Inferring topology of quantum phase space

Abstract

Does a semiclassical particle remember the phase space topology? We discuss this question in the context of the Berezin–Toeplitz quantization and quantum measurement theory by using tools of topological data analysis. One of its facets involves a calculus of Toeplitz operators with piecewise-constant symbols developed in an appendix by Laurent Charles.

Appendices

UMR 7586, Institut de Mathématiques de Jussieu-Paris Rive Gauche Sorbonne Universités, UPMC Univ Paris 06 F-75005, Paris, France. email: laurent.charles@imj-prg.fr.

Appendix: Toeplitz operators with piecewise constant symbol (by Laurent Charles)

UMR 7586, Institut de Mathématiques de Jussieu-Paris Rive Gauche Sorbonne Universités, UPMC Univ Paris 06 F-75005, Paris, France. email: laurent.charles@imj-prg.fr.

In Berezin–Toeplitz quantization, we consider a symplectic compact manifold M, a set $\Lambda \subset {\mathbb {R}}_{>0}$ having 0 as a limit point, and for any $\hbar \in \Lambda$, a Hermitian complex line bundle $L_\hbar$ and a finite-dimensional subspace $\mathcal {H}_{\hbar }$ of ${\mathcal {C}}^{\infty }(M, L_{\hbar })$. The space $\mathcal {H}_{\hbar }$ has a natural scalar product $\langle \Psi , \Psi ' \rangle _{\mathcal {H}_{\hbar }} = \int _M ( \Psi , \Psi ') \; d\nu$, where $(\Psi , \Psi ')$ is the pointwise scalar product and $\nu$ the Liouville measure⁴ of M. To any function $f \in {\mathcal {C}}^{\infty }(M)$, we associate an endomorphism $T_{\hbar } (f): \mathcal {H}_{\hbar } \rightarrow \mathcal {H}_\hbar$ such that

$$\begin{aligned} \langle T_{\hbar } (f) \Psi , \Psi ' \rangle _{\mathcal {H}_{\hbar }} = \int _M (\Psi , \Psi ' )(x) f(x) d \nu (x) , \qquad \Psi , \Psi ' \in \mathcal {H}_{\hbar }, \end{aligned}$$

(18)

When the spaces $\mathcal {H}_{\hbar }$ are conveniently defined, the family $(T_{\hbar }, \hbar \in {\Lambda })$ enjoys usual semi-classical properties. Typically, for a Kähler manifold M equipped with a positive line bundle L, we choose ${\Lambda }:= \{\hbar = 1/k, \; k \in {\mathbb {N}}^* \}$, $L_{\hbar } := L^{\otimes k}$ and define $\mathcal {H}_{\hbar }$ as the space of holomorphic sections of $L^k$. These definitions can be extended to any quantizable M, cf. Ma and Marinescu (2011) or Charles (2016) for instance. For the purpose of this appendix, we only need that the reproducing kernel of $\mathcal {H}_{\hbar }$ satisfies the two estimates (22) and (23).

In the definition (18) of $T_{\hbar } (f)$, instead of a smooth function f, we can more generally consider any distribution $f \in \mathcal {C}^{-\infty } (M)$. Indeed, in equation (18), the pointwise scalar product $(\Psi , \Psi ' )$ is a smooth function, so the integral of f against $(\Psi , \Psi ' ) d \nu$ still makes sense and defines an endomorphism $T_{\hbar } (f)$. The map $T_{\hbar } :\mathcal {C}^{-\infty }(M) \rightarrow {\text {End}} ( \mathcal {H}_{\hbar })$ is linear and positive in the sense that $T_\hbar (f) \geqslant 0$ when $f \geqslant 0$. The question is whether the asymptotic properties of $T_{\hbar } (f)$ still hold, in particular the estimates of the trace and the product.

Proposition A.1

For any $f \in \mathcal {C}^{-\infty } (M)$, we have

$$\begin{aligned} {\text {tr}} ( T_\hbar (f) ) = ( 2\pi \hbar )^{-n} \Bigg ( \int _M f d\nu \Bigg ) ( 1+ \mathcal {O}( \hbar )) \end{aligned}$$

The proof is an immediate generalization of the smooth case and will be given later. For the multiplicative properties of $T_{\hbar }$, the regularity is crucial. For instance, let us recall two estimates proved in Barron et al. (2014). Let $R_{\hbar } (f,g) = T_\hbar ( f) T_{\hbar } (g) - T_{\hbar } (fg)$. When f and g are both of class $\mathcal {C}^\ell$ with $\ell =1$ or 2, $\Vert R_{\hbar } (f,g) \Vert = \mathcal {O}( \hbar ^{\ell /2})$. When f and g are only assumed to be continuous, $\Vert R_\hbar ( f,g)\Vert$ tends to 0 in the semiclassical limit $\hbar \rightarrow 0$. It is not proved that these estimates are sharp, but we believe they are.

Our goal is to extend these multiplicative properties to a subalgebra of $L^{\infty } ( M)$ containing the characteristic functions of smooth domains. By a smooth domain, we mean a 0-codimensional smooth submanifold with boundary. For any endomorphism T of $\mathcal {H}_\hbar$, we introduce its Schatten norm normalized by the dimension $d(\hbar ) = \dim \mathcal {H}_{\hbar }$,

$$\begin{aligned} \Vert T \Vert _p := \Bigg ( \frac{ {\text {tr}} |T|^p}{d (\hbar )} \Bigg )^{1/p} , \qquad p \in [1,\infty ). \end{aligned}$$

If $(T_\hbar )$ is a family of endomorphisms depending on $\hbar$, we write $T_{\hbar } = \mathcal {O}_p (\hbar ^m)$ for $\Vert T_{\hbar } \Vert _p = \mathcal {O}(\hbar ^m)$. For any measurable set A of M, denote by $\chi _A \in L^{\infty } (M)$ its characteristic function. We say that A is a good set if

$$\begin{aligned} T_{\hbar } ( \chi _A ) ^2 = T_{\hbar } ( \chi _A ) + \mathcal {O}_{1} ( \hbar ^{1/2}) . \end{aligned}$$

(19)

We say that a function $f \in L^{\infty } (M)$ is a simple function if it has the form

$$\begin{aligned} f = \sum _{i=1}^{m} {\lambda }_i \chi _{A_i} \end{aligned}$$

(20)

where $m\in {\mathbb {N}}$, ${\lambda }_1$,...,${\lambda }_m$ are real numbers and $A_1$,...,$A_m$ are good sets.

Theorem A.2

• 1. Any smooth domain of M is a good set.

• 2. The good sets form an algebra, that is, they are closed under taking complement, finite intersection and finite union.

• 3. If $f, g \in L^{\infty }(M)$ are simple functions, then fg is simple and

  $$\begin{aligned} T_\hbar (f) T_\hbar (g) = T_{\hbar } (fg) + \mathcal {O}_2 ( \hbar ^{1/4}). \end{aligned}$$

• 4. If f is simple and takes only non-negative values, then $f^{1/2}$ is simple and

  $$\begin{aligned} T_\hbar (f)^{1/2} = T_{\hbar } ( f^{1/2} ) + \mathcal {O}_4 ( \hbar ^{1/8}) . \end{aligned}$$

Interestingly, only the first assertion relies on the estimates (22) and (23) of the Bergman kernel. The proof of the other assertions is independent and does not use any difficult result.

Corollary A.3

Let $f_1$,...,$f_m$ be m simple non-negative functions. Let $P_\hbar = T_\hbar (f_1)^{1/2}\cdots T_\hbar (f_m)^{1/2}$. Then

$$\begin{aligned} \frac{1}{d(\hbar )} {\text {tr}} (P_\hbar ^* P_{\hbar }) = \frac{1}{ \nu (M)}\int _M f_1 \cdots f_m \; d \nu + \mathcal {O}( \hbar ^{1/8}) . \end{aligned}$$

Remark

It is essential that we use a Schatten norm in the definition (19) of a good set. Indeed, for any measurable set A, $0 \leqslant T_\hbar ( \chi _A) \leqslant 1$, so $0 \leqslant T_{\hbar } ( \chi _A ) - T_{\hbar } ( \chi _A ) ^2 \leqslant 1/4$. When A is a good domain such that A and its complement are non-empty, we will prove in a forthcoming paper that $T_{\hbar } ( \chi _A)$ has an eigenvalue ${\lambda }( \hbar )$ converging to 1 / 2 when $\hbar \rightarrow 0$. Therefore,

$$\begin{aligned} \Vert T_{\hbar } ( \chi _A ) ^2 - T_{\hbar } ( \chi _A ) \Vert \rightarrow 1/4. \end{aligned}$$

The curious reader can think about the case where M is the two-sphere and A a hemisphere. In this case, we can explicitly compute the spectrum of $T_{\hbar } (\chi _A)$, as we learned from Douçot and Estienne (2017), cf. also Barron and Polterovich (2015).

$\square$

Proofs

Let $( \Psi _i)$ be an orthonormal basis of $\mathcal {H}_{\hbar }$ and define the Bergman kernel

$$\begin{aligned} K_{\hbar } (x,y) = \sum _{i =1}^{d( \hbar )} \Psi _i (x) \otimes \overline{\Psi }_i ( y) \in L_x^k \otimes \overline{L}_y^k , \qquad x,y \in M, \end{aligned}$$

(21)

where $\overline{L}$ is the conjugate line bundle of L. We will need the diagonal estimate

$$\begin{aligned} K_{\hbar } (x,x) = (2\pi \hbar )^{-n} ( 1 + \mathcal {O}( \hbar )) , \end{aligned}$$

(22)

where we identify $L_x^k \otimes \overline{L}_x^k$ with ${\mathbb {C}}$ by using the metric of L. The second estimate we need is

$$\begin{aligned} |K_{\hbar } (x,y)| \leqslant C_m {\hbar }^{-n} e^{-\hbar ^{-1} d(x,y)^2/C } + C_m \hbar ^{m} \end{aligned}$$

(23)

for any $m \in {\mathbb {N}}$, with some positive constants C and $C_m$ independent of x, y. Here d is any distance on M obtained by embedding M into an Euclidean space and restricting the Euclidean distance. In the Kähler case, (22) was first proved in Bouche (1990) and was subsequently extended in Zelditch (1998) to convergence in the ${\mathcal {C}}^{\infty }$-topology. Estimate (23) follows from Corollary 1 of Charles (2003).

Proof of Proposition A.1

We have

$$\begin{aligned} {\text {tr}} ( T_\hbar (f) )= & {} \sum _{i=1} ^{d(\hbar )} \langle T_{\hbar } (f) \Psi _i , \Psi _i \rangle = \int _M f(x) K_\hbar (x,x) d\nu (x) \\= & {} ( 2\pi \hbar )^{-n} \Bigg ( \int _M f d\nu \Bigg ) ( 1+ \mathcal {O}( \hbar )) \end{aligned}$$

by (22), which holds in the ${\mathcal {C}}^{\infty }$-topology. $\square$

In the sequel, to lighten the notation, we set $T_{A} := T_{\hbar } ( \chi _A)$ for any measurable set A of M. We denote by $A^c$ the complement of A.

Lemma A.4

A measurable set A of M is good if and only if

$$\begin{aligned} \int _{ A \times A^{c} }| K_{\hbar }(x,y)|^2 d\nu (x) \; d\nu (y) = \mathcal {O}( \hbar ^{-n + 1/2})\;. \end{aligned}$$

(24)

Proof

Since $0 \leqslant T_A \leqslant 1$, $T_A - T_A^2 \geqslant 0$. Hence

$$\begin{aligned} \Vert T_A^2 - T_A \Vert _1 = \frac{1}{d(\hbar )} {\text {tr}} ( T_A - T_A^2 ) = \frac{1}{d(\hbar )} {\text {tr}} ( T_A ( 1 - T_A)). \end{aligned}$$

Using that $d(\hbar ) = (2 \pi \hbar )^{-n} \nu (M) ( 1+ \mathcal {O}( \hbar ))$, we see that A is good if and only if ${\text {tr}} ( T_A ( 1 - T_A)) = \mathcal {O}( \hbar ^{-n +1/2})$. To conclude the proof, observe that for any measurable subsets A and B

$$\begin{aligned} {\text {tr}} (T_A T_B) = \int _{A \times B} | K_\hbar (x,y) |^2 d\nu (x) \; d\nu (y)\;. \end{aligned}$$

(25)

Indeed, computing the trace in the orthogonal basis $(\Psi _i)$, we have

$$\begin{aligned} {\text {tr}} (T_A T_B)= & {} \sum _{i,j} \langle T_A \Psi _i , \Psi _j \rangle \langle T_B \Psi _j, \Psi _i \rangle \\= & {} \sum _{i,j} \int _{A\times B} ( \Psi _i, \Psi _j ) (x) ( \Psi _j, \Psi _i )(y) \; d\nu (x) \; \nu (y)\;. \end{aligned}$$

By the definition (21) of the Bergman kernel,

$$\begin{aligned} |K_{\hbar } (x,y)|^2= & {} \sum _{i,j} \bigl ( \Psi _i ( x) \otimes \overline{\Psi }_i ( y) , \Psi _j ( x) \otimes \overline{\Psi }_j (y) \bigr ) \\= & {} \sum _{i,j} (\Psi _i,\Psi _j ) (x) (\Psi _j, \Psi _i )(y)\;, \end{aligned}$$

which proves (25). $\square$

Proof of assertion 1 of Theorem A.2

We will deduce from estimate (23) that any smooth domain A in M satisfies (24). Consider a finite cover $(U_{\alpha }, \; 1 \leqslant {\alpha }\leqslant N)$ of M such that each $U_{{\alpha }}$ is the domain of a coordinate system $(x_i)$ in which $A \cap U_{{\alpha }} = \{ x \in U_{\alpha }; \; x_1(x) \geqslant 0 \}$. Denote by $\Delta$ the diagonal of $M^2$. Let $(f_{\alpha }, \; 0\leqslant {\alpha }\leqslant N)$ be a partition of unity of $M^2$ subordinated to the cover $( \Delta ^c, U_1^2, \ldots , U_m^2 )$. It suffices to show that for each ${\alpha }$

$$\begin{aligned} \int _{A \times A^c} f_{{\alpha }}(x,y) |K_{\hbar }(x,y) |^2 \; d\nu (x) \; d\nu (y) = \mathcal {O}( {\hbar }^{-n+1/2})\;. \end{aligned}$$

(26)

For ${\alpha }=0$, this follows from the fact that ${\text {supp}} f_0 \cap \Delta = \emptyset$, so that $|K_{\hbar } (x,y) | = \mathcal {O}( \hbar ^{-\infty })$ uniformly on ${\text {supp}} f_0$. Let ${\alpha }\geqslant 1$ and choose a coordinate system $(x_i)$ on $U_{\alpha }$ as above. Introduce on $U_{\alpha }^2$ the coordinate system

$$\begin{aligned} s_i (x,y) = x_i (x), \qquad t_i ( x,y) = x_i ( x) - x_i ( y), \qquad x,y \in U_{{\alpha }}\;. \end{aligned}$$

Then by (23) there exists a constant C such that for any $(x,y) \in {\text {supp}} f_{\alpha }$, we have

$$\begin{aligned} | K_\hbar |^2 \leqslant C {\hbar }^{-2n} e^{-\hbar ^{-1} |t|^2 /C} + C \hbar ^{-n+ 1/2}\;, \end{aligned}$$

where $|t|^2 = \sum t_i^2$. Furthermore $(A \times \overline{A^c}) \cap U_{\alpha }^2 = \{ 0 \leqslant s_1 \leqslant t_1 \}$. So the integral in (26) is bounded above by

$$\begin{aligned} \hbar ^{-2n} \int _{ 0 \leqslant s_1 \leqslant t_1, \; |s'|_\infty \leqslant M } e^{-\hbar ^{-1} |t|^2 /C} \; ds \; dt + \mathcal {O}( \hbar ^{-n+1/2})\;, \end{aligned}$$

where $|s'|_{\infty } = \max ( |s_2|, \ldots , | s_{2n -1}|)$ and M is chosen so that the support of $f_{\alpha }$ is contained in $\{ |s'|_{\infty } \leqslant M \}$. Integrating with respect to the $s_i$’s and making the change of variable $t= t \hbar ^{-1/2}$, we obtain

$$\begin{aligned}&\hbar ^{-2n} \int _{ 0 \leqslant s_1 \leqslant t_1, \; |s'|_{\infty } \leqslant M} e^{-\hbar ^{-1} |t|^2 /C} \; ds \; dt = \hbar ^{-2n} (2M)^{2n-1} \int _{ 0 \leqslant t_1} t_1 e^{-\hbar ^{-1} |t|^2 /C} dt \\&\quad = \hbar ^{-n+1/2} (2M)^{2n-1} \int _{ {\mathbb {R}}_+ \times {\mathbb {R}}^{2n-1} } t_1 e ^{-|t|^2/C} dt\;. \end{aligned}$$

Hence, the estimate (26) holds. $\square$

Proof of assertion 2 of Theorem A.2

Since $T_{A^c} = 1 - T_A$, $T_A - T_A^2 = T_{A^c} - T_{A^c}^2$, so A is a good set if and only if $A^c$ is a good set. The intersection of two good sets A, B is good because

$$\begin{aligned} ( A\cap B) \times (A \cap B)^c \subset ( A \times A^c) \cup (B \times B^c) \end{aligned}$$

and thanks to Lemma A.4. $\square$

Proof of assertion 3 of Theorem A.2

It suffices to prove that for any good sets A and B,

$$\begin{aligned} T_A T_B = T_{A \cap B} + \mathcal {O}_2 ( \hbar ^{1/2})\;. \end{aligned}$$

(27)

We first show that

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} (T_A T_B) = \mu (A \cap B) + \mathcal {O}( \hbar ^{1/2})\;, \end{aligned}$$

(28)

with $\mu (A \cap B) = \nu ( A \cap B) /\nu (M)$. Introduce the good sets $a = A {\setminus } B$, $b = B {\setminus } A$ and $C = A \cap B$. Then A is the disjoint union of a and C, so $T_A = T_a + T_C$. In the same way, $T_B = T_b + T_C$. Therefore,

$$\begin{aligned} T_A T_B = T_a T_b + T_a T_C + T_C T_b + T_C^2. \end{aligned}$$

(29)

Since a and b are disjoint, we deduce from (25) that

$$\begin{aligned} 0 \leqslant {\text {tr}} (T_a T_b) \leqslant {\text {tr}}(T_a ( 1-T_a)). \end{aligned}$$

Further, since a is good, we obtain that $d(\hbar )^{-1} {\text {tr}} (T_a T_b) = \mathcal {O}( \hbar ^{1/2})$. By the same argument, $d(\hbar )^{-1} {\text {tr}} ( T_a T_C) = \mathcal {O}( \hbar ^{1/2})$ and $d(\hbar )^{-1} {\text {tr}} (T_C T_b) = \mathcal {O}( \hbar ^{1/2})$. Finally, since C is good,

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} T_C^2 =d(\hbar )^{-1} {\text {tr}} T_C + \mathcal {O}( \hbar ^{1/2}) = \mu (C) + \mathcal {O}( \hbar ^{1/2}) \end{aligned}$$

by Proposition A.1. Summing the various estimates, we get (28).

Second, we compute the Hilbert-Schmidt norm of $T_A T_B$. We will use the following consequence of the Hölder inequality: for any $\hbar$-dependent endomorphisms S, $S'$, T of $\mathcal {H}_{\hbar }$,

$$\begin{aligned} S= & {} S' + \mathcal {O}_p(\hbar ^m)\quad \text { and }\quad \Vert T \Vert = \mathcal {O}(1) \nonumber \\&\Rightarrow \quad \frac{1}{d(\hbar )} {\text {tr}} (ST) = \frac{1}{d(\hbar )} {\text {tr}} (S'T) + \mathcal {O}( \hbar ^m). \end{aligned}$$

(30)

We have:

$$\begin{aligned} \Vert T_A T_B \Vert ^2_2= & {} d(\hbar )^{-1} {\text {tr}} (T_A^2 T_B^2) \nonumber \\= & {} d(\hbar )^{-1} {\text {tr}} (T_A^2 T_B) + \mathcal {O}( \hbar ^{1/2}) \text { because}~ B~\text {is good and by }~(30) \nonumber \\= & {} d(\hbar )^{-1} {\text {tr}} (T_A T_B) + \mathcal {O}( \hbar ^{1/2}) \text { because}~ A~\text {is good and by}~ (30)\nonumber \\= & {} \mu ( A\cap B) + \mathcal {O}( \hbar ^{1/2}) \text { by}~ (28)\;. \end{aligned}$$

(31)

Now we come to the proof of (27). With $C = A \cap B$, we have

$$\begin{aligned} \Vert T_A T_B - T_C \Vert _2^2 = d(\hbar )^{-1} {\text {tr}} ( T_A^2T_B^2 + T_C^2 - T_A T_B T_C - T_C T_B T_A ). \end{aligned}$$

Next, by (31),

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} ( T_A^2T_B^2)= \mu (C) + \mathcal {O}( \hbar ^{1/2}), \quad d(\hbar )^{-1} {\text {tr}} T_C^2 = \mu (C) + \mathcal {O}( \hbar ^{1/2}). \end{aligned}$$

(32)

To estimate the trace of $T_A T_B T_C$, we use as above the sets $a = A {\setminus } B$ and $b = B {\setminus } A$. By (29),

$$\begin{aligned} T_A T_B T_C = T_a T_b T_C + T_a T_C^2 + T_C T_bT_C + T_C^3\;. \end{aligned}$$

Using that C is good and (30), we have

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} (T_C^3) = d(\hbar )^{-1} {\text {tr}}(T_C^2) + \mathcal {O}( \hbar ^{1/2}) = \mu ( C) + \mathcal {O}( \hbar ^{1/2}) \end{aligned}$$

by (28). Similarly, using that C is good, (30) and (28), we have

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} ( T_a T_C^2 ) = d(\hbar )^{-1} {\text {tr}} ( T_a T_C) + \mathcal {O}( \hbar ^{1/2}) = \mathcal {O}( \hbar ^{1/2})\;, \end{aligned}$$

because $a\cap C = \emptyset$. By the same argument, $d(\hbar )^{-1} {\text {tr}} ( T_b T_C^2 ) = \mathcal {O}( \hbar ^{1/2})$. Finally, C being good, $d(\hbar )^{-1} {\text {tr}} (T_a T_b T_C) = d(\hbar )^{-1} {\text {tr}} (T_a T_b T_C^2) + \mathcal {O}( \hbar ^{1/2})$, and by the Hölder inequality,

$$\begin{aligned} \Bigl | \frac{1}{d(\hbar )} {\text {tr}} (T_a T_b T_C^2) \Bigr | \leqslant \Vert T_CT_a \Vert _2 \Vert T_b T_C \Vert _2= \mathcal {O}( \hbar ^{1/4} ) \mathcal {O}( \hbar ^{1/4}) = \mathcal {O}( \hbar ^{1/2})\;, \end{aligned}$$

where we have applied (31) to C, a and b, C and used the fact that these sets are pairwise disjoint. Gathering these estimates we conclude that

$$\begin{aligned} d(\hbar )^{-1} {\text {tr}} ( T_A T_B T_C) = \mu (C) + \mathcal {O}( \hbar ^{1/2}). \end{aligned}$$

(33)

Exchanging A and B, we get the same estimate for $d(\hbar )^{-1} {\text {tr}} ( T_B T_a T_C)$. Now (33) and (32) imply that $\Vert T_A T_B - T_C \Vert _2^2 = \mathcal {O}( \hbar ^{1/2})$. $\square$

Proof of assertion 4 of Theorem A.2

Since the good sets are closed under taking the complement and finite intersections, we see that any simple function f can be written as a sum (20) with the $A_i$ being pairwise disjoint good sets. We can furthermore assume that these sets are non-empty. Assume that f is non-negative, then all the coefficients ${\lambda }_i$ are non-negative. Then $f^{1/2}$ is simple. Set $S_{\hbar } (f) = T_{\hbar }(f^{1/2})^2$. By the third assertion of Theorem A.2, we have

$$\begin{aligned} S_{\hbar } (f) = T_{\hbar } (f) + \mathcal {O}_2 ( \hbar ^{1/4})\;. \end{aligned}$$

(34)

Using that the square root is an operator monotone function, we have

$$\begin{aligned} \Vert \sqrt{S_{\hbar } (f) } - \sqrt{T_{\hbar } (f) } \Vert _4 \leqslant \bigl \Vert \sqrt{|S_\hbar (f) - T_{\hbar } (f) |} \bigr \Vert _4 = \Vert S_\hbar (f) - T_{\hbar } (f) \Vert _2^{1/2} \end{aligned}$$

and the right-hand side is a $\mathcal {O}( \hbar ^{1/8})$ by (34). In other words, $T_\hbar ( f^{1/2}) = T_{\hbar } ( f) ^{1/2} + \mathcal {O}_4 ( \hbar ^{1/8}).$ $\square$

Proof of Corollary A.3

By assertion 4 of Theorem A.2 and (30), we have

$$\begin{aligned} \frac{1}{d(\hbar )} {\text {tr}} ( P_\hbar ^* P_\hbar ) = \frac{1}{d(\hbar )} {\text {tr}} ( Q_\hbar ^* Q_{\hbar } ) + \mathcal {O}( \hbar ^{1/8})\;, \end{aligned}$$

with $Q_{\hbar } = T_{\hbar } (f_1) \cdots T_{\hbar } ( f_m)$. By assertion 3 of Theorem A.2,

$$\begin{aligned} Q_\hbar ^* Q_\hbar = T_{\hbar } ( f_1 \cdots f_m) + \mathcal {O}( \hbar ^{1/4}). \end{aligned}$$

Hence, by (30) and Proposition A.1,

$$\begin{aligned} \frac{1}{d(\hbar )} {\text {tr}} ( Q_\hbar ^* Q_\hbar ) = \frac{1}{\nu (M) } \int _M f_1 \cdots f_m \; d\nu + \mathcal {O}( \hbar ^{1/4})\;, \end{aligned}$$

and the result follows. $\square$