CHAPTER 6 The wave kernel

Let us return to the subject of “good distributions” as exemplified by Dirac delta ‘functions’ and the Schwartz kernels of pseudodifferential operators. In fact we shall associate a space of “conormal distributions” with any submanifold of a manifold.

Thus let X be a C^{∞} manifold and Y ⊂ X a closed embedded submanifold – we can easily drop the assumption that Y is closed and even replace embedded by immersed, but let’s treat the simplest case first! To say that Y is embedded means that each ȳ ∈ Y has a coordinate neighbourhood U, in X, with coordinate x_{1}, …, x_{n} in terms of which ȳ = 0 and

Y ∩ U = {x, = ⋯ = x_{k} = 0}. (6.1)

We want to define

I^{∗}(X, Y; Ω^{\frac{1}{2}} ) ⊂ C^{−∞}(X; Ω^{\frac{1}{2}} ) (6.2)

to consist of distributions which are singular only at Y and small “along Y.”

So if u ∈ C_{c}^{−∞}(U) then in local coordinates (6.1) we can identify u with u′ ∈ C_{c}^{−∞}(ℝ^{n}) so u′ ∈ H_{c}^{s}(ℝ^{n}) for some s ∈ ℝ. To say that u is ‘smooth along Y ’ means we want to have

D_{x_{k+1}}^{ℓ_{1}} … D_{x_{n}}^{ℓ_{n−k}}u′ ∈ H_{c}^{s′}(ℝ^{n}) ∀ ℓ_{1}, …, ℓ_{n−k} (6.3)

and a fixed s′, independent of ℓ (but just possibly different from the initial s); of course we can take s = s′. Now conditions like (6.3) do not limit the singular support of u′ at all! However we can add a requirement that multiplication by a function which vanishes on Y makes u′ smooth, by one degree, i.e.

x_{1}^{p_{1}} … x_{k}^{p_{k}}u′ ∈ H^{s+|p|}(ℝ^{n}), |p| = p_{1} + ⋯ + p_{k}. (6.4)

This last condition implies

D_{1}^{q_{1}} … D_{k}^{q_{k}}x_{1}^{p_{1}} … x_{k}^{p_{k}}u′ ∈ H^{s}(ℝ^{n}) if |q| ≤ |p|. (6.5)

Consider what happens if we rearrange the order of differentiation and multiplication in (6.5). Since we demand (6.5) for all p, q with |q| ≤ |p| we can show in tial that

∀ |q| ≤ |p| ≤ L (6.6)

⟹

∏_{i=1}^{L}(x_{j_{i}}D_{ℓ_{i}})u ∈ H^{s}(ℝ^{n}) ∀ pairs, (j_{i,ℓ_{i}}) ∈ (1, …, k)^{2}. (6.7)

Of course we can combine (6.3) and (6.7) and demand

\begin{matrix}∏_{i=1}^{L_{2}} D_{p_{i}} ∏_{i=1}^{L_{1}}(x_{j_{i}}D_{ℓ_{i}})u′ ∈ H_{c}^{s}(ℝ^{n})(j_{j}, ℓ_{i}) ∈ (1, …, k)^{2} \\ ∀ L_{1}, L_{2} p_{i} ∈ (k + 1, … u). \end{matrix} (6.8)

Problem 6.1. Show that (6.8) implies (6.3) and (6.4)

The point about (6.8) is that it is easy to interpret in a coordinate independent way. Notice that putting C^{∞} coefficients in front of all the terms makes no difference.

Lemma 6.1. The space of all C^{∞} vector fields on ℝ^{n} tangent to the submanifold {x_{1} = ⋯ = x_{k} = 0} is spanning over C^{∞}(ℝ^{n}) by

x_{i}D_{j}, D_{p} i, j ≤ k, p > k. (6.9)

Proof. A C^{∞} vector field is just a sum

V = ∑_{j≤k} a_{j}D_{j} + ∑_{p>k} b_{p}D_{p}.

Notice that the D_{p}, for p > k, are tangent to {x_{1} = ⋯ = x_{k} = 0}, so we can assume b_{p} = 0. Tangency is then given by the condition

V x)i = 0 and {x_{1} = ⋯ = x_{k} = 0}, i = 1, …, h

i.e. a_{j} = ∑_{ℓ=1} a_{jℓ}x_{ℓ}, 1 ≤ j ≤ h. Thus

V = ∑_{ℓ=1} a_{jℓ}x_{ℓ}D_{j}

which proves (6.9).

This allows us to write (6.8) in the compact form

V(ℝ^{n}, Y_{k})^{p}u′ ⊂ H_{c}^{s}(ℝ^{n}) ∀ p (6.10)

where V(ℝ^{n}, Y_{k}) is just the space of all C^{∞} vector fields tangent to Y_{k} = {x_{1} = ⋯ = x_{k} = 0}. Of course the local coordinate just reduce vector fields tangent to Y to vector fields tangent to Y_{k} so the invariant version of (6.10) is

V(X, Y)^{p}u ⊂ H^{s}(X; Ω^{\frac{1}{2}} ) ∀ p. (6.11)

To interpret (6.11) we only need recall the (Lie) action of vector fields on halfdensities. First for densities: The formal transpose of V is −V, so set

LV ϕ(ψ) = ϕ(−V ψ)

if ϕ ∈ C^{∞}(X; Ω), ψ ∈ C^{∞}(X). On ℝ^{n} then becomes

\begin{aligned}∫ ^{L}V ϕ & · ψ = − ∫ ϕ · V ψ \\ & = − ∫ ϕ(x)V ψ · dx \\ & = ∫ (V ϕ(x) + δ_{V} ϕ) ψ dx \\ δ_{V} & = ∑_{i=1}^{n} D_{i}a_{i} if V = Σa_{i}D_{i}. \end{aligned} (6.12)

6. THE WAVE KERNEL 91

i.e. L_{V} (ϕ|dx|) = (V ϕ)|dx| + δ_{V} ϕ. (6.13) Given the tensorial properties of density, set L_{V} (ϕ|dx|^{t}) = V ϕ|dx|^{t} + tδ_{V} ϕ. (6.14) This corresponds to the natural trivialization in local coordinates.

Definition 6.1. If Y ⊂ X is a closed embedded submanifold then IH^{s} (X, Y; Ω^{\frac{1}{2}} ) = \{u ∈ H^{s} (X; Ω^{\frac{1}{2}} ) satisfying (11)\}

I^{∗} (X, Y; Ω^{\frac{1}{2}} ) = ⋃_{s} IH^{s} (X, Y; Ω^{\frac{1}{2}} ) . (6.15)

Clearly u ∈ I^{∗}(X, Y; Ω^{\frac{1}{2}} ) ⟹ u ↾ X\Y ∈ C^{∞} (X\Y; Ω^{\frac{1}{2}} ) (6.16) and

⋂_{s} IH^{s} (X, Y; Ω^{\frac{1}{2}} ) = C^{∞} (X; Ω^{\frac{1}{2}} ) . (6.17)

Let us try to understand these distributions in some detail! To do so we start with a very simple case, namely Y = {p} is a point; so we only have one coordinate system. So construct p = 0 ∈ ℝ^{n}.

u ∈ I_{c}^{∗} (ℝ^{n}, \{0\}; Ω^{\frac{1}{2}} ) ⟹ u = u′|dx|^{\frac{1}{2}} when

x^{α}D_{x}^{β}u′ ∈ H_{c}^{s}(ℝ^{n}), s fixed ∀ |α| ≥ |β|.

(6.18)

Again by a simple commutative argument this is equivalent to D_{x}^{β}x^{α}u′ ∈ H_{c}^{s}(ℝ^{n}) ∀ |α| ≥ |β|. (6.19) We can take the Fourier transform of (6.19) and get ξ^{β}D_{ξ}^{α}û′ ∈ ⟨ξ⟩^{−s}L^{2}(ℝ^{n}) ∀ |α| ≥ |β|. (6.20) In this form we can just replace ξ^{β} by ⟨ξ⟩^{|β|}, i.e. (6.20) just says D_{ξ}^{α}û′(ξ) ∈ ⟨ξ⟩^{−s−|β|}L^{2}(ℝ^{n}) ∀ α. (6.21)

Notice that this is very similar to a symbol estimate, which would say D_{ξ}^{α}û′(ξ) ∈ ⟨ξ⟩^{m−|α|}L^{∞}(ℝ^{n}) ∀ α. (6.22)

Lemma 6.2. The estimate (6.21) implies (6.22) for any m > −s− \frac{n}{2}; conversely (6.22) implies (6.21) for any s < −m − \frac{n}{2}.

Proof. Let’s start with the simple derivative, (6.22) implies (6.21). This really reduces to the case α = 0. Thus

⟨ξ⟩^{M}L^{∞}(ℝ^{n}) ⊂ L^{2}(ℝ^{n}) ⟹ M < −\frac{n}{2}

is the inequality

(∫ |u|^{2}dξ)^{\frac{1}{2}} ≤ sup⟨ξ⟩^{−M}|u| (∫ ⟨ξ⟩^{2M}dξ)^{\frac{1}{2}}

and

∫ ⟨ξ⟩^{2M}dξ = ∫ (1 + |ξ|^{2})^{M} dξ < ∞ iff M < −\frac{n}{2}.

To get (6.22) we just show that (6.22) implies

⟨ξ⟩^{s+|α|}D_{ξ}^{α}û′ ∈ ⟨ξ⟩^{m+s}L^{∞} ⊂ L^{2} if m + s < −\frac{n}{2}. (6.23)

The converse is a little trickier. To really see what is going on we can reduce (6.21) to a one dimensional version. Of course, near ξ = 0, (6.21) just says û′ is C^{∞}, so we can assume that |ξ| > 1 on supp û′ and introduce polar coordinates:

ξ = tw, w ∈ S^{n−1}t > 1. (6.24)

Then Exercise 2. Show that (6.21) (or maybe better, (6.20)) implies that

D_{t}^{k}P û′(tw) ∈ t^{−s−k}L^{2}(ℝ^{+} × S^{n−1}; t^{n−1}dtdw) ∀ k (6.25)

for any C^{∞} differential operator on S^{n−1}.

In particular we can take P to be elliptic of any order, so (6.25) actually implies

sup_{w} D_{t}^{k}P û(t, w) ∈ t^{−s−k}L^{2}(ℝ^{+}; t^{n−1}dt) (6.26)

or, changing the meaning to dt,

sup_{w∈S^{n−1}} | D_{t}^{k}P û(t, w) |∈ t^{−s−k− \frac{n−1}{2}} L^{2} (ℝ^{+}, dt) . (6.27)

So we are in the one dimensional case, with s replaced by s + \frac{n−1}{2}. Now we can rewrite (6.27) as

D_{t}t^{q}D_{t}^{k}P û ∈ t^{r}L^{2}, ∀ k, r − q = −s − k − \frac{n − 1}{2} − 1. (6.28)

Now, observe the simple case:

f = 0t < 1, D_{t}f ∈ t^{r}L^{2} ⟹ f ∈ L^{∞} if r < −\frac{1}{2} (6.29)

since

sup |f| = ∫_{−∞}^{t} t^{r}g ≤ (∫ |g|^{2})^{\frac{1}{2}} · ( ∫_{−∞}^{t} t^{2r})^{\frac{1}{2}} .

Thus from (6.28) we deduce ≤ (∫ |g|^{2})^{\frac{1}{2}}

D_{t}^{k}P û ∈ t^{−q}L^{∞} if r < −\frac{1}{2}, i.e. − q > −s − k − \frac{n}{2}.

Finally this gives (6.22) when we go back from polar coordinates, to prove the lemma.

Definition 6.2. Set, for m ∈ ℝ,

I_{c}^{m}(ℝ^{n}, |[0\}) = \{u ∈ C_{c}^{−∞}(ℝ^{n}); û ∈ S^{m− \frac{n}{4}} (ℝ^{n})\}

with this definition,

IH^{s}(ℝ^{n}, {0}) ⊂ I_{c}^{m}(ℝ^{n}, {0}) ⊂ I_{c}^{s′}(ℝ^{n}, {0}) (6.30)

provided

s > −m − \frac{n}{4} > s′.

6. THE WAVE KERNEL 93

Exercise 3. Using Lemma 24, prove (6.30) carefully.

So now what we want to do is to define I_{c}^{m}(X, \{p\}; Ω^{\frac{1}{2}} ) for any p ∈ X by

\begin{matrix}u ∈ I_{c}^{m}(X, \{p\}; Ω^{\frac{1}{2}} ) ⟺ F^{∗}(ϕu) ∈ I_{c}^{m}(ℝ^{n}, \{0\}), \\ u ↾ X\\{p\} ∈ C^{∞}(X\\{p\}). \end{matrix} (6.31)

Here we have a little problem, namely we have to check that I^{m}(ℝ^{n}, {0}) is invariant under coordinate changes. Fortunately we can do this using (6.30).

Lemma 6.3. If F : Ω ⟶ ℝ^{n} is a diffeomorphism of a neighbourhood of 0 onto its range, with F (0) = 0, then F^{∗}{u ∈ I_{c}^{m}(ℝ^{n}, {0}; supp(u) ⊂ F (Ω)} ⊂ I_{c}^{m}(ℝ^{n}, {0}). (6.32)

Proof. Start with a simple case, that F is linear. Then

u = (2π)^{−n} ∫ e^{ixξ}a(ξ)dξ, a ∈ S^{m− \frac{n}{4}} (ℝ^{n}).

so

\begin{aligned}F^{∗}u & = (2π)^{−n} ∫ e^{iAx·ξ}a(ξ)dξ F x = Ax \\ & = (2π)^{−n} ∫ i^{ix·A^{t}ξ}a(ξ)dξ \\ & = (2π)^{−n} ∫ e^{ix·η}a((A^{t})^{−1}η)| det A|^{−1}dη. \end{aligned} (6.33)

Since a((A^{t})^{−1}η)| det A|^{−1} ∈ S^{m− \frac{n}{4}} ℝ^{n}) we have proved the result for linear transformations. We can always factorize F is F = G · A, A = (F_{∗}) (6.34) so that the differential of G at 0 is the identity, i.e. G(x) = x + O(|x|^{2}). (6.35)

Now (6.35) allows us to use an homotopy method, i.e. set G_{s}(x) = x + s(G(x) − x) s ∈ [0, 1) (6.36) so that G_{0} = Id, G_{s} = G. Such a 1-parameter family is given by integration of a vector field:

G_{s}^{∗}ϕ = ∫_{0}^{s} \frac{d}{ds}G_{s}^{∗}ϕdx (6.37)

= ∫_{0} s\frac{d}{ds}ϕ(G_{x}(x))ds (6.38)

= ∑_{1} ∫_{0}^{s} \frac{d}{G}_{s,i}^{ξ}ds (∂x_{j}ϕ) (G_{δ}(x)) ds (6.39)

= ∫_{0}^{s} G_{s}^{∗} (V_{s}ϕ) ds (6.40)

when the coefficients of V_{s} are

G_{s}^{∗}V_{s,j} = \frac{d}{ds}G_{s,i}. (6.41)

Now by (6.36) \frac{d}{ds}G_{s,i} = Σx_{i}x_{j}a_{ij}^{s}(x), so the same is true of the V_{s,i}, again using (6.36).

We can apply (6) to compute

G^{∗}u = ∫_{0}^{′} G_{s}^{∗} (V_{s}u) ds (6.42)

when u ∈ I_{c}^{m} (ℝ^{n}, {0}) has support near 0. Namely, by (6.30), u ∈ IH_{c}^{s} (ℝ^{n}, {0}) , with s < −m − \frac{n}{4}, but then

V_{s}u ∈ IH_{c}^{s+1}(ℝ^{n}, {0})

since V = ∑_{i,j=1}^{n} b_{ij}^{s}(x)x_{i}x_{j}D_{j}. Applying (6.30) again gives

G_{s}^{∗}(V_{s}u) ∈ I^{m′} (ℝ^{n}, {0}) , ∀ m′ > m − 1. (6.43)

This proves the coordinates invariance.

Last time we defined the space of conormal distributions associated to a closed embedded submanifold Y ⊂ X :

\begin{aligned}IH^{s}(X, Y) & = \{u ∈ H^{s}(X); V(X, Y)^{k}u ⊂ H^{s}(X) ∀ k\} & (6.44) \\ IH^{∗}(X, Y) & = I^{∗}(X, Y) = ⋃ sIH^{s}(X, Y). & (6.45) \end{aligned}

Here V(X, Y) is the space of C^{∞} vector fields on X tangent to Y. In the special case of a point in ℝ^{n}, say 0, we showed that

u ∈ I_{c}^{∗}(ℝ^{n}), {0}) ⟺ u ∈ C_{c}^{−∞}(ℝ^{n}) and û ∈ S^{M}(ℝ^{n}), M = M(u). (6.46)

In fact we then defined the “standard order filtration” by

u ∈ I_{c}^{m}(ℝ^{n}, \{0\}) = \{u ∈ C_{c}^{−∞}(ℝ^{n}); û ∈ S^{m− \frac{n}{4}} (ℝ^{n})\}, (6.47)

and found that

IH_{c}^{s}(ℝ^{n}, \{0\}) ⊂ I_{c}^{−s− \frac{n}{4}} (ℝ^{n}, \{0\}) ⊂ IH_{c}^{s′}(ℝ^{n}, \{0\}) ∀ s′ < s. (6.48)

Our next important task is to show that I_{c}^{m}(ℝ^{n}, {0}) is invariant under coordinate changes. That is, if F : U_{1} ⟶ ℝ^{n} is a diffeomorphism of a neighbourhood of 0 to its range, with F (0) = 0, then we want to show that

F^{∗}u ∈ I_{c}^{m}(ℝ^{n}, {0}) ∀ u ∈ I_{c}^{m}(ℝ^{n}, {0}), supp(u) ⊂ F (U_{1}). (6.49)

Notice that we already know the coordinate independence of the Sobolev-based space, so using (6.48), we deduce that

F^{∗}u ∈ I_{c}^{m′}(ℝ^{n}, {0}) ∀ u ∈ I_{c}^{m}(ℝ^{n}, {0}), n′ > m, supp(u) ⊂ F (U_{1}). (6.50)

In fact we get quite a lot more for our efforts:

Lemma 6.4. There is a coordinate-independent symbol map:

I^{m}(X, \{p\}; Ω^{\frac{1}{2}} )@ > σ_{Y}^{m} >> S^{m+ \frac{n}{4} −[J]} (T_{p}^{∗}ℝ^{n}; Ω^{\frac{1}{2}} ) (6.51)

given by the local prescription

σ_{Y}^{m}(u) = û(ξ)|dξ|^{\frac{1}{2}} (6.52)

where u = v|dx|^{\frac{1}{2}} is local coordinate based at 0, with ξ the dual coordinate in T_{p}^{∗}X.

Proof. Our definition of I^{m}(X, \{p\}; Ω^{\frac{1}{2}} ) is just that in any local coordinate based at p

u ∈ I^{m}(X, \{p\}; Ω^{\frac{1}{2}} ) ⟹ ϕu = v|dx|^{\frac{1}{2}} , v ∈ I_{c}^{m}(ℝ^{n}, \{0\}) (6.53)

and u ∈ C^{∞}(X\\{p\}; Ω^{\frac{1}{2}} ). So the symbol map is clearly supposed to be

σ^{m}(u)^{(ζ)} ≡_{↓} v̂(ξ)|dξ|^{\frac{1}{2}} ∈ S^{m+ \frac{n}{4} −[1]}(ℝ^{n}; Ω^{\frac{1}{2}} ) (6.54)

where ζ ∈ T_{p}^{∗}X is the 1-form ζ = ξ·dx in the local coordinates. Of course we have to show that (6.54) is independent of the choice of coordinates. We already know that a change of coordinates changes v̂ by a term of order m− \frac{n}{4} −1, which disappears in (6.54) so the residue class is determined by the Jacobian of the change of variables. From (6.33) we see exactly how v̂ transforms under the Jacobian, namely as a density on

T_{0}^{∗}ℝ^{n} : A ∈ GL(n, ℝ) ⟹ Â^{∗}v(η)|dη|^{\frac{1}{2}}

= v̂((A^{t})^{−1}η)| det A|^{−1}|dy|

so η = A^{t}ξ ⟹

Â^{∗}v(η)|dy| = v̂(ξ)|dξ|. (6.55)

However recall from (6.53) that u is a half-density, so actually in the new coordinates v′ = A^{∗}v · | det A|^{\frac{1}{2}} . This shows that (6.54) is well-defined.

Before going on to consider the general case let us note a few properties of I^{m}(X, \{p\}, Ω^{\frac{1}{2}} ) :

Exercise: Prove that

If P ∈ Diff^{m}(X; Ω^{\frac{1}{2}} ) then
P : I^{m}(X, \{p\}; Ω^{\frac{1}{2}} ) ⟶ I^{m+M}(X, \{p\}; Ω^{\frac{1}{2}} ) ∀ m
σ^{m+M}(P u) = σ^{M}(P) · σ^{m}(u).

(6.56)

To pass to the general case of Y ⊂ X we shall proceed in two steps. First let’s consider a rather ‘linear’ case of X = V a vector bundle over Y. Then Y can be identified with the zero section of V. In fact V is locally trivial, i.e. each p ∈ y has a neighbourhood U s.t.

π^{−1}(U) ≃ ℝ_{x}^{n} × U′_{y′}U′ ⊂ ℝ^{p} (6.57)

by a fibre-linear diffeomorphism projecting to a coordinate system on this base. So we want to define

I^{m}(V, Y; Ω^{\frac{1}{2}} ) = \{u ∈ I^{∗}(V, Y; Ω^{\frac{1}{2}} );

of ϕ ∈ C_{c}^{∞}(U) then under any trivialization (6.57)

\begin{matrix}ϕu(x, y) ≡ (2π)^{−n} ∫ e^{ix·ξ}a(y, ξ)dξ|dx|^{\frac{1}{2}} , mod C^{∞}, \\ a ∈ S^{m− \frac{n}{2} − \frac{p}{4}} (ℝ_{y}^{p}, ℝ_{ξ}^{n}). \end{matrix} (6.58)

Here p = dim Y, p+n = dim V. Of course we have to check that (6.58) is coordinateindependent. We can write the order of the symbol, corresponding to u having order m as

(6.59) m − \frac{dim V}{4} + \frac{dim Y}{2} = m + \frac{dim V}{4} − \frac{codim Y}{2} .

These additional shifts in the order are only put there to confuse you! Well, actually they make life easier later.

Notice that we know that the space is invariant under any diffeomorphism of the fibres of V, varying smoothly with the base point, it is also obvious that (6.58) in independent the choice of coordinates is U′, since that just transforms these variables. So a general change of variables preserving Y is

(y, x) ⟼ (f(y, x), X(y, x)) X(y, 0) = 0.

In particular f is a local diffeomorphism, which just changes the base variables in (6.58), so we can assume f(y) ≡ y. Then X(y, x) = A(y) · x + O(x^{2}). Since x ⟼ A(y) · x is a fibre-by-fibre transformation it leaves the space invariant too, So we are reduced to considering

G : (y, x) ⟼ (y, x + Σa_{ij}(x, y)x_{i}x_{j})y + Σb_{i}(x, y)x_{i}. (6.60)

To handle these transformations we can use the same homotopy method as before i.e.

G_{s}(x, y_{=}(y + s) ∑_{i} b_{i}(x, y)x_{i}, x + s ∑_{i,j} a_{ij}(x, y)x_{i}x_{j}) (6.61)

is a 1-parameter family of diffeomorphisms. Moreover

\frac{d}{ds}G_{s}^{∗}u = G_{s}^{∗}V_{s}k

where

V_{s} = ∑_{i,ℓ} β_{i,ℓ}(s, x, y)x_{i}∂_{y_{ℓ}} + ∑_{i,j,k} α_{i,j,k} + ∑_{i,j,k} α_{ijk}(α, y, s)ℓ_{i}, ℓ_{j} \frac{∂}{∂x_{k}} .

So all we really have to show is that

V_{s} : I^{M}(U′ × ℝ^{n}, U′ × {0}) ⟶ I^{M−1}(U′ × ℝ^{n}, U′ × {0}) ∀ M. (6.62)

Again the spaces are C^{∞}-modules so we only have to check the action of x_{i}∂_{y_{ℓ}} and x_{i}x + j∂_{x_{k}}. These change the symbol to

D_{ξ_{i}}∂_{y_{ℓ}}a and iD_{ξ_{i}}D_{ξ_{j}} · ξ_{k}a

respectively, all one order lower.

This shows that the definition (6.58) is actually a reasonable one, i.e. as usual it suffices to check it for any covering by coordinate partition.

Let us go back and see what the symbol showed before.

6. THE WAVE KERNEL 97

Lemma 6.5. If

u ∈ I^{m}(V, Y; Ω^{\frac{1}{2}} )u = v|dx|^{\frac{1}{2}} |dξ|^{\frac{1}{2}}

defines an element

σ^{m}(u) ∈ S^{m+ \frac{n}{4} + \frac{p}{4} −[1]}(V^{∗}; Ω^{\frac{1}{2}} )

independent of choices.

Last time we discussed the invariant symbol for a conormal distribution associated to the zero section of a vector bundle. It turns out that the general case is not any more complicated thanks to the “tubular neighbourhood” or “normal fibration” theorem. This compares Y ↪ X, a closed embedded submanifold, to the zero section of a vector bundle.

Thus at each point y ∈ Y consider the normal space: N_{y}Y = N_{y}{X, Y } = T_{y}x/T_{y}Y. (6.63) That is, a normal vector is just any tangent vector to X modulo tangent vectors to Y. These spaces define a vector bundle over Y :

NY = N{X; Y } = ⨆_{y∈Y} N_{y}Y (6.64)

where smoothness of a section is inherited from smoothness of a section of T_{y}X, i.e. NY = T_{y}X/T_{y}Y. (6.65)

Suppose Y_{i} ⊂ X_{i} are C^{∞} submanifolds for i = 1, 2 and that F : X_{1} ⟶ X_{2} is a C^{∞} map such that F (Y_{1}) ⊂ Y_{2}. (6.66) Then F_{∗} : T_{y}X_{1} ⟶ T_{F (y)}X_{2}, must have the property F_{∗} : T_{y}Y_{1} ⟶ T_{F (y)}Y_{2} ∀ y ∈ Y_{1}. (6.67) This means that F_{∗} defines a map of the normal bundles

\xymatrix{F_{∗} : NY_{1} \ar[r] \ar[d] & NY_{2} \ar[d] \\ Y_{1} \ar[r]_{F} & Y_{2}.} (6.68)

Notice the very special case that W ⟶ Y is a vector bundle, and we consider Y ↪ W as the zero section. Then N_{y}{W; Y } ⟷ W_{y} ∀ y ∈ Y (6.69) since T_{y}W = T_{y}Y ⊕ T_{y}(W_{y}) ∀ y ∈ W. (6.70) That is, the normal bundle to the zero section is naturally identified with the vector bundle itself.

So, suppose we consider C^{∞} maps f : B ⟶ N{X; Y } = NY (6.71) where B ⊂ X is an open neighbourhood of the submanifold Y. We can demand that f(y) = (y, 0) ∈ N_{y}Y ∀ y ∈ Y (6.72) which is to say that f induces the natural identification of Y with the zero section of NY and moreover we can demand

(6.73) f_{∗} : NY ⟶ NY is the identity.

Here f_{∗} is the map (6.68), so maps NY to the normal bundle to the zero section of NY, which we have just observed is naturally just NY again.

Theorem 6.1. For any closed embedded submanifold Y ⊂ X there exists a normal fibration, i.e. a diffeomorphism (onto its range) (6.71) satisfing (6.72) and (6.73); two such maps f_{1}, f_{2} are such that g = f_{2} ∘ f_{1}^{−1} is a diffeomorphism near the zero section of NY, inducing the identity on Y and inducing the identity (6.73).

Proof. Not bad, but since it uses a little Riemannian geometry I will not prove it, see [ ], [ ]. (For those who know a little Riemannian geometry, f^{−1} can be taken as the exponential map near the zero section of NY, identified as a subbundle of T_{Y} X using the metric.) Of course the uniqueness part is obvious.

Actually we do not really need the global aspects of this theorem. Locally it is immediate by using local coordinates in which Y = {x_{1} = ⋯ = x_{k} = 0}.

Anyway using such a normal fibration of X near Y (or working locally) we can simply define

I^{m}(X, Y; Ω^{\frac{1}{2}} ) = \{u ∈ C^{−∞}(X; Ω^{\frac{1}{2}} ); u is C^{∞} in X\Y and
(f^{−1})^{∗}(ϕu) ∈ I^{m}(NY, Y; Ω^{\frac{1}{2}} ) if ϕ ∈ C^{∞}(X), supp(ϕ) ⊂ B\}.

(6.74)

Naturally we should check that the definition doesn’t depend on the choice of f. This means knowing that I^{m}(NY, Y; Ω^{\frac{1}{2}} ) is invariant under g, as in the theorem, but we have already checked this. In fact notice that g is exactly of the type of (6.60). Thus we actually know that

σ^{m}(g^{∗}u) = σ^{m}(u) in S^{m+ \frac{n}{4} + \frac{p}{4} −[1]}(N^{∗}Y; Ω^{\frac{1}{2}} ).

So we have shown that there is a coordinate invariance symbol map

σ^{m} : I^{m}(X, Y; Ω^{\frac{1}{2}} ) ⟶ S^{m+ \frac{n}{4} + \frac{p}{4} −[1]}(N^{∗}Y; Ω^{\frac{1}{2}} ) (6.75)

giving a short exact sequence

0 ↪ I^{m−1}(X, Y; Ω^{\frac{1}{2}} ) ⟶ I^{m}(X, Y; Ω^{\frac{1}{2}} )@ > σ^{m} >> S^{m+ \frac{n}{4} + \frac{p}{4} −[1]}(N^{∗}Y; Ω^{\frac{1}{2}} ) ⟶ 0 (6.76)
where n = dim X − dim Y, p = dim Y.

Asymptotic completeness carries over immediately. We also need to go back and check the extension of (6.56):

Proposition 6.1. If Y ↪ X is a closed embedded submanifold and A ∈ Ψ_{c}^{m}(X; Ω^{\frac{1}{2}} ) then

A : I^{M}(X, Y; Ω^{\frac{1}{2}} ) ⟶ I^{M+m}(X, Y; Ω^{\frac{1}{2}} ) ∀ M (6.77)

and

σ^{m+M}(Au) = σ^{m}(A)σ^{m}(A) ↾ N^{∗}Y σ^{M}(u). (6.78)

Notice that σ^{m}(A) ∈ S^{m−[1]}(T^{∗}X) so the product here makes perfectly good sense.

Proof. Since everything in sight is coordinate-independent we can simply work in local coordinates where

X ∼ ℝ_{y}^{p} × ℝ_{x}^{n}, Y = {x = 0}. (6.79)

Then u ∈ I_{c}^{m}(X, Y; Ω^{\frac{1}{2}} ) means just

u = (2π)^{−n} ∫ e^{ix·ξ}a(y, ξ)dξ · |dx|^{\frac{1}{2}} , a ∈ S^{m− \frac{n}{4} + \frac{p}{4}} (ℝ^{p}, ℝ^{n}). (6.80)

Similarly A can be written in the form

A = (2π)^{−n−p} ∫ e^{i(x−x′)·ξ+i(y−y′)·η}b(x, y, ξ, η)dξdη. (6.81)

Using the invariance properties of the Sobolev based space if we write

A = A_{0} + Σx_{j}B_{j}, A_{0} = q_{L}(b(0, y, ξ, η))

we see that Au ∈ I^{m+M}(X, Y; Ω^{\frac{1}{2}} ) is equivalent to A_{0}u ∈ I^{m+M}(X, Y; Ω^{\frac{1}{2}} ). Then

A_{0}u = (2π)^{−n−p} ∫ e^{ix·ξ+i(y−y′)·η}b(0, y′, ξ, η)b(y′, ξ)dy′dηdξ,

where we have put A_{0} in right-reduced form. This means

A_{0}u = (2π)^{−n} ∫ e^{ix·ξ}c(y, ξ)dξ

where

c(y, ξ) = (2π)^{−p} ∫ e^{i(y−y′)·η}b(0, y′, ξ, η)a(y′, ξ)dy′dη.

Regarding ξ as a parameter, this is, before y′ integration, the kernel of a pseudodifferential operator is y. It can therefore be written in left-reduced form, i.e.

c(y, ξ) = (2π)^{−p} ∫ e^{i(y−y′)η}e(y, ξ, η)dηdy′ = e(y, ξ, 0) (6.82)

where e(y, ξ, η) = b(0, y, ξ, η)a(y, ξ) plus terms of order at most m + M − \frac{n}{4} + \frac{p}{4} − 1. This proves the formula (6.78).

Notice that if A is elliptic then Au ∈ C^{∞} implies u ∈ C^{∞}, i.e. there are no singular solutions. Suppose that P is say a differential operator which is not elliptic and we look for solutions of

P u ∈ C^{∞}(XΩ^{\frac{1}{2}} ). (6.83)

How can we find them? Well suppose we try

u ∈ I^{M}(X, Y; Ω^{\frac{1}{2}} ) (6.84)

for some submanifold Y. To know that u is singular we will want to have

(6.85) σ(u) is elliptic on N^{∗}Y

(which certainly implies that u ∉ C^{∞}).

The simplest case would be Y a hypersurface. In any case from (6.78) and (6.83) we deduce

σ^{m}(P) · σ^{M}(u) ≡ 0. (6.86)

So if we assume (6.85) then we must have

σ^{m}(P) ↾ N^{∗}Y = 0. (6.87)

100 6. THE WAVE KERNEL

Definition 6.3. A submanifold is said to be characteristic for a given operator P ∈ Diff^{m}(X; Ω^{\frac{1}{2}} ) if (6.87) holds.

Of course even if P is characteristic for y, and so (6.86) holds we do not recover (6.83), just P u ∈ I^{m+M−1}(X, Y; Ω^{\frac{1}{2}} ) (6.88) i.e., one order smoother than it “should be”. The task might seem hopeless, but let us note that these are examples, and important ones at that!!

Consider the (flat) wave operator

P = P_{t}^{2} − ∑_{i=1}^{n} D_{i}^{2} = D_{t}^{2} − Δ on ℝ^{n+1}. (6.89)

A hypersurface in ℝ^{n+1} looks like H = {h(t, x) = 0}, (dh ≠ 0 on H) . (6.90) The symbol of P is σ^{2}(P) = τ^{2} − |ξ|^{2} = τ^{2} − ξ_{1}^{2} − ⋯ − ξ_{n}^{2}, (6.91) where τ, ξ are the dual variables to t, x. So consider (6.87), N^{∗}Y = {(t, x; λdh(t, y)); h(t, x) = 0}. (6.92) Inserting this into (6.91) we find:

(λ\frac{∂h}{∂t})^{2} − (λ\frac{∂h}{∂x_{1}} )^{2} − ⋯ − (λ\frac{∂h}{∂x_{n}} )^{2} = 0 on h = 0 (6.93)

i.e. simply:

(\frac{∂h}{∂t})^{2} = |d_{x}h|^{2} on h(t, x) = 0. (6.94)

This is the “eikonal equation” for h (and hence H).

Solutions to (6.94) are easy to find – we shall actually find all of them (locally) next time. Examples are given by taking h to be linear: (6.95) H = {h = at + b · x = 0} is characteristic for P ⟺ a^{2} = |b|^{2}. Since h/a defines the same surface, all the linear solutions correspond to planes t = ω · x, ω ∈ 𝕊^{n−1}. (6.96)

So, do solutions of P u ∈ C^{∞} which are conormal with respect to such hypersurfaces exist? Simply take u = v(t − ω · x) v ∈ I^{∗}(ℝ, \{0\}; Ω^{\frac{1}{2}} ). (6.97) Then P u = 0, u ∈ I^{∗}(ℝ^{n+1}, H; Ω^{\frac{1}{2}} ). (6.98) For example v(s) = δ(s), u = δ(t − ω · x) is a “travelling wave”.

6.1. Hamilton-Jacobi theory

Let X be a C^{∞} manifold and suppose p ∈ C^{∞}(T^{∗}X\0) is homogeneous of degree m. We want to find characteristic hypersurfaces for p, namely hypersurfaces (locally) through x̄ ∈ X

H = {f(x) = 0} h ∈ C^{∞}(x)h(x̄) = 0, dh(x̄) ≠ 0 (6.99)

such that

p(x, dh(x)) = 0. (6.100)

Here we demand that (6.100) hold near x̄, not just on H itself. To solve (6.100) we need to impose some additional conditions, we shall demand

(6.101) p is real-valued

and

d_{fibre}p ≠ 0 or Σ(p) = {p = 0} ⊂ T^{∗}X\0. (6.102)

This second condition is actually stronger than really needed (as we shall see) but in any case it implies that

(6.103) Σ(P) ⊂ T^{∗}X\0 is a C^{∞} conic hypersurface

by the implicit function theorem.

The strategy for solving (6.100) is a geometric one. Notice that

Λ_{h} = {(x, dh(x)) ∈ T^{∗}X\0} (6.104)

actually determines h up to an additive constant. The first question we ask is – precisely which submanifold Λ ⊂ T^{∗}X\0 corresponds to graphs of differentials of C^{∞} functions? The answer to this involves the tautologous contact form.

\begin{matrix}α : T^{∗}X ⟶ T^{∗}(T^{∗}X) ⊄ π̃ ∘ α = Id \\ α(x, ξ) = π̃^{∗}ξ ∈ T_{(x,ξ)}^{∗}(T^{∗}X). \end{matrix} (6.105)

Here π̃ : T^{∗}(T^{∗}X) ⟶ T^{∗}X is the projection. Notice that if x_{1}, …, x_{n} are local coordinates in X then x_{1}, …, x_{n}, ξ_{1}, …, ξ_{n} are local coordinates T^{∗}X, where ξ ∈ T_{x}^{∗}X is written

ξ = ∑_{i=1}^{n} ξ_{i}dx_{i}. (6.106)

Since x_{1}, …, x_{n}, ξ_{1}, …, ξ_{n} are local coordinates in T^{∗}X they together with the dual coordinates Ξ_{1}, …, Ξ_{n}, X_{1}, …, X_{n} are local coordinates in T^{∗}(T^{∗}X) where

ζ ∈ T_{(x,ξ)}^{∗}(T^{∗}X) ⟹ ζ = ∑_{j=1}^{n} Ξ_{j}dx_{j} + ∑_{j=1}^{n} X_{j}dξ_{j}. (6.107)

In these local coordinates

α = ∑_{j=1}^{n} ξ_{j}dx_{j}! (6.108)

The first point is that α is independent of the original choice of coordinates, as is evident from (6.105).

Lemma 6.6. A submanifold Λ ⊂ T^{∗}X\0 is, near (x̄, ξ̄) ∈ Λ, of the form (6.104) for some h ∈ C^{∞}(X), if

(6.109) π : Λ ⟶ X is a local diffeomorphism

and

(6.110) α restricted to Λ is exact.

Proof. The first condition, (6.109), means that Λ is locally the image of a section of T^{∗}X :

Λ = {(x, ζ(x)), ζ ∈ C^{∞}(X; T^{∗}X)}.

Thus the section ζ gives an inverse Z to π in (6.109). It follows from (6.105) that

Z^{∗}α = ζ.

Thus if α is exact on Λ then ζ is exact on X, ζ = dh as required.

Of course if we are only working locally near some point (x̄, ξ̄) ∈ Λ then (6.110) can be replaced by the condition

ω = dα = 0 on X. (6.111)

Here ω = dα is the symplectic form on T^{∗}X :

ω = ∑_{j=1}^{n} dξ_{j} ∧ dx_{j}. (6.112)

Definition 6.4. A submanifold Λ ⊂ T^{∗}X of dimension equal to that of X is said to be Lagrangian if the fundamental 2-form, ω, vanishes when pulled back to Λ.

By definition a symplectic manifold is a C^{∞} manifold S with a C^{∞} 2-form ω ∈ C^{∞}(S; Λ^{2}) fixed satisfying two constraints

dω = 0 (6.113)

(6.114)  ω ∧ ⋯ ∧_{n factors} ω ≠ 0  dim S = 2n.

A particularly simple example of a symplectic manifold is a real vector space, necessarily of even dimension, with a non-degenerate antisymmetric 2-form:

\begin{cases}ω : E × E ⟶ ℝ \\ ω̃ : E ⟷ E^{∗}. \end{cases} (6.115)

Here ω̃(v)(w) = ω(v, w) ∀ w ∈ E. Now (6.113) is trivially true if we think of ω as a translation-invariant 2-form on E, thought of as a manifold.

Then a subspace V ⊂ E is Lagrangian if

\begin{aligned}ω(v, w) & = 0 ∀ v, w ∈ V & (6.116) \\ 2 dim V & = dim E. & (6.117) \end{aligned}

Of course the point of looking at symplectic vector spaces and Lagrangian subspaces is:

Lemma 6.7. If S is a symplectic manifold then T_{z}S is a symplectic vector space for each z ∈ S. A submanifold Λ ⊂ S is Lagrangian iff T_{z}Λ ⊂ T_{z}S is a Lagrangian subspace ∀ z ∈ Λ.

We can treat ω, the antisymmetric 2-form on E, as though it were a Euclidean inner product, at least in some regards! Thus if W ⊂ E is any subspace set

W^{ω} = {v ∈ E; ω(v, w) = 0 ∀ w ∈ W }. (6.118)

Lemma 6.8. If W ⊂ E is a linear subspace of a symplectic vector space then dim W^{ω} + dim W = dim E; W is Lagrangian if and only if

W^{ω} = W. (6.119)

Proof. Let W^{0} ⊂ E^{∗} be the usual annihilator:

W^{0} = {α ∈ E^{∗}; α(v) = 0 ∀ v ∈ W }.

Then dim W^{0} = dim E − dim W. Observe that

ω̃ : W^{ω} ⟷ W^{0}. (6.120)

Indeed if α ∈ W^{0} and ω̃(v) = α then

α(w) = ω̃(v)(w) = ω(v, w) = 0 ∀ w ∈ W

implies that v ∈ W^{ω}. Conversely if v ∈ W^{ω} then α = ω̃(v) ∈ W^{0}. Thus dim W^{ω} + dim W = dim E.

Now if W is Lagrangian then α = ω̃(w), w ∈ W implies

α(v) = ω̃(w)(v) = ω(w, v) = 0 ∀ v ∈ w.

Thus ω̃(W) ⊂ W^{0} ⟹ W ⊂ W^{ω}, by (6.120), and since dim W = dim W^{ω}, (6.119) holds. The converse follows similarly.

The “lifting” isomorphism ω̃ : E ⟷ E^{∗} for a symplectic vector space is like the Euclidean identification of vectors and covectors, but “twisted”. It is of fundamental importance, so we give it several names! Suppose that S is a symplectic manifold. Then

ω̃_{z} : T_{z}S ⟷ T_{z}^{∗}S ∀ z ∈ S. (6.121)

This means that we can associate (by the inverse of (6.121)) a vector field with each 1-form. We write this relation as

\begin{matrix}H_{γ} ∈ C^{∞}(S; T S) if γ ∈ C^{∞}(S; T^{∗}S) and \\ ω̃_{z}(H_{γ}) = γ ∀ z ∈ S. \end{matrix} (6.122)

Of particular importance is the case γ = df, f ∈ C^{∞}(S). Then H_{df} is written H_{f} and called the Hamilton vector field of f. From (6.122)

ω(H_{f}, v) = df(v) = vf ∀ v ∈ T_{z}S, ∀ z ∈ S. (6.123)

The identity (6.123) implies one important thing immediately:

H_{f}f ≡ 0 ∀ f ∈ C^{∞}(S) (6.124)

since

H_{f}f = df(H_{f}) = ω(H_{f}, H_{f}) = 0

by the antisymmetry of ω. We need a generalization of this:

Lemma 6.9. Suppose L ⊂ S is a Lagrangian submanifold of a symplectic manifold then for each f ∈ I(S) = {f ∈ C^{∞}(X); f ↾ {s = 0}, H_{f} is tangent to Λ.

Proof. H_{f} tangent to Λ means H_{f}(z) ∈ T_{z}Λ ∀ z ∈ Λ. If f = 0 on Λ then df = 0 on T_{z}Λ, i.e. df(z) ∈ (T_{z}Λ)^{0} ⊂ (T_{z}S) ∀ z ∈ Λ. By (6.119) the assumption that Λ is Lagrangian means ω̃_{z}(df(z)) ∈ T_{z}Λ, i.e. H_{f}(z) ∈ T_{ζ}Λ as desired.

This lemma gives us a necessary condition for our construction of a Lagrangian submanifold

Λ ⊂ Σ(P). (6.125)

Namely H_{p} must be tangent to Λ! We use this to construct Λ as a union of integral curves of H_{p}. Before thinking about this seriously, let’s look for a moment at the conditions we imposed on p, (6.101) and (6.102). If p is real then H_{p} is real (since ω is real). Notice that

(6.126) If S = T^{∗}X then each fibre T_{x}^{∗}X ⊂ T^{∗}X is Lagrangian .

Remember that on T^{∗}X, ω = dα, α = ξ · dx the canonical 1-form. Thus T_{x}^{∗}X is just x = const, so dx = 0, so α = 0 on T_{x}^{∗}X and hence in particular ω = 0, proving (6.126). This allows us to interpret (6.102) in terms of H_{p} as

(6.127) (6.102) ⟷ H_{p} is everywhere transversal to the fibres T_{x}^{∗}X.

Now we want to construct a little piece of Lagrangian manifold satisfying (6.125). Suppose z ∈ Σ(P) ⊂ T^{∗}X\0 and we want to construct a piece of Λ through z. Since π_{∗}(H_{p}(z)) ≠ 0 we can choose a local coordinate, t ∈ C^{∞}(X), such that

π_{∗}(H_{p}(z))t ≠ 0, i.e. H_{p}(π^{∗}t)(z) ≠ 0. (6.128)

Consider the hypersurface through π(z) ∈ X,

H = {t = t(z)} ⟹ π(z) ∈ H. (6.129)

Suppose f ∈ C^{∞}(H), df(π(z)) = 0. In fact we can choose f so that

f = f′ ↾ H, f′ ∈ C^{∞}(X), df′(π(z)) = z (6.130)

where z ∈ Ξ(P) was our chosen base point.

Theorem 6.2. (Hamilton-Jacobi) Suppose p ∈ C^{∞}(T^{∗}X\0) satisfies (6.101) and (6.102) near z ∈ T^{∗}X\0, H is a hypersurface through π(z) as in (6.128), (6.125) and f ∈ C^{∞}(H) satisfies (6.130), then there exists f̃ ∈ C^{∞}(X) such that

Λ = graph (df̃) ⊂ Σ(P) near z
f̃ ↾ H = f near π(z)
df̃(π(z)) = z

(6.131)

and any other such solution, f̃′, is equal to f̃ in a neighbourhood of π(z).

Proof. We need to do a bit more work to prove this important theorem, but let us start with the strategy. First notice that Λ ∩ π^{−1}(H) is already determined, near π(z).

To see this we have to understand the relationship between df(h) ∈ T^{∗}H and df̃(h) ∈ T^{∗}X, h ∈ H, f̃ ↾ H = f. Observe that H = {t = 0} lifts to T_{H}^{∗}X ⊂ T^{∗}X a hypersurface. By (6.124), H_{t} is tangent to T_{H}^{∗}X and non-zero. In local coordinates t, x, …, x_{n−1}, the x’s in H,

H_{t} = −\frac{∂}{∂τ}

where τ, ξ_{1}, …, ξ_{n} are the dual coordinates. Thus we see that

π_{H} : T_{H}^{∗}X ⟶ T^{∗}H π_{H}(β)(v) = β(v), v ∈ T_{h}H ⊂ T_{h}X, (6.132) is projection along ∂_{τ} . Now starting from f ∈ C^{∞}(H) we have

Λ_{f} ⊂ T^{∗}H.

Notice that if f̃ ∈ C^{∞}(X), f̃ | H = f then

(6.133) Λf̃ ∩ T_{H}^{∗}X has dimension n − 1

and

π_{H}(Λf̃ ∩ T_{H}^{∗}X) = Λ_{f}. (6.134)

The first follows from the fact that Λf̃ is a graph over X and the second from the definition, (6.132). So we find

Lemma 6.10. If z ∈ Σ(P) and H is a hypersurface through π(z) satisfying (6.128) and (6.129) then π_{H}^{P} : (Σ(P) ∩ T_{H}^{∗}X) ⟶ T^{∗}H is a local diffeomorphism in a neighbourhood z; if (6.130) is to hold then

Λf̃ ∩ T_{H}^{∗}X = (π_{H}^{P})^{−1}(Λ_{f}) near z. (6.135)

Proof. We know that H_{p} is tangent to Σ(P) but, by assumption (6.130) is not tangent to T_{H}^{∗}X at z. Then Σ(P)∩T_{H}^{∗}X does have dimension 2n−1−1 = 2(n−1). Moreover π_{H}is projection along ∂_{τ} which cannot be tangent to Σ(P) ∩ T_{H}^{∗}X (since it would be tangent to Σ(P)). Thus π_{H}^{P} has injective differential, hence is a local isomorphism.

So this is our strategy:

Start with f ∈ C^{∞}(H), look at Λ_{f} ⊂ T^{∗}H, lift to Λ ∩ T_{H}^{∗}X ⊂ Σ(P) by π_{H}^{P}. Now let

Λ = ⋃{H_{p} − curves through (π_{H}^{P})^{−1}(Λ_{f})}.

This we will show to be Lagrangian and of the form Λf̃, it follows that

p(x, df̃) = 0, f̃ ↾ H = f. (6.136)

6.2. Problems

Problem 6.2. Prove that if E is a symplectic vector space, with non-degenerate bilinear form ω then there is a basis v_{1}, …, v_{n}, w_{1}, …, w_{n} of E such that in terms of the dual basis of E^{∗}

ω = ∑_{j} v_{j}^{∗} ∧ w_{j}^{∗}. (6.137)

Hint: Construct the w_{j}, v_{j} successive. Choose v_{1} ≠ 0. Then choose w_{1} so that ω(v_{1}, w_{1}) = 1. Then choose v_{2} so ω(v_{1}, v_{2}) = ω(w_{1}, v_{2}) = 0 (why is this possible?) and w_{2} so ω(v_{2}, w_{2}) = 1, ω(v_{1}, w_{2}) = ψ(w_{1}, w_{2}) = 0. Then proceed and conclude that (6.137) must hold.

Deduce that there is a linear transformation T : E ⟶ ℝ^{2n} so that ω = T^{∗}ω_{D}, with ω_{D} given by (6.112).

Problem 6.3. Extend problem 6.2 to show that T can be chosen to map a given Lagrangian plane V ⊂ E to

{x = 0} ⊂ ℝ^{2n}

Hint: Construct the basis choosing v_{j} ∈ V ∀ j!

106 6. THE WAVE KERNEL

Problem 6.4. Suppose S is a symplectic manifold. Show that the Poisson bracket

{f, g} = H_{f}g

makes C^{∞}(S) into a Lie algebra.

6.3. Riemann metrics and quantization Metrics, geodesic flow, Riemannian normal form, Riemann-Weyl quantization.

6.4. Transport equation The first thing we need to do is to finish the construction of characteristic

hypersurfaces using Hamilton-Jacobi theory, i.e. prove Theorem XIX.37. We have already defined the submanifold Λ as follows:

1) We choose z ∈ Σ(P) and t ∈ C^{∞}(X) s.t. H_{p}π^{∗}(t) ≠ 0 at dz, then selected f ∈ C^{∞}(H), H = {t = 0} ∩ Ω, Ω ∋ πz s.t. z(v) = df(v) ∀ v ∈ T_{πz}H. (6.138) Then we consider Λ_{f} = graph{df} = {(x, df(x)), x ∈ H} ⊂ T^{∗}H (6.139) as our “initial data” for Λ. To move it into Σ(P) we noted that the map

Σ(P) ∩ T_{H}^{∗}X_{∥}_{{t=0 in T^{∗}X}} ⟶ T^{∗}H (6.140)

is a local diffeomorphism near z, df(π(z)) by (6.138). The inverse image of Λ_{f} in (6.140) is therefore a submanifold Λ̃_{f} ⊂ Σ(p) ∩ T_{H}^{∗}X of dimension dim X − 1 = dim H. We define (6.141) Λ = ⋃{H_{p} − curves of length ϵ starting on Λ̃_{f}}.

So we already know: (6.142) Λ ⊂ Σ(P) is a manifold of dimension n, and (6.143) π : Λ ⟶ X is a local diffeomorphism near n, What we need to know most of all is that (6.144) Λ is Lagrangian. That is, we need to show that the symplectic two form vanishes identically on T_{z′}Λ, ∀ z′ ∈ Λ (at least near z). First we check this at z itself! Now T_{z}Λ = T_{z}Λ̃_{f} + sp(H_{p}). (6.145) Suppose v ∈ T_{z}Λ̃_{f}, then ω(v, H_{p}) = −dp(v) = 0 since Λ̃_{f} ⊂ Σ(P). (6.146) Of course ω(H_{p}, H_{p}) = 0 so it is enough to consider ω|(T_{z}Λ̃_{f} × T_{z}Λ̃_{f}). (6.147)

6.4. TRANSPORT EQUATION 107

Recall from our discussion of the projection (6.140) that we can write it as projection along ∂_{τ} . Thus

\begin{matrix}ω_{X}(v, w) = ω_{H}(v′, w′) if v, w ∈ T_{z}(T_{H}X), \\ (c_{H}^{∗})_{∗}v = v′(c_{H}^{∗})_{∗}w = w′ ∈ T_{z}(T^{∗}H) \end{matrix} (6.148)

where z = df(π(z)). Thus the form (6.147) vanishes identically because Λ_{f} is Lagrangian.

In fact the same argument applies at every point of the initial surface Λ̃_{f} ⊂ Λ : (6.149) T_{z′}Λ is Lagrangian ∀ z′ ∈ Λ̃_{f}. To extend this result out into Λ we need to use a little more differential geometry. Consider the local diffeomorphisms obtained by exponentiating H_{p} : (6.150) exp(ϵH_{p})(Λ ∩ Ω) ⊂ Λ ∀ ϵ small, Ω ∋ z small. This indeed is really the definition of Λ_{j} more precisely, (6.151) Λ = ⋃_{ϵ small} exp(ϵH_{p})(Λ̃_{f}).

The main thing to observe is that, on T^{∗}H, the local diffeomorphisms exp(ϵH_{p}) are symplectic: exp(ϵH_{p})^{∗}ω_{X} = ω_{X}. (6.152) Clearly (6.152), (6.150) and (6.149) prove (6.144). The most elegant wary to prove (6.152) is to use Cartan’s identity (valid for H_{p} any vector field, ω any form)

\frac{d}{dϵ} exp(ϵH_{p})^{∗}ω = exp(ϵH_{p})^{∗}(L_{H_{p}}ω) (6.153)

where the Lie derivative is given explicitly by L_{V} = d ∘ ι_{V} + ι_{V} ∘ d, (6.154) c_{V} being contradiction with V (i.e. α(·, ·, …) ⟶ α(V, ·, ·, …)). Thus

L_{H_{p}}ω = d(ω(H_{p}, ·)) + ι_{V} (dω)_{∥}_{0} = d(dp) = 0. (6.155)

Thus from (6.142), (6.143) and (6.144) we know that (6.156) Λ = graph(df̃), f̃ ∈ C^{∞}(X), near π(z), must satisfy the eikonal equation p(x, df̃(x)) = 0 near π(z), Hf̃ ↾ H = f (6.157) where we may actually have to add a constant to f̃ to get the initial condition – since we only have df̃ = df on T H.

So now we can return to the construction of travelling waves: We want to find u ∈ I^{∗}(X, G; Ω^{\frac{1}{2}} ) G = \{f = 0\} (6.158) such that u is elliptic at z ∈ Σ(p) and P u ∈ C^{∞}(X). (6.159) So far we have noticed that σ_{m+M}(P u) = σ_{m}(P) ↾ N^{∗}G · σ(u) (6.160)

108 6. THE WAVE KERNEL

so that N^{∗}G ⊂ Σ(p) ⟺ p(x, df) = 0 on f = 0 (6.161) implies P u ∈ I^{m+M−1}(X, G; Ω^{\frac{1}{2}} ) near π(z) (6.162) which is one order smoother than without (6.161).

It is now clear, I hope, that we need to make the “next symbol” vanish as well, i.e. we want σ_{m+M−1}(P u) = 0. (6.163) Of course to arrange this it helps to know what the symbol is!

Proposition 6.2. Suppose P ∈ Ψ^{m}(X; Ω^{\frac{1}{2}} ) and G ⊂ X is a C^{∞} hypersurface characteristic for P (i.e. N^{∗}G ⊂ Σ(P)) then ∀ u ∈ I^{M}(X, G; Ω^{\frac{1}{2}} ) σ_{m+M−1}(P u) = (−iH_{p} + a)σ_{m}(u) (6.164) where a ∈ S^{m−1}(N^{∗}G) and H_{p} is the Hamilton vector field of p = σ_{m}(P).

Proof. Observe first that the formula makes sense since Λ = N^{∗}G is Lagrangian, Λ ⊂ Σ(p) implies H_{p} is tangent to Λ and if p is homogeneous of degree m (which we are implicitly assuming) then L_{H_{p}} : S^{r}(Λ; Ω^{\frac{1}{2}} ) ⟶ S^{r+m−1}(Λ; Ω^{\frac{1}{2}} ) ∀ m (6.165) where one can ignore the half-density terms. So suppose G = {x_{1} = 0} locally, which we can always arrange by choice of coordinates. Then

X = N^{∗}G = {(0, x′, ξ_{1}, 0) ∈ T^{∗}X}.

To say N^{∗}G ⊂ Σ(p) means p = 0 on Λ, i.e.

p = x_{1}q(x, ξ) + ∑_{j>1} ξ_{j}p_{j}(x, ξ) near z (6.166)

with q homogeneous of degree m and the p_{j} homogeneous of degree m−1. Working microlocally we can choose Q ∈ Ψ^{m}(X, Ω^{\frac{1}{2}} ), P_{j} ∈ Ψ^{m−1}(X, Ω^{\frac{1}{2}} ) with σ_{m}(Q) = q, σ_{m−1}(P_{j}) = p_{j} near z. (6.167) Then, from (6.166)

P = x_{1}Q + D_{x_{j}}P_{j} + R + P′, R ∈ Ψ^{m−1}(X; Ω^{\frac{1}{2}} )z ∉ W F′(P′), P′ ∈ Ψ^{m}(X, Ω^{\frac{1}{2}} ). (6.168)

Of course P′ does not affect the symbol near z so we only need observe that σ_{r−1}(x, u) = −d_{ξ_{1}}σ_{r}(u) (6.169) ∀ u ∈ I^{r}(X, G; Ω^{\frac{1}{2}} ) (6.170) σ_{r}(D_{x_{j}}u) = D_{x_{j}}σ_{r}(u). (6.171) This follows from the local expression u(x) = (2π)^{−1} ∫ e^{ix_{1}ξ_{1}}a(x′, ξ_{1})dξ_{1}. (6.172)

Then from (6.168) we get

\begin{matrix}σ_{m+M−1}(P u) = −D_{ξ_{1}}(qσ_{M}(u)) + ∑_{j} D_{x_{j}}(p_{j}σ_{M}(u)) + r · σ_{m}(u) \\ = −i (∑_{j>1} p_{j} ↾ Λ\frac{∂}{∂x_{j}} − q ↾ Λ\frac{∂}{∂ξ_{i}} ) σ_{M}(u) + a′σ_{M}(u). \end{matrix} (6.173)

Observe from (6.166) that the Hamilton vector field of p, at x_{1} = ξ′ = 0 is just the expression in parenthesis. This proves (6.164).

So, now we can solve (6.163). We just set

(6.174) σ_{M}(u)(exp(ϵH_{p})z′) = e^{iϵA} exp(ϵH_{p})^{∗}[b] ∀ z′ ∈ Λ̃_{f} = Λ ∩ {t = 0}.

where A is the solution of

H_{p}A = a, A ↾ t = 0 = 0 on Λ_{0} (6.175)

and b ∈ S^{r}(Λ_{0}) is a symbol defined on Λ_{0} = Λ ∩ {t = 0} near z.

Proposition 6.3. Suppose P ∈ Ψ^{m}(X; Ω^{\frac{1}{2}} ) has homogeneous principal symbol of degree m satisfying

p = σ_{m}(P) is real (6.176)

d _{fibre} p ≠ 0 on p = 0 (6.177)

and z ∈ Σ(p) is fixed. Then if H ∋ π(z) is a hypersurface such that π_{∗}(H_{p}) ∩ H and G ⊂ H is an hypersurface in H s.t.

z̄ = c_{H}^{∗}(z) ∈ H_{πz}^{∗}G

there exist a characteristic hypersurface G̃ ⊂ X for P such that G̃ ∩ H = G near π(z), z ∈ N_{πz}^{∗}G̃. For each

u_{0} ∈ I^{m+ \frac{1}{4}} (H, G; Ω^{\frac{1}{2}} ) with W F (u_{0}) ⊂ γ, (6.178)

γ a fixed small conic neighbourhood of z̄ n T^{∗}H there exists

(6.179) u ∈ I(X, G̃; Ω^{\frac{1}{2}} ) satisfying

u ↾ G = u_{0} near πz ∈ H (6.180)

P u ∈ C^{∞} near πz ∈ X. (6.181)

Proof. All the stuff about G and G̃ is just Hamilton-Jacobi theory. We can take the symbol of u_{0} to be the b in (6.174), once we think a little about halfdensities, and thereby expect (6.180) and (6.181) to hold, modulo certain singularities. Indeed, we would get

u_{1} ↾ G − u_{0} ∈ I^{r+ \frac{1}{4} −1}(H, G; Ω^{\frac{1}{2}} ) near πz ∈ H (6.182)

P u ∈ I^{r+m−2}(X, G̃; Ω^{\frac{1}{2}} ) near πz ∈ X. (6.183)

So we have to work a little to remove lower order terms. Let me do this informally, without worrying too much about (6.180) for a moment. In fact I will put (6.182) into the exercises!

All we really have to observe to improve (6.183) to (6.181) is that

\begin{matrix}g ∈ I^{r}(X, G̃; Ω^{\frac{1}{2}} ) ⟹ ∃ u ∈ I^{r+m−1}(X; G̃; Ω^{\frac{1}{2}} ) \\ s.t. P u − g ∈ I^{r−1}(X, G̃; Ω^{\frac{1}{2}} ) \end{matrix} (6.184)

which we can then iterate and asymptotically sum. In fact we can choose the solution so u ↾ H ∈ C^{∞}, near πz̄. To solve (6.184) we just have to be able to solve

−i(H_{p} + a)σ(u) = σ(g) (6.185)

which we can do by integration (duHamel’s principle).

The equation (6.185) for the symbol of the solution is the transport equation. We shall use this construction next time to produce a microlocal parametrix for P!

6.5. Problems

Problem 6.5. Let X be a C^{∞} manifold, G ⊂ X on C^{∞} hypersurface and t ∈ C^{∞}(X) a real-valued function such that

dt ≠ 0 on T_{p}G ∀ p ∈ L = G ∩ {t = 0}. (6.186)

Show that the transversality condition (6.186) ensures that H = {t = 0} and L = H ∩ G are both C^{∞} submanifolds.

Problem 6.6. Assuming (6.186) show that dt gives an isomorphism of line bundles

Ω^{\frac{1}{2}} (H) ≡ ΩH^{\frac{1}{2}}(X) ∼ ΩH^{\frac{1}{2}}(X)/|dt|^{\frac{1}{2}}

and hence one can define a restriction map,

C^{∞}(X; Ω^{\frac{1}{2}} ) ⟶ C^{∞}(H; Ω^{\frac{1}{2}} ).

Problem 6.7. Assuming 1 and 2, make sense of the restriction formula

↾ H : I^{m} (X, G; Ω^{\frac{1}{2}} ) ⟶ I^{m+ \frac{1}{4}} (H, L; Ω^{\frac{1}{2}} )

and prove it, and the corresponding symbolic formula

σ_{m+ \frac{1}{4}} (u ↾ H) = (ι_{H}^{∗})^{∗} (σ_{m}(u) ↾ N_{L}^{∗}G) /|dτ|^{\frac{1}{2}} .

NB. Start from local coordinates and try to understand restriction at that level before going after the symbol formula!

6.6. The wave equation

We shall use the construction of travelling wave solutions to produce a parametrix, and then a fundamental solution, for the wave equation. Suppose X is a Riemannian manifold, e.g. ℝ^{n} with a ‘scattering’ metrice:

g = ∑_{i,j=1}^{n} g_{ij}(x)dx^{i}dx^{j}, g_{ij} = δ_{ij}|x|R. (6.187)

Then the associates Laplacian, on functions, i.e.

Δu = − ∑_{i,j=1}^{n} \frac{1}{\sqrt{g}} \frac{∂}{∂x_{j}} (δgg^{ij}(x))\frac{∂}{∂x_{i}} u (6.188)

where g^{ij}(x) = (g_{ij}(x))^{−1} and g = det g_{ij}. We are interested in the wave equation

P u = (D_{t}^{2} − Δ)u = f on ℝ × X (6.189)

For simplicity we assume X is either compact, or X = ℝ^{n} with a metric of the form (6.187).

The cotangent bundle of ℝ × X is

T^{∗}(ℝ × X) ≃ T^{∗}ℝ × T^{∗}X

with canonical coordinates (t, x, τ, ξ). In terms of this

σ(P) = τ^{2} − |ξ|^{2}|ξ| = ∑_{i,j=1}^{n} g^{ij}(x)ξ_{i}ξ_{j}. (6.190)

Thus we certainly have an operator satisfying the conditions of (6.189) and (6.190), since

d _{fibre} p = (\frac{∂p}{∂τ} , \frac{∂p}{∂ξ}) = 0 ⟹ τ = 0 and g^{ij}(x)ξ_{i} = 0 ⟹ ξ = 0.

As initial surface we consider the obvious hypersurface {t = 0} (although it will be convenient to consider others). We are after the two theorems, one local and global, the other microlocal, although made to look global.

Theorem 6.3. If X is a Riemannian manifold, as above, then for every f ∈ C_{c}^{−∞}(ℝ × X) ∃! u ∈ C^{−∞}(ℝ × X) satisfying

(6.191) P u = f, u = 0 in t < inf{t̄; ∃(t̄, x) ∈ supp(f)}.

Theorem 6.4. If X is a Riemannian manifold, as above, then for every u ∈ C^{−∞}(ℝ × X),

W F (u)\W F (P u) ⊂ Σ(P)\W F (P u) (6.192)

is a union of maximally extended H_{o}-curves in the open subset Σ(P)\W F (P u) of Σ(P).

Let us think about Theorem 6.3 first. Suppose x̄X is fixed on δx̄ ∈ C^{−∞}(X; Ω) is the Dirac delta (g measure) at x̄. Ignoring, for a moment, the fact that this is not quite a generalized function we can look for the “forward fundamental solution” of P with pole at (0, x̄) :

\begin{matrix}P Ex̄(t, x) = δ(t)δx̄(x) \\ Ex̄ = 0 in t < 0. \end{matrix} (6.193)

Theorem 6.3 asserts its existence and uniqueness. Conversely if we can construct Ex̄ for each x̄, and get reasonable dependence on x̄ (continuity is almost certain once we prove uniqueness) then

K(t, x; t̄, x̄) = Ex̄(t − t̄, x) (6.194)

is the kernel of the operator f ↦ u solving (6.191).

So, we want to solve (6.193). First we convert it (without worrying about rigour) to an initial value problem. Namely, suppose we can solve instead

\begin{matrix}P Gx̄(t, x) = 0 in ℝ × X \\ Gx̄(0, x) = 0, D_{t}Gx̄(0, x) = δx̄(x) in X. \end{matrix} (6.195)

Note that

(g(t, x, τ, 0) ∉ Σ(P) ⟹ (t, x; τ, 0) ∉ W F (G). (6.196)

This means the restriction maps, to t = 0, in (6.195) are well-defined. In fact so is the product map:

Ex̄(t, x) = H(t)Gx̄(t, x). (6.197)

Then if G satisfied (6.195) a simple computation shows that Ex̄ satisfies (6.193). Thus we want to solve (6.195).

Now (6.195) seems very promising. The initial data, δx̄, is certainly conormal to the point {x̄}, so we might try to use our construction of travelling wave solutions. However there is a serious problem. We already noted that, for the wave equation, there cannot be any smooth characteristic surface other than a hypersurface. The point is that if H has codimension k then

N ∗ ; x̄H ⊂ T ∗ ; x̄(ℝ × X) has dimension k.

To be characteristic we must have

N ∗ ; x̄H ⊂ Σ(P) ⟹ k = 1 (6.198)

Since the only linear space contained in a (proper) cone is a line.

However we can easily ‘guess’ what the characteristic surface corresponding to the point (x, x̄) is – it is the cone through that point:

This certainly takes us beyond our conormal theory. Fortunately there is a way around the problem, namely the possibility of superposition of conormal solutions.

To see where this comes from consider the representation in terms of the Fourier transform:

δ(x) = (2π)^{−n} ∫ e^{ixξ}dξ. (6.199)

The integral of course is not quite a proper one! However introduce polar coordinates ξ = rω to get, at least formally

δ(x) = (2π)^{−n} ∫_{0}^{∞} ∫_{𝕊^{n−1}} e^{irx·ω}r^{n−1}dr dω. (6.200)

In odd dimensions r^{n−1} is even so we can write

δ(x) = \frac{1}{2(2π)^{n}} ∫_{𝕊^{n−1}} ∫_{−∞}^{∞} e^{irx·ω}r^{n−1}dr dω, n odd . (6.201)

Now we can interpret the r integral as a 1-dimensional inverse Fourier transform so that, always formally,

δ(x) = \frac{1}{2(2π)^{n−1}} ∫_{𝕊^{n−1}} f_{n}(x · ω)dω (6.202)

n odd (6.203)

f_{n}(s) = \frac{1}{(2π)} ∫ e^{irs}γ^{n−1}dr. (6.204)

In even dimensions we get the same formula with

f_{n}(s) = \frac{1}{2π} ∫ e^{irs}|r|^{n−1}dr. (6.205)

These formulas show that

f_{n}(s) = |D_{s}|^{n−1}δ(s). (6.206)

Here |S_{s}|^{n−1} is a pseudodifferential operator for n even or differential operator (= D_{s}^{n−1}) if n is odd. In any case

f_{n} ∈ I^{n−1+ \frac{1}{4}} (ℝ, \{0\})! (6.207)

Now consider the map

ℝ^{n} × 𝕊^{n−1} ∋ (x, ω) ↦ x · ω ∈ ℝ. (6.208)

Thus C^{∞} has different

ω · dx + x · dω ≠ 0 or x · ω = 0

So the inverse image of {0} is a smooth hypersurface R.

Lemma 6.11. For each n ≥ 2

f_{n}(x, ω) = \frac{1}{2π} ∫ e^{i(x·ω)r}|r|^{n−1}dr ∈ I^{\frac{n}{4} − \frac{1}{4}} (ℝ × 𝕊^{n−1}, R) . (6.209)

Proof. Replacing |r|^{n−1} by ρ(r)|r|^{n−1} + (1 − ρ(r))|r|^{n−1}, where ρ(r) = 0 n r < \frac{1}{2}, ρ(r) = 1 in r > 1, expresses f_{n} as a sum of a C^{∞} term and a conormal distribution. Check the order yourself!

Proposition 6.4. (Radon inversion formula). Under pushforward corresponding to ℝ^{n} × 𝕊^{n−1}@ > π_{1} >> ℝ^{n}

\begin{aligned}(π_{1})_{∗}f′_{n} & = 2(2π)^{n−1}δ(x), \\ f′_{n} & = f_{n}|dω||dx|. \end{aligned}

Proof. Pair with a test function ϕ ∈ S(ℝ^{n}) :

(π_{1})_{∗}f′_{n} = ∫∫ f_{n}(x · ω)ϕ(x)dx dω

by the Fourier inversion formula.

So now we have a superposition formula expressing δ(x) as an integral:

δ(x) = \frac{1}{2(2π)^{n−1}} ∫_{𝕊^{n−1}} f_{n}(x · ω)dω (6.210)

where for each fixed ω f_{n}(x · ω) is conormal with respect to x · ω = 0. This gives us a strategy to solve (6.195).

Proposition 6.5. Each x̄ ∈ X has a neighbourhood, Ux̄, such that for t̄ > 0 (independent of x̄) there are two characteristic hypersurfaces for each ω ∈ 𝕊^{n−1}

H_{x̄,ω)}^{±} ⊂ (−t̄, t̄) × Ux̄

depending on x̄, ω, and there exists

u^{±}(t, x; x̄, ω) ∈ I^{∗}((−t̄|t̄| × Ux̄, H_{(x̄,ω)}^{±})

such that

P u^{±} ∈ C^{∞} (6.211)

\begin{cases}u^{+} + ū ↾ t = 0 = δx̄(x · ω) & in Ux̄ \\ D_{t}(u^{+} + u^{−}) ↾ \{t = 0\} = 0 & in Ux̄. \end{cases} (6.212)

Proof. The characteristic surfaces are constructed through Hamilton-Jacobi theory:

\begin{matrix}N^{∗}H^{±} ⊂ Σ(P), \\ H_{0} = H^{±} ∩ \{t = 0\} = \{x · ω = 0\}. \end{matrix}

There are two or three because the conormal direction to H_{0} at 0; ωdx, has two Σ(P) :

τ = ±1, (τ, ω) ∈ T_{0}^{∗}(ℝ × X). (6.213)

With each of these two surfaces we can associate a microlocally unique conormal solution

\begin{matrix}P u^{±} = 0, u^{±} ↾ \{t = 0\} = u_{0}^{±} \\ u_{0}^{±} ∈ I^{∗}(ℝ^{n}, \{x · ω = 0\}) \end{matrix} (6.214)

Now, it is easy to see that there are unique choices

\begin{matrix}u_{δ}^{+} + u_{0}^{−} = δ(x · ω) \\ D_{t}u^{+} + D_{t}u^{−} ↾ \{t = 0\} = 0. \end{matrix} (6.215)

(See exercise 2.) This solves (6.212) and proves the proposition (modulo a fair bit of hard work!).

So now we can use the superposition principle. Actually it is better to add the variables ω to the problem and see that

\begin{matrix}u^{±}(t, x; ω, x̄) ∈ I^{∗}(ℝ × ℝ^{n} × 𝕊^{n−1} × ℝ^{n}; H^{±}) \\ H^{±} ⊂ ℝ × ℝ^{n} × 𝕊^{n−1} × ℝ^{n} \end{matrix} (6.216)

being fixed by the condition that

H^{±} ∩ ℝ × ℝ^{n} × {ω} × {x̄} = H_{x̄,ω}^{±}.

Then we set

G′ ; x̄(t, x) = ∫_{𝕊^{n−1}} (u^{+} + u^{−})(x, x; ω, x̄). (6.217)

This satisfies (6.195) locally near x̄ and modulo C^{∞}. i.e.

\begin{cases}P G′ ; x̄ ∈ C^{∞}((−t̄(t̄)) × Ux̄) \\ G′ ; x̄ ↾ \{t = 0\} = xv, \\ & v_{i} ∈ C^{∞} \\ D_{t}G′ ; x̄ = δx̄(x) + v_{2} \end{cases} (6.218)

Let us finish off by doing a calculation. We have (more or less) shown that u^{±} are conormal with respect to the hypersurfaces H^{±}. A serious question then

6.7. FORWARD FUNDAMENTAL SOLUTION 115

is, what is (a bound one) the wavefront set of G′ ; x̄? This is fairly easy provided we understand the geometry. First, since u^{±} are conormal,

W F (u^{±}) ⊂ N^{∗}H^{±}.

Then the push-forward theorem says

\begin{matrix}W F (G^{±}) ⊂ \{(t, x, τ, ξ); ∃ (t, x, τ, ξ, ω, w) ∈ W F (u^{±})\} \\ G^{±} = (π_{1})_{∗}u^{±} = ∫_{𝕊^{n−1}} u^{±}(t, s; ω, x̄)dω \end{matrix} (6.219)

so here

(t, x, τ, ξ, ω, w) ∈ T^{∗}(ℝ × ℝ^{n} × 𝕊^{n−1}) = T^{∗}(ℝ × ℝ^{n}) × T^{∗}𝕊^{n−1}.

We claim that the singularities of G′ ; x̄ lie on a cone:

W F (G′ ; x̄) ⊂ Λx̄ ⊂ T^{∗}(ℝ × ℝ^{n})

where Λx̄ is the conormal bundle to a cone:

\begin{matrix}Λx̄ = cl\{(t, x; τ, ξ); t ≠ 0, D(x, x̄) = ±t, \\ (τ, ξ) = τ(1, ∓d_{x}D(x, x̄)) \end{matrix} (6.220)

where D(x, x̄) is the Riemannian distance from x to x̄.

6.7. Forward fundamental solution Last time we constructed a local parametrix for the Cauchy problem:

\begin{cases}P G′ ; x̄ = f ∈ C^{∞}(Ω) & (0, x̄) ∈ Ω ⊂ ℝ × X \\ G′ ; x̄ ↾ t = 0 = u′ \\ D_{t}G′ ; x̄ ↾ \{t = 0\} = δx̄(x) + u′′ & u′, u′′ ∈ C^{∞}(Ω_{0}) \end{cases} (6.221)

where P = D_{t}^{2} − Δ is the wave operator for a Riemann metric on X. We also computed the wavefront set, and hence singular support of Gx̄ and deduced that (6.222) sing · supp .(Gx̄) ⊂ {(t, x); d(x, x̄) = |t|} in terms of the Riemannian distance. (6.223)

This allows us to improve (6.221) in a very significant way. First we can chop Gx̄ off by replacing it by

ϕ (\frac{t^{2} − d^{2}(x, x̄)}{ϵ^{2}}) . (6.224)

where ϕ ∈ C^{∞}(ℝ) has support near 0 and is identically equal to 1 in some neighbourhood of 0. This gives (6.221) again, with G′ ; x̄ now supported in say d^{2} < t^{2} +ϵ^{2}. (6.225)

Next we can improve (6.221) a little bit by arranging that u′ = u′′ = 0, D_{t}^{k}f |_{t=0}= 0 ∀ k. (6.226) This just requires adding to G′ a C^{∞}, v, function, so that v |_{t=0}= u′, D_{t}v |_{t=0}= −u′′, D_{t}^{k}(P u) |_{t=0}= −D_{t}^{k}f |_{t=0} k > 0. (6.227)

Once we have done this we consider

E′ ; x̄ = iH(t)G′ ; x̄

which now satisfies

\begin{matrix}P E′ ; x̄ = δ(t)δt̄(x) + Fx̄, Fx̄ ∈ C^{∞}(Ωx̄) \\ supp(E′_{x}) ⊂ \{d^{2}(x, x̄) ≤ t^{2} + ϵ^{2}\} ∩ \{t ≥ 0\}. \end{matrix} (6.228)

Here F vanishes in t < 0, so vanishes to infinite order at t = 0.

Next we remark that we can actually do all this with smooth dependence of x̄. This should really be examined properly, but I will not do so to save time. Thus we actually have

\begin{cases}E′(t, x, x̄) ∈ C^{−∞}(P (−∞, ϵ) × X × X) \\ P E′ = δ(t)σx̄(x) + F \\ supp E′ ⊂ \{d^{2}(x, x̄) ≥ t^{2} + ϵ^{2}\} ∩ \{t ≥ 0\}. \end{cases} (6.229)

We can, and later shall, estimate the wavefront set of E. In case X = ℝ^{n} we can take E to be the exact forward fundamental solution where |x| or x̄ ≥ R, so

\begin{matrix}supp(F) ⊂ \{t ≥ 0\} ∩ \{|x|, |x̄| ≤ R\} ∩ \{d^{2} ≤ t^{2} + ϵ^{2}\} \\ F ∈ C^{∞}((−∞, ϵ) × X × X). \end{matrix} (6.230)

Of course we want to remove F, the error term. We can do this because it is a Valterra operator, very similar to an upper triangular metric. Observe first that the operators of the form (6.230) form an algebra under t-convolution:

F = F_{1} ∘ F_{1}, F (t, x, x̄) = ∫_{0}^{t} ∫ F_{1}(t, −t′, x, x′)F_{2}(t^{1}, x^{1}, x̄)dx′dt′. (6.231)

In fact if one takes the iterates of a fixed operator

F^{(k)} = F^{(k−1)} ∘ F

One finds exponential convergence:

| D_{x}^{α}D_{t}^{p}F^{(k)}(t, x, x̄) |≤ \frac{C^{k+1}N, δ}{k!} |t|^{N} in t < ϵ − δ ∀ N. (6.232)

Thus if F is as in (6.230) then Id + F has inverse Id + F̃,

F̃ = ∑_{j≥1}(−1)^{j}F^{(j)} (6.233)

again of this form.

Next note that the composition of E′ with F̃ is again of the form (6.230), with R increased. Thus

E = E′ + E′ ∘ F (6.234)

is a forward fundamental solution, satisfying (6.229) with F ≡ 0.

In fact E is also a left parametrix, in an appropriate sense:

Proposition 6.6. Suppose u ∈ C^{−∞}((−∞, ϵ) × X) is such that

(6.235) supp(u) ∩ [−T, τ] × X is compact ∀ T and for τ < ϵ

then P u = 0 ⟹ u = 0.

Proof. The trick is to make sense of the formula

0 = E · P u = u. (6.236)

In fact the operators G with kernel G(t, x, x̄), defined in t < ϵ and such that G ∗ ϕ ⊂ C^{∞} ∀ ϕ ∈ C^{∞} and

{t ≥ 0} ∩ {d(x, x̄) ≤ R} ⊃ supp(G) (6.237)

act on the space (6.235) as t-convolution operators. For this algebra E ∗ P = Id so (6.236) holds!

We can use this proposition to prove that E itself is unique. Actually we want to do more.

Theorem 6.5. If X is either a compact Riemann manifold or ℝ^{n} with a scattering metric then P has a unique forward fundamental solution, ω.

supp(E) ⊂ {t ≥ 0}, P^{E} = Id (6.238)

and

supp(E) ⊂ {(t, x, x̄) ∈ ℝ × X × X; d(x, x̄) ≤ t} (6.239)

and further

W F′(E) ⊂ Id ∪F_{+} (6.240)

where F_{+} is the forward bicharacteristic relation on T^{∗}(ℝ × X)

ζ = (t, x, τ, ξ) ∉ Σ(P) ⟹ F_{+}(ζ) = ∅

ζ = (t, x, τ, ξ) ∈ Σ(P) ⟹ F_{+}(ζ) = {ζ′ = (t′, x′, τ′, ξ′)
t′ ≥ t × ζ′ = exp(T H_{p})ζ for some T }.

(6.241)

Proof. 1. Use E_{1} defined in (−∞, ϵ × X to continue E globally.

2. Use the freedom of choice of {t = 0} and uniqueness of E to show that (6.239)can be arranged for small, and hence all,

3. Then get (6.241) by checking the wavefront set of G.

As corollary we get proofs of (6.225) and (6.226).

Proof of Theorem XXI.5.

u(t, x) = ∫ E(t − t′, x, x′)f(t′, x′)dx′dt′.

Proof of Theorem XXI.6. We have to show that if both WF(P u) ∌ z and WF(u) ∌ z then exp(δH_{p})z ∉ W F (u) for small δ. The general case that follows from the (assumed) connectedness of H_{p} curves. This involves microlocal uniqueness of solutions of P u = f. Thus if ϕ ∈ C^{∞}(ℝ) has support in t > −δ, for δ > 0 small enough, π^{∗}t(z) = t̄

P (ϕ(t − t̄)u) = g has z ∉ W F (g),

and vanishes in t < δ. Then

ϕ(t − t̄)u = E × g
⟹ exp(τH_{p})(z) ∉ W F (u) for small τ.

118 6. THE WAVE KERNEL

6.8. Operations on conormal distributions I want to review and refine the push-forward theorem, in the general case,

to give rather precise results in the conormal setting. Thus, suppose we have a projection X × Y @ > x >> X (6.242) where we can view X × Y as compact manifolds or Euclidean spaces as desired, since we actually work locally. Suppose (6.243) Q ⊂ X × Y is an embeded submanifold. Then we know how to define and examine the conormal distribution associated to Q. If u ∈ I^{m}(X × Y, Q; Ω) (6.244) when is π_{∗}(u) ∈ C^{−∞}(X; Ω) conormal? The obvious thing we ned is a submanifold with respect to what it should be conormal! From our earlier theorem we know that W F (π_{∗}(u)) ⊂ {(x, ξ); ∃ (x, ξ, y, 0) ∈ W F (u) ⊂ N^{∗}Q}. (6.245) So suppose Q = {q_{j}(x, y) = 0, j = 1, …, k}, k = codim Q. Then we see that

(x̄, ξ̄, ȳ, 0) ∈ N^{∗}Q ⟺ (x̄, ȳ) ∈ Q, ξ̄ = ∑_{j=1}^{k} τ_{j}d_{x}q_{j}, ∑_{j=1}^{k} τ_{j}dyq_{j} = 0. (6.246)

Suppose for a moment that Q has a hypersurface, i.e. k = 1, and that (6.247) Q ⟶ π(Q) is a fibration then we expect

Theorem 6.6. π_{∗} : I^{m}(X × Y, Q, Ω) ⟶ I^{m′}(X, π(Q)).

Proof. Choose local coordinates so that

(6.248)  Q = {x_{1} = 0}
(6.249)  u = \frac{1}{2π} ∫ e^{ix_{1}ξ_{1}}a(x′, y, ξ_{1})dξ_{1}
(6.250)  π^{∗}u = \frac{1}{2π} ∫ e^{ix_{1}ξ_{1}}b(x′, ξ_{1})dξ_{1}
(6.251)  b = ∫ a(x′, y, ξ)dy.

Next consider the case of restriction to a submanifold. Again let us suppose Q ⊂ X is a hypersurface and Y ⊂ X is an embedded submanifold transversal to Q :

(6.252)  Q ⋔ Y = QY
(6.253)  i.e. T_{q}Q + T_{q}Y = T_{q}X ∀ q ∈ Qy
(6.254) ⟹ Q_{y} is a hypersurface in X.

Indeed locally we can take coordinates in which Q = {x_{1} = 0}, Y = {x′′ = 0}, x = (x_{1}, x′, x′′). (6.255)

Theorem 6.7.

C_{Y}^{∗} : I^{m}(X, Q) ⟶ I^{m+ \frac{k}{4}} (Y, Q_{Y})k = codim Y in X.

Proof. In local coordinates as in (6.255)

\begin{matrix}u = \frac{1}{2π} ∫ e^{ix_{1}ξ_{1}}a(x(x′, x′′, ξ_{1}))dξ, \\ c^{∗}u = \frac{1}{2π} ∫ e^{ix_{1}ξ_{1}}a(x′, 0, ξ_{1})dξ_{1}. \end{matrix}

Now let’s apply this to the fundamental solution of the wave equation. Well rather consider the solution of the initial value problem

\begin{cases}P G(t, x, x̄) = 0 \\ G(0, x, x̄) = 0 \\ D_{t}G(0, x, x̄) = δx̄(x). \end{cases} (6.256)

We know that G exists for all time and that for short time it is

G − ∫_{𝕊^{n−1}} (u_{+}(t, x, x̄; ω) + u_{−}(t, x, x̄; ω))dω + C^{∞} (6.257)

where u_{±} are conormal for the term characteristic hypersurfaces H_{p} satisfying

\begin{matrix}N^{∗}H_{±} ⊂ Σ(P) \\ H_{±} ∩ \{t = 0\} = \{(x − x̄) · ω = 0\} \end{matrix} (6.258)

Consider the 2 × 2 matrix of distribution

U(t) = (\begin{matrix}D_{t}G & G \\ D_{t}^{2}G & D_{t}G \end{matrix}) . (6.259)

Since W F U ⊂ Σ(P), in polar τ ≠ 0 we can consider this as a smooth function of t, with values in distribution on X × X.

Theorem 6.8. For each t ∈ ℝ U(t) is a boundary operator on L^{2}(X) ⊕ H′(X) such that

U(t) (\begin{matrix}u_{0} \\ u_{1} \end{matrix}) = (\begin{matrix}u(t) \\ D_{t}u(t) \end{matrix}) (6.260)

where u(t, x) is the unique solution of

\begin{matrix}(D_{t}^{2} − Δ) & u(t) = 0 \\ u(0) = u_{0} \\ D_{t} + u(0) = u_{1}. \end{matrix} (6.261)

Proof. Just check it!

Consider again the formula (6.257). First notice that at x = x̄, t = 0, dH^{±} = dt ± d(x − x̄)ω) (by construction). so

H_{±} ⋔ {x = x̄} = {t = 0} ⊂ ℝ × X ↪ ℝ × X × Y × 𝕊^{n−1}. (6.262)

Moreover the projection

ℝ × X × 𝕊^{n−1} ⟶ ℝ (6.263) clearly fibres {t = 0} over {t = 0} ∈= {0} ⊂ ℝ. Then we can apply the two theorems, on push-forward and pull-back, above to conclude that

T (t) = ∫_{X} G(t, x, x̄) ↾ x = x̄dx ∈ C^{−∞}(ℝ) (6.264)

is conormal near t = 0 i.e. C^{∞} in (−ϵ, ϵ)\{0} for some ϵ > 0 and conormal at 0. Moreover, we can, at least in principle, work at the symbol of T (t) at t = 0. We return to this point next time.

For the moment let us think of a more ‘fundamental analytic’ interpretation of (6.264). By this I mean

T (t) = trU(t). (6.265)

Remark 6.1. Trace class operators Δλ; Smoother operators are trace order, tr = ∫ K(x, x)

∫ U(t)ϕ(t) is smoothing

⟨T (t), ϕ(t)⟩ = tr⟨U(t), ϕ(t)⟩.

6.9. Weyl asymptotics

Let us summarize what we showed last time, and a little more, concerning the trace of the wave group

Proposition 6.7. Let X be a compact Riemann manifold and U(t) the wave group, so

U(t) : C^{∞}(X) × C^{∞}(X) ∋ (u_{0}, u_{1}) ↦ (u, (t), D + tu(t)) ∈ C^{∞}(X) × C^{∞}(X)

where u is the solution to

\begin{aligned}(D_{t}^{2} − Δ)u(t) & = 0 & (6.266) \\ u(0) & = u_{0} & (6.267) \\ D_{t}u(0) & = u_{1}. & (6.268) \end{aligned}

The trace of the wave group, T ∈ S′(ℝ), is well-defined by

T (ϕ) = Tr U(ϕ), U(ϕ) = ∫ U(t)ϕ(t)dt ∀ ϕ ∈ S(ℝ) (6.269)

and satisfies

T = Y ((1 + ∑_{j=1}^{∞} 2 cos(tλ_{j})) (6.270)

where 0 = λ_{0} < λ_{1}^{2} ≤ λ_{2}^{2} … λ_{j} ≥ 0

is the spectrum of the Laplacian repeated with multiplicity

(6.271) sing . supp(T) ⊂ L ∪ {0} ∪ −L

where L is the set of lengthes of closed geodesics of X and

\begin{matrix}if ψ ∈ C_{c}^{∞}(ℝ), ψ(t) = 0 if |t| ≥ inf L − ϵ, ϵ > 0, \\ ψT ∈ I(ℝ, \{0\}) \\ σ(ψT) = \end{matrix} (6.272)

Proof. We have already discussed (6.270) and the first part of (6.272) (given (6.271)). Thus we need to show (6.271), the Poisson relation, and compute the symbol of T as a cononormal distribution at 0 .

Let us recall that if G is the solution to

\begin{aligned}(D_{t}^{2} = Δ)G(t, x, x̄) & = 0 \\ G(0, x, x̄) & = 0 \\ D_{t}G(0, x, x̄) & = δx̄(x) \end{aligned} (6.273)

then

T = π_{∗}(ι_{Δ}^{∗}2D_{t}G), (6.274)

where

ι_{Δ} : ℝ × X ↪ ℝ × X × X (6.275)

is the embedding of the diagonal and

π : ℝ × X ⟶ ℝ (6.276)

is projective. We also know about the wavefront set of G. That is,

\begin{matrix}W F (G) ⊂ \{(t, x, x̄, τ, ξ, ξ̄); τ^{2} = |ξ|^{2} = |ξ̄|^{2}, \\ exp(sH_{p})(0, x̄, τ, ξ̄) = (t, x, τ, ξ), some s\}. \end{matrix} (6.277)

Let us see what (6.277) says about the wavefront set of T. First under the restriction map to ℝ × Δ

\begin{matrix}W F (ι_{Δ}^{∗}D_{t}G) ⊂ \{(t, y, τ, η); ∃ \\ (t, x, y, τ, ξ, ξ̄); η = ξ − ξ̄\}. \end{matrix} (6.278)

Then integration gives

W F (T) ⊂ {(t, τ); ∃ (t, y, τ, 0) ∈ W F (D_{t}G)}. (6.279)

Combining (6.278) and (6.279) we see t ∈ sing . supp(T) ⟹ ∃ (t, τ) ∈ W F (T)

⟹ ∃ (t, x, x, τ, ξ, ξ) ∈ W F (D_{t}G)
⟹ ∃ s s.t. exp(sH_{p})(0, x, τ, ξ) = (t, x, τ, ξ).

(6.280)

Now

p = τ^{2} − |ξ|^{2}, so H_{p} = 2τ∂_{t} − H_{g}, g = |ξ|^{2},

H_{g} being a vector field on T^{∗}X. Since W F is conic we can take |ξ| = 1 in the last condition in (6.280). Then it says

s = 2τt, exp(tH_{\frac{1}{2} g})(x, ξ) = (x, ξ), (6.281)

since τ^{2} = 1.

The curves in X with the property that their tangent vectors have unit length and the lift to T^{∗}X is an integral curve of H_{\frac{1}{2} g} are by definition geodesic, parameterized by arclength. Thus (6.281) is the statement that |t| is the length of a closed geodesic. This proves (6.271).

So now we have to compute the symbol of T at 0. We use, of course, our local representation of G in terms of conormal distributions. Namely

G = ∑_{j} ϕ_{j}G_{j}, ϕ_{j} ∈ C^{∞}(X), (6.282)

where the ϕ_{j} has support in coordinate particles in which

\begin{matrix}G_{j}(t, x, x̄) = ∫_{𝕊^{n−1}} (u_{+}(t, x, x̄; ω) + u_{−}(t, x, x̄; ω)) dω, \\ u_{p}m = \frac{1}{2π} ∫_{ξ} e^{ih_{±}(t,x,x̄,ω)ξ}a_{±}(x, x̄, ξ, ω)dξ. \end{matrix} (6.283)

Here h_{±} are solutions of the eikonal equation (i.e. are characteristic for P)

\begin{matrix}|∂_{t}h_{±}|^{2} = |h_{±}|^{2} \\ h_{±} |_{t=0}= (x − x̄) · ω \\ ±∂_{t}h_{±} > 0, \end{matrix} (6.284)

which fixes them locally uniquely. The a_{±} are chosen so that

(u_{+} + u_{±} |_{t=0}= 0, (D_{t}u_{+}D_{t}u_{−}) |_{t=0} δ((x − x̄) · ω)P u_{±} ∈ C^{∞}. (6.285)

Now, from (6.283)

u_{+} + u_{−} |_{t=0}= \frac{1}{2π} ∫ e^{((x−xx̄)·ω)ξ}(a_{+} + a_{−})(x, x̄, ξ, ω)dξ = 0

so a_{+} − a_{−}. Similarly

(6.286) D_{t}u_{+} + D_{t}u_{−} |_{t=0} = \frac{1}{2π} ∫ e^{i((x−x̄)·ω)ξ}[(D_{t}h_{+})a_{+} + (D_{t}h_{−})a_{−}]dξ ; \begin{aligned}& = \frac{1}{2(2π)^{n−1}} f_{n}((x − x̄) · ω) \\ (6.287) \end{aligned}

From (6.284) we know that D_{t}h_{±} = ∓i|d_{x}(x − x̄) · ω| = ∓i|ω| where the length is with respect to the Riemann measure. We can compute the symbols or both sides in (6.9) and consider that

−2i|ω|a_{+} ≡ \frac{1}{2(2π)^{n−1}} |ξ|^{n−1} = D_{t}h_{+}a_{+} + D_{t}h_{−}a_{−} |_{t=0} (6.288)

is necessary to get (6.285). Then

(6.289) T (t) = 2π_{∗}(ι_{Δ}^{∗}D_{t}G) ; \begin{aligned}& = \frac{1}{2π} ∑_{j,±} 2 ∫_{X} ∫_{𝕊^{n−1}} e^{ih_{±}(t,x,x,ω)ξ}(D_{t}h_{±}a_{±})(x, x̄, ω, ξ)dξdωdx. \\ (6.290) \end{aligned}

Here dx is really the Riemann measure on X. From (6.288) the leading part of this is

\frac{2}{2π} ∑_{j±} ∫_{X} ∫_{𝕊^{n−1}} e^{ih_{±}(t,x,x,ω)ξ} \frac{1}{4(2π)^{n−1}} |ξ|^{n−1}dξdωdx

since any term vanishes at t contributes a weaker singularity. Now

h_{±} = ±|ω|t + (x − x̄) · ω + 0(t^{2}).

From which we deduce that

\begin{matrix}ψ(t)T (t) = \frac{1}{2π} ∫ e^{itτ} a(τ)dτ \\ a(τ) ∼ C_{n} Vol(X)|τ|^{n−1}C_{n} = \end{matrix} (6.291)

where C_{n} is a universal constant depending only on dimension. Notice that if n is odd this is a “little” function.

The final thing I want to do is to show how this can be used to describe the asymptotic behaviour of the eigenvalue of Δ :

Proposition 6.8. (“Weyl estimates with optimal remainder”.) If N(λ) is the number of eigenvalues at Δ satisfying λ_{1}^{2} ≤ λ, counted with multiplicity, the

N(λ) = C_{n} Vol(X)λ^{n} + o(λ^{n−1}) (6.292)

The estimate of the remainder terms is the here – weaker estimates are easier to get.

Proof. (Tauberian theorem). Note that

T = F(μ) where N(λ) = ∫_{0}^{λ} μ(λ),

μ(λ) being the measure

μ(λ) = ∑_{λ_{j}^{2}∈spec(Δ)} δ(λ − λ_{j}).

Now suppose ρ ∈ S(ℝ) is even and ∫ ρ = 1, ρ ≥ 0. Then N_{ρ}(λ) = ∫ (λ′)ρ(λ − λ′) is a C^{∞} function. Moreover

{ \atop \frac{d}{dλ}N_{ρ}(λ)} = μ̂ · ρ̂. (6.293)

Suppose we can choose ρ so that

ρ ≥ 0, ∫ ρ = 1, ρ ∈ S, ρ̂(t) = 0, |t| > ϵ (6.294)

for a given ϵ > 0. Then we know μ̂ρ̂ is conormal and indeed

\frac{d}{dλ}Nρ(λ) ∼ C Vol(X)λ^{n−1} + …

⟹ N_{ρ}(λ) ∼ C′ Vol(X)λ^{n} + lots.

(6.295)

So what we need to do is look at the difference

N_{ρ}(λ) − N(λ) = ∫ N(λ − λ′)ρ(λ′) − N(λ)ρ(λ′). (6.296)

It follows that a bound for N

|N(λ + μ) − N(λ)| ≤ ((1 + |λ| + |μ|)^{n−1}(1 + |λ|) (6.297)

gives

N(λ) − N_{ρ}(λ) ≤ Cλ^{n−1}

which is what we want. Now (6.298) follows if we have

N(λ + 1) − N(λ) ≤ C(1 + |λ|) t/λ. (6.298)

This in turn follows from the positivity of ρ, since

∫ ρ(λ − λ′)μ(λ′) ≤ C(1 + |λ|)^{n−1}. (6.299)

124 6. THE WAVE KERNEL

Finally then we need to check the existence of ρ as in (6.294). If ϕ is real and even so is ϕ̂. Take ϕ with support in (−\frac{ϵ}{2}, \frac{ϵ}{2}) and construct ϕ∗ϕ, real and even with ϕ.