Asymptotic Expansion

In this appendix, we shall prove Prop. A.1 which, in particular, includes the proof of Lemma 4.7. The quantization is as in Subsection 4.3.

Proposition A.1

Let \(h_0\), \(E_}\) be as in Lemma 4.5. For \(h\in [0,h_0)\), \(|z|\le 2\Vert \mathscr \Vert _\infty \), we have

(1)

The symbol \(\frac}E_}\) has an asymptotic expansion in S: There are \(a_\in S\) such that

$$\begin & \frac} E_}(x_2,\xi _2;z,h)\sim \sum \limits _^\infty h^}E_(x_2,\xi _2;z) \ \text \ E_ \nonumber \\ & \quad = \sum \limits _^ a_(x_2,\xi _2)z^k, j\ge 1. \end$$

(A.1)

In particular, \(E_ = z - z_, \quad E_ = -z_,\quad E_ = - z_,\) where \(z_\) are given in Lemma A.2.

(2)

Let \(0<\delta <1/2\), if \(|\,}}z|\ge h^\), then \(\sqrtE_}^\) has an asymptotic expansions in \(S_\delta ^\delta \): There are \(b_, c_\in S\) such that in terms of \(\prod \limits _^k b_(x_2,\xi _2;z) = \sum \limits _^ z^\alpha c_(x_2,\xi _2)\) the expansion of

$$\begin \begin&\sqrt E_}^\sim \sum \limits _^\infty h^}F_ (x_2,\xi _2;z), \text F_ \\&= \sum _^ (z - z_)^ \prod \limits _^\left( b_(x_2,\xi _2;z)(z - z_)^\right) . \end \end$$

(A.2)

Thus \(h^} F_\in S^)+ \delta }\). In particular, we have

$$\begin \begin&F_ =(z - z_)^, \ F_ = F_z_F_,\ F_ \\&= F_\left( z_F_ + z_F_ - \frac, z - z_\}} \right) , \end \end$$

(A.3)

where \(\\) is the Poisson bracket.

(3)

Let \(0<\delta <1/2\), if \(|\,}}z|\ge h^\), then \(r_n\) has an asymptotic expansions in \(S_\delta ^\delta \): There are \(d_(x_2,\xi _2;z), e_(x_2,\xi _2)\in S\), such that in terms of \(\prod \limits _^k d_(x_2,\xi _2;z) = \sum \limits _^ z^\alpha e_(x_2,\xi _2)\)

$$\begin \begin&\quad r_n(x_2,\xi _2;z,h) \sim \sum \limits _^\infty h^} r_(x_2,\xi _2;z,h), \text r_ \\&= \sum \limits _^j(z - z_)^ \prod \limits _^k \left( d_(x_2,\xi _2;z)(z - z_)^\right) . \end \end$$

Thus \(h^}r_\in S_0^}\). In particular,

$$\begin r_ = F_,\quad r_ = F_,\quad r_ = F_ - (\partial _z z_)F_. \end$$

(4)

Finally, let \(\eta = x_2 + i\xi _2\), then the leading terms of \(\,}}_^2}(r_n)\) are:

$$\begin \begin } \ }_}}, n}:&\,}}_^2}(r_}}, n,0} + h^}r_}}, n,1} + hr_}},n,2}) = \frac + 0 + \frac\mathfrak U(\eta )h, \\ } \ }_}},n}^\theta :&\,}}_^2}(r_}}, n,0} + h^}r_}}, n,1}) = \frac + \frac\sqrt, \end \end$$

where \(\mathfrak U(\eta ) = \frac\left[ \alpha _1^2(|U_-(\eta )|^2 - |U(\eta )|^2)^2 + 4|\partial _}\overline-\partial _\eta U(\eta )|^2\right] \), \(\partial _\eta = \frac(\partial _ - i\partial _)\), \(s_n(\eta ) = \alpha _0 \sin (}) |V(\eta )| & n\ne 0 \\ \alpha _0 |V(\eta )| & n = 0, \end\right. }\) and \(c_n(\eta ) = \alpha _0 \cos (})|V(\eta )| & n \ne 0\\ \alpha _0 |V(\eta )| & n = 0. \end\right. }\)

We will prove Proposition A.1 in the rest of this appendix in two steps: First, we compute explicitly the leading terms (three terms for the chiral model, two for anti-chiral model) in the expansion of \(Z_n(x_2,\xi _2;z,h)\), the symbol of \(Z_n^W\), where \(E_}= \sqrt(z - Z_n^W)\) by (4.17). Then, we exhibit the z dependence for each term in the expansion of \(E_}\), from which we build up both the legitimacy of the existence of asymptotic expansions of \(E_}^\) and \(r_n\), and the z dependence of each term in the expansions.

Explicit leading terms Recall that by (4.17) and (4.20), \(E_}= \sqrt(z - Z_n^W)\) with

$$\begin \begin Z_n^W&(x_2,hD_;h) = R_n^+\tilde}}^W(I + \sqrtE_^\theta \tilde}}^W)^R_n^- \\ &= \sum \limits _^\infty h^}(-1)^k R_n^+\tilde}}^W(E_^\theta \tilde}}^W)^k R_n^- =: \sum \limits _^\infty h^}Q_^W(x_2,hD_;h), \end \end$$

(A.4)

where \(R_n^\), \(E_^\theta \), \(\tilde}}^W\) are given in (4.12), (4.15) and (4.20). Then we can express the asymptotic expansion of \(Z_n(x_2,\xi _2)\) in terms of \(Q_(x_2,\xi _2)\):

Proposition A.2

Let \(Q_^W(x_2,hD_;h) = (-1)^k R_n^+\tilde}}^W(E_^\theta \tilde}}^W)^k R_n^-\). Then symbols \(Q_\), \(Q_\), \(Q_\) have the following asymptotic expansions

$$\begin \begin&Q_(x_2,\xi _2;h) = Q_^(x_2,\xi _2) + \sqrtQ_^(x_2,\xi _2) + hQ_^(x_2,\xi _2) +}_(h^}),\\&Q_(x_2,\xi _2;h) = Q_^(x_2,\xi _2) +\sqrtQ_^(x_2,\xi _2) +}_(h),\\&Q_(x_2,\xi _2;h) = Q_^(x_2,\xi _2) +\mathcal O_(\sqrt). \end \end$$

For the chiral Hamiltonian, with \(\eta = x_2\ + i\xi _2\), \(D_\eta = \frac(D_ -i D_)\),

$$\begin \begin&Q_}},n,0}^} = Q_}},n,0}^} = Q_}},n,2}^} = 0, \quad Q_}},n,1}^} = -\frac\left[ |U|^2 - |U_-|^2\right] \sigma _3,\\&Q_}},n,0}^} = \frac \begin 0 & D_\eta U - D_}\overline\\ D_\eta U_- - D_} \overline & 0 \end,\\&Q_}},n,1}^} = -\frac\left[ 2|n|(|U|^2+|U_-|^2)}_ + (|U|^2 - |U_-|^2)\sigma _3\right] & n \ne 0, \\ -\frac\begin |U|^2 & 0 \\ 0 & |U_-|^2 \end&n = 0. \end\right. } \end \end$$

Analogously, for the anti-chiral Hamiltonian, when \(}^\theta = }_}}\), we have, when \(n = 0\),

$$\begin \begin&Q_}}, 0, 0}^= Q_}}, 0, 1}^= Q_}}, 0, 1}^= Q_}}, 0, 2}^= 0, \\&Q_}}, 0, 0}^= \alpha _0 \begin 0 & e^i}V \\ e^} V^* & 0 \end, Q_}}, 0, 0}^=\frac \begin 0 & e^i}\Delta _ V\\ e^i}\Delta _ \bar \end, \end \end$$

when \(n \ne 0\),

$$\begin \begin&Q_}}, n,0}^= 0, \ Q_}}, n,1}^=\frac)}}_, \ Q_}}, n,1}^= -\frac)}}_,\\&Q_}}, n,0}^= \alpha _0 \cos (\tfrac)\begin 0 & V^* \\ V & 0 \end, Q_}}, n,2}^= -\frac)\cos (\tfrac)}\begin 0 & V \\ V^* & 0 \end,\\&Q_}}, n,0}^= \frac\left( 2|n|\cos (\tfrac) - i\sigma _3\sin (\tfrac)\right) \begin 0 & \Delta _ V\\ \Delta _ \bar & 0 \end. \end \end$$

In particular, \(Z_n\) has an asymptotic expansion \(Z_n\sim \sum \limits _^\infty h^} z_\) in S with

$$\begin z_ = Q_^}, \ z_ = Q_^} + Q_^}, \ z_ = Q_^} + Q_^} + Q_^}. \end$$

Proof

\(Q_\) has the symbol \( Q_(x_2,\xi _2) = (-1)^k \int __}(K_n^\theta (x_1))^* \tilde}}^w \#(E_^\theta \tilde}}^w)^ K_n^\theta (x_1) dx_1.\) Recall that by(2.1), (4.13), and (4.15), we have

$$\begin K_n^\theta = \begin u_n^\theta & 0 \\ 0 & u_n^ \end,\quad }= \begin 0 & T \\ T^* & 0 \end, \quad T = \begin \alpha _0 V & \alpha _1 \overline\\ \alpha _1 U & \alpha _0 V \end, \quad ^\theta }= \begin ^\theta }& 0 \\ 0 & ^}\end. \end$$

Thus, inserting the above expressions into the definition of \(Q_\), we find for its symbol

$$\begin Q_ =\int \begin ^* & 0 \\ 0 & (})^* \end \begin 0 & \tilde^w \\ (\tilde^w)^* & 0 \end \left( \begin ^\theta }& 0 \\ 0 & ^}\end\begin 0 & \tilde^w \\ (\tilde^w)^* & 0 \end\right) ^k \begin ^* & 0 \\ 0 & }\end \frac, \end$$

where \(\tilde^w = T^w(}x_1,\xi _2 - h^}D_})\). In particular,

$$\begin \begin&Q_= \begin 0 & \int ()^* \tilde^w }dx_1\\ \int (})^* (\tilde^w)^* dx_1 \end,\\&Q_ = \begin -\int ()^* \tilde^w ^}(\tilde^w)^* dx_1 & 0 \\ 0 & -\int (})^* (\tilde^w)^* ^\theta }\tilde^w }dx_1 \end, \text \\&Q_ = \begin 0 & \int ()^* \tilde^w ^}(\tilde^w)^*^\theta }\tilde^w }dx_1\\ \int (})^* (\tilde^w)^*^\theta }\tilde^w ^}(\tilde^w)^*dx_1 & 0 \end. \end \end$$

(A.5)

Notice that since both \(\tilde^w\) and \(^\theta }\) depend on h, we need to further expand them in order to obtain asymptotic expansions of \(Q_\). Thus the proof of Proposition A.2 rests now on the following two lemmas.

Lemma A.3

(Expansion of \(\tilde^w\) and \(^\theta }\))

(1)

Let \(T\in C^\infty _b(_x^2)\). Recall the definition \(\tilde(x,\xi ):= T(x_2+h^}x_1, \xi _2 - h^} \xi _1)\in S(_^4)\). Then

$$\begin \begin \tilde^w(x,D_,\xi _2) =&T(x_2,\xi _2) + \sqrt\langle \nabla _T(x_2,\xi _2), (x_1,-D_) \rangle \\ +&\frac\langle (x_1,-D_), }T (x_2,\xi _2)(x_1,-D_)^T \rangle +\mathcal O__^2;\mathcal L(B^3_;B^0_))}(h^}) \end \end$$

(2)

Let \(^\theta }\) be as in (4.15). Then \(^\theta }(x,D_,\xi _2)\) has an asymptotic expansion \(^\theta }\sim \sum \limits _^\infty h^}\sigma _k(e_^\theta )\) where \( \sigma _k(^\theta }) = \sum \limits _ \frac}.\)

Lemma A.4

(Projections) Let \(S_n^\theta = }\^\theta \}\) with \(S_n:=S_n^\) and \(u_n:=u_n^.\) The following properties hold:

(1)

Reflection invariance with respect to \(\theta \) such that \(S_n^\theta = S_n^\), in particular \( = \cos \left( }\right) }+ i\sin \left( }\right) ^}.\)

(2)

Let \(M = \begin 0 & \alpha \\ \beta & 0 \end \in \mathbb C^\) then \(Mu_n\in S_\cup S_\), for any \(n\ge 0\). More specifically for \(\theta =0\)

$$\begin \begin Mu_ n}&= \frac(u_ - u_) \frac(u_ + u_), \text n \ge 2\\ Mu_ 1}&= \frac(u_2 - u_) \frac} u_0,\text Mu_0 = \frac}(u_1 - u_). \end \end$$

(3)

We have \(x_1\in S_^\theta \cup S_^\theta \), \(D_\in S_^\theta \cup S_^\theta \). More specifically

$$\begin \begin x_1u_ n}^&= \frac}[u_^\theta (\sqrt +\sqrt)+u_^\theta (\sqrt \mp \sqrt)\\&+u_^\theta (\sqrt +\sqrt) u_^\theta (\sqrt\mp \sqrt)],\text \vert n \vert \ge 2\\ x_1 u_ 1}^\theta&= \frac u_0^\theta + \frac} [u_2^\theta (\sqrt\sqrt) + u_^\theta (\sqrt \mp \sqrt)]\text x_1 u_0^\theta = \fraci}(u_1^\theta + u_^\theta ). \end \end$$

Proof

We omit the proof of this Lemma here as it follows from straightforward but lengthy basis expansions and the simple observation that \(\langle u_m^, u_n^\theta \rangle = \cos \left( }\right) \delta _ + i \sin \left( }\right) \delta _.\) \(\square \)

From the preceding Lemmas A.3 and A.4, we can compute the asymptotic expansion of each term of \(Q_\) in (A.5) and therefore prove Prop.A.2.

For the (1, 2)-entry of \(Q_\), by Lemma A.3, we have

$$\begin \begin \int ()^*\tilde^w}dx_1 =&\int ()^* T }dx_1 + \sqrt\int ()^* \langle \nabla _T, (x_1,-D_) \rangle }dx_1 \\&+ \frac\int ()^* \langle (x_1,-D_), } T(x_2,\xi _2)(x_1,-D_)^T \rangle }dx_1\\ =:&t_^+\sqrt t_^+h t_^+ }__^2;_)}(h^}). \end \end$$

Specializing now to the chiral case, in which case the \(\theta \)-dependence can be gauged away, we choose

$$\begin T(x_2,\xi _2) = \begin 0 & \alpha _1\overline \\ \alpha _1 U_-(x_2,\xi _2) & 0 \end \end$$

where in the chiral case, by Lemmas A.3 and A.4, we see that

$$\begin \begin&t_}}, n,0}^} = 0, \ t_}}, n,0}^} =\frac(\partial _}\overline - \partial _w U),\text t_}}, n,0}^} = 0, \end \end$$

while in the anti-chiral case, choosing \(T(x_2,\xi _2) =\alpha _0 V(x_2,\xi _2) }__}\)

$$\begin \begin&t_}}, n,0}^} = \alpha _0\cos (})V & n\ne 0,\\ \alpha _0 e^i} V & n = 0, \end\right. }, \quad t_}}, n,0}^} = 0, \text t_}}, n,0}^} \\ &= \frac(2|n|\cos (}) - i\sigma _3\sin (}))\Delta _V, & n\ne 0\\ \frace^}\Delta _V & n = 0. \end\right. } \end \end$$

Due to the conjugacy relation \(\int ()^* (\tilde^w)^*}dx_1 = (\int (})^* \tilde^w dx_1)^*\), the expansion of \(Q_^\theta \) follows by (A.5).

Similarly for the (1, 1)-entry \(Q_^\theta \), denote

$$\begin -\int ()^* \tilde^w ^}(\tilde^w)^* dx_1 =: t_^} + t_^\sqrt+\mathcal O__^2;_)}(h) \end$$

where, using Lemma , in the chiral case,

$$\begin \begin&t_}}, n,1}^ = -\frac(|U|^2 - |U_-|^2)\text t_}}, n,1}^} \\ &= -\frac[2|n|(|U|^2+|U_-|^2)+(|U|^2 - |U_-|^2)], & n \ne 0\\ -\frac|U|^2, & n = 0 \end\right. } \end \end$$

and in the anti-chiral case

$$\begin \begin&t_}}, n,1}^} = \frac})}, & n \ne 0\\ 0, & n = 0 \end\right. }\text t_}}, n,1}^} = -\frac})z}, & n \ne 0\\ 0, & n = 0. \end\right. } \end \end$$

In a similar fashion, the (2, 2)-entry of \(Q_\), defined in (A.5), can be obtained by precisely the same computations after only replacing \(\theta \) by \(-\theta \) and \(T^*\) by T, i.e., U switching with \(U_-\) and using \(V^*\) instead of V. Thus the asymptotic expansion of \(Q_^\theta \) follows.

Similarly for \(Q_^\theta \) we restrict us to the (1, 2) entry in (A.5). Then, we denote

$$\begin \int ()^* \tilde^w ^}(\tilde^w)^*^\theta }\tilde^w }dx_1 =: t_^} + \mathcal O__^2;_)}(\sqrt). \end$$

It follows then by Lemma , that in the chiral model, \(t_}}, n,2}^} = 0\) while in the anti-chiral model, \(t_^} = -\frac})\cos (})}V. \) By the conjugacy relation

$$\begin \int (})^* (\tilde^w)^*^\theta }\tilde^w ^}(\tilde^w)^*dx_1 = \Bigg (\int ()^* \tilde^w ^}(\tilde^w)^*^\theta }\tilde^w }dx_1\Bigg )^*, \end$$

this also yields directly the expansion of \(Q_}}, n,2}^\theta \). \(\square \)

Existence, derivation and z-dependence Now we prove the rest of Prop. A.1, which includes the existence and derivation of asymptotic expansion of \(E_}^\) and \(r_n\) and the z dependence of each terms in the expansions of \(E_}\), \(E_}^\) and \(r_n\).

Proof of Prop. A.1

By (A.4) and Prop. A.2, \(E_}= \sqrt(z - Z_n)\), and \(Z_n\) has an asymptotic expansion in S. Thus, \(\frac}E_}\) also has an asymptotic expansion in S: \(\frac}E_}\sim \sum \limits _j h^}E_\) with \(E_\in S\). To exhibit the z-dependence, we notice that only \(E_\) depends on z in (A.4). Thus, by (4.24), we have

$$\begin \begin Z_n^W&= R_n^+\tilde}}^W(}+ \sqrtE_\tilde}}^W)^ R_n^-= R_n^+ \tilde}}^W R_n^- + \sum \limits _^\infty R_n^+ \tilde}}^W (\sqrtE_\tilde}}^W)^\alpha R_n^-\\&= R_n^+ \tilde}}^W R_n^- + \sum \limits _^\infty h^} R_n^+ \tilde}}^W \left[ \sum \limits _ \frac \sum \limits _^\infty \left( \fracz}\right) ^\beta \right] ^\alpha R_n^-\\&= R_n^+ \tilde}}^W R_n^- + \sum \limits _^\infty \sum \limits _^\infty h^} z^ A_^W(x_2,hD_)\\&= R_n^+}}}^WR_n^- - \sum \limits _^\infty h^} \left( \sum \limits _^ z^k a_^W(x_2,hD_)\right) \end \end$$

for some appropriate \(A_(x_2,\xi _2)\in S\) and \(a_(x_2,\xi _2)\in S\). Thus we proved part (1).

We can formally derive (A.2) and (A.3) for \(\sqrtE_}^\), using a formal parametrix construction by using

$$\begin a\widetilde b \sim \sum _k\frac\left. \left( \left( \frac\sigma (D_,D_;D_y,D_\eta )\right) ^k\left( a(x_2,\xi _2)b(y,\eta )\right) \right) \right| _. \end$$

(A.6)

More specifically, there is a formal expansion of \(\sqrtE_}^\), which is denoted by \(\sqrtF_n \sim \sum \limits _j h^}F_\), such that \(\frac}E_}\widetilde \sqrtF_n= }_\). Denote \(\sigma (D_,D_;D_y,D_\eta )\) in (A.6) by \(\sigma \), we can solve for \(F_\) by considering

$$\begin \begin }_&= E_}\widetilde F_n^ \sim \sum \limits _^\infty \sum \limits _^\infty h^} E_ \widetilde F_\\&= \sum \limits _^\infty \sum \limits _^\infty h^} \sum \limits _^\infty h^\gamma \left. \left( \left( \frac\right) ^\gamma (E_(x_2,\xi _2)F_(y,\eta ))\right) \right| _\\&= \sum \limits _^\infty \sum \limits _^j \sum \limits _^ h^} \left. \left( \left( \frac\right) ^} (E_(x_2,\xi _2)F_(y,\eta ))\right) \right| _. \end \end$$

Then we compare the parameter of the term of \(h^}\) on both sides and get

$$\begin -E_F_ = \sum \limits _^ \sum \limits _^ \left. \left( \left( \frac\right) ^}(E_(x_2,\xi _2)F_(y,\eta ))\right) \right| _, \end$$

from which we can solve for \(F_\). Furthermore, by (A.1) and \(E_ = z - z_\), we can check inductively that for \(j\ge 0\), there are \(b_\), \(c_\) such that

$$\begin \begin&F_ = \sum _^ (z - z_)^ \prod \limits _^\left( b_(x_2,\xi _2;z)(z - z_)^\right) ,\\&\text \prod \limits _^k b_(x_2,\xi _2;z) = \sum \limits _^ z^\alpha c_(x_2,\xi _2), \text c_\in S. \end \end$$

Notice that \(\widetilde\) differs from the actual sharp product \(\#\):

$$\begin a\# b = e^\sigma (D_,D_;D_y,D_\eta )} \left( a(x_2,\xi _2)b(y,\eta )\right) |_. \end$$

(A.7)

Now we claim that this formal expansion for \(\sqrtF_n\) is legitimate as an asymptotic expansion in \(S_\delta ^\delta \) and in fact, it is exactly the asymptotic expansion of \(\sqrtE_}\) when \(|z|\le 2\Vert }\Vert _\infty \) and \(|\,}}z|\ge h^\delta \). In fact, \(\sqrt(E_}^ - F_n)\in S^\).

In fact, since |z| is bounded and \(|\,}}z|\ge h^\delta \) and \(F_\) is a rational function in z, thus \(h^}F_ \in S_^) +\delta }\). Since \(j(\delta - \frac) +\delta \rightarrow -\infty \), (A.2) is not only a formal expansion but is indeed an asymptotic expansion of \(F_n\) in the symbol class \(S_\delta ^\delta \).

Furthermore, comparing (A.6) with (A.7), we see that \(F_n\# E_}= 1 - R_n\) with \(R_n \in S^\). By Beal’s lemma, there is \(\tilde_n\in S^\) such that \((1 - R_n^W)^ = 1-\tilde_n^W\). Thus \(\sqrtE_}^ = F \# (1 - \tilde_n^W)\in S_\delta ^\delta \) and have exactly the same asymptotic expansion as \(F_n\) in (A.2) since \(\tilde_n\in S_\delta ^\). Thus part (2) is proved.

It follows that \(r_n:= \partial _z E_}\# E_}^\) is also well-defined with an asymptotic expansion in \(S_\delta ^\delta \). Since

$$\begin \begin&r_n \sim \sum \limits _^\infty h^}\partial _z E_ \# \sum \limits _^\infty h^} F_\\&= \sum \limits _^\infty \sum \limits _^\infty h^} \sum \limits _^\infty h^\gamma \left. \left( \left( \frac\right) ^\gamma (E_(x_2,\xi _2;z)F_(y,\eta ;z))\right) \right| _\\&= \sum _^\infty \sum \limits _^j \sum \limits _^ h^} r_ \left. \left( \left( \frac\right) ^}(E_(x_2,\xi _2;z)F_(y,\eta ;z))\right) \right| _. \end \end$$

Combining it with part (1) and (2) and the fact that \(\sigma \) is linear in \(D_\), \(D_\), we get part (3). Part (4) follows directly from parts (1), (2), (3) with Prop. A.2. \(\square \)

For the Proof of Lemma 4.8

In this subsection, we provide several lemmas that together complete the proof of Lemma 4.8. We start with a proposition that expresses the Hilbert-Schmidt norm of the quantization in terms of its operator-valued symbol.

Proposition B.1

Let \(}_1\), \(}_2\) be two Hilbert spaces. Let \(P: ^2 \rightarrow \mathcal (}_1;}_2)\) be an operator-valued symbol in the symbol class \(S(^2_;\mathcal (}_1;}_2))\). Furthermore, let \(\Vert \cdot \Vert _\,}}}\) denote the Hilbert-Schmidt norm of maps \(}_1\) to \(}_2\) or \(L^2(_y;}_1)\) to \(L^2(_y;}_2)\). Then

$$\begin \Vert P^W(y,hD_y)\Vert _\,}}}^2=\frac\int _^2}\Vert P(y,\eta )\Vert _\,}}}^2 \ dy \ d\eta . \end$$

In particular, if \(}_1 = }_2 = \), for the scalar-valued symbol P, we have

$$\begin \Vert P^W(y,hD_y)\Vert _\,}}}^2 = \frac^;)}^2}. \end$$

(B.1)

The next Lemma allows us to interchange the order of trace and integration.

Lemma B.2

Let \(E_\), \(E_\) be as in (4.16). Let \(\tilde}}_R^W\), \(\bar}}_R^W\) be as in the proof of Lemma 4.7. Then, there exists a constant \(C>0\) such that

$$\begin \begin&\Vert \bar}}_R^WE_\Vert _}(L^2(^2_x), L^2(_))}\le Ch^ R\text \\ &\Vert E_\tilde}}_R^W\Vert _}(L^2(^2_x), L^2(_))}\le Ch^ R. \end \end$$

Proof

The first equation follows from (B.1). For the second equation, we first recall that \(\square \)

Claim 1

If \(a\in S(^;\mathcal (X,Y);m_1)\), \(b\in S(^;\,}}(Y,Z);m_2)\) and \(m_1m_2\in L^2(^_) \), where \(m_1,m_2\) are order functions, then

$$\begin b \# a \in S(^;\,}}(X,Z);m_1m_2) \text (b \# a)^W = b^Wa^W \in \,}}(L^2(^n_x;X);L^2(^n_x;Y)). \end$$

Similar to Lemma 1 in [39], we can show that

Claim 2

For any \(k'\) such that \(1<k'\), we have

(1)

\(E_(x_2,\xi _2)\in S(^2_;\mathcal (B^_;^2))\),

(2)

\(\tilde}}_R^w(x,D_,\xi _2)\in S(^2_;\,}}(L^2_;B^_);m)\), where \(m(x_2,\xi _2) = (1+ (|(x_2,\xi _2)| - R)_+)^\) is the order function.

Then it follows that, by Claim 1, we have \(E_\# \tilde}}_R^w\in S(^2_;\,}}(L^2_);m)\), i.e.,

$$\begin \Vert E_\# \tilde}}_R^w(x_2,\xi _2)\Vert _\,}}(L^2_)} \le m(x_2,\xi _2) = (1+(|(x_2,\xi _2)| - R)_+)^. \end$$

Thus by Prop. B.1, since for all \(k>0\),

$$\begin \int _^2}[1+(|(x_2,\xi _2)|-R)_+]^dxd\xi =\pi R^2+\mathcal O(R^)=}(R^2), \end$$

we get \(\Vert E_\tilde}}_R^W\Vert _\,}}(L^2(_;L^2(_;^4));L^2(_;^2))} \le Ch^R\) and the Lemma is proved.

Lemma B.3

Let \(E_,E_,E_}\) be as in (4.16). For \(\,}}z\ne 0\), both operators

$$\begin\tilde}}_R^WE_E_}^E_\tilde}}_R^W\text \bar}}_R^WE_E_E_}^\bar}}_R^W\end$$

are trace class as bounded linear operators \(\mathcal (L^2(_;L^2(_;^4)))\) and \(\mathcal (L^2(_;^2))\), respectively.

Proof

By Lemma B.2, the fact that \(\tilde}}_R^WE_\) is the adjoint of \(E_\tilde}}_R^W\) and boundedness of \(E_}\) from (4.19), we have

$$\begin \begin&\,}}_1(\tilde}}_R^WE_E_}^E_\tilde}}_R^W)\le \frac}|\,}}z|}\text \,}}_2(\bar}}_R^WE_E_E_}^\bar}}_R^W)\le \frac}|\,}}z|}. \end \end$$

\(\square \)

The second proposition allows us to change the position of \(E_\) in the averaging and limiting process in the proof of Lemma 4.8.

Lemma B.4

Let \(E_,E_,E_}\) be as in (4.16), then

$$\begin \,}}_^2_;^4)} (}_R^WE_E_}^E_}_R^W) - \,}}__;^2)}(\bar}}_R^WE_E_E_}^\bar}}_R^W) \le \frac}}\,}}z|}. \end$$

Proof

Since \(\,}}(AB) = \,}}(BA)\) when AB and BA are both of trace class.

$$\begin \begin&\,}}_1 (\tilde}}_R^WE_E_}^E_\tilde}}_R^W) - \,}}_2(\bar}}_R^WE_E_E_}^\bar}}_R^W)\\ =&\,}}_2 (E_(\tilde}}_R^W)^2E_E_}^) - \,}}_2((\bar}}_R^W)^2E_E_E_}^)\\ =&\,}}_2\left[ (E_\tilde}}_R^W- \bar}}_R^WE_)\tilde}}_R^WE_E_}^\right] + \,}}_2\left[ \bar}}_R^W(E_\tilde}}_R^W- \bar}}_R^WE_)E_E_}^\right] \\ =:&\,}}_2\left[ [E_, }_R]_w\tilde}}_R^WE_E_}^\right] + \,}}_2\left[ \bar}}_R^W[E_, }_R]_wE_E_}^\right] \\ =:&\,}}_2(A_1) + \,}}_2(A_2) \end \end$$

where \([E_, }_R]_W:= E_\tilde}}_R^W- \bar}}_R^WE_\). Then the following claim completes the proof. \(\square \)

Claim 3

For \(\,}}z\ne 0\), \(A_1\), \(A_2\) are trace class operators and there is a \(C>0\) such that

$$\begin \,}}_2(A_1),\,}}_2(A_2)\le C h^|\,}}z|^R^. \end$$

Proof of Claim 3

From Lemma B.2, we already know

$$\begin \Vert [E_,}_R]_W \Vert _\,}}^\,}}}} \le Ch^ R, \quad \end$$

where \(\,}}^\,}}} = \,}}(L^2(_;L^2(_;^4));L^2(_;^2))\). We will improve the upper bound from \(Ch^R\) to \(Ch^R^\).

Let \(\bar_R^c=1-\bar_R\), \(}}}_R^c = 1 - }}}_R\). First notice that from the proof of Lemma B.2, and replacing \(\bar_R\) by \(\bar_R^c\), we have

$$\begin \begin&\Vert [E_, }_R]_w(x_2,\xi _2)\Vert _\,}}} \\&\le \tfrac}\text \Vert [E_, }_R^c]_w(x_2,\xi _2)\Vert _\,}}}\\&\le \tfrac} \end \end$$

where \([E_, }_R]_w(x_2,\xi _2) = E_\# \tilde}}_R^w- }}}_R \# E_\) is the symbol in \((x_2,\xi _2)\) of \([E_, }_R]_W\) and \(\,}}= \,}}(L^2(_;^4);^2)\). Since \([E_, }_R]_w = -[E_, }_R^c]_w\), we have

$$\begin \Vert [E_, }_R]_w(x_2,\xi _2)\Vert _\,}}} \le C_k[1+||(x_2,\xi _2)| - R |]^. \end$$

Thus by Prop. B.1 and a straightforward computation of the following integral

$$\begin \int _^2_} [1+||(x_2,\xi _2)| - R |]^dx_2d\xi _2 = \frac + \frac = }(R), \end$$

we find that \( \Vert [E_, }_R]_W\Vert _\,}}^\,}}}}\le Ch^R^.\) Since \(\tilde}}_R^WE_\) is the adjoint of \(E_\tilde}}_R^W\), this yields that

$$\begin \,}}(A_1) \le Ch^R^,\quad \,}}(A_2)\le Ch^R^. \end$$

\(\square \)

In next Lemma, we state the averaging property of periodic symbols to reduce the regularized trace to a fundamental cell.

Lemma B.5

Let \(E_,E_,E_}\), \(}}}_R\) be as in (4.16). Then

$$\begin & \lim \limits _ \frac \int _^2}\,}}_^2}(\bar}}_R\#\partial _z E_}\#E_}^\#\bar}}_R)~d~d \\ & = \frac \int _ \partial _}\tilde \,}}_^2}(\partial _z E_}\#E_}^) ~d ~d. \end$$

The proof of this Lemma can be found in [39, Prop. 3].