Proof of Product and Commutator Estimates

Abstract

The aim of this chapter is to prove Theorems 5.2.2 and 5.2.3.

The aim of this chapter is to prove Theorems 5.2.2 and 5.2.3.

8.1 Corrected Berezin–Toeplitz Operators

Given a function $f \in \mathcal {C}^{\infty }(M, \mathbb {R})$, we introduce the corrected Berezin–Toeplitz quantisation of f:

$$\begin{aligned} T_k^{\mathrm {c}}(f) = \varPi _k \biggl (f + \frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ):\mathcal {H}_k \rightarrow \mathcal {H}_k \end{aligned}$$

(8.1)

where $X_f$ is the Hamiltonian vector field associated with f. The operator

$$ P_k(f) = f + \frac{1}{\text {i}k} \nabla ^k_{X_f}:\mathcal {C}^{\infty }(M, L^k) \rightarrow \mathcal {C}^{\infty }(M, L^k) $$

is called the Kostant–Souriau operator associated with f. The Kostant–Souriau operators satisfy the following nice properties.

Lemma 8.1.1.

For any $f, g \in \mathcal {C}^{\infty }(M, \mathbb {R})$,

$$ P_k(fg) = P_k(f) P_k(g) - \frac{1}{\text {i}k}\biggl (\{f, g\} + \frac{1}{\text {i}k} \nabla _{X_f}^k \nabla _{X_g}^k\biggr ). $$

Proof.

Since $X_{fg} = f X_g + g X_f$, we have that

$$ P_k(fg) = f\biggl (g + \frac{1}{\text {i}k} \nabla ^k_{X_g}\biggr ) + g \biggl (\frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ) = f P_k(g) + g \biggl (\frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ). $$

We can rewrite this as

$$ P_k(fg) = P_k(f) P_k(g) - \frac{1}{\text {i}k} \nabla ^k_{X_f}P_k(g) + g \biggl (\frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ). $$

Let us simplify the second term of the right-hand side; for $\phi \in \mathcal {C}^{\infty }(M, L^k)$, one has

$$ \nabla ^k_{X_f}(P_k(g) \phi ) = (\mathcal {L}_{X_f}g) \phi + g \nabla _{X_f}^k \phi + \frac{1}{\text {i}k} \nabla _{X_f}^k \nabla _{X_g}^k \phi . $$

Using that $\mathcal {L}_{X_f}g = \{f, g\}$, this implies that

$$\begin{aligned} P_k(fg) = P_k(f) P_k(g) - \frac{1}{\text {i}k}\bigl (\{f, g\} + \frac{1}{\text {i}k} \nabla _{X_f}^k \nabla _{X_g}^k\bigr ). \end{aligned}$$

$\square$

This shows that $P_k(fg)$ differs from $P_k(f) P_k(g)$ by a remainder “of order $k^{-1}$”. It turns out that for commutators, however, there is an exact (i.e. without remainder) correspondence principle for Kostant–Souriau operators.

Lemma 8.1.2.

For any $f, g \in \mathcal {C}^{\infty }(M, \mathbb {R})$,

$$ [P_k(f), P_k(g) ] = \frac{1}{\text {i}k} P_k(\{f, g\}). $$

Proof.

Since $P_k(gf) = P_k(fg)$, the previous lemma yields

$$ [P_k(f), P_k(g) ] = \frac{1}{\text {i}k} \biggl (\{f, g\} + \frac{1}{\text {i}k} \nabla _{X_f}^k \nabla _{X_g}^k - \{g, f\} - \frac{1}{\text {i}k} \nabla _{X_g}^k \nabla _{X_f}^k\biggr ). $$

This can be rewritten as

$$ [P_k(f), P_k(g) ] = \frac{1}{\text {i}k} \biggl (2\{f, g\} + \frac{1}{\text {i}k} [\nabla _{X_f}^k,\nabla _{X_g}^k ]\biggr ). $$

Moreover, by definition of the curvature, we have that

$$ [\nabla ^k_{X_f},\nabla ^k_{X_g} ] = {{\mathrm{curv}}}(\nabla ^k)(X_f, X_g) + \nabla ^k_{[X_f, X_g ]}, $$

which yields, since ${{\mathrm{curv}}}(\nabla ^k) = -\text {i}k\omega$, and since $[X_f, X_g ]$ is the Hamiltonian vector field associated with $\{f, g\}$,

$$ [\nabla ^k_{X_f},\nabla ^k_{X_g} ] = -\text {i}k \{f, g\} + \nabla ^k_{X_{\{f, g\}}}. $$

Putting all these equalities together, we finally obtain that

$$ [P_k(f), P_k(g) ] = \frac{1}{\text {i}k} \biggl (\{f, g\} + \frac{1}{\text {i}k} \nabla ^k_{X_{\{f, g\}}}\biggr ), $$

which was to be proved. $\square$

The idea behind the proof of Theorems 5.2.2 and 5.2.3 is to derive from the properties above some estimates for the corrected Berezin–Toeplitz operators and to take profit of these estimates by comparing the corrected operator $T_k^{\mathrm {c}}(f)$ with the usual Berezin–Toeplitz operator $T_k(f)$. In order to do so, we will need a result due to Tuynman [46], but let us first introduce some notation. Let $g = \omega (\cdot , j\cdot )$ be the Kähler metric on M, let $\mu _g$ be the associated volume form, let ${{\mathrm{grad}}}_g$ be the associated gradient, and let $\varDelta$ be the associated Laplacian. We recall that for any $f \in \mathcal {C}^{2}(M)$,

$$ \varDelta f = {{\mathrm{div}}}_g({{\mathrm{grad}}}_g f) $$

where the divergence ${{\mathrm{div}}}_g(X)$ of a vector field X on M is the function defined by the equality

$$ \mathcal {L}_X \mu _g = {{\mathrm{div}}}_g(X) \mu _g. $$

Proposition 8.1.3

(Tuynman’s lemma). Let $X \in \mathcal {C}^{1}(M, TM \otimes \mathbb {C})$. Then

$$ \varPi _k \nabla ^k_X \varPi _k = -\varPi _k {{\mathrm{div}}}_g(X^{1, 0}) \varPi _k, $$

where we recall that $X^{1, 0} = (X-\text {i}jX)/2$. Furthermore, if $f \in \mathcal {C}^{2}(M, \mathbb {R})$, then

$$ \varPi _k \biggl (\frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ) \varPi _k = - \frac{1}{2k}\varPi _k (\varDelta f) \varPi _k. $$

The following corollary is immediate.

Corollary 8.1.4.

For every $X \in \mathcal {C}^{1}(M, TM \otimes \mathbb {C})$,

$$ \left\| {\varPi _k \biggl (\frac{1}{\text {i}k} \nabla ^k_X\biggr ) \varPi _k}\right\| = O\bigl (k^{-1}\bigr ) \left\| {X}\right\| _1. $$

In particular, for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left\| {\varPi _k \biggl (\frac{1}{\text {i}k} \nabla ^k_{X_f}\biggr ) \varPi _k}\right\| = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _2. $$

Consequently, for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$, $\left\| {T_k^{\mathrm {c}}(f) - T_k(f)}\right\| = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _2$.

Proof of Proposition 8.1.3. Set $Y = X^{1, 0}$. By virtue of Lemma 8.1.5 below, proving the first statement amounts to showing that for every $\phi \in \mathcal {H}_k$,

$$ \langle \varPi _k \bigl (\nabla ^k_X \phi \bigr ),\phi \rangle _k = -\langle \varPi _k({{\mathrm{div}}}_g(Y) \phi ),\phi \rangle _k. $$

Using the facts that $\varPi _k$ is self-adjoint and that $\varPi _k \phi = \phi$ whenever $\phi$ belongs to $\mathcal {H}_k$, we only need to prove that

$$\begin{aligned} \forall \phi \in \mathcal {H}_k, \quad \langle \nabla ^k_X\phi ,\phi \rangle _k = - \langle {{\mathrm{div}}}_g(Y)\phi ,\phi \rangle _k. \end{aligned}$$

(8.2)

Recall that $\mu _g = \mu$ the Liouville measure on M. We have that

$$\begin{aligned} \langle {{\mathrm{div}}}_g(Y)\phi ,\phi \rangle _k = \int _M {{\mathrm{div}}}_g(Y) h_k(\phi , \phi ) \, \mu _g = \int _M h_k(\phi , \phi ) \, \mathcal {L}_Y\mu _g. \end{aligned}$$

(8.3)

Now, by integrating the equality

$$ \mathcal {L}_Y(h_k(\phi , \phi ) \mu _g) = \mathcal {L}_Y \bigl (h_k(\phi , \phi )\bigr ) \mu _g + h_k(\phi , \phi ) \mathcal {L}_Y \mu _g, $$

we obtain that

$$ \int _M h_k(\phi , \phi ) \, \mathcal {L}_Y\mu _g = - \int _M \mathcal {L}_Y \bigl (h_k(\phi , \phi )\bigr ) \mu _g. $$

Indeed, by Cartan’s formula, and using the fact that $h_k(\phi , \phi ) \mu _g$ is closed, we have that $\mathcal {L}_Y(h_k(\phi , \phi ) \mu _g) = \,\mathrm{d}\bigl (i_Y(h_k(\phi , \phi ) \mu _g)\bigr )$, thus its integral on M vanishes. Coming back to (8.3), this yields

$$ \langle {{\mathrm{div}}}_g(Y)\phi ,\phi \rangle _k = - \int _M \mathcal {L}_Y \bigl (h_k(\phi , \phi )\bigr ) \mu _g = - \int _M \Bigl (h_k\bigl (\nabla ^k_Y \phi , \phi \bigr ) + h_k\bigl (\phi , \nabla ^k_{\overline{Y}}\phi \bigr )\Bigr ) \mu _g, $$

where the second equality comes from the fact that $\nabla ^k$ and $h_k$ are compatible. But $\overline{Y}$ is a section of $T^{0, 1}M$, and $\phi$ is a holomorphic section of $L^k$, so $\nabla ^k_{\overline{Y}}\phi = 0$, which implies that $\nabla ^k_X\phi = \nabla ^k_Y\phi$ since $X = Y + \overline{Y}$, and (8.2) is proved.

We now want to apply this to $X_f$ where f belongs to $\mathcal {C}^{2}(M, \mathbb {R})$. Observe that

$$ {{\mathrm{div}}}_g\bigl (X_f^{1, 0}\bigr ) = \tfrac{1}{2} \bigl ({{\mathrm{div}}}_g(X_f) -\text {i}{{\mathrm{div}}}_g(jX_f)\bigr ). $$

We claim that ${{\mathrm{div}}}_g(X_f) = 0$; indeed, since $\mu _g = \mu$, we have that

$$ {{\mathrm{div}}}_g(X_f) \mu _g = \mathcal {L}_{X_f} \mu _g = \mathcal {L}_{X_f} \mu = 0. $$

Consequently, ${{\mathrm{div}}}_g(X_f^{1, 0}) = -(\text {i}/2) {{\mathrm{div}}}_g(jX_f)$. Thanks to Lemma 2.6.1, this yields

$$ {{\mathrm{div}}}_g(X_f^{1, 0}) = \frac{\text {i}}{2} {{\mathrm{div}}}({{\mathrm{grad}}}_g f) = \frac{\text {i}}{2} \varDelta f, $$

and the second statement follows. $\square$

Lemma 8.1.5.

Let T be a bounded operator acting on a complex Hilbert space $\mathcal {H}$. If $\langle T \xi ,\xi \rangle = 0$ for every $\xi \in \mathcal {H}$, then $T = 0$.

Proof.

This is a standard exercise but we still prove it. Let $\xi , \eta \in \mathcal {H}$. Then

$$ 0 = \langle T(\xi + \eta ),\xi + \eta \rangle = \langle T\xi ,\xi \rangle + \langle T\xi ,\eta \rangle + \langle T\eta ,\xi \rangle + \langle T\eta ,\eta \rangle $$

which yields

$$ \langle T\xi ,\eta \rangle = - \langle T\eta ,\xi \rangle . $$

Replacing $\eta$ by $\text {i}\eta$, this implies that

$$ -\text {i}\langle T\xi ,\eta \rangle = -\text {i}\langle T\eta ,\xi \rangle , $$

and combining these two equalities yields $\langle T \xi ,\eta \rangle = 0$. $\square$

8.2 Unitary Evolution of Kostant–Souriau Operators

The goal of this section is to give an alternate, more geometric proof of Lemma 8.1.2, and to use this as an excuse to address the topic of the Schrödinger equation for these operators. More precisely, given a function $f \in \mathcal {C}^{\infty }(M, \mathbb {R})$, we want to look for solutions of

$$\begin{aligned} \frac{\,\mathrm{d}\varPsi _t}{\,\mathrm{d}t} = -\text {i}k P_k(f) \varPsi _t, \quad t \in \mathbb {R}, \end{aligned}$$

(8.4)

where $\varPsi _t$ is a smooth section of $L^k \rightarrow M$ and $\varPsi _0 \in \mathcal {C}^{\infty }(M, L^k)$ is a given initial condition. We can solve this equation as follows. Given a path $\gamma :[0,T ] \rightarrow M$, let

$$ \mathcal {T}^k_{\gamma }:L_{\gamma (0)}^k \rightarrow L_{\gamma (T)}^k $$

be the parallel transport operator in $L^k$ with respect to $\nabla ^k$. Moreover, let $\phi ^t$ be the Hamiltonian flow of f at time t.

Proposition 8.2.1.

Given $\varPsi _0 \in \mathcal {C}^{\infty }(M, L^k)$, the family of sections $\varPsi _t \in \mathcal {C}^{\infty }(M, L^k)$ defined as

$$ \varPsi _t\bigl (\phi ^t(m)\bigr ) = \exp \bigl (-\text {i}ktf(m)\bigr ) \mathcal {T}^k_{({\phi ^s(m)})_{s \in [0,t ]}}\bigl (\varPsi _0(m)\bigr ) $$

for every $m \in M$, is a solution of (8.4) with initial condition $\varPsi _0$.

This defines an operator $U_k(t):\mathcal {C}^{\infty }(M, L^k) \rightarrow \mathcal {C}^{\infty }(M, L^k)$ sending $\varPsi _0$ to $\varPsi _t$, which describes the prequantum evolution of the system.

Proof.

We fix $m \in M$ and $\varPsi _0 \in \mathcal {C}^{\infty }(M, L^k)$. We claim that it is enough to prove the proposition for t so small that for every $s \in [-t, t ]$, the point $\phi ^s(m)$ belongs to a trivialisation open set V for L. This is because the operator $U_k(t)$ satisfies the semigroup relation $U_k(t_1 + t_2) = U_k (t_2) U_k(t_1)$.

Let u be a local non-vanishing section of L over V, and let $\varphi = h u^k$ for some $h \in \mathcal {C}^{\infty }(V, \mathbb {R})$. Moreover, let $\alpha$ be the differential form such that $\nabla s = -\text {i}\alpha \otimes s$. Then we can write $P_k(f) \varphi = (\widetilde{P}_k(f) h) u^k$ with

$$\begin{aligned} \widetilde{P}_k(f) h = (f-i_{X_f}\alpha ) h + \frac{1}{\text {i}k} \mathcal {L}_{X_f} h. \end{aligned}$$

(8.5)

Moreover, a standard computation yields

$$ \mathcal {T}^k_{({\phi ^s(m)})_{s \in [0,t ]}}\bigl (\varphi (m)\bigr ) = \exp \biggl (\text {i}k \int _0^t (\phi ^s)^*(i_{X_f}\alpha )\,\mathrm{d}s\biggr ) h(m) u^k\bigl (\phi ^t(m)\bigr ), $$

and consequently, if $\varPsi _0 = h_0 s^k$ on V, then $\varPsi _t = h_t s^k$ on V where

$$ h_t(m) = \exp \Biggr (\text {i}k \biggl (\int _{-t}^0 (\phi ^s)^*(i_{X_f}\alpha )\,\mathrm{d}s - t f(m)\biggr )\Biggr ) h_0\bigl (\phi ^{-t}(m)\bigr ). $$

for every $m \in M$. We only need to compare the time derivative of $h_t$ and $\widetilde{P}_k(f) h_t$. To simplify notation, we will write

$$ \theta (t, m) = \int _{-t}^0 (\phi ^s)^*(i_{X_f}\alpha )(m)\,\mathrm{d}s - t f(m). $$

On the one hand,

$$ \frac{\,\mathrm{d}h_t}{\,\mathrm{d}t} = \exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \bigl (-(\phi ^{-t})^*(\mathcal {L}_{X_f} h_0) + \text {i}k \bigl ((\phi ^{-t})^*(i_{X_f}\alpha ) - f\bigr ) (\phi ^{-t})^*h_0\bigr ). $$

On the other hand, we have that

$$ \mathcal {L}_{X_f} h_t= \exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \biggl ((\phi ^{-t})^*(\mathcal {L}_{X_f} h_0) + \text {i}k \bigl ((\phi ^{-t})^*h_0\bigr ) \int _{-t}^0 \mathcal {L}_{X_f}\bigl ((\phi ^s)^*(i_{X_f}\alpha )\bigr )\,\mathrm{d}s\biggr ). $$

Using Cartan’s formula, we have that

$$ \,\mathrm{d}\bigl ((\phi ^s)^*(i_{X_f}\alpha )\bigr ) = (\phi ^s)^*\bigl (\,\mathrm{d}(i_{X_f}\alpha )\bigr ) = (\phi ^s)^*(\mathcal {L}_{X_f}\alpha ) - (\phi ^s)^*(i_{X_f}\,\mathrm{d}\alpha ). $$

Since $\,\mathrm{d}\alpha = \text {i}{{\mathrm{curv}}}(L) = \omega$ and $(\phi ^s)^*(\mathcal {L}_{X_f}\alpha ) = \mathrm{d}(\phi ^s)^*\alpha /\mathrm{d}s$, we can write

$$ \int _{-t}^0 \mathcal {L}_{X_f}\bigl ((\phi ^s)^*(i_{X_f}\alpha )\bigr )\,\mathrm{d}s = i_{X_f}\alpha - (\phi ^{-t})^*(i_{X_f}\alpha ), $$

therefore we finally obtain that

$$ \widetilde{P}_k(f) h_t = \exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \biggl (\bigl (f - (\phi ^{-t})^*(i_{X_f}\alpha )\bigr ) (\phi ^{-t})^*h_0 + \frac{1}{\text {i}k} (\phi ^{-t})^*(\mathcal {L}_{X_f} h_0)\biggr ), $$

which yields the desired formula $-\text {i}k \widetilde{P}_k(f) h_t = (\mathrm{d}h_t/\mathrm{d}t)$. $\square$

One can check that $U_k(t)$ extends to a unitary operator on $L^2(M, L^k)$. It turns out that the Kostant–Souriau operators satisfy an exact version of Egorov’s theorem (Theorem 5.3.2).

Proposition 8.2.2.

Let $f \in \mathcal {C}^{\infty }(M, \mathbb {R})$ and let $U_k(t)$ be the evolution operator associated with $P_k(f)$. Then

$$ U_k(t)^* P_k(g) U_k(t) = P_k (g \circ \phi ^t) $$

for every $g \in \mathcal {C}^{\infty }(M, \mathbb {R})$, where $\phi ^t$ is the Hamiltonian flow of f at time t.

Proof.

Again, we can work in a trivialisation open set for L, since

$$\begin{aligned}&U_k(t_1 + t_2)^* P_k(g) U_k(t_1 + t_2) = U_k(t_2)^* U_k(t_1)^* P_k(g) U_k(t_1) U_k(t_2) , \\&\qquad \qquad \qquad \quad g \circ \phi ^{t_1 + t_2} = g \circ \phi ^{t_1} \circ \phi ^{t_2}. \end{aligned}$$

Hence we keep the same notation as in the proof of the previous proposition. If $U_k(t) \varPsi _0 = h_t u^k$ on V, the computations performed in this proof yield

$$\begin{aligned} \,\mathrm{d}h_t=&\exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \\&\times \biggl ((\phi ^{-t})^*(dh_0) + \text {i}k \biggl (\alpha - (\phi ^{-t})^*\alpha - \int _{-t}^0 (\phi ^s)^*(i_{X_f}\,\mathrm{d}\alpha )\,\mathrm{d}s- t \,\mathrm{d}f\biggr ) (\phi ^{-t})^*h_0\biggr ). \end{aligned}$$

We can simplify this further because

$$ (\phi ^s)^*(i_{X_f}\,\mathrm{d}\alpha ) = (\phi ^s)^*(i_{X_f}\omega ) = -(\phi ^s)^*(\,\mathrm{d}f) = -\,\mathrm{d}\bigl ((\phi ^s)^*f\bigr ) = -\,\mathrm{d}f, $$

hence we obtain that

$$ \mathcal {L}_{X_g} h_t = \exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \Bigl ((\phi ^{-t})^*(\mathcal {L}_{X_g} h_0) + \text {i}k \Bigl (i_{X_g}\alpha -i_{X_g}\bigl ((\phi ^{-t})^*\alpha \bigr )\Bigr ) (\phi ^{-t})^*h_0\Bigr ). $$

Therefore, (8.5) yields

$$ \widetilde{P}_k(g)h_t = \exp \bigl (\text {i}k \theta (t, \cdot )\bigr ) \biggl (\frac{1}{\text {i}k} (\phi ^{-t})^*(\mathcal {L}_{X_g} h_0) + \Bigl (g -i_{X_g}\bigl ((\phi ^{-t})^*\alpha \bigr )\Bigr ) (\phi ^{-t})^*h_0\biggr ). $$

Consequently, if $U_k(t)^* P_k(g) U_k(t) = q_t u^k$ on V, we finally obtain that

$$\begin{aligned} q_t = \frac{1}{\text {i}k} \mathcal {L}_{X_{g \circ \phi ^t}} h_0 + \bigl (g \circ \phi ^t - i_{X_{g \circ \phi ^t}}\alpha \bigr ) h_0 = \widetilde{P}_k(g \circ \phi ^t) h_0. \end{aligned}$$

$\square$

In order to reprove Lemma 8.1.2 with the help of these two results, it suffices to write the time derivative of $\phi _k(t) = U_k(t)^* P_k(g) U_k(t) \varPsi _0$, for $\varPsi _0 \in \mathcal {C}^{\infty }(M, L^k)$, in two different ways. On the one hand, by definition of $U_k$,

$$ \frac{\,\mathrm{d}\phi _k}{\,\mathrm{d}t}\biggr |_{t=0} = \text {i}k [P_k(f), P_k(g) ] \varPsi _0. $$

On the other hand, since $\phi _k(t) = P_k (g \circ \phi ^t) \varPsi _0$, Lemma 5.3.3 implies that

$$ \frac{\,\mathrm{d}\phi _k}{\,\mathrm{d}t}\biggr |_{t=0} = P_k (\{f, g\}) \varPsi _0, $$

and we conclude by comparing these two equalities that the Kostant–Souriau operators satisfy the exact correspondence principle.

8.3 Product Estimate

We will need the following result, of which we will give a proof in Section 8.5.

Theorem 8.3.1.

There exists $C > 0$ such that for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left\| {[P_k(f),\varPi _k ]}\right\| \le C k^{-1} \left\| {f}\right\| _2. $$

This estimate is fundamental and allows us to obtain product and commutator estimates. We now use it to prove Theorem 5.2.2. We compute the difference

$$ T_k(f)T_k(g) - T_k(fg) = \varPi _k f [\varPi _k,g ] \varPi _k = \varPi _k f [\varPi _k, P_k(g) ] \varPi _k - \varPi _k f \left[ {\varPi _k},{\frac{1}{\text {i}k}\nabla _{X_g}^k}\right] \varPi _k. $$

Thanks to Theorem 8.3.1, we know that $\left\| {\varPi _k f [\varPi _k, P_k(g)] \varPi _k}\right\| = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _0 \left\| {g}\right\| _2$. The other term can be estimated by writing it as

$$ \varPi _k f \left[ {\varPi _k},{\frac{1}{\text {i}k}\nabla _{X_g}^k}\right] \varPi _k = \varPi _k f \varPi _k \biggl (\frac{1}{\text {i}k}\nabla _{X_g}^k\biggr ) \varPi _k - \varPi _k \biggl (\frac{1}{\text {i}k} \nabla _{fX_g}^k\biggr ) \varPi _k. $$

Both terms can be estimated using Corollary 8.1.4. The first one satisfies

$$ \left\| {\varPi _k f \varPi _k \biggl (\frac{1}{\text {i}k}\nabla _{X_g}^k\biggr ) \varPi _k}\right\| = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _0 \left\| {g}\right\| _2, $$

whereas the second one satisfies

$$ \left\| {\varPi _k \biggl (\frac{1}{\text {i}k} \nabla _{fX_g}^k\biggr ) \varPi _k}\right\| = O\bigl (k^{-1}\bigr ) \left\| {fX_g}\right\| _1 = O\bigl (k^{-1}\bigr ) (\left\| {f}\right\| _0 \left\| {g}\right\| _2 + \left\| {f}\right\| _1 \left\| {g}\right\| _1). $$

This proves the first estimate of the theorem. To derive the second one, observe that $T_k(fg)$ is self-adjoint and that the adjoint of $T_k(f)T_k(g)$ is $T_k(g) T_k(f)$, and use the fact that the operator norm of the adjoint of an operator is the same as the norm of the operator.

8.4 Commutator Estimate

We first prove commutator estimates for corrected Berezin–Toeplitz operators.

Proposition 8.4.1.

For any $f, g \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left\| {[T_k^{\mathrm {c}}(f), T_k^{\mathrm {c}}(g) ] - \frac{1}{\text {i}k} T_k^{\mathrm {c}}(\{f, g\})}\right\| = O\bigl (k^{-2}\bigr ) \left\| {f}\right\| _2 \left\| {g}\right\| _2. $$

Proof.

We will compare $[T_k^{\mathrm {c}}(f), T_k^{\mathrm {c}}(g) ]$ with $[P_k(f), P_k(g) ]$. In order to do so, we compute:

$$ \varPi _k [\varPi _k,P_k(f) ] [\varPi _k,P_k(g) ] \varPi _k = \varPi _k P_k(f) [\varPi _k,P_k(g) ] \varPi _k - \varPi _k P_k(f) \varPi _k [\varPi _k, P_k(g) ] \varPi _k. $$

Expanding the first term on the right-hand side of this equality, we get

$$ \varPi _k P_k(f) [\varPi _k, P_k(g) ] \varPi _k = \varPi _k P_k(f) \varPi _k P_k(g) \varPi _k - \varPi _k P_k(f) P_k(g) \varPi _k $$

and the second term satisfies

$$ \varPi _k P_k(f) \varPi _k [\varPi _k, P_k(g) ] \varPi _k = \varPi _k P_k(f) \varPi _k P_k(g) \varPi _k - \varPi _k P_k(f) \varPi _k P_k(g) \varPi _k = 0. $$

Therefore, we have that

$$ \varPi _k [\varPi _k,P_k(f) ] [\varPi _k, P_k(g) ] \varPi _k = T_k^{\mathrm {c}}(f) T_k^{\mathrm {c}}(g) - \varPi _k P_k(f) P_k(g) \varPi _k. $$

Thanks to Theorem 8.3.1, the left-hand side is a $\mathcal {O}(k^{-2}) \left\| {f}\right\| _2 \left\| {g}\right\| _2$, thus

$$ [T_k^{\mathrm {c}}(f),T_k^{\mathrm {c}}(g) ] = \varPi _k [P_k(f), P_k(g) ] \varPi _k + \mathcal {O}(k^{-2}) \left\| {f}\right\| _2 \left\| {g}\right\| _2 $$

which yields, using Lemma 8.1.2,

$$\begin{aligned}&[T_k^{\mathrm {c}}(f), T_k^{\mathrm {c}}(g) ]&= \frac{1}{\text {i}k} T_k^{\mathrm {c}}(\{f, g\}) + \mathcal {O}(k^{-2}) \left\| {f}\right\| _2 \left\| {g}\right\| _2.&\end{aligned}$$

$\square$

We now prove Theorem 5.2.3. Thanks to Proposition 8.1.3, we have that

$$ T_k(f) = T_k^{\mathrm {c}}(f) + \frac{1}{2k} T_k(\varDelta f), $$

and similarly for g. Consequently, $[T_k(f),T_k(g) ] = [T_k^{\mathrm {c}}(f), T_k^{\mathrm {c}}(g) ] + R_k$, with

$$ R_k = \frac{1}{2k}[T_k(\varDelta f), T_k^{\mathrm {c}}(g) ] + \frac{1}{2k}[T_k^{\mathrm {c}}(f), T_k(\varDelta g) ] + \frac{1}{4k^2} [T_k(\varDelta f), T_k(\varDelta g) ]. $$

Let us estimate $R_k$. Firstly, we have that

$$ [T_k(\varDelta f),T_k^{\mathrm {c}}(g) ] = [T_k(\varDelta f), T_k(g) ] - \frac{1}{2k} [T_k(\varDelta f), T_k(\varDelta g) ]. $$

Applying Theorem 5.2.2 to $\varDelta f \in \mathcal {C}^{1}(M, \mathbb {R})$ and $g \in \mathcal {C}^{3}(M, \mathbb {R})$, we obtain that

$$ [T_k(\varDelta f), T_k(g) ] = O\bigl (k^{-1}\bigr )(\left\| {f}\right\| _2 \left\| {g}\right\| _2 + \left\| {f}\right\| _3 \left\| {g}\right\| _1) = O\bigl (k^{-1}\bigr ) \left\| {f, g}\right\| _{1, 3}. $$

Moreover, Lemma 5.1.2 implies that

$$\begin{aligned}{}[T_k(\varDelta f), T_k(\varDelta g) ] = O\bigl (1\bigr ) \left\| {\varDelta f}\right\| _0 \left\| {\varDelta g}\right\| _0 = O\bigl (1\bigr ) \left\| {f}\right\| _2 \left\| {g}\right\| _2. \end{aligned}$$

(8.6)

It follows from these estimates that

$$ \frac{1}{2k}[T_k(\varDelta f), T_k^{\mathrm {c}}(g) ] = O\bigl (k^{-2}\bigr ) \left\| {f, g}\right\| _{1, 3}. $$

A similar reasoning leads to

$$ \frac{1}{2k}[T_k^{\mathrm {c}}(f), T_k(\varDelta g) ] = O\bigl (k^{-2}\bigr ) \left\| {f, g}\right\| _{1, 3}. $$

These two results combined with (8.6) imply that $R_k = O\bigl (k^{-2}\bigr ) \left\| {f, g}\right\| _{1, 3}$. Now, thanks to the previous proposition, we have that

$$ [T_k^{\mathrm {c}}(f), T_k^{\mathrm {c}}(g) ] = \frac{1}{\text {i}k} T_k^{\mathrm {c}}(\{f, g\}) + O\bigl (k^{-2}\bigr ) \left\| {f, g}\right\| _{1, 3}. $$

Therefore,

$$ [T_k(f), T_k(g) ] = \frac{1}{\text {i}k} T_k(\{f, g\}) + \frac{\text {i}}{2k^2} T_k(\varDelta \{f, g\}) + O\bigl (k^{-2}\bigr ) \left\| {f, g}\right\| _{1, 3}, $$

and we conclude thanks to the estimate

$$ T_k(\varDelta \{f, g\}) = O\bigl (1\bigr ) \left\| {\varDelta \{f, g\}}\right\| _0 = O\bigl (1\bigr ) \left\| {f, g}\right\| _{1, 3}, $$

which follows from Lemma 5.1.2.

8.5 Fundamental Estimates

This section, which follows the same lines as in the article [20], is devoted to the proof of Theorem 8.3.1; this strongly relies on the asymptotic expansion of the Schwartz kernel of the projector given by Theorem 7.2.1. Let be as in this theorem, that is, satisfying the properties stated in Proposition 7.1.1. Let $U \subset M^2$ be the open set where E does not vanish; observe that U contains the diagonal $\varDelta _M$ of $M^2$. Define as before a function $\varphi _E \in \mathcal {C}^{\infty }(U)$ and a differential form $\alpha _E \in \Omega ^1(U) \otimes \mathbb {C}$ by the formulas

$$ \varphi _E = -2\log \left\| {E}\right\| , \quad \widetilde{\nabla }E = -\text {i}\alpha _E \otimes E, $$

where we recall that $\widetilde{\nabla }$ is the connection induced by $\nabla$ on . The function $\varphi _E$ vanishes along $\varDelta _M$ and is positive outside $\varDelta _M$. We derived the following properties of $\varphi _E$ and $\alpha _E$ in Lemmas 7.1.3 and 7.1.4:

• (1) $\alpha _E$ vanishes along $\varDelta _M$,

• (2) $\varphi _E$ vanishes to second order along $\varDelta _M$,

• (3) for every $x \in M$, the kernel of the Hessian of $\varphi _E$ at (x, x) is equal to $T_{(x, x)} \varDelta _M$, and this Hessian is positive definite on the complement of $T_{(x, x)}\varDelta _M$.

In what follows, we will need the following additional property.

Lemma 8.5.1.

Let $f \in \mathcal {C}^{2}(M, \mathbb {R})$, and let $g \in \mathcal {C}^{2}(U, \mathbb {R})$ be defined by the formula $g(x, y) = f(x) - f(y)$. Then the function

$$ u = g - \alpha _E(X_f, X_f) $$

vanishes to second order along $\varDelta _M$.

Proof.

It is clear that u vanishes along $\varDelta _M$ since g and $\alpha _E$ do. Now, let Y and Z be two vector fields on M; we compute for $(y, z) \in U$

$$ \bigl (\mathcal {L}_{(Y, Z)}u\bigr )(y, z) = \bigl (\mathcal {L}_Y f\bigr )(y) - \bigl (\mathcal {L}_Z f\bigr )(z) - \mathcal {L}_{(Y, Z)}\bigl (\alpha _E(X_f, X_f)\bigr )(y, z). $$

As before, set $\widetilde{\omega } = p_1^* \omega - p_2^*\omega$ with $p_1, p_2$ the natural projections $M^2 \rightarrow M$. Therefore the first two terms in the above equation satisfy

$$ \bigl (\mathcal {L}_Y f\bigr )(y) - \bigl (\mathcal {L}_Z f\bigr )(z) = \widetilde{\omega }\bigl ((Y, Z), (X_f, X_f)\bigr )(x, y). $$

Moreover, since $\,\mathrm{d}\alpha _E = \text {i}{{\mathrm{curv}}}(\widetilde{\nabla }) = \widetilde{\omega }$, the last term in the previous equation can be written as

$$\begin{aligned}&\mathcal {L}_{(Y, Z)}\bigl (\alpha _E(X_f, X_f)\bigr ) = \widetilde{\omega }\bigl ((Y, Z), (X_f, X_f)\bigr ) + \alpha _E([(Y, Z),(X_f, X_f) ]) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad {}+ \mathcal {L}_{(X_f, X_f)}\bigl (\alpha _E(Y, Z)\bigr ). \end{aligned}$$

Thus we finally obtain that

$$ \mathcal {L}_{(Y, Z)}u = \alpha _E([(X_f, X_f),(Y, Z) ]) - \mathcal {L}_{(X_f, X_f)}\bigl (\alpha _E(Y, Z)\bigr ). $$

The first term vanishes along $\varDelta _M$ because $\alpha _E$ does. The second term vanishes along $\varDelta _M$ because $\alpha _E$ vanishes along $\varDelta _M$ and $(X_f, X_f)$ is tangent to $\varDelta _M$. $\square$

These properties yield the following result. For $u \in \mathcal {C}^{0}(M^2, \mathbb {R})$, let $Q_k(u)$ be the operator acting on $\mathcal {C}^{0}(M, L^k)$ with Schwartz kernel $F_k(u) = \bigl (k/(2\pi )\bigr )^n E^k u$.

Lemma 8.5.2.

Taking a smaller U still containing $\varDelta _M$ if necessary, for every compact subset $K \subset U$ and for every $p \in \mathbb {N}$, there exists a constant $C_{K, p} > 0$ such that for any $u \in \mathcal {C}^{0}(M^2, \mathbb {R})$ with support contained in K, and for every $k \ge 1$,

$$ \left\| {Q_k(u)}\right\| \le C_{K, p} \left| {u}\right| _{K, p} k^{-p/2} $$

where $\left| {u}\right| _{K, p}$ is the supremum of $\left| {u}\right| \varphi _E^{-p/2}$ on $K \setminus \varDelta _M$, which may be infinite.

Proof.

Assume first that $K \subset V^2$, where $V \subset M$ is a trivialisation open set for M, with coordinates $x_1, \dots , x_{2n}$, such that $V^2 \subset U$. So we may identify V with a subset of $\mathbb {R}^{2n}$ and assume that we are working in a subset of $\mathbb {R}^{4n}$. Since $\varphi _E$ vanishes to second order along $\varDelta _M$, Taylor’s formula with integral remainder yields

$$ \varphi _E(x, y) = \frac{1}{2}\,\mathrm{d}^2\varphi _E(x, x)(v, v) + \int _0^1 \frac{(1-t)^2}{2} \,\mathrm{d}^3\varphi _E\bigl ((1-t)(x, x) + t (x, y)\bigr )(v, v, v) \,\mathrm{d}t $$

with $v = (0, y-x)$. The last term is a $O\bigl (\left| {x-y}\right| ^3\bigr )$ uniformly on K. Since $\,\mathrm{d}^2\varphi _E(x, x)$ is positive definite on the orthogonal of $\{x=y\} \subset \mathbb {R}^{4n}$, we have that

$$ \lambda _{\min }(x) \left\| {v}\right\| ^2 \le \,\mathrm{d}^2\varphi _E(x, x)(v, v) \le \lambda _{\max }(x) \left\| {v}\right\| ^2 $$

whenever $y \ne x$, where $\lambda _{\min }(x)$ (respectively $\lambda _{\max }(x)$) is the smallest (respectively largest) positive eigenvalue of $\,\mathrm{d}^2\varphi _E(x, x)$. Therefore, there exists $C > 0$ such that

$$\begin{aligned} \frac{\left\| {x-y}\right\| ^2}{C} \le \varphi _E(x, y) \le C \Vert x-y\Vert ^2 \end{aligned}$$

(8.7)

for every $(x, y) \in K$. Now, let $u \in \mathcal {C}^{0}(M^2, \mathbb {R})$ be compactly supported in K. The previous estimate shows that for every $(x, y) \in K$, $x \ne y$,

$$ \frac{\left| {u(x, y)}\right| }{\varphi _E(x, y)^{p/2}} \ge \frac{\left| {u(x, y)}\right| }{C^{p/2} \left\| {x-y}\right\| ^p}, $$

thus $\left| {u(x, y)}\right| \le C^{p/2} \left| {u}\right| _{K, p} \left\| {x-y}\right\| ^p$ on $V^2$. If $\left| {u}\right| _{K, p}$ is infinite, the result is obvious. If not, since $\left\| {E}\right\| = \exp (-\varphi _E/2)$, we have that

$$ \int _{M} \left\| {F_k(u)(x, y)}\right\| \,\mathrm{d}x \le \biggl (\frac{k}{2\pi }\biggr )^n C^{p/2} \left| {u}\right| _{K, p} \int _V \exp \biggl (-\frac{k\left\| {x-y}\right\| ^2}{2C}\biggr ) \left\| {x-y}\right\| ^p \,\mathrm{d}x. $$

The integral on V is smaller that the integral on $\mathbb {R}^{2n}$ of the same integrand. The change of variable $v = \sqrt{k/C} (x-y)$ yields

$$ \int _{M} \left\| {F_k(u)(x, y)}\right\| \,\mathrm{d}x \le \frac{C^{p+n}}{(2\pi )^n} k^{-p/2} \left| {u}\right| _{K, p} \int _{\mathbb {R}^{2n}} \exp \biggl (-\frac{\left\| {v}\right\| ^2}{2}\biggr ) \left\| {v}\right\| ^p \,\mathrm{d}v, $$

which implies that $\int _{M} \left\| {F_k(u)(x, y)}\right\| \,\mathrm{d}x \le C_{K, p}^1 k^{-p/2} \left| {u}\right| _{K, p}$. A similar computation leads to $\int _{M} \left\| {F_k(u)(x, y)}\right\| \,\mathrm{d}y \le C_{K, p}^2 k^{-p/2} \left| {u}\right| _{K, p}$ for some $C_{K, p}^2 > 0$. It follows from the Schur test that

$$ \left\| {Q_k(u)}\right\| \le C_{K, p} k^{-p/2} \left| {u}\right| _{K, p} $$

for some $C_{K, p} > 0$.

Let us now turn to the general case. Taking a smaller U, still containing the diagonal, if necessary, let $(V_i)_{1 \le i \le d}$ be a finite family of trivialisation sets of M such that $K \subset \bigcup _{i=1}^d V_i^2 \subset U$. Choose a partition of unity $\eta , (\eta _i)_{1 \le i \le d}$ subordinate to the open cover $M^2 \subset (M^2 \setminus K) \cup (\bigcup _{i=1}^d V_i^2)$. Let $u \in \mathcal {C}^{0}(M^2, \mathbb {R})$ be compactly supported in K; then

$$ u = \sum _{i=1}^d \eta _i u, \quad Q_k(u) = \sum _{i=1}^d Q_k(\eta _i u). $$

It follows from the first part of the proof that

$$ \left\| {Q_k(\eta _i u)}\right\| \le C_{K, p, i} k^{-p/2} \left| {\eta _i u}\right| _{K, p} \le C_{K, p, i} k^{-p/2} \left| {u}\right| _{K, p} $$

for some constants $C_{K, p, i} > 0$. We conclude by applying the triangle inequality. $\square$

Proposition 8.5.3.

For every $p \in \mathbb {N}$, for every $u \in \mathcal {C}^{\infty }(M^2, \mathbb {R})$ supported in U and vanishing to order p along $\varDelta _M$, there exists $C_u > 0$ such that for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left\| {Q_k(u)}\right\| \le C_u k^{-p/2}, \quad \left\| {[P_k(f), Q_k(u) ]}\right\| \le C_u k^{-p/2-1} \left\| {f}\right\| _2 , $$

where $P_k(f) = f + \bigl (1/(\text {i}k)\bigr ) \nabla ^k_{X_f}:\mathcal {C}^{\infty }(M, L^k) \rightarrow \mathcal {C}^{\infty }(M, L^k)$ is the Kostant–Souriau operator associated with f.

Before proving this result, let us state several lemmas.

Lemma 8.5.4.

Let $u \in \mathcal {C}^{\infty }(M^2, \mathbb {R})$ be compactly supported in U, and let $f \in \mathcal {C}^{2}(M, \mathbb {R})$. Let $g \in \mathcal {C}^{2}(M^2, \mathbb {R})$ be defined by the formula $g(x, y) = f(x) - f(y)$ as before, and define the vector field $Y_f = (X_f, X_f)$ on $M^2$. Then

$$ [P_k(f), Q_k(u) ] = Q_k\Bigl (\bigl (g - \alpha _E(Y_f)\bigr )u\Bigr ) + \frac{1}{\text {i}k} Q_k\bigl (\mathcal {L}_{Y_f}u\bigr ). $$

Proof.

We start by writing

$$ [P_k(f), Q_k(u) ] = f Q_k(u) - Q_k(u) f + \frac{1}{\text {i}k} \bigl (\nabla _{X_f}^k \circ Q_k(u) - Q_k(u) \circ \nabla _{X_f}^k\bigr ). $$

The Schwartz kernel of $f Q_k(u) - Q_k(u) f$ is equal to $f(x) F_k(u)(x, y) - F_k(u)(x, y) f(y)$. By Lemma 6.4.3, the Schwartz kernel of $\nabla _{X_f}^k \circ Q_k(u)$ is equal to . By Lemma 6.4.4, the Schwartz kernel of $Q_k(u) \circ \nabla _{X_f}^k$ is equal to since ${{\mathrm{div}}}(X_f) = 0$. Therefore, the Schwartz kernel of $[P_k(f), Q_k(u) ]$ is given by

Remembering the definition of $\alpha _E$, and since u has support in U, we have that

$$ \widetilde{\nabla }_{(X_f, X_f)}^k(E^ku) = u \widetilde{\nabla }_{Y_f}^k E^k + (\mathcal {L}_{Y_f}u) E^k = \bigl (-\text {i}k \alpha _E(Y_f) u + \mathcal {L}_{Y_f}u\bigr ) E^k. $$

Consequently, the Schwartz kernel of $[P_k(f), Q_k(u) ]$ is equal to

$$ F_k\Bigl (\bigl (g - \alpha _E(Y_f)\bigr )u\Bigr ) + \frac{1}{\text {i}k} F_k\bigl (\mathcal {L}_{Y_f}u\bigr ); $$

in other words, $[P_k(f), Q_k(u) ] = Q_k\left( \bigl (g - \alpha _E(Y_f)\bigr )u\right) + \bigl (1/(\text {i}k)\bigr ) Q_k(\mathcal {L}_{Y_f}u)$. $\square$

In order to prove Proposition 8.5.3, we will investigate the two terms in the right-hand side of the equality obtained in this lemma. The following result will help us dealing with the first term.

Lemma 8.5.5.

Let K be a compact subset of U. Then there exists $C > 0$ such that for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left| {g - \alpha _E(Y_f)}\right| \le C \left\| {f}\right\| _2 \varphi _E $$

on K, with $g(x, y) = f(x) - f(y)$ and $Y_f = (X_f, X_f)$ as above.

Proof.

Assume first that $K \subset V^2$ where V is a trivialisation open set for M such that $V^2 \subset U$. Introduce some coordinates $x_1, \dots , x_{2n}$ on V. By Taylor’s formula and (8.7), there exist some functions $g_i \in \mathcal {C}^{1}(V, \mathbb {R})$, $1 \le i \le 2n$, such that for $x, y \in V$

$$\begin{aligned} g(x, y) = \sum _{i=1}^{2n} g_i(y)(y_i - x_i) + O\bigl (\varphi _E\bigr ) \left\| {f}\right\| _2, \end{aligned}$$

(8.8)

and the $O\bigl (\varphi _E\bigr )$ is uniform on K. Now, write

$$ \alpha _E(x, y) = \sum _{j=1}^{2n} (\mu _j(x, y) \,\mathrm{d}x_j + \nu _j(x, y) \,\mathrm{d}y_j) $$

for some functions $\mu _j, \nu _j \in \mathcal {C}^{\infty }(V^2)$. Since $\alpha _E$ vanishes along $\varDelta _M$, so does $\mu _j$. Therefore, by Taylor’s formula, there exist some functions $\mu _{ji} \in \mathcal {C}^{\infty }(V)$, $1 \le i \le 2n$, such that

$$ \mu _j(x, y) = \sum _{i=1}^{2n} \mu _{ji}(y) (y_i - x_i) + O\bigl (\varphi _E\bigr ) . $$

Similarly, there exist some functions $\nu _{ji} \in \mathcal {C}^{\infty }(V)$, $1 \le i \le 2n$, such that

$$ \nu _j(x, y) = \sum _{i=1}^{2n} \nu _{ji}(y) (y_i - x_i) + O\bigl (\varphi _E\bigr ) . $$

Consequently, we have that

$$ \alpha _E(x, y) = \sum _{i=1}^{2n} \Biggl (\sum _{j=1}^{2n} \mu _{ji}(y) \,\mathrm{d}x_j + \nu _{ji}(y) \,\mathrm{d}y_j\Biggr ) (y_i - x_i) + O\bigl (\varphi _E\bigr ) \sum _{j=1}^{2n}(\,\mathrm{d}x_j + \,\mathrm{d}y_j). $$

Now, by Taylor’s formula, $\,\mathrm{d}x_j(X_f)(x) = \,\mathrm{d}x_j(X_f)(y) + O\bigl (\varphi _E^{1/2}\bigr ) \left\| {f}\right\| _2$. Thus, the previous formula implies that

$$\begin{aligned} \alpha _{E}(Y_f)(x, y) = \sum _{i=1}^{2n} \kappa _i(y) (y_i - x_i) + O\bigl (\varphi _E\bigr ) \left\| {f}\right\| _2 \end{aligned}$$

(8.9)

for some smooth functions $\kappa _i$, and the $O\bigl (\varphi _E\bigr )$ is uniform on K. Since, by Lemma 8.5.1, the function $g - \alpha _E(Y_f)$ vanishes to second order along $\varDelta _M$, it follows from (8.8) and (8.9) that $g_i - \kappa _i = 0$ for every $i \in \llbracket 1,2n \rrbracket$. Therefore

$$ g - \alpha _E(Y_f) = O\bigl (\varphi _E\bigr ) \left\| {f}\right\| _2 $$

uniformly on K.

To handle the general case, we use the same partition of unity argument that we have used at the end of the proof of Lemma 8.5.2. $\square$

Finally, the following lemma will take care of the second term in the equality displayed in Lemma 8.5.4.

Lemma 8.5.6.

Let $u \in \mathcal {C}^{\infty }(M^2, \mathbb {R})$ be a function vanishing to order p along $\varDelta _M$. Then there exists $C > 0$ such that for any vector field X of $M^2$ of class $\mathcal {C}^{1}$ and tangent to $\varDelta _M$, we have that

$$ |\mathcal {L}_X u| \le C \left\| {X}\right\| _1 \varphi _E^{p/2}. $$

Proof.

We start by proving the lemma for vector fields which are compactly supported in $V^2$, where V is a trivialisation open set of M, endowed with coordinates $x_1, \dots , x_{2n}$. Write

$$ \,\mathrm{d}u = \sum _{i=1}^{2n} \biggl (\frac{\partial u}{\partial x_i} \,\mathrm{d}x_i + \frac{\partial u}{\partial y_i} \,\mathrm{d}y_i\biggr ) = \sum _{i=1}^{2n} \biggl (\frac{\partial u}{\partial y_i} (dy_i - \,\mathrm{d}x_i) + \biggr (\frac{\partial u}{\partial x_i} + \frac{\partial u}{\partial y_i}\biggr ) \,\mathrm{d}x_i\biggr ). $$

Since u vanishes to order p along $\varDelta _M$ and the vector field $\partial _{x_i} + \partial _{y_i}$ is tangent to $\varDelta _M$, the function ${\partial u/\partial x_i} + {\partial u/\partial y_i}$ vanishes to order p along $\varDelta _M$, so by Taylor’s formula, it is a $O\bigl (\varphi _E^{p/2}\bigr )$. Moreover, there exists $C_{1} > 0$ such that for any $\mathcal {C}^{1}$ vector field X compactly supported in $V^2$, $\left| {\,\mathrm{d}x_i(X)}\right| \le C_{1} \left\| {X}\right\| _0$. Furthermore, ${\partial u/\partial y_i}$ vanishes to order $p-1$ along $\varDelta _M$, so it is a $O\bigl (\varphi _E^{(p-1)/2}\bigr )$. We claim that there exists $C_{2} > 0$ such that for any $\mathcal {C}^{1}$ vector field X compactly supported in $V^2$ and tangent to $\varDelta _M$,

$$ \left| {(\,\mathrm{d}y_i - \,\mathrm{d}x_i)(X)}\right| \le C_{2} \left\| {X}\right\| _1 \varphi _E^{1/2}. $$

Indeed, take any such vector field X and write it as

$$ X = \sum _{i=1}^{2n} \alpha _i(x, y) \partial _{x_i} + \beta _i(x, y) \partial _{y_i}, $$

where $\alpha _i(x, x) = \beta _i(x, x)$ since X is tangent to $\varDelta _M$. Now

$$ (\,\mathrm{d}y_i - \,\mathrm{d}x_i)(X) = \beta _i(x, y) - \alpha _i(x, y) = \int _0^1 \,\mathrm{d}(\beta _i - \alpha _i)\bigl ((1-t)(x, x) + t(x, y)\bigr )v \,\mathrm{d}t $$

with $v = (0, y-x)$, by Taylor’s formula. This last term is smaller than a constant not depending on X times $\left\| {X}\right\| _1 \varphi _E^{1/2}$.

Combining all of the above estimates, we obtain the result for vector fields which are compactly supported in $V^2$. We prove the general case by using a partition of unity argument. $\square$

Let us now show how to apply all of the above.

Proof of Proposition 8.5.3. Let K denote the support of u. Since u vanishes to order p along the diagonal, it follows from Taylor’s formula, (8.7) and a partition of unity argument that $\left| {u}\right| _{K, p}$ is finite. Consequently, the first estimate follows from Lemma 8.5.2.

To prove the second estimate, recall that it follows from Lemma 8.5.4 that

$$ [P_k(f), Q_k(u) ] = Q_k\Bigl (\bigl (g - \alpha _E(Y_f)\bigr )u\Bigr ) + \frac{1}{\text {i}k} Q_k\bigl (\mathcal {L}_{Y_f}u\bigr ). $$

It follows from Lemma 8.5.5 that $\left| {g - \alpha _E(Y_f)}\right| \le C \left\| {f}\right\| _2 \varphi _E$ for some constant $C > 0$ not depending on f. Moreover, since u vanishes to order p along $\varDelta _M$, u is a $O\bigl (\varphi _E^{p/2}\bigr )$. Thus, $\bigl (g - \alpha _E(Y_f)\bigr )u = O\bigl (\varphi _E^{(p+2)/2}\bigr )$, and by Lemma 8.5.2,

$$ \left\| {Q_k\Bigr (\bigl (g - \alpha _E(Y_f)\bigr )u\Bigr )}\right\| = O\bigl (k^{-p/2-1}\bigr ) \left\| {f}\right\| _2. $$

Similarly, it follows from Lemma 8.5.6 that $\left| {\mathcal {L}_{Y_f}u}\right| \le C' \left\| {f}\right\| _2 \varphi _E^{p/2}$ for some $C' > 0$ not depending on f. Therefore, Lemma 8.5.2 yields

$$ \left\| {Q_k\bigl (\mathcal {L}_{Y_f}u\bigr )}\right\| = O\bigl (k^{-p/2}\bigr ) \left\| {f}\right\| _2, $$

and the result follows. $\square$

We are now ready to prove Theorem 8.3.1. Write as in Theorem 7.2.1

$$ \varPi _k(x, y) = \biggl (\frac{k}{2 \pi }\biggr )^n E^k(x, y) u(x, y, k) + R_k(x, y), $$

and let $u \sim \sum _{\ell \le 0} k^{-\ell } u_\ell$ be the asymptotic expansion of $u(\,\cdot \,,\cdot \,, k)$. Choose a compactly supported function $\chi \in \mathcal {C}^{\infty }(M^2, \mathbb {R})$ such that ${{\mathrm{supp}}}(\chi ) \subset U$ and equal to one near $\varDelta _M$. Fixing $m \in \mathbb {N}$, we write

$$ \varPi _k = \sum _{\ell = 0}^m k^{-\ell } Q_k(\chi u_\ell ) + \sum _{\ell = 0}^m k^{-\ell } Q_k\bigl ((1-\chi )u_\ell \bigr ) + Q_k\Biggl (u - \sum _{\ell = 0}^m k^{-\ell } u_\ell \Biggr ) + R_k, $$

where $R_k$ is the operator with Schwartz kernel $R_k(\,\cdot \,, \cdot \,)$. We only need to estimate the commutator of each of these terms with $P_k(f)$. Since $\chi u_\ell$ is compactly supported in U, it follows from Proposition 8.5.3 that $[P_k(f), Q_k(\chi u_\ell ) ] = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _2$, so

$$ \left[ {P_k(f)},{\sum _{\ell = 0}^m k^{-\ell }Q_k(\chi u_\ell )}\right] = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _2. $$

For the second term, we use the following fact. Let V be a neighbourhood of $\varDelta _M$, and let $r = \sup _{M^2 \setminus V} \left\| {E}\right\| < 1$; then for any $v \in \mathcal {C}^{0}(M^2)$ vanishing in V, we have that

$$ \left\| {F_k(v)}\right\| \le C k^n r^k \left\| {v}\right\| _0 $$

for some $C > 0$ not depending on v. Therefore this Schwartz kernel is a $O\bigl (k^{-\infty }\bigr ) \left\| {v}\right\| _0$ uniformly on $M^2$, and by Proposition 6.4.1, $Q_k(v) = O\bigl (k^{-\infty }\bigr ) \left\| {v}\right\| _0$. Since $1-\chi$ vanishes in a neighbourhood of $\varDelta _M$, combining this fact with the equality

$$ [P_k(f), Q_k\bigl ((1-\chi )u_\ell \bigr ) ] = Q_k\bigl ((1-\chi )\bigl (g - \alpha _E(Y_f)\bigr )u_\ell \bigr ) + \frac{1}{\text {i}k} Q_k\Bigl (\mathcal {L}_{Y_f}\bigl ((1-\chi )u_\ell \bigr )\Bigr ), $$

coming from Lemma 8.5.4, we obtain that

$$ \left[ {P_k(f)},{\sum _{\ell = 0}^m k^{-\ell }Q_k\bigl ((1 - \chi ) u_\ell \bigr )}\right] = O\bigl (k^{-1}\bigr ) \left\| {f}\right\| _2. $$

It only remains to estimate the commutator $[P_k(f), S_k ]$ where

$$ S_k = Q_k\Biggl (u - \sum _{\ell = 0}^m k^{-\ell } u_\ell \Biggr ) + R_k. $$

The Schwartz kernel $S_k(\,\cdot \,, \cdot \,)$ of $S_k$ is a $O\bigl (k^{n-(m+1)}\bigr )$. We conclude the proof by taking m large enough and using the following lemma.

Lemma 8.5.7.

There exists $C > 0$ such that for every $f \in \mathcal {C}^{2}(M, \mathbb {R})$,

$$ \left\| {[P_k(f), S_k ]}\right\| \le C k^{n-(m+1)} \left\| {f}\right\| _2. $$

Proof.

By computing $\widetilde{\nabla }^k\bigl (F_k(u - \sum _{\ell = 0}^m k^{-\ell } u_\ell )\bigr )$, we obtain that for every vector field X on $M^2$ of class $\mathcal {C}^{0}$, there exists $C_X > 0$ such that $\left\| {\widetilde{\nabla }^k_X S_k}\right\| \le C_X k^{n-m}$. This implies that there exists $C > 0$ such that for every vector field X on $M^2$ of class $\mathcal {C}^{0}$, the inequality $\left\| {\widetilde{\nabla }^k_X S_k}\right\| \le C k^{n-m} \left\| {X}\right\| _0$ holds. Indeed, let $(\eta _i)_{1 \le i \le q}$ be a partition of unity subordinate to an open cover $(U_i)_{1 \le i \le q}$ of $M^2$ by trivialisation open sets for $TM^2$, with a local basis $(Y_{ij})_{1 \le j \le 4n}$, and write

$$ X = \sum _{i=1}^q \eta _i X = \sum _{i=1}^q \sum _{j=1}^{4n} \lambda _{ij} Y_{ij}, $$

where $\lambda _{ij}$ is a continuous function, which satisfies $\Vert \lambda _{ij} \Vert _0 \le C' \left\| {X}\right\| _0$ for some $C' > 0$. Consequently,

$$ \left\| {\widetilde{\nabla }^k_X S_k}\right\| = \left\| {\sum _{i=1}^q \sum _{j=1}^{4n} \lambda _{ij} \widetilde{\nabla }^k_{Y_{ij}} S_k}\right\| \le C' (\max _{i, j} C_{Y_{ij}}) \left\| {X}\right\| _0. $$

To finish the proof, we obtain as in the proof of Lemma 8.5.4 that the Schwartz kernel of $[P_k(f), S_k ]$ is equal to

By the above estimate, $\left\| {\widetilde{\nabla }_{(X_f, X_f)}^k S_k}\right\| \le C k^{n-m} \left\| {f}\right\| _1$, and the result follows. $\square$