}(da_{y}))

= \frac{1}{2} Tr_{Res}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} ∧ D_{H}(a_{y}^{−1}da_{y})).

[1.2.2000.391] (3.140)

Using the trace property and the commutativity of a 2-form with other forms the last expression can be written

\frac{1}{6} Tr_{Res}(D_{H}(a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}d_{y}a_{y} ∧ a_{y}^{−1}da_{y})) = 0^{[1.2.2000.392]} (3.141)

by property (3.121) of the residue trace.

Now, suppose that a smooth perturbation as in (3.138) does exist. We can replace A_{y} by A_{y} + E_{y} without affecting β, since the residue trace vanishes on the ideal of smoothing operators. Thus we can assume that A_{y} itself is invertible. Then consider the 1-form defined using the regularized trace

\overline{α} = \overline{Tr}_{H}(A_{y}^{−1}d_{y}A_{y}).^{[1.2.2000.393]} (3.142)

This is an extension of the 1-form d log det_{F} on G_{iso}^{−∞}(ℝ^{n}; ℂ^{N}). The extension is not in general closed, because the regularized trace does not satisfy the trace condition. Using the stanadard formula for the variation of the inverse, dA_{y}^{−1} = −A_{y}^{−1}dA_{y}A_{y}^{−1}, the exterior derivative is the 2-form

d\overline{α} = − \overline{Tr}_{H}(A_{y}^{−1}(d_{y}A)A_{y}^{−1}d_{y}A_{y}).^{[1.2.2000.394]} (3.143)

The 2-form argument is a commutator. Indeed, in terms of local coordinates we can write

A_{y}^{−1}(d_{y}A)A_{y}^{−1}d_{y}A_{y} = ∑_{j,k=1}^{p} A_{y}^{−1}(\frac{∂A}{∂y_{j}})A_{y}^{−1}(\frac{∂A}{∂y_{k}})dy_{j} ∧ dy_{k}

= \frac{1}{2} ∑_{j,k=1}^{p} (A_{y}^{−1}(\frac{∂A}{∂y_{j}})A_{y}^{−1}(\frac{∂A}{∂y_{k}}) − A_{y}^{−1}(\frac{∂A}{∂y_{k}})A_{y}^{−1}(\frac{∂A}{∂y_{j}})) dy_{j} ∧ dy_{k}

= \frac{1}{2} ∑_{j,k=1}^{p}[A_{y}^{−1}(\frac{∂A}{∂y_{j}}), A_{y}^{−1}(\frac{∂A}{∂y_{k}})]dy_{j} ∧ dy_{k}

Applying the trace defect formula (3.119) shows that

d\overline{α} = −\frac{1}{2} Tr_{Res} (A_{y}^{−1}d_{y}A_{y} ∧ D_{H}(A_{y}^{−1}d_{y}A_{y})) ,^{[1.2.2000.395]} (3.144)

locally and hence globally.

Expanding the action of the derivation D_{H} gives

(3.145) d\overline{α} = β − \frac{1}{2} Tr_{Res} (A_{y}^{−1}d_{y}A_{y} ∧ A_{y}^{−1}d_{y}(D_{H}A_{y})) = β − dγ, where

γ = \frac{1}{2} Tr_{Res} (A_{y}^{−1}d_{y}A_{y} ∧ .A_{y}^{−1}D_{H}A_{y}) .^{[1.2.2000.396]}

We conclude that if A_{y} has an invertible lift then β is exact.

Note that the form γ in (3.145) is well-defined as a form on E_{iso,0}^{0}(ℝ^{n}; ℂ^{N}), and is independent of the perturbation. Thus the cohomology class which we have constructed as the obstruction to perturbative invertibility can be written

[β] = [β − dγ] ∈ H^{2}(E_{iso,0}^{0}(ℝ^{n}; ℂ^{N})).^{[1.2.2000.397]} (3.146)

3.25. Determinant bundle

To better explain the topological origin of the cohomology class (3.146) we construct the determinant bundle. This was originally introduced for families of Dirac operators by Quillen [10]. Recall that the Fredholm determinant is a character

(3.147) det_{Fr} : Id +Ψ_{iso}^{−2n−1}(ℝ^{n}; ℂ^{N}) ⟶ ℂ,

det_{Fr}(AB) = det_{Fr}(A) det_{Fr}(B)∀ A, B ∈ Id +Ψ_{iso}^{−2n−1}(ℝ^{n}; ℂ^{N}).^{[1.2.2000.398]}

As we shall see, it is not possible to extend the Fredholm determinant as a multiplicative function to G_{iso}^{0}(ℝ^{n}; ℂ^{N}), essentially because of the non-extendibility of the trace.

However in trying to extend the determinant we can consider the possible values it would take on a point A ∈ G_{iso}^{0}(ℝ^{n}; ℂ^{N}) as the set of pairs (A, z), z ∈ ℂ. Thus we simple consider the product

D^{0} = G^{0} × ℂ,^{[1.2.2000.399]} (3.148)

where from now on we simplify the notation and write G^{0} = G_{iso}(ℝ^{n}; ℂ^{N}) etc. Although it is not reasonable to expect full multiplicative of the determinant, it is more reasonable to expect the determinant of A(Id +B), B ∈ Ψ^{−2n−1} to be related to the product of determinants. Thus it is natural to identify pairs in D^{0},

(A, z) ∼_{p} (A′, z′) if
A, A′ ∈ G^{0}, A′ = A(Id +B), z′ = det_{Fr}(Id +B)z, B ∈ Ψ^{p}, p < −2n.

[1.2.2000.400] (3.149)

The equivalence relations here are slightly different, depending on p. In all cases the action of the determinant is linear, so the quotient is a line bundle.

[1.2.2000.401]

Lemma 3.18. For any integer p < −2n, and also p = −∞, the quotient

D_{p}^{0} = D^{0}/ ∼_{p} ^{[1.2.2000.402]} (3.150)

is a smooth line bundle over G_{p}^{0} = G^{0}/G^{p}.

3.31. PROBLEMS^{[S.P ROBLEMS.3]} 101

Proof. The projection is just the quotient in the first factor and this clearly defines a commutative square

\xymatrix{D^{0} \ar[r]^{[∼_{p}]} \ar[d]^{π} & D_{p}^{0} \ar[d]^{π} \\ G^{0} \ar[r]^{/G^{p}} & G_{p}^{0}.}^{[1.2.2000.403]} (3.151)

3.26. Index bundle

3.27. Index formulæ

susceptible

3.28. Isotropic essential support

3.29. Isotropic wavefront set

3.30. Isotropic FBI transform

3.31. Problems^{[S.P roblems.3]}

[21.2.1998.121]

Problem 3.1. Define the isotropic Sobolev spaces of integral order by

H_{iso}^{k}(ℝ^{n}) = \begin{cases}\{u ∈ L^{2}(ℝ^{n}); x^{α}D_{x}^{β}u ∈ L^{2}(ℝ^{n}) ∀ |α| + |β| ≤ k\} & k ∈ ℕ \\ \{u ∈ S′(ℝ^{n}); u = ∑_{|α|+|β|≤−k} x^{α}D_{x}^{β}u_{α,β}, u_{α,β} ∈ L^{2}(ℝ^{n})\} & k ∈ −ℕ. \end{cases} ^{[21.2.1998.122]} (3.152)

Show that if A ∈ Ψ_{iso}^{p}(ℝ^{n}) with p an integer, then A : H_{iso}^{k}(ℝ^{n}) ⟶ H_{iso}^{k−p}(ℝ^{n}) for any integral k. Deduce (using the properties of elliptic isotropic operators) that the general definition

H_{iso}^{m}(ℝ^{n}) = {u ∈ S′(ℝ^{n}); Au ∈ L^{2}(ℝ^{n}), ∀ A ∈ Ψ_{iso}^{−m}(ℝ^{n})} , m ∈ ℝ^{[21.2.1998.123]} (3.153)

is consistent with (3.152) and has the properties

(3.154) A ∈ Ψ_{iso}^{M}(ℝ^{n}) ⟹ A : H_{iso}^{m}(ℝ^{n}) ⟶ H_{iso}^{m−M}(ℝ^{n}), ^{[21.2.1998.124]}

⋂_{m} H_{iso}^{m}(ℝ^{n}) = S(ℝ^{n}), ⋃_{m} H_{iso}^{m}(ℝ^{n}) = S′(ℝ^{n})^{[21.2.1998.126]} (3.155)

A ∈ Ψ_{iso}^{m}(ℝ^{n}), u ∈ S′(ℝ^{n}), Au ∈ H^{m′}(ℝ^{n}) ⟹ u ∈ H^{m′−m}(ℝ^{n}),^{[21.2.1998.125]} (3.156)

[21.2.1998.127]

Problem 3.2. Show that if ϵ > 0 then

H_{iso}^{ϵ}(ℝ^{n}) ⊊ (1 + |x|)^{−ϵ}L^{2}(ℝ^{n}) ∩ H^{ϵ}(ℝ^{n})

Deduce that H_{iso}^{ϵ}(ℝ^{n}) ↪ L^{2}(ℝ^{n}) is a compact inclusion (i.e. the image of a bounded set is precompact).

[21.2.1998.128]

Problem 3.3. Using Problem 3.2, or otherwise, show that each element of Ψ_{iso}^{−ϵ}(ℝ^{n}), ϵ > 0, defines a compact operator on L^{2}(ℝ^{n}).

102 3. ISOTROPIC AND SCATTERING CALCULI^{[C.IANDS.CALCULI]}

[21.2.1998.129]

Problem 3.4. Show that if E ∈ Ψ_{iso}^{−∞}(ℝ^{n}) then there exists F ∈ Ψ_{iso}^{−∞}(ℝ^{n}) such that

(Id +E)(Id +F) = Id_{G} with G ∈ Ψ_{iso}^{−∞}(ℝ^{n}) of finite rank,

that is, G · S(ℝ^{n}) is finite dimensional.

[21.2.1998.130]

Problem 3.5. Using Problem 3.4 show that an elliptic element A ∈ Ψ_{iso}^{m}(ℝ^{n}) has a parametrix B ∈ Ψ_{iso}^{−m}(ℝ^{n}) up to finite rank error; that is, such that A∘B −Id and B ∘ A − Id are finite rank elements of Ψ_{iso}^{−∞}(ℝ^{n}). Deduce that such an elliptic A defines a Fredholm operator

A : H_{iso}^{M}(ℝ^{n}) ⟶ H_{iso}^{M−m}(ℝ^{n})

for any M. [The requirements for an operator A between Hilbert spaces to be Fredholm are that it be bounded, have finite-dimensional null space and closed range with a finite-dimensional complement.]

[21.2.1998.111]

Problem 3.6. [The harmonic oscillator] Show that the ‘harmonic oscillator’

H = |D|^{2} + |x|^{2}, Hu = ∑_{j=1}^{n} D_{j}^{2}u + |x|^{2}u,

is an elliptic element of Ψ_{iso}^{2}(ℝ^{n}). Consider the ‘creation’ and ‘annihilation’ operators

C_{j} = D_{j} + ix_{j}, A_{j} = D_{j} − ix_{j} = C_{j}^{∗},^{[21.2.1998.131]} (3.157)

and show that

(3.158) H = ∑_{j=1}^{n} C_{j}A_{j} + n = ∑_{j=1}^{n} A_{j}C_{j} − n,

[A_{j}, H] = 2A_{j}, [C_{j}, H] = −2C_{j}, [C_{l}, C_{j}] = 0, [A_{l}, A_{j}] = 0, [A_{l}, C_{j}] = 2δ_{lk} Id,^{[21.2.1998.132]}

where [A, B] = A ∘ B − B ∘ A is the commutator bracket and δ_{lk} is the Kronecker symbol. Knowing that (H − λ)u = 0, for λ ∈ ℂ and u ∈ S′(ℝ^{n}) implies u ∈ S(ℝ^{n}) (why?) show that

E_{λ} = {u ∈ S′(ℝ^{n}); (H − λ)u = 0} ≠ {0} ⟺ λ ∈ n + 2ℕ_{0}^{[21.2.1998.133]} (3.159)

and E_{−n+2k} = { ∑_{|α|=k} c_{α}C^{α} exp(−|x|^{2}/2), c_{α} ∈ ℂ} , k ∈ ℕ_{0}.^{[21.2.1998.134]} (3.160)

[1.2.2000.406]

Problem 3.7. [Definition of determinant of matrices.]

[iml.6]

Problem 3.8. [Proof that dα = 0 in (3.93).] To prove that the 1-form is closed it suffices to show that it is closed when restricted to any 2-dimensional submanifold. Thus we may suppose that A = A(s, t) depends on 2 parameters. In terms of these parameters

α = Tr(A(s, t)^{−1} \frac{dA(s, t)}{ds})ds + Tr(A(s, t)^{−1} \frac{dA(s, t)}{dt})dt.^{[iml.7]} (3.161)

Show that the exterior derivative can be written

dα = Tr([A(s, t)^{−1} \frac{dA(s, t)}{dt}, A(s, t)^{−1} \frac{dA(s, t)}{ds}])ds ∧ dt^{[iml.8]} (3.162)

and hence that it vanishes.

[21.2.1998.137]

Problem 3.9. If E and F are vector spaces, show that the space of operators Ψ_{iso}^{m}(ℝ^{n}; E, F) from S′(ℝ^{n}; E) to S′(ℝ^{n}; F) is well-defined as the matrices with entries in Ψ_{iso}^{m}(ℝ^{n}) for any choice of bases of E and F.

[1.2.2000.349]

Problem 3.10. Necessity of ellipticity for a psuedodifferential operator to be Fredholm on the isotropic Sobolev spaces.

1. Reduce to the case of operators of order 0.

2. Construct a sequence in L^{2} such that ∥u_{n}∥ = 1, u_{n} → 0 weakly and Au_{n} → 0

strongly in L^{2}.

[21.2.1998.135]

Problem 3.11. [Koszul complex] Consider the form bundles over ℝ^{n}. That is Λ^{k}ℝ^{n} is the vector space of dimension \binom{n}{k} consisting of the totally antisymmetric k-linear forms on ℝ^{n}. If e_{1}, e_{2}, …, e_{n} is the standard basis for ℝ^{n} then for a k-tuple α e^{α} defined on basis elements by

e^{α}(e_{i_{1}}, …, e_{i_{k}}) = ∏_{j=1}^{k} δ_{1_{j}α_{j}}

extends uniquely to a k-linear map. Elements dx^{α} ∈ Λ^{k}ℝ^{n} are defined by the total antisymmetrization of the e^{α}. Explicitly,

dx^{α}(v_{1}, …, v_{k}) = ∑_{π} sgn πe^{α}(v_{π_{1}}, …, v_{π_{n}})

where the sum is over permutations π of {1, …, n} and sgn π is the parity of π. The dx^{α} for strictly increasing k-tuples α of elements of {1, …, n} give a basis for Λ^{k}ℝ^{n}. The wedge product is defined by dx^{α} ∧ dx^{β} = dx^{α,β}.

Now let S′(ℝ^{n}; Λ^{k}) be the tensor product, that is u ∈ S′(ℝ^{n}; Λ^{k}) is a finite sum

u = ∑_{α} u_{α}dx^{α}.^{[21.2.1998.136]} (3.163)

The annihilation operators in (3.157) define an operator, for each k,

D : S′(ℝ^{n}; Λ^{k}) ⟶ S′(ℝ^{n}; Λ^{k+1}), Du = ∑_{j=1}^{n} A_{j}u_{α}dx^{j} ∧ dx^{α}.

Show that D^{2} = 0. Define inner products on the Λ^{k}ℝ^{n} by declaring the basis introduced above to be orthonormal. Show that the adjoint of D, defined with respect to these inner products and the L^{2} pairing is

D^{∗} : S′(ℝ^{n}; Λ^{k}) ⟶ S′(ℝ^{n}; Λ^{k−1}), D^{∗}u = ∑_{j=1}^{n} C_{j}u_{α}ι_{j}dx^{α}.

Here, ι_{j} is ‘contraction with e_{j};’ it is the adjoint of dx^{j} ∧ . Show that D + D^{∗} is an elliptic element of Ψ_{iso}^{1}(ℝ^{n}; Λ^{∗}). Maybe using Problem 3.6 show that the null space of D + D^{∗} on S′(ℝ^{n}; Λ^{∗}ℝ^{n}) is 1-dimensional. Deduce that

(3.164) {u ∈ S′(ℝ^{n}); Du = 0} = ℂ exp(−|x|^{2}/2),

{u ∈ S′(ℝ^{n}; Λ^{k}); Du = 0} = (S′(ℝ^{n}; Λ^{k−1}), k ≥ 1.^{[21.2.1998.138]}

Observe that, as an operator from S′(ℝ^{n}; Λ^{odd}) to S′(ℝ^{n}; Λ^{even}), D + D^{∗} is an elliptic element of Ψ_{iso}^{1}(ℝ^{n}; Λ^{odd}, Λ^{even}) and has index 1.

[22.2.1998.141]

Problem 3.12. [Isotropic essential support] For an element of S^{m}(ℝ^{n}) define (isotropic) essential support, or operator wavefront set, of A ∈ Ψ_{iso}^{m}(ℝ^{n}) by

(3.165) WF_{iso}(A) = cone supp(σ_{L}(A)) ⊂ ℝ^{2n} \ {0}.^{[22.2.1998.145]}

Show that WF_{iso}(A) = cone supp(σ_{L}(A)) and check the following

(3.166) WF′_{iso}(A + B) ∪ WF′_{iso}(A ∘ B) ⊂ WF′_{iso}(A) ∩ WF′_{iso}(B),^{[22.2.1998.149]}

(3.167) WF′_{iso}(A) = ∅ ⟺ A ∈ Ψ_{iso}^{−∞}(ℝ^{n}).^{[22.2.1998.146]}

[22.2.1998.150]

Problem 3.13. [Isotropic partition of unity] Show that if U_{i} ⊂ 𝕊^{n−1} is an open cover of the unit sphere and Ũ_{i} = \{Z ∈ ℝ^{2n} \ \{0\}; \frac{Z}{|Z|} ∈ U_{i}\} is the corresponding conic open cover of ℝ^{2n} \ {0} then there exist (finitely many) operators A_{i} ∈ Ψ_{iso}^{0}(ℝ^{n}) with WF′_{iso}(A_{i}) ⊂ Ũ_{i}, such that

Id − ∑_{i} A_{i} ∈ Ψ_{iso}^{−∞}(ℝ^{n}).^{[22.2.1998.151]} (3.168)

[22.2.1998.152]

Problem 3.14. Suppose A ∈ Ψ_{iso}^{m}(ℝ^{n}), is elliptic and has index zero as an operator on S′(ℝ^{n}). Show that there exists E ∈ Ψ_{iso}^{−∞}(ℝ^{n}) such that A + E is an isomorphism of S′(ℝ^{n}).

[22.2.1998.147]

Problem 3.15. [Isotropic wave front set] For u ∈ S′(ℝ^{n}) define

(3.169) WF_{iso}(u) = ⋂ {WF′_{iso}(A); A ∈ Ψ_{iso}^{0}(ℝ^{n}), Au ∈ S(ℝ^{n})} .^{[22.2.1998.148]}