We will now construct the resolvent of $X$ at $z$ from the contribution of each term in (5.7), except for the last term. For this last term, we observe that

$$ \mathrm{Id} - \mathrm{Op}_{\Sigma}(\mathrm{Id}_{\mathcal{F}}) \approx 0. \qquad (4.56) $$

Since $\sigma$ is fixed in (5.5) we have that

$$ (5.9) \qquad \lim_{|\omega| \to \infty} \left\| (\mathrm{Id} - \mathrm{Op}_{\Sigma}(\mathrm{Id}_{\mathcal{F}})) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \stackrel{(2.37)}{=} 0. $$

**5.1. Contribution of $\Omega_0$.** Recall that $\mathrm{Op}(1_{\Omega_0})$ defined from $\Omega_0$ in (5.5) depends on $\omega$ defined in (5.2).

**Lemma 5.1.** We have (for negative time) $\forall \epsilon > 0, \exists C_{\epsilon} > 0, \forall t \le 0, \exists \omega_t > 0, \forall \omega > \omega_t,$

$$ (5.10) \qquad \left\| e^{tX} \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \le C_{\epsilon} e^{t(\gamma_{\bar{K}} - \epsilon)}, $$

and (for positive time) $\forall \epsilon > 0, \exists C_{\epsilon} > 0, \forall t \ge 0, \exists \omega_t > 0, \forall \omega > \omega_t,$

$$ (5.11) \qquad \left\| e^{tX_F} \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \le C_{\epsilon} e^{t(\gamma_0^+ + \epsilon)}, $$

$$ (5.12) \qquad \left\| e^{tX} \mathrm{Op}_{\Sigma}(\tilde{T}_{\ge (K+1)}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \le C_{\epsilon} e^{t(\gamma_{K+1}^+ + \epsilon)}. $$

*Proof.* Let $t \le 0$. We have

$$ (5.13) \qquad \begin{aligned} e^{tX} \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) &\approx_{(4.56)} \mathrm{Op}_{\Sigma}(e^{tX_F}) \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \\ &\approx_{(4.65)} \mathrm{Op}_{\Sigma}(e^{tX_F} \tilde{T}_{[0,K]}), \end{aligned} $$

Now we use the cutoff in frequency $\mathrm{Op}(1_{\Omega_0})$. We have

$$ (5.14) \qquad \left\| \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \le C, $$

and

$$ \begin{aligned} & \left\| e^{tX} \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \\ & \le_{(5.13),(2.37)} \left\| \mathrm{Op}_{\Sigma}(e^{tX_F} \tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} + r(\omega) \\ & \le \left\| \mathrm{Op}_{\Sigma}(e^{tX_F} \tilde{T}_{[0,K]}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} \left\| \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_{\mathcal{W}}(M)} + r(\omega) \\ & \le_{(4.63),(5.14)} C_{K,\epsilon'} e^{t(\gamma_{\bar{K}} - \epsilon')} + r(\omega), \end{aligned} $$

with some function $r(\omega)$ that satisfies $\lim_{|\omega| \to \infty} r(\omega) = 0$. This gives (5.10). Similarly, using (4.62) we get (5.11) and using (4.61) we get (5.12). $\square$
**Approximate resolvent $R_{in}^{(0)}(z)$ and $R_{out}^{(0)}(z)$:** For $T \ge 0$ we define

$$ (5.15) \qquad R_{in}^{(0)}(z) := -\left(\int_{-T}^{0} e^{-tz} e^{tX} dt\right) \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}). $$

Then

$$ (5.16) \qquad \left\| R_{in}^{(0)}(z) \right\|_{\mathcal{H}_W(M) (5.15),(5.10)} \le \int_0^T e^{t\mathrm{Re}(z)} C_{\varepsilon'} e^{-t(\gamma_K - \varepsilon')} dt \le C_{\varepsilon'} \int_0^T e^{-(\varepsilon-\varepsilon')} dt \le C_{\varepsilon}, $$

where we have taken $\varepsilon' = \frac{1}{2}\varepsilon$. We also have

$$ (z-X) R_{in}^{(0)}(z) = (\mathrm{Id} - e^{Tz} e^{-TX}) \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) $$

and

$$ (5.17) \qquad \left\| (z-X) R_{in}^{(0)}(z) - \mathrm{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_W(M)} \le e^{T\mathrm{Re}(z)} C_{\varepsilon'} e^{-T(\gamma_K - \varepsilon')} $$

$$ (5.18) \qquad \le (5.1) \qquad C_{\varepsilon'} e^{-(\varepsilon-\varepsilon')T} \le C_{\varepsilon} e^{-\frac{1}{2}\varepsilon T} $$

with $\varepsilon = \frac{1}{2}\varepsilon'$. Similarly we define

$$ (5.19) \qquad R_{out}^{(0)}(z) := \left( \int_0^T e^{-tz} e^{tX} dt \right) \mathrm{Op}_{\Sigma}(\tilde{T}_{\ge (K+1)}) \mathrm{Op}(1_{\Omega_0}) $$

and get

$$ (5.20) \qquad \left\| R_{out}^{(0)}(z) \right\|_{\mathcal{H}_W(M) (5.12),(5.1)} \le C_{\varepsilon} $$

and

$$ (5.21) \qquad \left\| (z-X) R_{out}^{(0)}(z) - \mathrm{Op}_{\Sigma}(\tilde{T}_{\ge (K+1)}) \mathrm{Op}(1_{\Omega_0}) \right\|_{\mathcal{H}_W(M) (5.12),(5.1)} \le C_{\varepsilon} e^{-\frac{1}{2}\varepsilon T}. $$

### 5.2. Contribution of $\Omega_1$. 
Recall that $\mathrm{Op}(1_{\Omega_1})$ defined from $\Omega_1$ in (5.6) depends on $\sigma > 0$. Let $\sigma' > 0$.

$$ e^{tX} \mathrm{Op}(1_{\Omega_1}) = e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma,\sigma'})) \mathrm{Op}(1_{\Omega_1}) + e^{tX} \mathrm{Op}(\chi_{\Sigma,\sigma'}) \mathrm{Op}(1_{\Omega_1}). $$

For the first term, from Theorem 4.6, we have $\forall \Lambda > 0, \exists C > 0, \forall t \ge 0, \exists \sigma_t, \forall \sigma' > \sigma_t$,

$$ \left\| e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma, \sigma'})) \mathrm{Op}(1_{\Omega_1}) \right\|_{\mathcal{H}_W(M) (4.22)} \le \frac{1}{2} C e^{-\Lambda t}. $$

For the second term, since the support of the symbols is disjoint, we have $\forall \Lambda > 0, \exists C > 0, \forall t \ge 0, \forall \sigma', \exists \sigma > \sigma', \exists \omega_t > 0, \forall |\omega| > \omega_t$,

$$ \left\| \mathrm{Op}(\chi_{\Sigma, \sigma'}) \mathrm{Op}(1_{\Omega_1}) \right\|_{\mathcal{H}_W(M)} \le \frac{1}{2} C e^{-\Lambda t}. $$

Taking $\Lambda > 0$ large enough, we deduce that $\exists C > 0, \forall t \ge 0, \exists \sigma_t, \forall \sigma > \sigma_t, \exists \omega_t > 0, \forall |\omega| > \omega_t$,

$$ (5.22) \qquad \left\| e^{tX} \mathrm{Op}(1_{\Omega_1}) \right\|_{\mathcal{H}_W(M)} \le C e^{t \gamma_{K+1}^+}. $$
**Approximate resolvent:** For $T \ge 0$ we define

$$ (5.23) \qquad R^{(1)}(z) := \int_{0}^{T} e^{-tz} e^{tX} \mathrm{Op}(1_{\Omega_1}) dt $$

and similarly to (5.16) and (5.17) we get

$$ (5.24) \qquad \left\| R^{(1)}(z) \right\|_{\mathcal{H}_W(M) (5.22), (5.1)} \le C_{\epsilon}, $$

and

$$ (5.25) \qquad \left\| (z-X) R^{(1)}(z) - \mathrm{Op}(1_{\Omega_1}) \right\|_{\mathcal{H}_W(M) (5.22), (5.1)} \le C e^{-\epsilon T}. $$

**5.3. Contribution of $\Omega_2$.** We define

$$ R^{(2)}(z) := \mathrm{Op}\left(\frac{1}{z - i\omega(\cdot)} 1_{\Omega_2}\right). $$

We have the following properties. Recall that $\delta > 0$ enters in the Definition (5.3) of $J'_{\omega, \delta}$, hence for $\Omega_2$.

**Lemma 5.2** ([20, Lemma 5.17]). There exists $C > 0$ such that for any $\omega \in \mathbb{R}$, $\delta > 0$,

$$ (5.26) \qquad \left\| R^{(2)}(z) \right\|_{\mathcal{H}_W(M)} \le \frac{C}{\delta}, $$

and

$$ (5.27) \qquad \left\| (z-X) R^{(2)}(z) - \mathrm{Op}(\chi_{\Omega_2, \sigma}) \right\|_{\mathcal{H}_W(M)} \le \frac{C}{\delta}. $$

**5.4. Sum of contributions.** The final approximate resolvent is defined by

$$ R(z) := R_{\mathrm{in}}^{(0)}(z) + R_{\mathrm{out}}^{(0)}(z) + R^{(1)}(z) + R^{(2)}(z). $$

From previous estimates we have

$$ \left\| R(z) \right\|_{\mathcal{H}_W(M) (5.16), (5.20), (5.24), (5.26)} \le 3C_{\epsilon} + \frac{C}{\delta}, $$

and

$$ \left\| (z-X) R(z) - \mathrm{Id} \right\|_{\mathcal{H}_W(M) (5.7), (5.9), (5.17), (5.21), (5.25), (5.27)} \le r(\omega) + 2C_{\epsilon}e^{-\frac{1}{2}\epsilon T} + C e^{-\epsilon T} + \frac{C}{\delta}, $$

with $\lim_{|\omega| \to \infty} r(\omega) = 0$. Hence we can take in this order $T \gg 1$, $\sigma \gg 1$, $\delta \gg 1$, $|\omega| \gg 1$, large enough so that

$$ (z-X) R(z) = \mathrm{Id} - r_2(z) $$

with

$$ \|r_2(z)\|_{\mathcal{H}_W(M)} < \frac{1}{2}. $$

Hence $(\mathrm{Id} - r_2(z))^{-1}$ is bounded. We set $\tilde{R}(z) := R(z)(\mathrm{Id} - r_2(z))^{-1}$ and get that
$$ (z-X)\tilde{R}(z) = \mathrm{Id}. $$

With a similar construction on the left $\tilde{R}_l(z)(z-X) = \mathrm{Id}$, we deduce that

$$ \|(z-X)^{-1}\| \le C_{\epsilon} $$

is bounded by $C_{\epsilon} > 0$ that does not depend on $\omega$. We have shown (1.15) and (1.16). This finishes the proof of Theorem 1.7.
6. PROOF OF THEOREM 1.5 (EMERGENCE OF QUANTUM DYNAMICS)

Recall that the space $\mathcal{H}_W(M)$ in (4.18) depends on the weight $W$ in (4.15) that itself depends on a parameter $R \in \mathbb{R}$. Let $K \in \mathbb{N}$ and $\gamma_{K+1}^+$ defined in (4.45). Using Theorem 4.8, we can take $R$ large enough such that the generator $X$ in $\mathcal{H}_W(M)$ has discrete Ruelle spectrum on $\mathbb{R}(z) > \gamma_{K+1}^+$.

We will use the approximate projector $\text{Op}_\Sigma(\tilde{T}_{[0,K]})$ defined from the symbol (4.39) and (4.55), the operator $\text{Op}(\chi_\omega)$ in (1.25) that selects the low frequencies. We have the decomposition

$$
(6.1) \qquad e^{tX} = \operatorname{Op}(\chi_{\Sigma,\sigma}) e^{tX} \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma,\sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega}))
$$

$$
(6.2) \qquad + \operatorname{Op}(\chi_{\Sigma, \sigma}) e^{tX} (\operatorname{Id} - \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]})) \operatorname{Op}(\chi_{\Sigma, \sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega}))
$$

$$
(6.3) \qquad + (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma, \sigma})) e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\omega})) + \mathrm{Op}(\chi_{\Sigma, \sigma}) e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma, \sigma})) (\mathrm{Id} - \mathrm{Op}(\chi_{\omega}))
$$

$$
(6.4) \qquad + e^{tX} \operatorname{Op}(\chi_{\omega}).
$$

For the first term in the right hand side of (6.1) we have

$$
(6.5) \qquad e^{tX} \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \underset{(4.56)}{\approx} \operatorname{Op}_{\Sigma}(e^{tX_f}) \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \\
\qquad \qquad \approx_{(4.65)} \operatorname{Op}_{\Sigma}(e^{tX_f} \tilde{T}_{[0,K]}),
$$

hence

$$
(6.6) \qquad \operatorname{Op}(\chi_{\Sigma, \sigma}) e^{tX} \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma, \sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega}))
$$

$$
(6.7) \qquad = \operatorname{Op}(\chi_{\Sigma, \sigma}) \operatorname{Op}_{\Sigma}(e^{tX_f} \tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma, \sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega})) + r_{t,\omega}
$$

with

$$
(6.8) \qquad \lim_{|\omega| \to \infty} \left\| r_{t,\omega} \right\|_{\mathcal{H}_W(M) \quad (2.37)} = 0.
$$

For the second term in the right hand side of (6.1), we proceed exactly as for the proof of Eq. (5.12) in Lemma 5.1, with the difference that we replace $\operatorname{Op}(1_{\Omega_0})$ by $\operatorname{Op}(\chi_{\Sigma,\sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega}))$. We get that $\forall \epsilon > 0, \exists C_{\epsilon} > 0, \forall t \ge 0, \exists \omega_t > 0, \forall \omega > \omega_t,$

$$
\left\| \operatorname{Op}(\chi_{\Sigma, \sigma}) e^{tX} (\operatorname{Id} - \operatorname{Op}_{\Sigma}(\tilde{T}_{[0,K]})) \operatorname{Op}(\chi_{\Sigma, \sigma}) (\operatorname{Id} - \operatorname{Op}(\chi_{\omega})) \right\|_{\mathcal{H}_W(M)} \le C_{\epsilon} e^{t(\gamma_{K+1}^{+} + \epsilon)}.
$$

Let $\Lambda \ge -\gamma_{K+1}^+$. For the terms in line (6.3), we have $\exists C > 0, \forall t \ge 0, \exists \sigma_t, \forall \sigma > \sigma_t,$

$$
\left\| (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma, \sigma})) e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\omega})) \right\|_{\mathcal{H}_W(M) \quad (4.23)} \le C e^{-\Lambda t} \le C e^{t \gamma_{K+1}^{+}},
$$
$$
\left\| \mathrm{Op}(\chi_{\Sigma, \sigma}) e^{tX} (\mathrm{Id} - \mathrm{Op}(\chi_{\Sigma, \sigma})) (\mathrm{Id} - \mathrm{Op}(\chi_{\omega})) \right\|_{\mathcal{H}_W(M) \quad (4.23)} \le C e^{-\Lambda t} \le C e^{t \gamma_{K+1}^{+}}.
$$

The operator $\text{Op}(\chi_\omega)$ is compact hence we can decompose the last term in the right
hand side of (6.1) and the last term in (6.5) as

$$
(6.9) \quad (e^{tX} - \operatorname{Op}(\chi_{\Sigma,\sigma}) \operatorname{Op}_{\Sigma}(e^{tX_f} \tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma,\sigma})) \operatorname{Op}(\chi_{\omega}) = R_t + r'',
$$
where $R_t$ is finite rank and $\|r''\|_{H_W(M)} \le C_\epsilon e^{t(\gamma_{K+1}^+ + \epsilon)}$. We get

$$e^{tX}_{(6.1),(6.6),(6.9)} = \operatorname{Op}(\chi_{\Sigma,\sigma}) \operatorname{Op}_{\Sigma}(e^{tX_{\tau}} \tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma,\sigma}) + R_t + r'' + r_{t,\omega},$$

where $|\omega|$ is taken large enough in (6.8) so that $\|r_{t,\omega}\|_{H_W(M)} \le Ce^{t(\gamma_{K+1}^+ + \epsilon)}$ also. We have obtained $\exists C > 0, \forall t > 0, \exists \sigma_t > 0, \forall \sigma > \sigma_t, \exists R_{\sigma,t}$ finite rank operator,

$$\|e^{tX} - \operatorname{Op}(\chi_{\Sigma,\sigma}) \operatorname{Op}_{\Sigma}(e^{tX_{\tau}} \tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma,\sigma}) + R_{\sigma,t}\|_{H_W(M)} \le C e^{t(\gamma_{K+1}^+ + \epsilon)},$$

giving Theorem 1.5.

**Remark 6.1.** By the effects of weight $W$ we can replace $\operatorname{Op}(\chi_{\Sigma,\sigma}) \operatorname{Op}_{\Sigma}(e^{tX_{\tau}} \tilde{T}_{[0,K]}) \operatorname{Op}(\chi_{\Sigma,\sigma})$ by $\operatorname{Op}_{\Sigma}(e^{tX_{\tau}} \tilde{T}_{[0,K]})$ in (1.13).

## 7. PROOF OF THEOREM 1.11 (WEYL LAW)

The aim of this section is to prove Theorem 1.11. We do this in three steps.

### 7.1. Approximate projector.
For $\omega \in \mathbb{R}$, $\delta > 0$ and $k \in \mathbb{N}$, we define an approximate projector $P$ for the union of band $k' \in [0, k]$ and frequency interval $[\omega - \delta, \omega + \delta]$ as follows. We will relate its trace to a symplectic volume. We denote $1_{[\omega-\delta,\omega+\delta]} : C^\beta(\Sigma; \mathcal{F}_{[0,k]}) \to C^\beta(\Sigma; \mathcal{F}_{[0,k]})$ the multiplication operator by the characteristic function for $\rho \in \Sigma$,

$$1_{[\omega-\delta,\omega+\delta]}(\rho) := \begin{cases} 1 & \text{if } \omega(\rho) \in [\omega-\delta, \omega+\delta] \\ 0 & \text{if not} \end{cases}.$$

We define the operator

$$(7.1) \qquad P := \operatorname{Op}_{\Sigma}(1_{[\omega-\delta,\omega+\delta]} \tilde{T}_{[0,k]}) : C^{\infty}(M) \to C^{\infty}(M).$$

Lemma 7.1 shows that the trace of $P$ is related to the symplectic volume of the support of function $1_{[\omega-\delta,\omega+\delta]}$ on the trapped set $\Sigma$.

**Lemma 7.1.** The operator $P$ is trace class in $H_W(M)$ and $\exists C > 0, \forall \omega \ge \delta \ge 1$,

$$(7.2) \qquad \|P\|_{\mathrm{Tr}} \le C\omega^d\delta, \qquad \|P\|_{H_W} \le C,$$

$$(7.3) \qquad \left| \mathrm{Tr}(P) - \mathrm{rank}(\mathcal{F}_{[0,k]}) \frac{\mathrm{Vol}(M) \omega^d}{(2\pi)^{d+1}} (2\delta) \right| \le C\omega^d\delta \left( |\omega|^{-(\beta-\mu)/2} + \frac{\delta}{\omega} \right) + C\omega^d.$$

**Remark 7.2.** Later in (7.39) we will use (7.3) as follows. For any small $c' > 0$ we will take $\delta \gg 1$ and $\omega \gg \delta$ large enough so that the right hand side of (7.3) is smaller than $c'\omega^d\delta$.

*Proof.* We have

$$\|P\|_{\mathrm{Tr}} = \|\operatorname{Op}_{\Sigma}(1_{[\omega-\delta,\omega+\delta]} \tilde{T}_{[0,k]})\|_{\mathrm{Tr}} \le C\omega^d\delta,$$

$$\|P\|_{H_W(M)} = \|\operatorname{Op}_{\Sigma}(1_{[\omega-\delta,\omega+\delta]} \tilde{T}_{[0,k]})\|_{H_W(M)} \le C.$$

Eq. (7.3) comes from (4.87) because $T_{[0,k]}(m)$ is a projector onto $\mathrm{Pol}_{[0,k]}(E_s(m))$ hence

$$\mathrm{Tr}(T_{[0,k]}(m)) = \dim(\mathrm{Pol}_{[0,k]}(E_s(m))) = \mathrm{rank}(\mathcal{F}_{[0,k]}).$$
and

$$ \int_M \mathrm{Tr}(T_{[0,k]}(m)) dm = \mathrm{Vol}(M) \mathrm{rank}(\mathcal{F}_{[0,k]}). \quad \square $$

## 7.2. Truncated resolvent.

**Definition 7.3.** For $z \in \mathbb{C}$, $\Lambda > 0$, let

$$ (7.4) \qquad \mathcal{R}(z) := (z - (X - \Lambda P))^{-1}, $$

$$ (7.5) \qquad D(z) := (z - X)^{-1} - \mathcal{R}(z). $$

*Remark 7.4.* A similar construction of truncated resolvent $\mathcal{R}(z)$ has been used in [20, Proof of Prop. 5.14].

Let $\delta' > 0$ and define the rectangle in the spectral plane

$$ (7.6) \qquad \tilde{R}(\omega, \delta, \delta') := [\gamma_k^- - 2\epsilon, \gamma_0^+ + 2\epsilon] \times i[\omega - \delta + \delta', \omega + \delta - \delta'], $$

Lemma 7.5 is similar to [20, Lemma 5.16] so we omit the proof.

**Lemma 7.5.** There exists $C > 0$ such that, if $\delta' > 0$ in (7.6) and $\Lambda > 0$ in (7.4) are sufficiently large, then for any $\omega \in \mathbb{R}$, $\delta > \delta' > 0$, $z \in \tilde{R}(\omega, \delta, \delta')$,

$$ (7.7) \qquad \|\mathcal{R}(z)\|_{\mathcal{H}_W(M)} \le C. $$

*Remark 7.6.* A consequence of Lemma 7.5 and expression (7.5) is that the poles of $D(z)$ in $\tilde{R}(\omega, \delta, \delta')$ coincide with the poles of the resolvent $(z - X)^{-1}$, i.e. Ruelle resonances.

Let us consider the following union of two horizontal bands (see Figure 7.1).

<figure><img src="image_20.png" /><figcaption>FIGURE 7.1. Picture in the spectral plane for the rectangular domain $\tilde{R}(\omega, \delta, \delta')$ defined in (7.6), the subdomains $B_{\pm}(\omega, \delta, \delta')$ defined in (7.8) and $R(\omega, \delta, \delta')$ defined later in (7.30)</figcaption></figure>

$$ (7.8) \qquad B_{\pm}(\omega, \delta, \delta') := [\gamma_k^- - \epsilon, \gamma_0^+ + \epsilon] \times i[\omega \pm (\delta - 2\delta'), \omega \pm (\delta - \delta')], $$

$$ (7.9) \qquad B(\omega, \delta, \delta') := B_+(\omega, \delta, \delta') \cup B_-(\omega, \delta, \delta'). $$
We will need later (in the proof of Lemma 7.11), Proposition 7.7.

**Proposition 7.7.** The operator $D(z) : \mathcal{H}_W(M) \to \mathcal{H}_W(M)$ is meromorphic w.r.t. $z \in \tilde{R}(\omega, \delta, \delta')$ and trace class outside of its poles of first order. In particular $\text{Tr}(D(z)) \in L_{\text{loc}}^1$ is locally integrable and with setting $z := x + iy$, we have $\exists C > 0, \forall \delta > 1, \forall \omega > \delta$,

$$ (7.10) \qquad \int_{B(\omega, \delta, \delta')} |\text{Tr}(D(z))| \, dxdy \le C\delta'\omega^d \ln \delta. $$

*Proof.* We first show few lemmas.

**Lemma 7.8.** $D(z)$ is trace class outside of its poles and

$$ (7.11) \qquad \text{Tr}(D(z)) = \partial_z \ln F(z) $$

with

$$ (7.12) \qquad A(z) := \text{Id} - \mathcal{R}(z) \Lambda P, $$

$$ (7.13) \qquad F(z) := \det(A(z)). $$

$D(z)$ is meromorphic because both $(z-X)^{-1}$ and $\mathcal{R}(z)$ are meromorphic, see [20, Lemma 5.16]. $\mathcal{R}(z)$ is even holomorphic on the rectangle $\tilde{R}(\omega, \delta, \delta')$, from Lemma 7.5. Then

$$ \begin{align} D(z) &= (z-X)^{-1} - \mathcal{R}(z) \tag{7.14} \\ &= (z - (X - \Lambda P))^{-1} \left( (z - (X - \Lambda P)) - (z-X) \right) (z-X)^{-1} \nonumber \\ &= \mathcal{R}(z) \Lambda P (z-X)^{-1}. \end{align} $$

In this last expression $P$ is trace class. $(z-X)^{-1}$ and $\mathcal{R}(z)$ are bounded outside of their poles hence $D(z)$ is trace class outside the poles of $(z-X)^{-1}$ and $\mathcal{R}(z)$. We have

$$ (7.15) \qquad D(z) = (z-X)^{-1} - \mathcal{R}(z) = (z - (X - \Lambda P) + \Lambda P)^{-1} - \mathcal{R}(z) \\ \qquad = \mathcal{R}(z) (\text{Id} - \mathcal{R}(z) \Lambda P)^{-1} - \mathcal{R}(z) = \mathcal{R}(z) (A(z)^{-1} - \text{Id}). $$

We compute

$$ (7.16) \qquad \partial_z A(z) = \mathcal{R}(z)^2 \Lambda P = \mathcal{R}(z) (\text{Id} - A(z)) = D(z) A(z), $$

then we have

$$ \begin{align*} \partial_z \ln A(z) &= (\partial_z A(z)) A(z)^{-1} = D(z), \tag{7.16} \\ \ln F(z) &= \ln \det(A(z)) = \text{Tr}(\ln A(z)), \tag{7.13} \\ \partial_z \ln F(z) &= \text{Tr}(\partial_z \ln A(z)) = \text{Tr}(D(z)). \end{align*} $$

We have obtained (7.11).

<figure><img src="image_21.png" /></figure>

For later use, we also notice that

$$ (7.17) \qquad \mathcal{R}(z) = A(z)(z-X)^{-1}. $$
**Lemma 7.9.** There exists $C > 0$, such that for every $z \in B(\omega, \delta, \delta')$,

$$ (7.18) \qquad |F(z)| \le e^{C(\ln \delta)\omega^d}, $$

and for every $z \in B(\omega, \delta, \delta')$ with $\mathrm{Re}(z) = \gamma_k^- - \epsilon$,

$$ (7.19) \qquad |F(z)| \ge e^{-C(\ln \delta)\omega^d}. $$

*Proof.* For $z \in B_-(\omega, \delta, \delta')$, writing $\lambda_j(z)$, $j = 1, 2, \dots, J$, for the eigenvalues of the trace class operator $\mathcal{R}(z) \Lambda P$, we have by Lidskii's theorem

$$ \begin{gathered} \log|F(z)| = \log|\det(\mathrm{Id} - \mathcal{R}(z) \Lambda P)| = \sum_j \log|1 - \lambda_j(z)| \\ (7.20) \qquad \le \sum_j \log(1 + |\lambda_j(z)|) \le \sum_j |\lambda_j(z)| \le \|\mathcal{R}(z) \Lambda P\|_{\mathrm{Tr}}. \end{gathered} $$

For the last inequality, see e.g. [24, p.64]. Notice that $\|\mathcal{R}(z) \Lambda P\|_{\mathrm{Tr}} \stackrel{(7.2),(7.7)}{\le} C\delta\omega^d$ but this is not enough to get (7.18). We will improve the last bound as follows. We decompose the frequency interval $[\omega - \delta, \omega + \delta]$ as the union of $[\delta]$ (the integer part of $\delta$) intervals of bounded length $l = \frac{2\delta}{|\delta|}$, so that

$$ (7.21) \qquad [\omega - \delta, \omega + \delta] = \bigcup_{w \in \{1, 2, \dots, \lfloor \delta \rfloor\}} I_w $$

with

$$ (7.22) \qquad I_w := [\omega - \delta + (w - 1)l, \omega - \delta + wl]. $$

Parallel to (7.1) we define

$$ P_w := \mathcal{J}_\sigma^\Delta \mathcal{O} p_\Sigma (T_{[0,k]}) \chi_{I_w} \mathcal{J}_\sigma : C^\infty(M) \to C^\infty(M), $$
i.e. an approximate projector on frequency interval of size $l$ and we write

$$ (7.23) \qquad P = \sum_w P_w. $$

Similarly to (7.2) we have

$$ \|P_w\|_{\mathrm{Tr}} \le C\omega^d $$

and

$$ (7.24) \qquad \|\mathcal{R}(z) \Lambda P_w\|_{\mathrm{Tr}} \le \frac{1}{w} C\omega^d $$

uniformly for $1 \le w \le |\delta|$. Hence

$$ \|\mathcal{R}(z) \Lambda P\|_{\mathrm{Tr}} \le \sum_{w=1}^{\lfloor \delta \rfloor} \|\mathcal{R}(z) \Lambda P_w\|_{\mathrm{Tr}} \le C\omega^d \ln \delta. $$

We get

$$ |F(z)| \stackrel{(7.20)}{\le} e^{\|\mathcal{R}(z)\Lambda P\|_{\mathrm{Tr}}} \le e^{C\omega^d \ln \delta}. $$

We have obtained (7.18). To prove the second claim (7.19), we consider an arbitrary $z \in B(\omega, \delta, \delta')$ with $\mathrm{Re}(z) = \gamma_k^- - \epsilon$. From Theorem 1.7 we have $\|(z-X)^{-1}\| \le C$. Then

$$ (\mathrm{Id} - \mathcal{R}(z) \Lambda P)^{-1} = A(z)^{-1} = (z - X)^{-1} \mathcal{R}(z)^{-1} = (\mathrm{Id} + (z - X)^{-1} \Lambda P)^{-1} $$
is uniformly bounded from (7.2). Hence $|1 - \lambda_j(z)| > c$ with some $c > 0$ independent on $\omega$ and therefore we have $|\log|1 - \lambda_j(z)|| < C' |\lambda_j(z)|$ for some $C' > 0$. Thus

$$ -\log|F(z)| = -\sum_j \log|1 - \lambda_j(z)| \le C' \sum_j |\lambda_j(z)| \le C' \|\mathcal{R}(z) \Lambda P\|_{\text{Tr}} \le C' \omega^d \ln \delta, $$

giving (7.19).

<figure><img src="image_22.png" /></figure>

Let $\text{spec}(X)$ denote the discrete Ruelle spectrum of $X$.

**Lemma 7.10.** We have $\exists C > 0, \forall \omega, \forall z \in B(\omega, \delta, \delta')$,

$$ (7.25) \qquad \left| \partial_z \ln F(z) - \sum_{z_j \in \text{spec}(X) \cap B(\omega, \delta, \delta')} \frac{1}{z-z_j} \right| \le C (\ln \delta) \omega^d, $$

where the sum over $z_j \in \text{spec}(X) \cap B(\omega, \delta, \delta')$ is counted with multiplicity of the Ruelle eigenvalues.

*Proof.* This is lemma $\alpha$ in [44, p.56]. We reproduce the proof here. From (7.7), $F(z)$ is a well defined holomorphic function in the domains $B(\omega, \delta, \delta')$ and from (7.17), the zeroes of $F(z)$ coincide up to multiplicities with the poles of $(z-X)^{-1}$, i.e. Ruelle eigenvalues $z_j \in \text{spec}(X) \cap B(\omega, \delta, \delta')$. Using Weyl upper bound $O(\omega^d)$ on the density in [20, Thm.3.6], note that the number of such Ruelle eigenvalues is bounded by $C\delta'\omega^d \le C\delta\omega^d$. From (7.19), we can fix some $z_0 \in B(\omega, \delta, \delta')$ with $\text{Re}(z_0) = \gamma_k^- - \epsilon$ so that $|F(z_0)| \stackrel{(7.19)}{\ge} e^{-C'\delta\omega^d}$. The function

$$ (7.26) \qquad G(z) := \frac{F(z) \prod_{z_j} (z_0 - z_j)}{F(z_0) \prod_{z_j} (z - z_j)}. $$

$G(z)$ is holomorphic on $B(\omega, \delta, \delta')$ and from estimate (7.18) on the boundary of $B(\omega, \delta, \delta')$ and maximum modulus principle we get

$$ (7.27) \qquad \forall z \in B(\omega, \delta, \delta'), \quad |G(z)| \stackrel{(7.18)}{\le} e^{C\omega^d \ln \delta}. $$

Moreover $G(z)$ has no zero on the simply connected domain $B(\omega, \delta, \delta')$ hence

$$ (7.28) \qquad g(z) := \ln G(z) $$

is well defined as a holomorphic function on $B(\omega, \delta, \delta')$ with $g(z_0) = 0$ and satisfies $\text{Reg}(z) \stackrel{(7.27)}{\le} C\omega^d \ln \delta$. By Borel-Carathéodory theorem<fnref n="14" /> this implies

$$ |g(z)| < C'\omega^d \ln \delta $$

on a region slightly larger than $B(\omega, \delta, \delta')$ and therefore by Cauchy integral formula

$$ |\partial_z g(z)| \stackrel{(7.28),(7.26)}{=} \left| \partial_z \ln F(z) - \sum_{z_j} \frac{1}{z-z_j} \right| < C'' \omega^d \ln \delta $$

<footnote n="14">¹⁴Here we do not need to give the constants and the proof is simple: $g$ maps $B(\omega, \delta, \delta')$ on the half-plane $\text{Re}(z) < C\delta\omega^d$. We post-compose by a Moebius transformation $f$ which maps this half-plane to the unit disk and maps $g(z_0) = 0$ to 0. Then the image of $B(\omega, \delta, \delta')$ by $f \circ g$ is contained in a disk $|z| < r < 1$. This implies that the image $g(B(\omega, \delta, \delta'))$ is contained in a (large) disk $|z| < C'\delta\omega^d$.</footnote>
giving the conclusion. Now we finish with the proof of Lemma 7.7. Using Weyl upper bound on the density in [20, Thm. 2.3], we have, with $z = x + iy$, that

$$ (7.29) \quad \sum_{z_j \in \text{spect}(X) \cap B(\omega, \delta, \delta')} \int_{B(\omega, \delta, \delta')} \frac{1}{|z - z_j|} dxdy < C (\delta' \omega^d) \ln \delta' $$

Then

$$ \int_{B(\omega, \delta, \delta')} |\text{Tr}(D(z))| \, dxdy = \int_{B(\omega, \delta, \delta')} \frac{|\partial_z \ln F(z)| \, dxdy}{(7.11)} \le \int_{(7.25),(7.29)} C \delta' \omega^d \ln \delta + C \delta' \omega^d \ln \delta' \le 2C \delta' \omega^d \ln \delta. $$

We have obtained (7.10). □

<figure><img src="image_23.png" /></figure>

**7.3. Counting resonances by argument principle.** From Lemma 7.5, the truncated resolvent $\mathcal{R}(z)$ is bounded on the rectangle $\tilde{R}(\omega, \delta, \delta')$ and hence by the Definition (7.5) the poles of $z \to \text{Tr} D(z)$ coincide with the poles of $(z - X)^{-1}$, i.e., the Ruelle resonance. In the following, we count the number of Ruelle resonance in the rectangle $\tilde{R}(\omega, \delta, \delta')$ by using a slightly generalized version of the argument principle. We learned this method from the paper of S. Dyatlov [9, Section 11].

In the following, we suppose that $\delta > 0, \delta' > 0$ is fixed so that $2\delta' < \delta$ and we consider the limit $|\omega| \to \infty$. We take a $C^\infty$ function $f_{\delta,\delta'} : \mathbb{R} \to [0, 1]$ that is $f_{\delta,\delta'}(\omega') = 1$ for $|\omega'| < \delta - 2\delta'$ and $f_{\delta,\delta'}(\omega') = 0$ for $|\omega'| > \delta - \delta'$ and $|\partial_{\omega'} f| < C\delta'^{-1}$. Then for $\omega \in \mathbb{R}, z \in \mathbb{C}$, let

$$ f_{\omega, \delta, \delta'}(z) := f_{\delta, \delta'}(\text{Im}(z - i\omega)). $$

Let us define the rectangle in the spectral plane (see Figure 7.1)

$$ (7.30) \quad R(\omega, \delta, \delta') := [\gamma_k^- - \epsilon, \gamma_k^+ + \epsilon] \times i[\omega - \delta + 2\delta', \omega + \delta - 2\delta'], $$

In the following we suppose that the $k$-th band is separated from $(k+1)$-st band, i.e.

$$ \gamma_{k+1}^{+} < \gamma_{k}^{-}. $$

Note that $R(\omega, \delta, \delta')$ is a proper subset of $\tilde{R}(\omega, \delta, \delta')$ defined in (7.6). We have that $\forall z \in R(\omega, \delta, \delta')$, $f_{\omega,\delta,\delta'}(z) = 1$ and $\forall z \in \tilde{R}(\omega, \delta, \delta')$,

$$ (7.31) \quad |\partial_{\bar{z}} f_{\omega, \delta, \delta'}(z)| < C \delta'^{-1}. $$

We consider the following paths $\Gamma := \Gamma_- \cup \Gamma_+$ on $\mathbb{C}$, parameterized as

$$ \Gamma_- : \omega \in \mathbb{R} \to z = \gamma_k^- - \epsilon + i\omega \in \mathbb{C}, $$

$$ \Gamma_+ : \omega \in \mathbb{R} \to z = \gamma_k^+ + \epsilon + i\omega \in \mathbb{C}. $$

From (7.7) and (1.16), there is no pole of $D(z)$ on $\Gamma \cap \text{supp}(f_{\omega,\delta,\delta'})$ supposing that $|\omega|$ is sufficiently large. We know that $D(z)$ is in trace class. Hence the integral $\int_{\Gamma_+ - \Gamma_-} f_{\omega,\delta,\delta'}(z) \text{Tr}(D(z)) dz$ is well defined. Lemma 7.11 shows that this integral is close to the number of Ruelle resonances in the rectangle $R(\omega, \delta, \delta')$.

**Lemma 7.11.** We have $\exists C > 0, \forall \delta > 1, \forall \omega > \delta, \forall 0 < \delta' < \delta$, (7.32)

$$ \left| \left( \frac{1}{2\pi i} \int_{\Gamma_+ - \Gamma_-} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz \right) - \#(\operatorname{spect}(X) \cap \tilde{R}(\omega, \delta, \delta')) \right| < C\omega^d \max(\ln \delta, \delta'). $$
*Proof.* From the Definition of $f_{\omega,\delta,\delta'}$ above, we have $f_{\omega,\delta,\delta'}(z) = 0$ if $|\mathrm{Im}(z) - \omega| > \delta - \delta'$, so we can close the contour integral on horizontal segments $\mathrm{Im}(z) = \omega \pm (\delta - \delta')$. The residues of the poles of $D(z)$ inside the contour are the spectral projector onto eigenspaces. Hence Cauchy integral formula for smooth functions gives

$$ \frac{1}{2\pi i} \int_{\Gamma_+ - \Gamma_-} f_{\omega, \delta, \delta'}(z) \mathrm{Tr}(D(z)) dz = \sum_{z_j \in \mathrm{spect}(X) \cap \tilde{\mathbb{R}}(\omega, \delta, \delta')} f_{\omega, \delta, \delta'}(z_j) + \frac{1}{\pi} \int_{\tilde{\mathbb{R}}(\omega, \delta, \delta')} (\partial_{\bar{z}} f_{\omega, \delta, \delta'})(z) \mathrm{Tr}(D(z)) dxdy $$

with $z = x + iy$. We have $\exists C > 0, \forall \delta > 0, \forall \delta' < \frac{1}{2}\delta, \forall \omega > \delta$,

$$ (7.33) \qquad \left| \int_{\tilde{\mathbb{R}}(\omega, \delta, \delta')} (\partial_{\bar{z}} f_{\omega, \delta, \delta'})(z) \mathrm{Tr}(D(z)) dxdy \right| \le \delta'^{-1} \int_{B(\omega, \delta, \delta')} |\mathrm{Tr}(D(z))| dxdy \\ \le C \delta'^{-1} (\ln \delta) \delta' \omega^d = C \omega^d \ln \delta. \quad (7.31), (7.10) $$

From properties of $f_{\omega,\delta,\delta'}$ and using Weyl upper bound on the density in [20, Thm.2.3] we have

$$ \left| \sum_{z_j \in \mathrm{spect}(X) \cap \tilde{\mathbb{R}}(\omega, \delta, \delta')} f_{\omega, \delta, \delta'}(z_j) - \#\{z_j \in \mathrm{spect}(X) \cap \tilde{\mathbb{R}}(\omega, \delta, \delta')\} \right| \le C \omega^d \delta'. $$

We have obtained (7.32). □

7.4. **Relation with the symplectic volume.** The last step is to relate the integral $\frac{1}{2\pi i} \int_{\Gamma_+ - \Gamma_-} f_{\omega, \delta, \delta'}(z) \mathrm{Tr}(D(z)) dz$ in (7.32) to the trace of $P$ in (7.3). Recall Definition (7.1) of $P$. We define the operator

$$ P_0 := \mathrm{Op}_{\Sigma} \left( 1_{[\omega - \delta + \delta'/2, \omega + \delta - \delta'/2]} T_{[0, k]} \right). $$

For $z \in R(\omega, \delta, \delta')$, let

$$ d(z) = \min\{\mathrm{Im}(z) - (\omega + \delta + \delta'/2), \mathrm{Im}(z) + (\omega - \delta - \delta'/2)\} \ge \delta'/2. $$

**Lemma 7.12** (Approximate expressions of $D(z)$). There exists $C > 0$ such that for any $T > 0$, $\omega > \delta > \delta' > 1$ and any $z \in \mathbb{C}$ with $\mathrm{Im}(z) \in [\omega - \delta + \delta', \omega + \delta - \delta']$ and $\mathrm{Re}(z) = \gamma_0^+ + \varepsilon$, we have
(7.34)

$$ \left\| D(z) - \left( \int_{0}^{T} e^{-tz} e^{t(X-\Lambda)} dt \right) (\Lambda P_{0}) \left( \int_{0}^{T} e^{-tz} e^{tX} dt \right) \right\|_{\mathrm{Tr}} \le C \omega^{d} \left( e^{-\varepsilon T} + \frac{\delta'}{(d(z))^{2}} \right). $$

If instead $\mathrm{Re}(z) = \gamma_k^- - \varepsilon$, we have
(7.35)

$$ \left\| D(z) - \left( \int_{0}^{T} e^{-tz} e^{t(X-\Lambda)} dt \right) (\Lambda P_0) \left( \int_{-T}^{0} e^{-tz} e^{tX} dt \right) \right\|_{\mathrm{Tr}} \le C \omega^d \left( e^{-\varepsilon T} + \frac{\delta'}{(d(z))^2} \right). $$

*Proof.* Let $z \in \mathbb{C}$ with $\mathrm{Im}(z) \in [\omega - \delta + \delta', \omega + \delta - \delta']$. In the following we let $T > 0$ be sufficiently large. Suppose $\mathrm{Re}(z) = \gamma_0^+ + \varepsilon$. As in (7.23), we decompose $P =$
$\sum_{w \in W} P_w$ with $W := \{1, 2, \dots, |\delta|\}$. As in (7.21) with interval $I_w$ defined in (7.22) we decompose

$$[\omega - \delta + \delta'/2, \omega + \delta - \delta'/2] = \bigcup_{w \in W_0} I_w$$

with a subset $W_0 \subset W$. So we can write

$$P_0 = \sum_{w \in W_0} P_w.$$

We have

$$D(z) = \sum_{w \in W} D_w(z)$$

with

$$D_w(z) := \Lambda \mathcal{R}(z) P_w(z-X)^{-1}.$$

For $w \in W_0$, let

$$\tilde{D}_w(z) := \Lambda \left( \int_0^T e^{-tz} e^{t(X-\Lambda)} dt \right) P_w \left( \int_0^T e^{-tz} e^{tX} dt \right).$$

Then

$$(z - X + \Lambda) \tilde{D}_w(z) (z - X) = \Lambda [e^{-tz} e^{t(X-\Lambda)}]_0^T P_w [e^{-tz} e^{tX}]_0^T \\ = \Lambda P_w + \Lambda r_w,$$

with

$$r_w = e^{T(-z+X-\Lambda)} P_w + P_w e^{T(-z+X)} + e^{T(-z+X-\Lambda)} P_w e^{T(-z+X)}.$$

We deduce that

$$\tilde{D}_w(z) = D_w(z) + \Lambda \mathcal{R}(z) r_w (z-X)^{-1}$$

with

$$\mathcal{R}(z) r_w (z-X)^{-1} = \mathcal{R}(z) e^{T(-z+X-\Lambda)} P_w (z-X)^{-1} + \mathcal{R}(z) P_w (z-X)^{-1} e^{T(-z+X)} \\ + \mathcal{R}(z) e^{T(-z+X-\Lambda)} P_w (z-X)^{-1} e^{T(-z+X)}.$$

For the right hand side of the last equality, we have the following estimates. We use the notation $\mathcal{O}_{\text{Tr}}(*)$ for an operator whose trace norm is bounded by a constant multiple of *. If $w \in W_0$ then as in (7.24), we have

$$\mathcal{R}(z) r_w (z-X)^{-1} = \mathcal{O}_{\text{Tr}} \left( \omega^d e^{-\epsilon T} \frac{1}{\text{dist}(I_w, \text{Im}(z))^2} \right),$$

where $\text{dist}(I_w, \text{Im}(z))$ is the distance between $I_w$ and $\text{Im}(z)$. If $w \in W \setminus W_0$ then

$$D_w(z) = \mathcal{O}_{\text{Tr}} \left( \omega^d \frac{1}{(d(z))^2} \right).$$

Taking the sum with respect to $w \in W_0$, we see

$$\sum_{w \in W_0} \left\| \mathcal{R}(z) r_w (z-X)^{-1} \right\|_{\text{Tr}} = \sum_{w \in W_0} \mathcal{O} \left( \omega^d e^{-\epsilon T} \frac{1}{\text{dist}(I_w, \text{Im}(z))^2} \right) = \mathcal{O} (\omega^d e^{-\epsilon T}),$$

$$\sum_{w \in W \setminus W_0} \mathcal{O}_{\text{Tr}} \left( \omega^d \frac{1}{(d(z))^2} \right) = \mathcal{O}_{\text{Tr}} \left( \omega^d \frac{\delta'}{(d(z))^2} \right).$$
Hence

$$ \left\| D(z) - \left( \int_0^T e^{-tz} e^{t(X-\Lambda)} dt \right) (\Lambda P_0) \left( \int_0^T e^{-tz} e^{tX} dt \right) \right\|_{\text{Tr}} = O \left( \omega^d \left( e^{-\epsilon T} + \frac{\delta'}{(d(z))^2} \right) \right). \quad \square $$

Now we give the key estimate.

**Lemma 7.13.** For any arbitrarily small $c > 0$, if we take $\delta$ large enough, $\delta' > 1$ large enough and $\delta/\delta'$ large enough, we have

$$ (7.36) \qquad \left| \left( \frac{1}{2\pi i} \int_{\Gamma_+ - \Gamma_-} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz \right) - \operatorname{Tr}(P) \right| < c\omega^d\delta. $$

*Proof.* We first consider the integration on $\Gamma_+$. From Lemma 7.12, we have

$$ (7.37) \qquad \begin{aligned} \int_{\Gamma_+} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz &= \int_{\mathbb{R}} f_{\delta, \delta'}(\omega' - \omega) \operatorname{Tr}(D(\gamma_0^+ + \epsilon + i\omega')) d(i\omega') \\ &= i \operatorname{Tr} \left[ \int_{\omega' \in \mathbb{R}} f_{\delta, \delta'}(\omega' - \omega) \left( \int_{t'=0}^T e^{-t'(\gamma_0^+ + \epsilon + i\omega')} e^{t'(X-\Lambda)} dt' \right) (\Lambda P_0) \left( \int_{t=0}^T e^{-t(\gamma_0^+ + \epsilon + i\omega')} e^{tX} dt \right) \right] d\omega' \\ &\quad + \mathcal{O}_{\operatorname{Tr}}(\omega^d (\delta e^{-\epsilon T} + \delta')). \end{aligned} $$

Let

$$ \tilde{f}_{\omega, \delta, \delta'}(t) := \int_{\mathbb{R}} e^{-t(\gamma_0^+ + \epsilon + i\omega')} f_{\delta, \delta'}(\omega' - \omega) d\omega' = e^{-t(\gamma_0^+ + \epsilon + i\omega)} \hat{f}_{\delta, \delta'}(t), $$

where $\hat{f}_{\delta, \delta'}(t) := \int_{\mathbb{R}} e^{-it\omega'} f_{\delta, \delta'}(\omega') d\omega'$ satisfies a fast decay for $|t| \gg \delta'^{-1}$ as follows: $\forall N > 0, \exists C_N > 0, \forall t,$

$$ (7.38) \qquad |\hat{f}_{\delta, \delta'}(t)| \le C_N \langle \delta' t \rangle^{-N} \quad \text{and} \quad \int_{\mathbb{R}} \hat{f}_{\delta, \delta'}(t) dt = 2\pi f_{\delta, \delta'}(0) = 2\pi. $$

Then, with the change of variables $(t, t') \to (t, s = t + t')$, we see

$$ \begin{aligned} & \int_{\Gamma_+} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz \\ &= i \int_0^T \int_0^T \tilde{f}_{\omega, \delta, \delta'}(t+t') \operatorname{Tr}(e^{t'(X-\Lambda)} P_0 \Lambda e^{tX}) dt' dt + \mathcal{O}_{\operatorname{Tr}}(\omega^d (\delta e^{-\epsilon T} + \delta')) \\ &= i \int_0^{2T} \left( \int_{t=\max\{0, s-T\}}^{\min\{T, s\}} e^{-s(\gamma_0^+ + \epsilon + i\omega)} \hat{f}_{\delta, \delta'}(s) \operatorname{Tr}(e^{(s-t)(X-\Lambda)} P_0 \Lambda e^{tX}) dt \right) ds \\ &\quad + \mathcal{O}_{\operatorname{Tr}}(\omega^d (\delta e^{-\epsilon T} + \delta')). \end{aligned} $$

From (7.38), by letting $\delta'$ be large (and $\hat{f}_{\delta, \delta'}(t)$ be more concentrate to 0), we get

$$ \left| \int_{\Gamma_+} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz \right| = \mathcal{O}_{\operatorname{Tr}}(\delta'^{-1} \omega^d \delta) + \mathcal{O}_{\operatorname{Tr}}(\omega^d (\delta e^{-\epsilon T} + \delta')). $$
Next we consider the integration on $\Gamma_-$ Similarly to the argument above, we see

$$
\begin{align*}
\int_{-\Gamma_{-}} f_{\omega, \delta, \delta'}(z) \operatorname{Tr}(D(z)) dz &= i \int \int_{\Gamma_{\delta, \delta'}} (\omega' - \omega) \operatorname{Tr}(D(\gamma_{0}^{-} - \epsilon - i\omega')) d\omega' \\
&= i \operatorname{Tr} \left[ \int_{(7.34)} \int_{\Gamma_{\delta, \delta'}} (\omega' - \omega) \left( \int_{0}^{T} e^{-t'(\gamma_{0}^{-} - \epsilon - i\omega')} e^{t'(X-\Lambda)} dt' \right) (\Lambda P_{0}) \left( \int_{-T}^{0} e^{-t(\gamma_{0}^{-} - \epsilon - i\omega')} e^{tX} dt \right) \right] d\omega' \\
&\quad + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')) \\
&= i \int_{s=-T}^{T} \left( \int_{t=\max\{-T, s-T\}}^{\min\{0, s\}} e^{-s(\gamma_{0}^{-} - \epsilon - i\omega')} \hat{f}_{\delta, \delta'}(s) \operatorname{Tr}(e^{s(X-\Lambda)} e^{-t(X-\Lambda)} \Lambda P_{0} e^{tX}) dt \right) ds \\
&\quad + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')) \\
&= 2\pi i \int_{t=-T}^{0} \operatorname{Tr}(e^{-t(X-\Lambda)} \Lambda P_{0} e^{tX}) dt + \mathcal{O}_{\operatorname{Tr}}(\delta'^{-1}\omega^{d}\delta) + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')) \\
&= 2\pi i \operatorname{Tr}([e^{t\Lambda} P_{0}]_{t=-T}^{0}) + \mathcal{O}_{\operatorname{Tr}}(\delta'^{-1}\omega^{d}\delta) + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')) \\
&= 2\pi i \operatorname{Tr}(P_{0}) + \mathcal{O}_{\operatorname{Tr}}(e^{-\epsilon T}\omega^{d}) + \mathcal{O}_{\operatorname{Tr}}(\delta'^{-1}\omega^{d}\delta) + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')) \\
&= 2\pi i \operatorname{Tr}(P) + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}\delta') + \mathcal{O}_{\operatorname{Tr}}(e^{-\epsilon T}\omega^{d}) + \mathcal{O}_{\operatorname{Tr}}(\delta'^{-1}\omega^{d}\delta) + \mathcal{O}_{\operatorname{Tr}}(\omega^{d}(\delta e^{-\epsilon T} + \delta')).
\end{align*}
$$

Therefore, by letting $T$ large $\delta \gg \delta' \gg 1$ and then letting $\omega$ large, we obtain the required estimate. $\square$

7.5. **Weyl law.** Using the previous estimates we get that

$$
\begin{equation} \tag{7.39}
\begin{split}
\left| \# (\operatorname{spect}(X) \cap \tilde{R}(\omega, \delta, \delta')) - \operatorname{rank}(\mathcal{F}_{[0,k]}) \frac{\operatorname{Vol}(M) \omega^d}{(2\pi)^{d+1}} (2\delta) \right|
\end{split}
\end{equation}
\[
\begin{aligned}
& <_{(7.32),(7.36),(7.3)} C \omega^d \max\{\delta', \ln \delta\} + c \omega^d \delta + C \omega^d \delta (|\omega|^{-\beta/2} + \frac{\delta}{\omega}) \\
& \quad + C \omega^d \left(C_N \sigma^{-N} + |\omega|^{-\beta/2} \sigma + \frac{\delta}{\omega}\right)
\end{aligned}
\]
\[
\begin{split}
& < c' \omega^d \delta,
\end{split}
\]
$$

with arbitrarily small $c' > 0$ and $\omega \gg \delta \gg \delta' \gg 1$. Comparing this estimate for adjacent $k$'s, we obtain Theorem 1.11.

8. PROOF OF THEOREM 1.13 (ACCUMULATION OF EIGENVALUES ON NARROWER BANDS)

The aim of this section is to prove Theorem 1.13. We consider the operator $X_F$ defined in Section 2.2 where $X$ is a contact Anosov flow. For $k \in \mathbb{N}$, consider the vector bundle $\pi : \mathcal{F}_k \to \Sigma$, with $\mathcal{F}_k = |\det N_s|^{-1/2} \otimes \operatorname{Pol}_k(N_s) \otimes F$ defined in (4.43), twisted by $F$, and the flow induced from $X_F$,

$$
\phi_k^t : \mathcal{F}_k \to \mathcal{F}_k, \quad t \in \mathbb{R},
$$

that is a linear bundle map over $\tilde{\phi}^t : \Sigma \to \Sigma$. For a point $\rho \in \Sigma$, $t \in \mathbb{R}$, let

$$
g_{k,t}(\rho) := \ln\left(\left(\max_{\nu \in \pi^{-1}(\rho)} \frac{\|\phi_k^t(\nu)\|}{\|\nu\|}\right)^{1/t}\right),
$$
that defines a continuous function $g_{k,t}$ on $\Sigma$. Let $d\varrho$ be the Liouville measure (3.8) on $\Sigma$ and

$$ \mu_{k,t} := (g_{k,t}^{\circ})^{\dagger} (d\varrho) $$

be the push-forward measure on $\mathbb{R}$ by the map $g_{k,t}$. We have already defined $\gamma_k^{\pm}$ in (4.45) that can be written as

$$ \gamma_k^+ := \lim_{t \to +\infty} \max_{\rho \in \Sigma} g_{k,t}(\rho) = \lim_{t \to +\infty} \max(\text{supp}(\mu_{k,t})) $$

$$ \gamma_k^- := \lim_{t \to -\infty} \min_{\rho \in \Sigma} g_{k,t}(\rho) = \lim_{t \to -\infty} \min(\text{supp}(\mu_{k,t})). $$

Now we define the **maximal and minimal Lyapunov exponents with respect to the Liouville measure** by

$$ \begin{aligned} (8.1) \qquad & \dot{\gamma}_k^+ = \max \left( \supp \left( \lim_{t \to +\infty} \mu_{k,t} \right) \right), \\ & \dot{\gamma}_k^- = \min \left( \supp \left( \lim_{t \to -\infty} \mu_{k,t} \right) \right). \end{aligned} $$

One has

$$ \gamma_k^- \le \dot{\gamma}_k^- \le \dot{\gamma}_k^+ \le \gamma_k^+. $$

From the definition of $\dot{\gamma}_k^-$, $\dot{\gamma}_k^+$, for any small $\epsilon > 0$, if $I_\epsilon := [\dot{\gamma}_k^- - \epsilon, \dot{\gamma}_k^+ + \epsilon]$ and

$$ \Sigma_{\omega, \delta, I_\epsilon, t} := (g_{k,t}^{-1}(I_\epsilon) \cap [\omega, \omega + \delta]) \subset \Sigma, $$

then we have that for any $c > 0$, there exists $T_c > 0$ such that

$$ (8.2) \qquad (d\varrho)(\Sigma \setminus \Sigma_{\omega, \delta, I_\epsilon, T_c}) \le c\delta\omega^d. $$

For the proof of Theorem 1.13, we follow [18, Thm 1.3.11 and Chapter 8,]. Let $c > 0$. Take $T_c > 0$ large enough such that (8.2) holds true. Then similarly to (4.62) one gets $\forall \epsilon > 0, \exists C_\epsilon > 0, \forall t \ge 0, \exists \sigma_l > 0, \forall \sigma > \sigma_l, \exists \omega_\sigma > 0, \forall \omega > \omega_\sigma$,

$$ \begin{aligned} \|e^{tX} \text{Op}_{\Sigma} (1_{\Sigma_{\omega,\delta,I_{\epsilon},T_c}} T_k)\|_{\mathcal{H}_{W}(M)} &\le C_{\epsilon} e^{t(\dot{\gamma}_k^+ + \epsilon)}, \\ \|e^{-tX} \text{Op}_{\Sigma} (1_{\Sigma_{\omega,\delta,I_{\epsilon},T_c}} T_k)\|_{\mathcal{H}_{W}(M)} &\le C_{\epsilon} e^{-t(\dot{\gamma}_k^- - \epsilon)}, \\ \|\text{Op}_{\Sigma} (1_{\Sigma \setminus \Sigma_{\omega,\delta,I_{\epsilon},T_c}} T_k)\|_{\text{Tr}} &= c\delta\omega^d + C\omega^d. \end{aligned} $$

Using arguments with perturbation of the resolvent as in previous proofs, we deduce the Theorem.

## 9. PROOF OF THEOREM 1.18 (HOROCYCLE OPERATORS)

Suppose $s \in C^\beta(M; E_s)$ is a Hölder continuous section of the stable bundle over $M$ and $u \in C^\beta(M; E_u)$ a section of the unstable bundle. Let $\pi : T^*M \to M$ be the canonical projection. We have seen in (4.25) that for $\rho \in \Sigma$, the differential $d\pi$ gives a linear isomorphism $d\pi : N(\rho) = N_s(\rho) \oplus N_u(\rho) \to E_u(m) \oplus E_s(m)$ with $m = \pi(\rho)$. By pull back we get sections $\tilde{s} := (d\pi)^{-1}(s) \in C(\Sigma; N_s)$ and $\tilde{u} := (d\pi)^{-1}(u) \in C(\Sigma; N_u)$. Recall from Section 4.1.2 that $dA$ is a symplectic form on $E_u \oplus E_s$ and from Definition 3.4 that $N(\rho) = N_s(\rho) \oplus N_u(\rho)$ is an $\Omega$-symplectic vector space and that $\Omega_\rho(\tilde{u}, \tilde{s}) = \omega(\rho) \cdot (dA)(u, s)$, hence $\tilde{u}$ (so then $u$) defines a dual vector

$$ (9.1) \qquad \tilde{u}^* := \check{\Omega} \tilde{u} = \Omega(\tilde{u}, \cdot)_{(A.1)} = \omega(\cdot)_{(A.9)} (dA)(u, d\pi(\cdot)) \in N_s^*(\rho). $$
For $k \in \mathbb{N}$, let $p \in C(\Sigma; \mathcal{F}_k)$ a continuous section of $\mathcal{F}_k(N_s) \underset{(4.27)}{=} |\det N_s|^{-1/2} \otimes \text{Pol}_k(N_s) \otimes F$ over $\Sigma$ where $\text{Pol}_k(N_s) \equiv \text{Sym}^k(N_s^* \otimes \dots \otimes N_s^*)$ is equivalent to the symmetric tensor algebra of degree $k$.

**Definition 9.1.** For $s \in C^\beta(M; E_s)$ we define

$$ (9.2) \qquad \iota_s : \begin{cases} C^\beta(\Sigma; \mathcal{F}_k) & \to C^\beta(\Sigma; \mathcal{F}_{k-1}), \\ p & \to \iota_s p = k p(\bar{s}, \dots, \dots) \end{cases} $$

as the interior product, i.e. the pointwise (i.e. independently in every fiber) contraction in the first entry of $p$ by $\bar{s} := (d\pi)^{-1}(s) \in C^\beta(\Sigma; N_s)$, but with a factor $k$. For $u \in C^\beta(M; E_u)$ we define

$$ u \odot : \begin{cases} C^\beta(\Sigma; \mathcal{F}_k) & \to C^\beta(\Sigma; \mathcal{F}_{k+1}), \\ p & \to \tilde{u}^* \odot p = \text{Sym}(\tilde{u}^* \otimes p) \end{cases} $$

as the symmetric point-wise tensor product of $p$ by $\tilde{u}^* := \omega(dA)(u, d\pi(.)) \in C^\beta(\Sigma; N_s^*)$.

**Lemma 9.2 (Weyl algebra).** For $s \in C^\beta(M; E_s)$, $u \in C^\beta(M; E_u)$ and $k \in \mathbb{N}$ we have

$$ (9.3) \qquad [\iota_s, u \odot] = \omega(.) (dA)(s, u) \text{Id}_{\mathcal{F}_k} : C^\beta(\Sigma; \mathcal{F}_k) \to C^\beta(\Sigma; \mathcal{F}_k), $$

where $\omega$ is the frequency function (2.7) on $\Sigma$ and $\text{Id}_{\mathcal{F}_k}$ is the identity operator in fibers.

*Proof.* This is point wise relation so we consider a point $\rho \in \Sigma$ with frequency $\omega \underset{(2.7)}{=} \omega(\rho)$ and the vector space $N_s(\rho)$ defined in (4.25). We use the isomorphism $\text{Pol}_k(N_s) \equiv \text{Sym}^k(N_s^* \otimes \dots \otimes N_s^*)$ to derive (9.3) that goes as follows. If $(e_i)_i$ is a basis of $N_s(\rho)$, we write $\bar{s} = \sum_{i=1}^d \sigma_i e_i \in N_s(\rho)$ with components $\sigma = (\sigma_i)_i \in \mathbb{R}^d$. For $p \in \text{Sym}^k(N_s^* \otimes \dots \otimes N_s^*)$, we associate a degree $k$ polynomial $P \in \text{Pol}_k(\mathbb{R}^d)$ by $P(\sigma) := p(\bar{s}, \bar{s}, \dots, \bar{s})$. We define $e_i^* \odot p := \text{Sym}(e_i^* \otimes p)$ and observe that $(e_i^* \odot p)(\bar{s}) = \sigma_i P(\sigma)$ and $(\iota_{e_i} p)(\bar{s}) = (\partial_{\sigma_i} P)(\sigma)$. Then the “Weyl algebra relation” on polynomials $[\partial_{\sigma_i}, \sigma_j] = \delta_{i=j} \text{Id}$ gives $[\iota_{e_i}, e_j^* \odot] = \delta_{i=j} \text{Id}$ hence for $\bar{s} \in N_s$, $\tilde{u}^* \in N_s^*$ we have $[\iota_{\bar{s}}, \tilde{u}^* \odot] = \tilde{u}^* (\bar{s}) \text{Id}$. With (9.1), we get (9.3). $\square$

*Proof of Theorem 1.18.* Let $K \in \mathbb{N}$. We will use $\tilde{T}_{[0,K]}$ to restrict to the finite rank bundle of homogeneous polynomials of degree less than $K$. For $k < K$, let $\tilde{T}_k$ be the projector on homogeneous polynomials of degree $k$. We have

$$ \begin{aligned} [\text{Op}_\Sigma(\iota_s \tilde{T}_{[0,K]}), \text{Op}_\Sigma((u \odot) \tilde{T}_{[0,K]})] \text{Op}_\Sigma(\tilde{T}_k) & \underset{(4.65)}{\approx} \text{Op}_\Sigma([\iota_s, u \odot] \tilde{T}_k) \\ & = \text{Op}_\Sigma(\omega(.) (dA)(s, u) \tilde{T}_k) \end{aligned} $$

that gives (1.36). We have

$$ \begin{aligned} & e^{tX_F} \text{Op}_\Sigma(\iota_s \tilde{T}_k) \underset{(4.56)}{\approx} \text{Op}_\Sigma(e^{tX_F}) \text{Op}_\Sigma(\iota_s \tilde{T}_k) \underset{(4.65)}{\approx} \text{Op}_\Sigma(e^{tX_F} \iota_s \tilde{T}_k) \\ & = \text{Op}_\Sigma(\iota_d \phi^t s e^{tX_F} \tilde{T}_k) \underset{(4.42),(4.65)}{\approx} \text{Op}_\Sigma(\iota_d \phi^t s \tilde{T}_k) \text{Op}_\Sigma(e^{tX_F}) \\ & \approx \underset{(4.56)}{\approx} \text{Op}_\Sigma(\iota_d \phi^t s \tilde{T}_k) e^{tX_F}, \end{aligned} $$

giving (1.37) and similarly to get (1.38). $\square$

<figure><img src="image_24.png" /></figure>
## APPENDIX A. GENERAL NOTATIONS USED IN THIS PAPER

* **Dual** *. If $E$ is a finite dimensional vector space and $E^*$ its dual space, $x \in E, \xi \in E^*$ we denote the duality by $\langle \xi | x \rangle \in \mathbb{R}$. If $E_1, E_2$ are vector spaces and $A: E_1 \to E_2$ is a linear map, we will denote $A^*: E_2^* \to E_1^*$ the induced dual map on the dual spaces defined by $\langle A^* \xi | x \rangle = \langle \xi | Ax \rangle, \forall x \in E_1, \xi \in E_2^*$.

* **Bilinear map.** Let $\Omega: E \times E \to \mathbb{R}$ be a bilinear map on a vector space $E$. It defines a linear map $\check{\Omega}: E \to E^*$ by

$$ (A.1) \qquad \check{\Omega}(x)(.) = \Omega(x, .). $$

- If $\check{\Omega}$ is invertible ($\Omega$ is said to be non-degenerated), it induces a bilinear form on $E^*$ denoted $\Omega^{-1} : E^* \times E^* \to \mathbb{R}$ and defined by $\Omega^{-1}(\xi_1, \xi_2) := \Omega(\check{\Omega}^{-1}\xi_1, \check{\Omega}^{-1}\xi_2) = \xi_1(\check{\Omega}^{-1}\xi_2)$.

- The bilinear map $\Omega: E \times E \to \mathbb{R}$ is anti-symmetric iff

$$ (A.2) \qquad \check{\Omega}^* = -\check{\Omega}. $$

The bilinear map $\Omega: E \times E \to \mathbb{R}$ is symmetric iff $\check{\Omega}^* = \check{\Omega}$.

* **Adjoint** †. Suppose $A: E_1 \to E_2$ is a linear map and $\Omega_1$ (respect. $\Omega_2$) is a non-degenerated bilinear form on $E_1$ (respect. $E_2$). The $\Omega$-adjoint of $A$ is $A^{\dagger \Omega}: E_2 \to E_1$ defined by

$$ (A.3) \qquad A^{\dagger \Omega} = \check{\Omega}_1^{-1} A^* \check{\Omega}_2 $$

equivalently given by

$$ \Omega_1 (A^{\dagger \Omega} x_1, x_2) = \Omega_2 (x_1, A x_2), \forall x_1 \in E_1, \forall x_2 \in E_2. $$

In this paper bilinear forms will be either a symplectic form $\Omega(.,.)$ or a metric $g(.,.)$.

* **Pull-back** ◦. If $f: M \to N$ is a smooth map between two manifolds,

- We denote $f^\circ$ the **pull back** operator

$$ (A.4) \qquad f^\circ: \begin{cases} C^\infty(N) & \to C^\infty(M), \\ u & \to u \circ f \end{cases} $$

and if $f$ is a diffeomorphism, we denote $f^{-\circ} := (f^{-1})^{\circ}$ the **push forward** operator, so that $f^{-\circ}f^{\circ} = \text{Id}$.

- We denote $df: TM \to TN$ the differential of $f$ that is a linear bundle map and denote $(df)^*: T^*N \to T^*M$ its dual.

## APPENDIX B. MORE INFORMATION ABOUT FLOWS

B.1. **Transfer operator.** Suppose that $X$ is a vector field on $M$ and $(\phi^t)^\circ = e^{tX}$ is the pull back operator defined in (2.2). For any smooth measure $d\mu$ on $M$, the $L^2(M, d\mu)$ scalar product is defined by $\langle v|u \rangle_{L^2} := \int_M \bar{u}v d\mu$ with $u, v \in C_c^\infty(M)$. Then the $L^2$-adjoint $(\phi^t)^{\dagger}$ of $(\phi^t)^\circ$, defined by $\langle v| (\phi^t)^{\dagger} u \rangle_{L^2} = \langle (\phi^t)^\circ v|u \rangle_{L^2}, \forall u, v \in C_c^\infty(M)$ is given by

$$ (\phi^t)^{\dagger} = |\det(d\phi^{-t})| \cdot (\phi^{-t})^\circ = e^{tX^\dagger}, \quad \text{with } X^\dagger = -X - \text{div}_\mu X. $$
($\phi^t$)$^{\circ\dagger}$ is called the **Ruelle-Perron-Frobenius operator** or **transfer operator**. Observe that $(\phi^t)^{\circ\dagger}$ **pushes forward probability distributions** because

$$ \int_M (\phi^t)^{\circ\dagger} u d\mu = \langle 1 | (\phi^t)^{\circ\dagger} u \rangle_{L^2} = \langle \underbrace{(\phi^t)^\circ}_{1} 1 | u \rangle_{L^2} = \int_M u d\mu. $$

In particular the evolution of Dirac measures gives $(\phi^t)^{\circ\dagger} \delta_m = \delta_{\phi^t(m)}$ and is equivalent to evolution of points under the flow map $\phi^t$.

B.2. **More general pull back operators $e^{tX_F}$**. Suppose that $X$ is a smooth vector field on $M$ and $F \to M$ is a smooth vector bundle over $M$. Suppose that $X_F : C^\infty(M;F) \to C^\infty(M;F)$ is a derivation over $X$ fulfilling (2.4).

B.2.1. *Expression in local frames.* For a general vector bundle $F$ of rank $r$, with respect to a local frame $(e_1, \dots, e_r)$, a section $u \in C^\infty(M;F)$ is expressed as $u(m) = \sum_{j=1}^r u_j(m) e_j(m)$ with $m \in M$ and components $u_j(m) \in \mathbb{C}$. We introduce a matrix of potential functions $V_{k,j}(m) \in \mathbb{C}$ defined by

$$ (B.1) \qquad (X_F e_k)(m) = \sum_j V_{k,j}(m) e_j(m). $$

Then the operator $X_F$ in (2.3) is expressed as

$$ (B.2) \qquad (X_F u)(m) = \sum_{j=1}^{r} \left( (X u_j)(m) + \sum_{k=1}^{r} V_{k,j}(m) u_k(m) \right) e_j(m). $$

We see that the expression (B.2) generalizes (2.5).

B.2.2. *Expression of $e^{tX_F}$ from a flow $\tilde{\phi}_F^t$ on $F$.* An equivalent way to present the operator $X_F$ in (2.4) is to consider a vector field $\tilde{X}_F$ on $F$, a lift of $X$, i.e. $(d\pi)\tilde{X}_F = X$, with the projection $\pi : F \to M$, and such that the flow map generated by $\tilde{X}_F$

$$ \tilde{\phi}_F^t : F \to F, $$

i.e. $(\tilde{\phi}_F^t)^\circ = e^{t\tilde{X}_F} : S(F) \to S(F)$, is a smooth linear bundle map over $\phi^t$ which means that for every $m \in M, t \in \mathbb{R}$, $\tilde{\phi}_F^t(m) : F(m) \to F(\phi^t(m))$ is a linear map. See Figure B.1.

<figure><img src="image_25.png" /><figcaption>FIGURE B.1. Pull back operator $e^{tX_F}$</figcaption></figure>

This defines a **group of pull back operators** with generator $X_F$ acting on a section $u \in C^\infty(M;F)$ of the vector bundle $F$:

$$ e^{tX_F} : C^\infty(M;F) \to C^\infty(M;F), \quad t \in \mathbb{R}, $$
by

$$ (B.3) \qquad e^{tX_F} u := \tilde{\phi}_F^{-t} (u \circ \phi^t). $$

We get $e^{tX_F}(fu) = \tilde{\phi}_F^{-t}((e^{tX}f)(u \circ \phi^t)) = (e^{tX}f)(e^{tX_F}u)$ and deduce (2.4) by derivation w.r.t. $t$. Conversely the flow map $\tilde{\phi}_F^t$ and its generator $\tilde{X}_F$ is defined from $X_F$ by (B.3) and use of any section $u$.

## APPENDIX C. BARGMANN TRANSFORM AND METAPLECTIC OPERATORS

In this section we collect results that concern the Bargmann transform on a vector space $E$, the quantization of affine and linear symplectic map, that are the unitary representations of the Heisenberg group and the metaplectic operators. All these definitions and results are well known and important in many fields of mathematics and physics. They are at the core of micro-local analysis. We present these definitions and results in a form that is adapted to our paper that relies strongly on them. References are [22], [29], [32, appendix 4.4], [18, chap.3, chap.4].

### C.1. Weyl Heisenberg group.
Let $E$ be a real vector space, $n = \dim E$. For $x \in E$, $\xi \in E^*$ we denote $\langle \xi | x \rangle \in \mathbb{R}$ the duality. Let $dx \in |\Lambda^n(E)|$ be a density on $E$, and let $d\xi$ be the induced density on $E^*$. Let us denote the Fourier transform

$$ \mathcal{F} : \left\{ \begin{array}{ll} S(E) & \to S(E^*), \\ u & \to v(\xi) = \frac{1}{(2\pi)^{n/2}} \int e^{-i\langle \xi | x \rangle} u(x) dx. \end{array} \right. $$

Its $L^2$-adjoint is

$$ \mathcal{F}^{\dagger} : \left\{ \begin{array}{ll} S(E^*) & \to S(E), \\ v & \to u(x) = \frac{1}{(2\pi)^{n/2}} \int e^{i\langle \xi | x \rangle} v(\xi) d\xi \end{array} \right. $$

and we have that $\mathcal{F}^{\dagger}\mathcal{F} = \text{Id}_{S(E)}$ hence $\mathcal{F}^{-1} = \mathcal{F}^{\dagger}$.

For $x \in E$, $\xi \in E^*$ we define the translation maps

$$ T_x : \left\{ \begin{array}{ll} E & \to E \\ y & \to y + x \end{array} \right. , \quad T_\xi : \left\{ \begin{array}{ll} E^* & \to E^* \\ \xi' & \to \xi' + \xi \end{array} \right. $$

And denote $T_x^{-\circ}$, $T_\xi^{-\circ}$ the push forward operators. We will denote

$$ (C.1) \qquad \hat{T}_{x, \xi} := T_x^{-\circ} (\mathcal{F}^{-1} T_\xi^{-\circ} \mathcal{F}). $$

**Lemma C.1.** On $S(E)$, one has

$$ (C.2) \qquad (\mathcal{F}^{-1} T_{\xi}^{-\circ} \mathcal{F}) T_{x}^{-\circ} = e^{i\langle \xi | x \rangle} T_{x}^{-\circ} (\mathcal{F}^{-1} T_{\xi}^{-\circ} \mathcal{F}). $$ $$ \hat{T}_{x, \xi} \hat{T}_{x', \xi'} = e^{i\langle \xi | x' \rangle} \hat{T}_{x+x', \xi+\xi'}. $$

In particular $\hat{T}_{x,\xi}$ is a unitary operator in $L^2(E, dx)$ and

$$ (C.3) \qquad \hat{T}_{x, \xi}^{-1} = \hat{T}_{x, \xi}^{\dagger} = \hat{T}_{-(x, \xi)} e^{i \xi x}. $$

**Remark C.2.** The family of operators $\hat{T}_{x,\xi}$ with $(x, \xi) \in T^*E = E \oplus E^*$ are called the Schrödinger unitary representation of the Weyl Heisenberg group.
*Proof.* Let $x \in E$. Then

$$ (\mathcal{F}\delta_x)(\xi') = \frac{1}{(2\pi)^{n/2}}e^{-i\xi'x}, \quad (T_{\xi}^{-\circ}\mathcal{F}\delta_x)(\xi') = \frac{1}{(2\pi)^{n/2}}e^{-i(\xi'-\xi)x}, \\ (\mathcal{F}^{-1}T_{\xi}^{-\circ}\mathcal{F}\delta_x)(x') = \frac{1}{(2\pi)^n} \int e^{i\xi'x}e^{-i(\xi'-\xi)x}d\xi' = e^{i\xi x}\delta_x(x'). $$

Hence $\mathcal{F}^{-1}T_{\xi}^{-\circ}\mathcal{F}\delta_x = e^{i\xi x}\delta_x$. Hence

$$ T_x^{-\circ}(\mathcal{F}^{-1}T_{\xi}^{-\circ}\mathcal{F})\delta_{x'} = e^{i\xi x'}\delta_{x+x'}, \\ (\mathcal{F}^{-1}T_{\xi}^{-\circ}\mathcal{F})T_x^{-\circ}\delta_{x'} = e^{i(\xi|x')}\delta_{x+x'} = e^{i(\xi|x)}T_x^{-\circ}(\mathcal{F}^{-1}T_{\xi}^{-\circ}\mathcal{F})\delta_{x'}. $$

Then

$$ \begin{aligned} \hat{T}_{x, \xi} \hat{T}_{x', \xi'} &= T_x^{-\circ} (\mathcal{F}^{-1} T_{\xi}^{-\circ} \mathcal{F}) T_{x'}^{-\circ} (\mathcal{F}^{-1} T_{\xi'}^{-\circ} \mathcal{F}) \\ &= e^{i\langle\xi|x'\rangle} T_x^{-\circ} T_{x'}^{-\circ} (\mathcal{F}^{-1} T_{\xi}^{-\circ} \mathcal{F}) (\mathcal{F}^{-1} T_{\xi'}^{-\circ} \mathcal{F}) \\ &= e^{i\langle\xi|x'\rangle} \hat{T}_{x+x', \xi+x'}. \end{aligned} $$

□

**C.2.** **Bargmann transform on a Euclidean vector space (E, g).** Let E be a vector space with a Euclidean metric g. We have the canonical identification

$$ T^*E = E \oplus E^*, $$

with the **canonical symplectic form** $\Omega$ on $E \oplus E^*$ given by: for $(x_1, \xi_1), (x_2, \xi_2) \in E \oplus E^*$,

$$ (C.4) \qquad \Omega((x_1, \xi_1), (x_2, \xi_2)) = \langle \xi_1 | x_2 \rangle - \langle \xi_2 | x_1 \rangle. $$

We denote $dx$ the density on $E$ induced by $g$.

**Definition C.3.** A Gaussian wave packet in vertical gauge (V) with parameters $(x, \xi) \in E \oplus E^*$, is the function $\varphi_{x, \xi}^{(V)} \in S(E)$ given by: for $y \in E$,

$$ (C.5) \qquad \varphi_{x, \xi}^{(V)}(y) := \pi^{-\frac{n}{4}} e^{i\langle\xi|y-x\rangle} e^{-\frac{1}{2}\|y-x\|^2_g}. $$

Similarly a Gaussian wave packet in radial gauge (R) is

$$ (C.6) \qquad \varphi_{x, \xi}^{(R)}(y) := e^{\frac{i}{2}\langle\xi|x\rangle} \varphi_{x, \xi}^{(V)}(y) = \pi^{-\frac{n}{4}} e^{i\langle\xi|y-\frac{1}{2}x\rangle} e^{-\frac{1}{2}\|y-x\|^2_g}. $$

*Remark C.4.* We have

* The terms 'vertical gauge' and 'radial gauge' come from a geometrical construction of these waves packets using prequantization that we do not describe here. But we can explain the terminology as follows. The wave packet is seen as a map $\varphi^{(V)}: (x, \xi) \in E \oplus E^* \to \mathcal{H} = L^2(E)$ that induces a map to the projective space $[\varphi^{(V)}]: E \oplus E^* \to \mathbb{P}(\mathcal{H})$ (i.e. ignoring phase) and a pull back line bundle $L \to E \oplus E^*$ with Hermitian connection (Levi Civita). The map $\varphi^{(V)}$ can be seen as a trivialization of this line bundle $L$. Its covariant derivative is given by $D\varphi^{(V)} = Pd\varphi^{(V)} = \eta^{(V)}\varphi^{(V)}$ [43, (1.5) p.540], with projector onto the fiber $P = \varphi^{(V)}\langle\varphi^{(V)}|,\rangle$ hence connection one-form

$$ \eta^{(V)} = \langle \varphi^{(V)} | d\varphi^{(V)} \rangle_{(C.5)} = -i\xi dx. $$
We observe that the kernel of this one-form is along $x$ = cste on phase space $E \oplus E^*$, i.e. 'vertical'. Similarly for the wave packet (C.6), considered as another trivialization, we compute the connection one-form

$$ \eta^{(R)} = \langle \varphi^{(R)} | d\varphi^{(R)} \rangle_{(C.6)} = \frac{i}{2} (-\xi dx + x d\xi), $$

for which the kernel at point $(x, \xi)$ contains the 'radial' vector $(x, \xi)$.

• $\|\varphi_{x,\xi}^{(V)}\|_{L^2(E,dx)} = \|\varphi_{x,\xi}^{(R)}\|_{L^2(E,dx)} = 1$ with $dx$ the density on $E$ associated to the metric g.

• We have

$$ (C.7) \qquad \varphi_{x,\xi}^{(V)} \Big|_{(C.1)} = \hat{T}_{x,\xi} \varphi_{0,0}^{(V)}. $$

**Definition C.5.** The Bargmann transform in vertical gauge is the continuous operator

$$ (C.8) \qquad \mathcal{B}_{E,(V)} : \begin{cases} S(E) & \to S(E \oplus E^*), \\ u & \to (x, \xi) \to \langle \varphi_{x,\xi}^{(V)} | u \rangle_{L^2(E;dx)}. \end{cases} $$

**Lemma C.6.** With respect to the density $\frac{1}{(2\pi)^n} dxd\xi$ on $E \oplus E^*$, the Bargmann transform $\mathcal{B}_{E,(V)} : L^2(E;dx) \to L^2(E \oplus E^*; \frac{1}{(2\pi)^n} dxd\xi)$ is an isometry. Its $L^2$-adjoint operator $\mathcal{B}_{(V)}^\dagger : L^2(E \oplus E^*; \frac{1}{(2\pi)^n} dxd\xi) \to L^2(E;dx)$ is a smoothing operator given by

$$ (C.9) \qquad (\mathcal{B}_{(V)}^{\dagger} \nu)(y) = \int_{E \oplus E^{*}} \varphi_{x, \xi}^{(V)}(y) \nu(x, \xi) \frac{dxd\xi}{(2\pi)^n} $$

and satisfies the resolution of identity

$$ (C.10) \qquad \mathrm{Id}_{L^2(E)} = \mathcal{B}_{(V)}^{\dagger} \mathcal{B}_{(V)} = \int_{E \oplus E^{*}} \pi(x, \xi) \frac{dxd\xi}{(2\pi)^n} $$

with $\pi(x, \xi) := \varphi_{x,\xi}^{(V)} \langle \varphi_{x,\xi}^{(V)} | \cdot \rangle$ that is the rank one orthogonal projector in $L^2(E)$ onto $\mathcal{C}\varphi_{x,\xi}^{(V)}$.

*Proof.* We first check (C.9). For $u \in S(E)$ and $v \in S(E \oplus E^*)$ we have

$$ \begin{aligned} \langle u | \mathcal{B}_{(V)}^{\dagger} v \rangle_{L^2} &= \langle \mathcal{B}_{(V)} u | v \rangle_{L^2} = \int_{E \oplus E^{*}} \overline{\langle \varphi_{x, \xi}^{(V)} | u \rangle_{L^2}} v(x, \xi) \frac{dxd\xi}{(2\pi)^n} \\ &= \int_{E} \overline{u(y)} \int_{E \oplus E^{*}} \varphi_{x, \xi}^{(V)}(y) v(x, \xi) \frac{dxd\xi}{(2\pi)^n} dy. \end{aligned} $$

Now we check that $\|\mathcal{B}_{(V)}u\|^2 = \|u\|^2$ that is equivalent to (C.10). We have

$$ \begin{aligned} \|\mathcal{B}_{(V)}u\|^2 &= \int_{E \oplus E^*} |\langle \varphi_{x, \xi}^{(V)} | u \rangle_{L^2}|^2 \frac{dxd\xi}{(2\pi)^n} \\ &= \pi^{-\frac{n}{2}} \int e^{-i\langle \xi | y-x \rangle} e^{-\frac{1}{2}\|y-x\|^2} g(u) dy e^{i\langle \xi | y'-x \rangle} e^{-\frac{1}{2}\|y'-x\|^2} \overline{g(u(y'))} dy' \frac{dxd\xi}{(2\pi)^n}. \end{aligned} $$
But

$$
\int e^{i\langle\xi|y'-y\rangle} d\xi = (2\pi)^n \delta(y' - y) \\
\int_E e^{-\|y\|^2} dy = \pi^{n/2},
$$

hence

$$
\| \mathcal{B}_{(V)} u \|^2 = \pi^{-n/2} \int e^{-\|y-x\|^2_g} |u(y)|^2 dy dx = \int |u(y)|^2 dy = \|u\|^2.
$$

□

By using $\varphi_{x,\xi}^{(R)}$ in radial gauge instead of $\varphi_{x,\xi}^{(V)}$, we can define similarly operators $\mathcal{B}_{(R)}$, $\mathcal{B}_{(R)}^{\dagger}$.

Let

$$
(C.11) \qquad \mathcal{P}_{(V)} := \mathcal{B}_{(V)} \mathcal{B}_{(V)}^{\dagger} : S(E \oplus E^{*}) \to S(E \oplus E^{*}),
$$

$$
(C.12) \qquad \mathcal{P}_{(R)} := \mathcal{B}_{(R)} \mathcal{B}_{(R)}^{\dagger} : S(E \oplus E^{*}) \to S(E \oplus E^{*})
$$

be the orthogonal projector in $L^2((E \oplus E^*)^\perp)$ onto their closed images $\text{Im}(\mathcal{B}_{(V)})$, $\text{Im}(\mathcal{B}_{(R)})$
respectively, called the **Bergman projector**. Remark that Lemma C.6 holds true for
$\mathcal{B}_{(R)}$ as well.

The metric $g$ on $E$ induces a canonical metric on $E \oplus E^*$ denoted $g := g \oplus g^{-1}$.

**Lemma C.7.** The Schwartz kernels of $\mathcal{P}_{(V)}$, $\mathcal{P}_{(R)}$ are given by

$$
(C.13) \quad \langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \delta_{x, \xi} \rangle = \exp \left( \frac{i}{2} (\xi' + \xi) (x' - x) - \frac{1}{4} \|(x', \xi') - (x, \xi)\|_{g \oplus g^{-1}}^2 \right),
$$

$$
(C.14) \quad \langle \delta_{x', \xi'} | \mathcal{P}_{(R)} \delta_{x, \xi} \rangle = \exp \left( -\frac{i}{2} \Omega \left( (x', \xi'), (x, \xi) \right) - \frac{1}{4} \|(x', \xi') - (x, \xi)\|_{g \oplus g^{-1}}^2 \right).
$$

*Proof.* We have

$$
\begin{align*}
\langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \delta_{x, \xi} \rangle &= \langle \varphi_{x', \xi'}^{(V)} | \varphi_{x, \xi}^{(V)} \rangle = \pi^{-n/2} \int e^{-i\langle\xi'|y-x'\rangle} e^{-\frac{1}{2}\|y-x'\|^2_g} e^{i\langle\xi|y-x\rangle} e^{-\frac{1}{2}\|y-x\|^2_g} dy \\
&= \pi^{-n/2} e^{i\left(\frac{1}{2}\|\xi'|x'\rangle - \langle\xi|x\rangle\right) - \frac{1}{2}\|x'\|^2_g - \frac{1}{2}\|x\|^2_g} \int e^{-\|y\|^2_g + \langle b|y\rangle} dy
\end{align*}
$$

with $b = \check{g}(x' + x) + i(\xi - \xi') \in E^*$. We use the Gaussian integral

$$
(C.15) \qquad \int_E e^{-\alpha \|y\|_g^2 + \langle b | y \rangle} dy = \left(\frac{\pi}{\alpha}\right)^{\frac{n}{2}} e^{\frac{1}{4\alpha} \|b\|_g^2 - 1},
$$

with $\alpha = 1$, giving

$$
\begin{align*}
\langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \delta_{x, \xi} \rangle &= e^{i\left(\langle \xi' | x' \rangle - \langle \xi | x \rangle\right) - \frac{1}{2}\|x'\|^2 - \frac{1}{2}\|x\|^2} e^{\frac{1}{4}\left(\|x' + x\|^2 - \|\xi - \xi'\|^2 + 2i\langle \xi - \xi' | x' + x \rangle\right)} \\
&= e^{i\left(\langle \xi' | x' \rangle - \langle \xi | x \rangle + \frac{1}{2}\langle \xi - \xi' | x' + x \rangle\right)} e^{-\frac{1}{4}\left(\|x' - x\|^2 + \|\xi - \xi'\|^2\right)} \\
&= e^{i\frac{1}{2}\langle \xi' + \xi | x' - x \rangle} e^{-\frac{1}{4}\left\| (x', \xi') - (x, \xi) \right\|^2_{g+g^{-1}}}
\end{align*}
$$
We have

$$
\begin{align*}
\langle \delta_{x', \xi'} | \mathcal{P}_{(R)} \delta_{x, \xi} \rangle &= e^{i \frac{1}{2} (\langle \xi | x \rangle - \langle \xi' | x' \rangle)} \langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \delta_{x, \xi} \rangle \\
&= e^{i \frac{1}{2} (-\langle \xi' | x \rangle + \langle \xi | x' \rangle)} e^{-\frac{1}{4} \|(x', \xi') - (x, \xi)\|_{g+g-1}^2} \\
&= e^{-i \frac{1}{2} \Omega((x', \xi'), (x, \xi))} e^{-\frac{1}{4} \|(x', \xi') - (x, \xi)\|_{g+g-1}^2}.
\end{align*}
$$

□

**Lemma C.8.** Let $\pi : E \oplus E^* \to E$ denote the projector on the first component, and $\pi^\circ : S(E) \to S(E \oplus E^*)$ the pull back operator. We have

$$
\mathcal{B}_{(V)} = \pi^{\frac{n}{4}} \mathcal{P}_{(V)} \pi^{\circ}.
$$

Proof. We compute the Schwartz kernel of both sides for $y \in E, (x', \xi') \in E \oplus E^*$. We
have $\langle \delta_{x', \xi'} | \mathcal{B}_{(V)} \delta_y \rangle = \varphi^{(V)}_{x', \xi'}(y)$ and

$$
\begin{align*}
\langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \pi^\circ \delta_y \rangle &= \int e^{\frac{i}{2}(\xi' + \xi)(x' - x)} - \frac{1}{4} \left( \|x' - x\|^2 + \|\xi - \xi'\|^2 \right) \delta(x-y) \frac{dx d\xi}{(2\pi)^n} \\
&= (2\pi)^{-n} e^{\frac{i}{2}\xi'(x' - y)} - \frac{1}{4}\|x' - y\|^2 - \frac{1}{4}\|\xi'\|^2 \int e^{\frac{i}{2}\xi(x' - y) + \frac{1}{2}\xi'\xi' - \frac{1}{4}\|\xi\|^2} d\xi.
\end{align*}
$$

We use the Gaussian integral (C.15) with $\alpha = \frac{1}{4}$, $b = \frac{i}{2}(x' - y) + \frac{1}{2}\xi'$,

$$
\frac{1}{4\alpha} \|b\|^2 = \frac{1}{4} (\|\xi'\|^2 - \|x' - y\|^2) + \frac{i}{2} (x' - y) \xi',
$$

$$
\begin{align*}
\langle \delta_{x', \xi'} | \mathcal{P}_{(V)} \pi^\circ \delta_y \rangle &= (2\pi)^{-n} e^{\frac{i}{2}\xi'(x'-y)} - \frac{1}{4}\|x'-y\|^2 - \frac{1}{4}\|\xi'\|^2 \left(\frac{\pi}{\alpha}\right)^{\frac{n}{2}} e^{\frac{1}{4\alpha}\|b\|^2} \\
&= (4\pi)^{\frac{n}{2}} (2\pi)^{-n} \pi^{\frac{n}{4}} \pi^{-\frac{n}{4}} e^{i\xi'(x'-y)} - \frac{1}{2}\|x'-y\|^2 \\
&= \pi^{-\frac{n}{4}} \varphi^{(V)}_{x', \xi'}(y).
\end{align*}
$$

□

C.3. **Linear map** $\phi : (E_2, g_2) \to (E_1, g_1)$. Suppose $\phi : (E_2, g_2) \to (E_1, g_1)$ is a linear
invertible map between two metric spaces. Let

$$
\Phi := \phi^{-1} \oplus \phi^* : E_1 \oplus E_1^* \to E_2 \oplus E_2^*
$$

be the induced map (i.e. pull back of differential forms $\Phi : T^*E_1 \to T^*E_2$). We have
defined in (C.8), $\mathcal{B}_{E_1} : S(E_1) \to S(E_1 \oplus E_1^*)$ and $\mathcal{B}_{E_2} : S(E_2) \to S(E_2 \oplus E_2^*)$ where
$\mathcal{B}_{E_j}$ is either $\mathcal{B}_{(V)}$ or $\mathcal{B}_{(R)}$.

**Lemma C.9.** Suppose $\phi : (E_2, g_2) \to (E_1, g_1)$ is a linear invertible map between two Euclidean vector spaces. We have that

$$
(C.16) \qquad \phi^\circ = Y(\phi) \mathcal{B}_{E_2}^\dagger \Phi^{-\circ} \mathcal{B}_{E_1}
$$

with

$$
(C.17) \qquad Y(\phi) := \left( \det \left( \frac{1}{2} \left( \mathrm{Id} + (\phi^{-1})^{\dagger} \phi^{-1} \right) \right) \right)^{1/2},
$$
where $(\phi^{-1})^{\dagger}$ is the metric-adjoint of $\phi^{-1}$ defined by $\langle x_1 | (\phi^{-1})^{\dagger} x_2 \rangle_{g_1} = \langle \phi^{-1} x_1 | x_2 \rangle_{g_2}$ for any $x_1 \in E_1$, $x_2 \in E_2$.

*Remark C.10.* If $\phi$ is an isometry then $\phi^{\dagger}\phi = \mathrm{Id}$ and $Y(\phi) = 1$.

*Proof.* We compute the Schwartz kernels. For $y' \in E_1$, $z' \in E_2$,

$$ \langle \delta_{z'} | \mathcal{B}_{E_2}^{\dagger} \Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'} \rangle = \langle \mathcal{B}_{E_2} \delta_{z'} | \Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'} \rangle. $$

For $y \in E_1$, $\eta \in E_1^*$,

$$ (\mathcal{B}_{E_1} \delta_{y'}) (y, \eta) = \langle \varphi_{y,\eta}^{(1)} | \delta_{y'} \rangle = \pi^{-\frac{n}{4}} e^{-i\langle\eta|y'-y\rangle} e^{-\frac{1}{2}\|y'-y\|_{g_1}^2}. $$

For $z \in E_2$, $\xi \in E_2^*$,

$$ (\Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'}) (z, \xi) = \pi^{-\frac{n}{4}} e^{-i\langle\Phi^{-1}\xi|y'-\Phi z\rangle} e^{-\frac{1}{2}\|y'-\Phi z\|_{g_1}^2}, $$

$$ (\mathcal{B}_{E_2} \delta_{z'}) (z, \xi) = \pi^{-\frac{n}{4}} e^{-i\langle\xi|z'-z\rangle} e^{-\frac{1}{2}\|z'-z\|_{g_2}^2}. $$

Hence

$$ \langle \mathcal{B}_{E_2} \delta_{z'} | \Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'} \rangle = \pi^{-\frac{n}{2}} \int e^{i\langle\xi|z'-z\rangle} e^{-\frac{1}{2}\|z'-z\|_{g_2}^2} e^{-i\langle\Phi^{-1}\xi|y'-\Phi z\rangle} e^{-\frac{1}{2}\|y'-\Phi z\|_{g_1}^2} \frac{dz d\xi}{(2\pi)^n}. $$

We do the symplectic change of variables $(z, \xi) \in E_2 \oplus E_2^* \to (y, \eta) = \Phi^{-1}(z, \xi) = (\phi(z), \phi^{*-1}\xi) \in (E_1 \oplus E_1^*)$, i.e. $\xi = \phi^*\eta$, $z = \phi^{-1}y$, and get

$$ \begin{aligned} \langle \delta_{z'} | \mathcal{B}_{E_2}^{\dagger} \Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'} \rangle &= \pi^{-\frac{n}{2}} \int e^{i\langle\Phi^{*}\eta|z' - \Phi^{-1}y\rangle} e^{-\frac{1}{2}\|z' - \Phi^{-1}y\|_{g_2}^2} e^{-i\langle\eta|y' - y\rangle} e^{-\frac{1}{2}\|y' - y\|_{g_1}^2} \frac{dy d\eta}{(2\pi)^n} \\ &= \pi^{-\frac{n}{2}} \int e^{i\langle\eta|\Phi z' - y'\rangle} e^{-\frac{1}{2}\|\Phi^{-1}(\Phi z' - y)\|_{g_2}^2} e^{-\frac{1}{2}\|y' - y\|_{g_1}^2} \frac{dy d\eta}{(2\pi)^n}. \end{aligned} $$

We have

$$ \int e^{i\langle\eta|\Phi z' - y'\rangle} \frac{d\eta}{(2\pi)^n} = \delta(\Phi z' - y') = \langle\delta_{z'} | \Phi^{\circ} \delta_{y'}\rangle, $$

hence

$$ \begin{aligned} \langle \delta_{z'} | \mathcal{B}_{E_2}^{\dagger} \Phi^{-\circ} \mathcal{B}_{E_1} \delta_{y'} \rangle &= \langle \delta_{z'} | \Phi^{\circ} \delta_{y'} \rangle \pi^{-\frac{n}{2}} \int e^{-\frac{1}{2} \| \Phi^{-1}(y' - y) \|_{g_2}^2} e^{-\frac{1}{2} \| y' - y \|_{g_1}^2} dy \\ &= \langle \delta_{z'} | \Phi^{\circ} \delta_{y'} \rangle \pi^{-\frac{n}{2}} \int_F e^{-\frac{1}{2} \| \Phi^{-1} Y \|_{g_2}^2} e^{-\frac{1}{2} \| Y \|_{g_1}^2} dY \\ &= \langle \delta_{z'} | \Phi^{\circ} \delta_{y'} \rangle \pi^{-\frac{n}{2}} \int e^{-\frac{1}{2} \langle Y | (\mathrm{Id} + (\Phi^{-1})^{\dagger} \Phi^{-1}) Y \rangle} dY \\ &= \langle \delta_{z'} | \Phi^{\circ} \delta_{y'} \rangle \pi^{-\frac{n}{2}} \left( \frac{(2\pi)^n}{\det(\mathrm{Id} + (\Phi^{-1})^{\dagger} \Phi^{-1})} \right)^{1/2} \\ &= \langle \delta_{z'} | \Phi^{\circ} \delta_{y'} \rangle Y(\Phi)^{-1}, \end{aligned} $$

giving (C.16).

□
Later we will use Lemma C.11, where the determinant is measured with respect to local densities.

**Lemma C.11.** With $Y(\phi)$ defined in (C.17), we have

$$ (C.18) \qquad Y(\phi^{-1}) = |\det\phi| Y(\phi). $$

If $\phi = \phi_1 \oplus \phi_2$ on $E = E_1 \oplus E_2$, then

$$ (C.19) \qquad Y(\phi) = Y(\phi_1) Y(\phi_2). $$

*Proof.* Notice that $\det(\phi^\dagger\phi) = |\det\phi|^2$. Then

$$ \begin{aligned} Y(\phi)^2 &= \det\left(\frac{1}{2}(\mathrm{Id} + \phi^{-1}\dagger\phi^{-1})\right) \\ &= (\det(\phi^\dagger\phi))^{-1} \det\left(\frac{1}{2}(\phi^\dagger\phi + \mathrm{Id})\right) = |\det\phi|^{-2} Y(\phi^{-1})^2. \end{aligned} $$

□

**Corollary C.12.** If $E$ is endowed with two metrics $g_1, g_2$ then

$$ \mathrm{Id}_{S(E)} = Y_{g_2,g_1} \mathcal{B}_{g_2}^{\dagger g_2} \mathcal{B}_{g_1} $$

with

$$ (C.20) \qquad Y_{g_2,g_1} = \left( \det\left(\frac{1}{2}(\mathrm{Id} + \mathrm{Id}^{\dagger})\right) \right)^{1/2} = \prod_{j=1}^{n} \left(\frac{1+I_j}{2}\right)^{1/2}. $$

$(I_j)_j \in \mathbb{R}^n$ are inertial moments of $g_2$ with respect to $g_1$, i.e. in suitable coordinates, $g_1 = \sum_j dy_j^2, g_2 = \sum_j I_j dy_j^2$. Consequently

$$ \mathcal{P}_{g_1,g_2} := Y_{g_2,g_1} \mathcal{B}_{g_1} \mathcal{B}_{g_2}^{\dagger g_2} : S(E \oplus E^*) \to S(E \oplus E^*) $$

is a projector, with $\mathrm{Im}(\mathcal{P}_{g_1,g_2}) = \mathrm{Im}(\mathcal{B}_{g_1})$, $\mathrm{Ker}(\mathcal{P}_{g_1,g_2}) = (\mathrm{Im}(\mathcal{B}_{g_2}))^{\perp}$.

*Proof.* We apply Lemma C.9 with $\phi = \mathrm{Id} : (E, g_2) \to (E, g_1)$ and $\Phi = \mathrm{Id}$. We have

$$ \langle u | (\phi^{-1})^{\dagger} v \rangle_{g_1} = \langle \phi^{-1} u | v \rangle_{g_2} = \langle u | v \rangle_{g_2} $$

in coordinates such that $g_1 = \sum_j dy_j^2, g_2 = \sum_j I_j dy_j^2$,

$$ (\mathrm{Id})_{j,k}^{\dagger} = (\phi^{-1})_{j,k}^{\dagger} = I_j \delta_{j=k} $$

giving (C.20).

□

**C.4. Compatible triple** $g, \Omega, J$. Let us recall the definition of a compatible triple $g, \Omega, J$. Let $F$ be a vector space endowed with a symplectic bilinear form $\Omega$ and a Euclidean metric $g$. Let

$$ (C.21) \qquad J := \check{g}^{-1} \check{\Omega} : F \to F $$

with $\check{\Omega} : F \to F^*$, $\check{g} : F \to F^*$ defined in Section A.

**Definition C.13.** We say that $\Omega, g$ are compatible structures on $F$ if $J := \check{g}^{-1} \check{\Omega} : F \to F$ is an almost complex structure on $F$, i.e.

$$ (C.22) \qquad J^2 = -\mathrm{Id}. $$
**Proposition C.14.** *From* (C.21), we have that for any $u, v \in F$

$$
(C.23) \qquad \Omega(u, v) = g(Ju, v).
$$

If moreover (C.22) holds true, we have that for any $u, v \in F$,

$$
(C.24) \qquad g(Ju, Jv) = g(u, v), \quad \Omega(Ju, Jv) = \Omega(u, v).
$$

*Proof.* We have

$$
g(Ju, v) \underset{(C.21)}{=} \langle \check{g} \check{g}^{-1} \check{\Omega} u | v \rangle = \Omega(u, v).
$$

Suppose $J^2 = \check{g}^{-1}\check{\Omega}\check{g}^{-1}\check{\Omega} = -\text{Id}$ hence $\check{g} = -\check{\Omega}\check{g}^{-1}\check{\Omega}$. Then

$$
\begin{align*}
g(Ju, Jv) &= \langle \check{g} \check{g}^{-1} \check{\Omega} u | \check{g}^{-1} \check{\Omega} v \rangle = \langle -\check{\Omega} \check{g}^{-1} \check{\Omega} u | v \rangle = g(u, v), \\
\Omega(Ju, Jv) &= \langle \check{\Omega} \check{g}^{-1} \check{\Omega} u | \check{g}^{-1} \check{\Omega} v \rangle = \langle -\check{\Omega} \check{g}^{-1} \check{\Omega} \check{g}^{-1} \check{\Omega} u | v \rangle = \Omega(u, v).
\end{align*}
$$

□

**Example C.15.** Let $(E, g)$ be a vector space with Euclidean metric $g$. On $E \oplus E^*$, the canonical symplectic form $\Omega$ in (C.4) is compatible with the induced metric $g := g \oplus g^{-1}$.

**Lemma C.16.** Let $(F, \Omega, g)$ be a vector space with a symplectic structure $\Omega$ and compatible metric $g$. If $E \subset F$ is a linear Lagrangian subspace, let $E^{\perp g}$ be the orthogonal subspace with respect to $g$. Then $E^{\perp g} = J(E)$ is also Lagrangian and $E = J(E^{\perp g})$. Hence we have an isomorphism $\Psi$ that is symplectic (for $\Omega$) and isometric (for $g$)

$$
(C.25) \qquad \Psi : \begin{cases} F = E \oplus E^{\perp g} & \to E \oplus E^*, \\ (v, u) & \mapsto (v, \check{\Omega}u), \end{cases}
$$

where $E \oplus E^*$ is endowed with the canonical symplectic form (C.4) and with the metric
$g = g \oplus g^{-1}$.

*Proof.* Let $v_1, v_2 \in E$ Lagrangian. Then

$$
g(Jv_1, v_2) \underset{(C.23)}{=} \Omega(Jv_1, Jv_2) \underset{(C.24)}{=} \Omega(v_1, v_2) \underset{\text{Lag.}}{=} 0.
$$

Considering dimensions, $\dim E = \dim E^{\perp g} = \frac{1}{2}\dim F$, we deduce $E^{\perp g} = J(E)$ and $E = J(E^{\perp g})$, since $J^2 = -\text{Id}$. From compatibility $J$ is a symplectic map hence $E^{\perp g}$ is Lagrangian as well. $\square$

C.5. **Bergman projector on a symplectic vector space** $(F, \Omega, g)$. Let $(F, \Omega, g)$ be a vector space with a symplectic structure $\Omega$ and compatible metric $g$.

**Definition C.17.** Let $\mathcal{P}_F : S(F) \to S(F)$ be the operator defined by its Schwartz kernel: for any $\rho, \rho' \in F$,

$$
(C.26) \qquad \langle \delta_{\rho'} | \mathcal{P}_F \delta_{\rho} \rangle = \exp \left( -\frac{i}{2} \Omega(\rho', \rho) - \frac{1}{4} \|\rho' - \rho\|_g^2 \right).
$$

**Lemma C.18.** $\mathcal{P}_F$ is an orthogonal projector in $L^2(F, \frac{d\rho}{(2\pi)^n})$, called the **Bergman projector**.

*Proof.* We choose any $E \subset F$ Lagrangian linear subspace. Then $\mathcal{P}_F = \Psi^\circ \circ \mathcal{P}_{E \oplus E^*, (R)} \circ \Psi^{-\circ}$ is conjugated to the Bergman projector in radial gauge defined in (C.14). $\square$
### C.6. The metaplectic decomposition of $F \oplus F^*$.
Let $(F, \Omega)$ be a linear symplectic space. Let us denote $\Omega$ the canonical symplectic form on $F \oplus F^*$ as in (C.4). Recall $\check{\Omega} : F \to F^*$ defined in Section A and that $\tilde{\Omega}^* = -\check{\Omega}$.

**Lemma C.19 (Metaplectic decomposition).** Let $(F, \Omega)$ be a linear symplectic space and

$$ (C.27) \qquad K := \operatorname{graph}(\check{\Omega}), \quad N := \operatorname{graph}(-\check{\Omega}). $$

We have

$$ (C.28) \qquad F \oplus F^* = K^{\perp \Omega} \oplus N, $$

where the right hand side is an $\Omega$-orthogonal decomposition into linear symplectic subspaces. See Figure C.1. Explicitly for any $(x, \xi) \in F \oplus F^*$, $(x, \xi) = (\nu, \check{\Omega}(\nu)) + (\zeta, -\check{\Omega}(\zeta)) \in K \oplus N$ is given by

$$ (C.29) \qquad \nu = \frac{1}{2} (x + \check{\Omega}^{-1}(\xi)), $$

$$ (C.30) \qquad \zeta = \frac{1}{2} (x - \check{\Omega}^{-1}(\xi)). $$

<figure><img src="image_26.png" /><figcaption>FIGURE C.1. Picture for the orthogonal decomposition $F \oplus F^* = K^{\perp} \oplus N$ in (C.28)</figcaption></figure>

**Remark C.20.** The metaplectic decomposition (C.28) is central in this paper. It has also been used in [13, Prop.6] and in [18, Prop.2.2.9].

**Proof.** If $(x, \xi) = (x, \check{\Omega}(x)) \in \operatorname{graph}(\check{\Omega})$ and $(x', \xi') = (x', -\check{\Omega}(x')) \in \operatorname{graph}(-\check{\Omega})$ then

$$ \begin{aligned} \Omega((x, \xi), (x', \xi')) &= \langle \xi | x' \rangle - \langle x | \xi' \rangle = \langle \check{\Omega}(x) | x' \rangle + \langle x | \check{\Omega}(x') \rangle \\ &= \Omega(x, x') + \Omega(x', x) = 0. \end{aligned} $$

Moreover if $(x, \xi) = (x, \check{\Omega}(x)) = (x, -\check{\Omega}(x)) \in \operatorname{graph}(\check{\Omega}) \cap \operatorname{graph}(-\check{\Omega})$ then $\check{\Omega}(x) = -\check{\Omega}(x)$ hence $\check{\Omega}(x) = 0$ hence $x = 0$. This orthogonality and transversality implies that $\operatorname{graph}(\check{\Omega})$, $\operatorname{graph}(-\check{\Omega})$ are symplectic. Let us prove (C.29). We write (C.31)

$$ \left\{ \begin{array}{l} x = \nu + \zeta, \\ \xi = \check{\Omega}(\nu) - \check{\Omega}(\zeta) = \check{\Omega}(\nu - \zeta), \end{array} \right. \iff \left\{ \begin{array}{l} \nu = \frac{1}{2}(x + \check{\Omega}^{-1}(\xi)), \\ \zeta = \frac{1}{2}(x - \check{\Omega}^{-1}(\xi)). \end{array} \right. \qquad \square $$

<figure><img src="image_27.png" /></figure>
**Lemma C.21.** Let $(F, \Omega, g)$ be a vector space with a symplectic structure $\Omega$ and a compatible metric $g$. Then (C.28) is orthogonal for the metric $g = g \oplus g^{-1}$ as well. The maps

$$ (C.32) \qquad \sqrt{2\pi_K} : (x, \breve{\Omega}(x)) \in (K, \Omega, g) \to x \in (F, \Omega, g), $$

$$ (C.33) \qquad \sqrt{2\pi_N} : (x, \breve{\Omega}^*(x)) \in (N, \Omega, g) \to x \in (F, -\Omega, g) $$

are isomorphism for the respective bilinear forms.

*Proof.* $K \oplus N$ is also orthogonal for the metric $g$ because

$$ \begin{aligned} g((x, \xi), (x', \xi')) &= g(x, x') + g^{-1}(\xi, \xi') = g(x, x') + g^{-1}(\breve{\Omega}(x), -\breve{\Omega}(x')) \\ &= g(x, x') - g^{-1}(\breve{\Omega}(x), \breve{\Omega}(x')) = 0 \end{aligned} $$

because $g^{-1}(\breve{\Omega}(x), \breve{\Omega}(x')) = g(x, x')$ from compatibility between $\Omega$ and $g$.

If $(x, \xi) = (x, \breve{\Omega}(x)) \in K$ and $(x', \xi') = (x', \breve{\Omega}(x')) \in K$ then

$$ \begin{aligned} \Omega((x, \xi), (x', \xi')) &= \langle \xi | x' \rangle - \langle x | \xi' \rangle = \langle \breve{\Omega}(x) | x' \rangle - \langle x | \breve{\Omega}(x') \rangle = \langle \breve{\Omega}(x) | x' \rangle - \langle \breve{\Omega}^*(x) | x' \rangle \\ &= \langle \breve{\Omega}(x) | x' \rangle + \langle \breve{\Omega}(x) | x' \rangle = 2\Omega(x, x'), \end{aligned} $$

$$ g((x, \xi), (x', \xi')) = g(x, x') + g^{-1}(\xi, \xi') = g(x, x') + g^{-1}(\breve{\Omega}(x), \breve{\Omega}(x')) = 2g(x, x'), $$

and a similar computation on $N$, if $\breve{\Omega}$ is replaced by $-\breve{\Omega}$. $\square$

Let $\mathcal{B}_F : S(F) \to S(F \oplus F^*)$ the Bargman transform in radial Gauge. Let $\mathcal{P}_{F \oplus F^*} := \mathcal{B}_F \mathcal{B}_F^\dagger$. Let $\mathcal{P}_K, \mathcal{P}_N$ defined as in (C.26) for the respective spaces $(K, \Omega_K, g_K)$, $(N, \Omega_N, g_N)$ defined in (C.27) with metrics induced from $\Omega, g$. From (C.28) we have the natural identification

$$ S(F \oplus F^*) = S(K \oplus N) = S(K) \otimes S(N). $$

**Lemma C.22.** We have

$$ (C.34) \qquad \mathcal{P}_{F \oplus F^*} = \mathcal{P}_K \otimes \mathcal{P}_N. $$

*Proof.* Since $\Omega_{F \oplus F^*} = \Omega_K \oplus \Omega_N$ and $g = g_K \oplus g_N$, (C.34) follows from the expression of the kernels (C.14) and (C.26) that coincide for both sides. $\square$

**Lemma C.23.** From the linear maps $\pi_K, \pi_N$ in (C.32), we have $L^2$-unitary operators

$$ (C.35) \qquad \begin{aligned} U_K &:= (\sqrt{2\pi_K})^\circ : \operatorname{Im}(\mathcal{P}_F) \to \operatorname{Im}(\mathcal{P}_K), \\ U_N &:= (\sqrt{2\pi_N})^\circ : \operatorname{Im}(C\mathcal{P}_F C) \to \operatorname{Im}(\mathcal{P}_N), \end{aligned} $$

where $C : u \in S(F) \to \bar{u} \in S(F)$ is the conjugation operator.

*Proof.* This is a direct consequence of Lemma C.21 and expression (C.26). $\square$

### C.7. Metaplectic decomposition of a linear symplectic map.

$$ \Phi : (F_1, \Omega_1) \to (F_2, \Omega_2) $$

a linear invertible symplectic map and $g_j$ a compatible metric on $F_j$ for $j = 1, 2$.

Let us first give a definition of metaplectic operator that will be useful. The metaplectic correction $Y(\Phi) > 0$ has been defined in (C.17). The reader may consult [45, section 13.2] for a general reference about metaplectic operators.
**Definition C.24.**

• We define the **metaplectic operator (or quantization) of** $\Phi$ by

$$ (C.36) \qquad \tilde{\text{Op}}(\Phi) := (Y(\Phi))^{1/2} \mathcal{P}_{F_2} \Phi^{-\circ} \mathcal{P}_{F_1} : \text{Im}(\mathcal{P}_{F_1}) \to \text{Im}(\mathcal{P}_{F_2}) $$

with the metaplectic correction $Y(\Phi) > 0$ defined in (C.17).

• If $E_1 \subset F_1$, $E_2 \subset F_2$ are Lagrangian linear subspaces, and using the identification (C.25), we define

$$ (C.37) \qquad \text{Op}(\Phi) := (Y(\Phi))^{1/2} \mathcal{B}_{E_2}^{\dagger} \Phi^{-\circ} \mathcal{B}_{E_1} : S(E_1) \to S(E_2) $$

that is conjugated to $\tilde{\text{Op}}(\Phi)_{/\text{Im}\mathcal{B}_{E_1} \to \text{Im}\mathcal{B}_{E_2}}$ since

$$ (C.38) \qquad \tilde{\text{Op}}(\Phi) = \mathcal{B}_{E_2} \text{Op}(\Phi) \mathcal{B}_{E_1}^{\dagger}. $$

*Remark C.25.* We will show in Proposition C.29 that $\tilde{\text{Op}}(\Phi) : L^2(F_1) \to L^2(F_2)$ is $L^2$-unitary (hence $\text{Op}(\Phi) : L^2(E_1) \to L^2(E_2)$ is also unitary).

Let

$$ (C.39) \qquad \tilde{\Phi} := \Phi^{-1} \oplus \Phi^* : F_2 \oplus F_2^* \to F_1 \oplus F_1^* $$

the induced map on cotangent spaces. Since $\Phi$ is symplectic we have that $\tilde{\Phi}$ preserves the decompositions (C.28), $F_j \oplus F_j^* = K_j \oplus N_j$, with $j = 1, 2$ and we denote its components by

$$ (C.40) \qquad \tilde{\Phi} = \Phi_K \oplus \Phi_N. $$

Beware that $\Phi_K, \Phi_N$ are conjugated to $\Phi^{-1}$, i.e. $\Phi^{-1}\pi_{N_2} = \pi_{N_1}\Phi_N : N_2 \to F_1$ and $\Phi^{-1}\pi_{K_2} = \pi_{K_1}\Phi_K : K_2 \to F_1$.

**Theorem C.26** ([13], [18]). We have

$$ (C.41) \qquad \mathcal{B}_{F_1} \Phi^{\circ} \mathcal{B}_{F_2}^{\dagger} = \tilde{\text{Op}}(\tilde{\Phi}) = \tilde{\text{Op}}(\Phi_K) \otimes \tilde{\text{Op}}(\Phi_N) : \text{Im}(\mathcal{P}_{F_2 \oplus F_2^*}) \to \text{Im}(\mathcal{P}_{F_1 \oplus F_1^*}). $$

*Proof.* We have

$$ (C.42) \qquad Y(\tilde{\Phi})_{(C.39),(C.19)} = Y(\Phi^{-1}) Y(\Phi^*)_{(C.18)} = (Y(\Phi))^2. $$

Hence

$$ \begin{aligned} \mathcal{B}_{F_1} \Phi^{\circ} \mathcal{B}_{F_2}^{\dagger} &= Y(\Phi) \mathcal{P}_{F_1 \oplus F_1^*} \tilde{\Phi}^{-\circ} \mathcal{P}_{F_2 \oplus F_2^*} \\ &= Y(\Phi) (Y(\tilde{\Phi}))^{-1/2} \tilde{\text{Op}}(\tilde{\Phi})_{(C.36)} = \tilde{\text{Op}}(\tilde{\Phi})_{(C.42)}, \end{aligned} $$

and

$$ \begin{aligned} \tilde{\text{Op}}(\tilde{\Phi})_{(C.36)} &= (Y(\tilde{\Phi}))^{1/2} \mathcal{P}_{F_2 \oplus F_2^*} \tilde{\Phi}^{-\circ} \mathcal{P}_{F_1 \oplus F_1^*} \\ &= (Y(\Phi_K) Y(\Phi_N))^{1/2} (\mathcal{P}_{K_1}^{\dagger} \Phi_K \tilde{\Phi}^{-\circ} \mathcal{P}_{K_2}) \otimes (\mathcal{P}_{N_1}^{\dagger} \Phi_N \tilde{\Phi}^{-\circ} \mathcal{P}_{N_2}) \\ &= \tilde{\text{Op}}(\Phi_K) \otimes \tilde{\text{Op}}(\Phi_N)_{(C.36)}. \end{aligned} $$

□

*Remark C.27.* Theorem C.26 is central to the analysis in this paper. It is a factorization formula for the map $\Phi$ and somehow gives a square root of $\Phi^{\circ}$.
**Proposition C.28.** For any linear symplectic maps $\Phi_{2,1} : (F_1, \Omega_1) \to (F_2, \Omega_2)$, $\Phi_{3,2} : (F_2, \Omega_2) \to (F_3, \Omega_3)$ with compatible metrics $g_j$ on $F_j$, $j = 1, 2, 3$, we have

$$ (C.43) \qquad \tilde{\text{Op}}(\Phi_{3,2}\Phi_{2,1}) = \tilde{\text{Op}}(\Phi_{3,2})\tilde{\text{Op}}(\Phi_{2,1}). $$

*Proof.* We have

$$ \begin{align*} \tilde{\text{Op}}((\Phi_{3,2}\Phi_{2,1})_K) \otimes \tilde{\text{Op}}((\Phi_{3,2}\Phi_{2,1})_N) &\stackrel{(C.41)}{=} \mathcal{B}_{F_3} (\Phi_{3,2}\Phi_{2,1})^{-\circ} \mathcal{B}_{F_1}^{\dagger} \\ &= \mathcal{B}_{F_3} \Phi_{3,2}^{-\circ} \Phi_{2,1}^{-\circ} \mathcal{B}_{F_1}^{\dagger} = \mathcal{B}_{F_3} \Phi_{3,2}^{-\circ} \mathcal{B}_{F_2}^{\dagger} \mathcal{B}_{F_2} \Phi_{2,1}^{-\circ} \mathcal{B}_{F_1}^{\dagger} \\ &\stackrel{(C.41)}{=} \mathcal{B}_{F_3} (\Phi_{3,2})_K \mathcal{B}_{F_2} (\Phi_{2,1})_K \otimes \mathcal{B}_{F_3} (\Phi_{3,2})_N \mathcal{B}_{F_2} (\Phi_{2,1})_N. \end{align*} $$

Considering each factor gives (C.43). $\square$

**Proposition C.29.** The operator $\tilde{\text{Op}}(\Phi) : \text{Im}(\mathcal{P}_{F_1}) \to \text{Im}(\mathcal{P}_{F_2})$ defined in (C.36) is $L^2$-unitary.

*Remark C.30.* Since $\mathcal{B}_{E_j} : L^2(E_j) \to \text{Im}\mathcal{B}_{E_j} = \text{Im}\mathcal{P}_{F_j}$ is unitary this implies that $\text{Op}(\Phi) : L^2(E_1) \to L^2(E_2)$ is also unitary.

*Proof.* We have $Y(\text{Id}) = 1$ hence $\tilde{\text{Op}}(\Phi)(\text{Id}) = \mathcal{P} = \text{Id}_{/\text{Im}\mathcal{P}}$. Also

$$ (C.44) \qquad Y(\Phi^{-1}) \stackrel{(C.18)}{=} |\det\Phi| Y(\Phi) = Y(\Phi). $$

We have

$$ \tilde{\text{Op}}(\Phi^{-1}) \tilde{\text{Op}}(\Phi) \stackrel{(C.43)}{=} \tilde{\text{Op}}(\text{Id}) = \text{Id}_{/\text{Im}\mathcal{P}}, $$

so $\tilde{\text{Op}}(\Phi)^{-1} = \tilde{\text{Op}}(\Phi^{-1})$ and

$$ \begin{align*} \tilde{\text{Op}}(\Phi)^{\dagger} &\stackrel{(C.36)}{=} (Y(\Phi))^{1/2} \mathcal{P}\Phi^{\circ}\mathcal{P} \stackrel{(C.44)}{=} (Y(\Phi^{-1}))^{1/2} \mathcal{P}(\Phi^{-1})^{-\circ}\mathcal{P} \\ &\stackrel{(C.36)}{=} \tilde{\text{Op}}(\Phi^{-1}) = \tilde{\text{Op}}(\Phi)^{-1}. \end{align*} $$

Lemma C.31 considers a special case for the symplectic map $\Phi$.

**Lemma C.31** ([18, proof of Prop 4.3.1 p.79]). Let $(E_1, g_1), (E_2, g_2)$ be Euclidean vector spaces, and $\phi : E_2 \to E_1$ an invertible linear map. We denote $\phi^{\circ} : S(E_1) \to S(E_2)$ the pull-back operator and $\Phi := \phi^{-1} \oplus \phi^{*} : E_1 \oplus E_1^{*} \to E_2 \oplus E_2^{*}$ the induced symplectic map on cotangent spaces. $\text{Op}(\Phi)$ has been defined in (C.37) and $\tilde{\text{Op}}(\Phi)$ in (C.36). We have

$$ (C.45) \qquad \text{Op}(\Phi) = |\det\phi|^{1/2} \phi^{\circ}. $$

$$ (C.46) \qquad \tilde{\text{Op}}(\Phi) = |\det\phi|^{1/2} \mathcal{B}_{E_2}\phi^{\circ}\mathcal{B}_{E_1}^{\dagger}. $$

*Proof.* We have

$$ (C.47) \qquad Y(\Phi) \stackrel{(C.19)}{=} Y(\phi)Y(\phi^{-1}) \stackrel{(C.18)}{=} |\det\phi|(Y(\phi))^{2}. $$

Then

$$ \text{Op}(\Phi) \stackrel{(C.37)}{=} (Y(\Phi))^{1/2} \mathcal{B}_{E_2}^{\dagger} \Phi^{-\circ} \mathcal{B}_{E_1} \stackrel{(C.16)}{=} (Y(\Phi))^{1/2} (Y(\phi))^{-1} \phi^{\circ} \stackrel{(C.47)}{=} |\det\phi|^{1/2} \phi^{\circ}. \square $$
### C.8. Symplectic spinors.
The previous constructions correspond to what is sometimes called **symplectic spinors** in the literature. We summarize what we have obtained and comment on this.

We start from a linear symplectic space $(F, \Omega)$. We have the decomposition $F \oplus F^* = K \oplus_{\Omega} N$ in (C.28). We take an $\Omega$-compatible metric $g$ on $F$, so that we can define $\mathcal{B}_F : L^2(F) \to \text{Im}\mathcal{B}_F \subset L^2(F \oplus F^*)$ in radial Gauge from (C.8), but also the operators $\mathcal{P}_F, \mathcal{P}_{F \oplus F^*}, \mathcal{P}_K, \mathcal{P}_N$.

Recall from Lemma C.23, we have $L^2$-unitary isomorphisms

$$U_K : \text{Im}\mathcal{P}_F \to \text{Im}\mathcal{P}_K, \quad U_N : \text{Im}(\mathcal{C}\mathcal{P}_F\mathcal{C}) \to \text{Im}\mathcal{P}_N,$$

and from (C.34), we have a $L^2$-unitary isomorphism

$$\mathcal{B}_F : L^2(F) \to \text{Im}\mathcal{P}_K \otimes \text{Im}\mathcal{P}_N.$$

This suggests to give Definition C.32 and Theorem C.33.

**Definition C.32.** For a linear symplectic space $(F, \Omega)$ with $\Omega$-compatible metric $g$, we define the space of **symplectic spinors** $\text{Spin}_{\pm}(F)$ by

$$\text{Spin}_{+}(F) := \text{Im}\mathcal{P}_{F} \subset L^{2}(F), \quad \text{Spin}_{-}(F) := \text{Im}(\mathcal{C}\mathcal{P}_{F}\mathcal{C}) \subset L^{2}(F).$$

We define the $L^2$-unitary operator

$$(C.48) \qquad U_{\text{spin}} := (U_K^\dagger \otimes U_N^\dagger) \circ \mathcal{B}_F : L^2(F) \to \text{Spin}_+(F) \otimes \text{Spin}_-(F).$$

**Theorem C.33.** Let $\Phi : (F_1, \Omega_1) \to (F_2, \Omega_2)$ a linear invertible symplectic map. We have the commutative diagram with unitary operators

$$(C.49) \qquad \begin{tikzcd} L^2(F_1) \arrow[r, "\Phi^{-\circ}"] \arrow[d, "U_{\text{spin}}"] & L^2(F_2) \arrow[d, "U_{\text{spin}}"] \\ \text{Spin}_+(F_1) \otimes \text{Spin}_-(F_1) \arrow[r, "\mathcal{O}\text{p}(\Phi) \otimes (\mathcal{C}\mathcal{O}\text{p}(\Phi)\mathcal{C})"] & \text{Spin}_+(F_2) \otimes \text{Spin}_-(F_2) \arrow[d, "U_{\text{spin}}"] \end{tikzcd}$$

*Remark C.34.* The operator $\Phi^{-\circ}$ in the upper line is the push forward of the symplectic map and can be interpreted as the “classical dynamics”. The lower line is the product of two metaplectic operators or quantum operators. So this diagram shows that

* In Eq. (C.48), the classical space $L^2(F)$ is identified as the product of two quantum spinor spaces.

* In Eq. (C.49), the “classical dynamics” can be factorized as the product of the quantum dynamics times its complex conjugation.

*Proof.* Since $\Phi\pi_{N_1} = \pi_{N_2}\Phi_N^{-1} : N_1 \to F_2$, we have $\Phi^\circ = U_{N_1}^\dagger\Phi_N^{-\circ}U_{N_2} = U_{K_1}^\dagger\Phi_K^{-\circ}U_{K_2} : L^2(F_2) \to L^2(F_1)$. From Lemma C.23, we have

$$\mathcal{C}\mathcal{P}_F\mathcal{C} = U_N^\dagger\mathcal{P}_N U_N.$$

Hence

$$\begin{align*} U_1^\dagger \mathcal{O} \text{p}(\Phi_N) U_{N_2} &\stackrel{(C.36)}{=} (Y(\Phi))^{1/2} U_{N_1}^\dagger \mathcal{P}_{N_1} \Phi_N^{-\circ} \mathcal{P}_{N_2} U_{N_2} \\ &= (Y(\Phi))^{1/2} U_{N_1}^\dagger \mathcal{P}_{N_1} U_{N_1} \Phi^\circ U_{N_2}^\dagger \mathcal{P}_{N_2} U_{N_2} = (Y(\Phi))^{1/2} \mathcal{C}\mathcal{P}_{F_1} \mathcal{C}\Phi^\circ \mathcal{C}\mathcal{P}_{F_2} \mathcal{C} \\ &\stackrel{(C.44)}{=} \mathcal{C}(Y(\Phi^{-1}))^{1/2} \mathcal{P}_{F_1} \Phi^\circ \mathcal{P}_{F_2} \mathcal{C} = \mathcal{C}\mathcal{O}\text{p}(\Phi^{-1}) \mathcal{C} \end{align*}$$
and similarly $U_{K_1}^{\dagger} \tilde{\text{Op}}(\Phi_K) U_{K_2} = \tilde{\text{Op}}(\Phi^{-1})$. So we get

$$
\begin{align*}
\Phi^{\circ}_{(C.41)} &= B_{F_1}^{\dagger} (\tilde{\text{Op}}(\Phi_K) \otimes \tilde{\text{Op}}(\Phi_N)) B_{F_2} : L^2(F_2) \to L^2(F_1) \\
&= B_{F_1}^{\dagger} \left( (U_{K_1} \tilde{\text{Op}}(\Phi^{-1}) U_{K_2}^{\dagger}) \otimes (U_{N_1} c \tilde{\text{Op}}(\Phi^{-1}) c U_{N_2}^{\dagger}) \right) B_{F_2} \\
&= U_{\text{spin}}^{\dagger} (\tilde{\text{Op}}(\Phi^{-1}) \otimes (c \tilde{\text{Op}}(\Phi^{-1}) c)) U_{\text{spin}}
\end{align*}
$$

giving (C.49).

□

### C.9. Some useful decompositions.
Here we give a relation that is simple here in the linear setting and is used in the main text as an approximation, in the proof of Theorem 3.15. Let us denote $\tilde{F} = F \oplus F^* = K \oplus N$ and $\exp : T\tilde{F} \to \tilde{F}$ the exponential map. Now we restrict the base to $K \subset \tilde{F}$ and consider $N \subset T_K \tilde{F}$ as a sub-bundle $N \to K$ of $T_K \tilde{F} \to K$. We denote $\exp_N : N \to \tilde{F}$ the exponential map that is an isomorphism.

As before, we have the operators $\mathcal{B}_F : S(F) \to S(\tilde{F})$, $\mathcal{B}_F^\dagger : S(\tilde{F}) \to S(F)$, $\widetilde{\exp_N^\circ} : S(\tilde{F}) \to S(N)$, $(\widetilde{\exp_N^{-1}})^\circ : S(N) \to S(\tilde{F})$.

Let $\Phi : F \to F$ be a linear symplectic map. We have $\Phi^\circ : S(F) \to S(F)$. The map $\tilde{\Phi} : \tilde{F} \to \tilde{F}$ that can be seen as the bundle map $\tilde{\Phi}_N : N \to N$ over $\Phi_K$. Then $\tilde{\text{Op}}(\tilde{\Phi}_N) : S(N) \to S(N)$ is a bundle map over $\Phi_K$. The metaplectic correction $(Y(\Phi_K))^{1/2} > 0$ has been defined in (C.17) and in fact we have $Y(\Phi_K) = Y(\Phi_N) = Y(\Phi)$.

**Lemma C.35.** We have

$$
(C.50) \qquad \Phi^{\circ} = \mathcal{B}_{F}^{\dagger} (\widetilde{\exp_{N}^{-1}})^{\circ} (Y(\Phi_{K}))^{1/2} \tilde{\text{Op}}(\tilde{\Phi}_{N}) \widetilde{\exp_{N}^{\circ}} \mathcal{B}_{F}.
$$

Equivalently

$$
(C.51) \qquad \tilde{\text{Op}}(\tilde{\Phi})_{(C.41)} = \mathcal{P}_F(\widetilde{\exp_N^{-1}})^\circ (Y(\Phi_K))^{1/2} \tilde{\text{Op}}(\tilde{\Phi}_N) \widetilde{\exp_N^\circ} \mathcal{P}_F.
$$

**Remark C.36.** As the proof below shows, the metaplectic correction $(Y(\Phi_K))^{1/2} > 0$ in (C.50) is due to the fact that $\tilde{\text{Op}}(\tilde{\Phi}_N)$ over $\Phi_K$ does not preserve the space $\text{Im}\mathcal{P}_K$.

*Proof.* We have

$$
\begin{align*}
\Phi^{\circ}_{(C.10)} &= \mathcal{B}_{F}^{\dagger} \mathcal{B}_{F} \Phi^{\circ} \mathcal{B}_{F}^{\dagger} \mathcal{B}_{F}_{(C.41)} = \mathcal{B}_{F}^{\dagger} (\tilde{\text{Op}}(\Phi_{K}) \otimes \tilde{\text{Op}}(\Phi_{N})) \mathcal{B}_{F} \\
&= \mathcal{B}_{F}^{\dagger} \left( \left( (Y(\Phi_{K}))^{1/2} \mathcal{P}_{K} \Phi_{K}^{-\circ} \mathcal{P}_{K} \right) \otimes \tilde{\text{Op}}(\Phi_{N}) \right) \mathcal{B}_{F}.
\end{align*}
$$

We use that $\mathcal{B}_F = \mathcal{P}_F \mathcal{B}_F = (\mathcal{P}_K \otimes \mathcal{P}_N) \mathcal{B}_F$ and similarly $\mathcal{B}_F^\dagger = \mathcal{B}_F^\dagger (\mathcal{P}_K \otimes \mathcal{P}_N)$. Hence we can remove the operators $\mathcal{P}_K$ and get

$$
\begin{align*}
\Phi^{\circ} &= \mathcal{B}_{F}^{\dagger} (Y(\Phi_{K}))^{1/2} \Phi_{K}^{-\circ} \otimes \tilde{\text{Op}}(\Phi_{N}) \mathcal{B}_{F} \\
&= (Y(\Phi_{K}))^{1/2} \mathcal{B}_{F}^{\dagger} \widetilde{\exp_{N}^{-1}} \tilde{\text{Op}}(\tilde{\Phi}_{N}) \widetilde{\exp_{N}^{\circ}} \mathcal{B}_{F},
\end{align*}
$$

where the last line expresses that $\tilde{\text{Op}}(\tilde{\Phi}_N) : S(N) \to S(N)$ is a bundle map over $\Phi_K$. Using (C.41) and (C.12) we deduce (C.51). □
In Lemma C.37, we use the (twisted) operator $\widetilde{\exp}^\circ : S(\tilde{F}) \to S(T\tilde{F})$. We also use the restriction operators $r_0 : S(T\tilde{F}) \to S(\tilde{F})$, $r_{/K} : S(T\tilde{F}) \to S(T_K\tilde{F})$, $r_N : S(T_K\tilde{F}) \to S(N)$. We have

$$ \mathrm{Id}_{S(\tilde{F})} = r_0 \widetilde{\exp}^\circ = (\widetilde{\exp_N^{-1}})^\circ r_N r_{/K} \widetilde{\exp}^\circ, $$

where the second equality can be seen as a generalization of the first equality for the respective decompositions $\tilde{F} = \{0\} \oplus \tilde{F} = K \oplus N$. More generally we have Lemma C.37.

**Lemma C.37.** For the symplectic map $\tilde{\Phi} = \Phi_K \oplus \Phi_N : \tilde{F} \to \tilde{F}$, using $\tilde{\mathrm{Op}}(\Phi)$ : $S(T\tilde{F}) \to S(T\tilde{F})$ and $\tilde{\mathrm{Op}}_{/K}(\Phi) : S(T_K\tilde{F}) \to S(T_K\tilde{F})$ as bundle map operators, we have the decomposition

$$ (C.52) \qquad r_0 \tilde{\mathrm{Op}}(\Phi) \widetilde{\exp}^\circ = (\widetilde{\exp_N^{-1}})^\circ r_N \tilde{\mathrm{Op}}_{/K}(\Phi) r_{/K} \widetilde{\exp}^\circ. $$

**C.10. Taylor operators** $T_k$ on $S(E)$. Let $E$ a Euclidean vector space, $n = \dim E$. If $(e_1, \dots, e_n)$ is a basis of $E$, we write $x = \sum_{i=1}^n x_i e_i \in E$. A polynomial on $E$ is $P(x) := p(x_1, \dots, x_n)$ with $p \in \mathbb{C}[x_1, \dots, x_n]$ a polynomial on $\mathbb{R}^n$. For $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^n$, we write $x^\alpha := x_1^{\alpha_1} \dots x_n^{\alpha_n}$ a monomial of degree $|\alpha| := \alpha_1 + \dots + \alpha_n$. For $k \in \mathbb{N}$, we denote $\mathrm{Pol}_k(E)$ the space of homogeneous polynomials of degree $k$, that is independent on the basis. $P \in \mathrm{Pol}_k(E)$ can be written $P(x) = \sum_{\alpha \in \mathbb{N}^k, |\alpha|=k} P_\alpha x^\alpha$ with components $P_\alpha \in \mathbb{C}$. We denote $\alpha! := \alpha_1! \dots \alpha_n!$ and $\delta^{(\alpha)} := \delta_0^{(\alpha_1)}(x_1) \dots \delta_0^{(\alpha_n)}(x_n)$ the Dirac distribution on $E$ (with $\alpha$-derivatives). We have that for any $\alpha, \alpha' \in \mathbb{N}^n$, $\langle \frac{1}{\alpha'!} \delta_0^{(\alpha')}|x^\alpha\rangle = \delta_{\alpha'=\alpha}$. Hence the set $(x^\alpha)_\alpha$ forms a basis of $\mathrm{Pol}(E)$ and $((\delta^{(\alpha)}|.))_\alpha$ is the dual basis. See Section 9 for comments about the equivalence $\mathrm{Pol}_k(E) \equiv \mathrm{Sym}((E^*)^{\otimes k})$.

**Definition C.38.** For $k \in \mathbb{N}$, let

$$ T_k : S'(E) \to \mathrm{Pol}_k(E) \subset S'(E) $$

be the projector onto $\mathrm{Pol}_k(E)$ with kernel $\oplus_{k' \neq k} \mathrm{Pol}_k(E)$. We have

$$ \forall k, k', \quad T_{k'} T_k = T_k \delta_{k'=k}, $$

and explicitly, with respect to a basis of $E$,

$$ (C.53) \qquad T_k = \sum_{\alpha \in \mathbb{N}^k, |\alpha|=k} x^\alpha \langle \frac{1}{\alpha!} \delta_0^{(\alpha)} | . \rangle. $$

Lemma C.39 shows how this relation is changed by a small norm operator if we insert a truncation at large distance $\sigma$ from the origin. We put $\chi_\sigma(\rho) = 1$ for $|\rho|_g \le \sigma$, $\chi_\sigma(\rho) = 0$ for $|\rho|_g > \sigma$ and $\mathrm{Op}(\chi_\sigma) := \mathcal{B}^\dagger \mathcal{M}_{\chi_\sigma} \mathcal{B}$.

**Lemma C.39.** For $k, k' \in \mathbb{N}$, we have $\forall N, \exists C_N > 0, \forall \sigma > 0$,

$$ R_\sigma := \mathrm{Op}(\chi_\sigma)(T_{k'} \mathrm{Op}(\chi_\sigma) T_k - T_k \delta_{k'=k}) \mathrm{Op}(\chi_\sigma) $$

satisfies

$$ |\langle \delta_{\rho'} | \mathcal{B}^\dagger R_\sigma \mathcal{B} \delta_\rho \rangle| \le \langle \mathrm{dist}_{g/\sigma^2}(\rho', \rho) \rangle^{-N} C_N \sigma^{-N} \langle \mathrm{dist}_{g/\sigma^2}(\rho, 0) \rangle^{-N}. $$

Consequently $\|R_\sigma\| \le C_N \sigma^{-N}$ for any $N > 0$, with $C_N > 0$.
Proof. We have

(C.54)

$$ \langle \delta_{\rho'} | \mathcal{B} \mathcal{R} \delta_{\rho} \rangle_{(C.53)} = \chi_{\sigma}(\rho') \chi_{\sigma}(\rho) $$

$$ (C.55) \qquad \sum_{\alpha' \in \mathbb{N}^k, |\alpha'| = k', \alpha \in \mathbb{N}^k, |\alpha| = k,} \langle \varphi_{\rho'} | x^{\alpha'} \rangle \left( \left\langle \frac{1}{\alpha'!} \delta_0^{(\alpha')} | \mathrm{Op}(\chi_{\sigma}) x^{\alpha} \rangle - \delta_{\alpha'} = \alpha \right\rangle \left\langle \frac{1}{\alpha!} \delta_0^{(\alpha)} | \varphi_{\rho} \rangle \right. $$

We have $\forall N, \exists C_N, \forall \sigma > 0$,

$$ \left| \left\langle \frac{1}{\alpha'!} \delta_0^{(\alpha')} | \mathcal{B}^{\dagger} \mathcal{M}_{\chi_{\sigma}} \mathcal{B} x^{\alpha} \rangle - \delta_{\alpha'} = \alpha \right| \le C_N \sigma^{-N}, $$

and due to truncations $\chi_{\sigma}(\rho') \chi_{\sigma}(\rho)$ in (C.54) we deduce that

$$ |\langle \delta_{\rho'} | \mathcal{B}^{\dagger} \mathcal{R} \delta_{\rho} \rangle| \le \langle \mathrm{dist}_{g/\sigma^2}(\rho', \rho) \rangle^{-N} C_N \sigma^{-N} \langle \mathrm{dist}_{g/\sigma^2}(\rho, 0) \rangle^{-N}, $$

and $||R_{\sigma}|| \le C_N \sigma^{-N}$ from example (2.37). $\square$

<figure><img src="image_28.png" /></figure>

C.11. **Analysis on $T(E \oplus E^*)$.** In this section we “lift” the analysis from $E \oplus E^*$ to $T(E \oplus E^*)$. This might seem a bit artificial for vector spaces, but it will be useful in this paper as a linearized model for manifolds.

C.11.1. **Bargman transform.** Let $(E, g)$ be a linear Euclidean space. Let $F := E \oplus E^*$ with induced metric $g = g \oplus g^{-1}$. One has $TF = T(E \oplus E^*) = (E \oplus E^*) \oplus (E \oplus E^*)$. The exponential map is

$$ \exp : \begin{cases} TF & \to F, \\ (\rho, \rho') & \to \rho + \rho'. \end{cases} $$

Let us define the “twisted pull back operator” $\widetilde{\exp}^o : S(F) \to S(TF)$ by its Schwartz kernel as follows. For $(\rho_1, \rho'_1) \in TF, \rho_2 \in F, \rho_1 = (x_1, \xi_1), \rho'_1 = (x'_1, \xi'_1)$,

$$ \langle \delta_{\rho_1, \rho'_1} | \widetilde{\exp}^o \delta_{\rho_2} \rangle = \delta_{\rho_2 - (\rho_1 + \rho'_1)} e^{-i\xi_1 x'_1}. $$

We define the “restriction” operator $r_0 : S(TF) \to S(F)$ by its Schwartz kernel

$$ \langle \delta_{\rho_2} | r_0 \delta_{\rho_1, \rho'_1} \rangle = \delta_{\rho_2 - \rho_1} \delta_{\rho'_1}. $$

Let the “horizontal space” be

$$ (C.56) \qquad H := F \oplus E. $$

If $\mathcal{B} : S(E) \to S(F)$ is the Bargman transform defined in (C.8), we define the partial Bargman transform

$$ B : S(H) \to S(TF) $$

by $\mathcal{B}(u(\rho) \otimes v(x')) = u(\rho) \otimes (\mathcal{B}v)(x')$, i.e. action on the second part only.

Let $\sigma > 0$ and let us introduce the cutoff function that truncates at distance $\sigma$ in the fiber $F$:

$$ \chi_{\sigma} : \begin{cases} S(TF) & \to S'(TF), \\ u(\rho, \rho') & \to \begin{cases} u(\rho, \rho') & \text{if } \|\rho'\|_g \le \sigma, \\ 0 & \text{otherwise.} \end{cases} \end{cases} $$

Let

$$ B_{\chi}^{\Delta} := B^{\dagger} \chi_{\sigma} \widetilde{\exp}^{\circ} : S(F) \to S(H), $$

$$ B := r_0 B : S(H) \to S(F). $$
C.11.2. Linear map. Let $A : E \to E$ a linear invertible map and $\tilde{A} := A^{-1} \oplus A^*$ the induced map on $F := E \oplus E^*$. Let $A^\circ : S(E) \to S(E)$ the pull back operator and $A_H := \tilde{A}^{-1} \oplus A : H \to H$ the induced map on $H$ in (C.56).

**Lemma C.40.** Let us consider the difference operator

$$R := BA_H^\circ B_\chi^\Delta - BA^\circ B^\dagger : S(F) \to S(F).$$

Then $\forall N > 0, \exists C_N > 0$ such that for any $\sigma > 0, \rho, \rho'$,

$$|\langle\delta_{\rho'}|R\delta_{\rho}\rangle| \le C_N \langle|\rho - \tilde{A}^{-1}\rho'|\rangle^{-N} \sigma^{-N}.$$

*Remark C.41.* In particular, for $\sigma = \infty$, i.e. no cutoff, then $R = 0$. Another interesting particular case is $A = \text{Id}$.

*Proof.* We compute the Schwartz kernel.

$$ \begin{aligned} \langle \delta_{\rho'} | BA_H^\circ B_\chi^\Delta \delta_\rho \rangle &= \langle \delta_{\rho'} | r_0 BA_H^\circ B^\dagger \chi_\sigma \widetilde{\exp}^\circ \delta_\rho \rangle \\ &= \int \langle \delta_\rho | r_0 \delta_{\rho_1, \rho_1'} \rangle \langle \delta_{\rho_1, \rho_1'} | BA_H^\circ B^\dagger \delta_{\rho_2, \rho_2'} \rangle \chi_\sigma(\rho_2') \langle \delta_{\rho_2, \rho_2'} | \widetilde{\exp}^\circ \delta_\rho \rangle \\ &= \int (\delta_{\rho'-\rho_1} \delta_{\rho_1'}) (\delta_{\tilde{A}\rho_2-\rho_1} \langle \varphi_{\rho_1'} | A^\circ \varphi_{\rho_2'} \rangle) \chi_\sigma(\rho_2') (\delta_{\rho-(\rho_2+\rho_2')} e^{-i\xi_2 x_2'}) \\ &= \langle \varphi_0 | A^\circ \varphi_{\rho-\tilde{A}^{-1}\rho'} \rangle e^{-i((A^*)^{-1}\xi')(x-Ax')} \chi_\sigma(\rho-\tilde{A}^{-1}\rho') \end{aligned} $$

because Dirac measures gave $\rho_2' = \rho - \rho_2 = \rho - \tilde{A}^{-1}\rho'$, $\xi_2 = (A^*)^{-1} \xi'$ and $x_2' = x - Ax'$.
Then

$$ \begin{aligned} \langle \delta_{\rho'} | BA_H^\circ B_\chi^\Delta \delta_\rho \rangle &= \langle \varphi_0 | A^\circ \varphi_{\rho-\tilde{A}^{-1}\rho'} \rangle e^{i\xi'(x'-A^{-1}x')} \chi_\sigma(\rho-\tilde{A}^{-1}\rho') \\ &= \langle \varphi_0 | A^\circ T_{x-Ax'}^{-\circ} (\mathcal{F} T_{\tilde{\xi}-A^{-1}\xi'}^{-\circ}, \mathcal{F}) \varphi_0 \rangle e^{i\xi'(x'-A^{-1}x')} \chi_\sigma(\rho-\tilde{A}^{-1}\rho') \\ &= \langle \varphi_0 | A^\circ (\mathcal{F} T_{-A^{-1}\xi'}^{-\circ}, \mathcal{F}) T_{-Ax'}^{-\circ} T_x^{-\circ} (\mathcal{F} T_{\tilde{\xi}}^{-\circ}, \mathcal{F}) \varphi_0 \rangle \\ &= e^{-i(-A^{-1}\xi')x} e^{i\xi'(-A^{-1}x')} \chi_\sigma(\rho-\tilde{A}^{-1}\rho') \\ &= \langle \varphi_0 | (\mathcal{F} T_{-\tilde{\xi}}^{-\circ}, \mathcal{F}) T_{-x'}^{-\circ} A^\circ T_x^{-\circ} (\mathcal{F} T_{\tilde{\xi}}^{-\circ}, \mathcal{F}) \varphi_0 \rangle \chi_\sigma(\rho-\tilde{A}^{-1}\rho') \\ &= \langle \varphi_{\rho'} | A^\circ \varphi_\rho \rangle \chi_\sigma(\rho-\tilde{A}^{-1}\rho') = \langle \delta_{\rho'} | BA^\circ B^\dagger \delta_\rho \rangle \chi_\sigma(\rho-\tilde{A}^{-1}\rho'). \end{aligned} $$

$$ \text{If } |\rho - \tilde{A}^{-1}\rho'| \ge \sigma \text{ then } \langle \delta_{\rho'} | R\delta_{\rho} \rangle = 0 \text{ otherwise} \\ \forall N > 0, \exists C_N > 0, \quad |\langle \varphi_0 | A^\circ \varphi_{\rho-\tilde{A}^{-1}\rho'} \rangle| \le C_A C_N \sigma^{-N}. \quad \square $$

C.11.3. Taylor projectors. For $\rho \in F$, $\hat{T}_\rho$ has been defined in (C.1). Let us define

$$T_H^{(k)} := \hat{T}_{-\rho} T_k \hat{T}_{\rho} \quad : S(H) \to S(H),$$

where $\rho \in F$ denotes the first variable and the operators act on the second variable $x' \in E$ only.

**Lemma C.42.** Let us consider the difference operator

$$R := BT_H^{(k)} B_\chi^\Delta - BT_k B^* : S(F) \to S(F).$$

Then for $||\rho - \rho'|| \le \sigma$ we have $\langle \delta_{\rho'} | R \delta_{\rho} \rangle = 0$.
*Remark C.43.* In particular for $\sigma = \infty$, i.e. no cutoff, then $R = 0$.

*Proof.* We repeat the lines of proof of Lemma C.40.

$$
\begin{align*}
\langle \delta_{\rho'}, |BT_H^{(k)} B_{\chi}^{\Delta} \delta_{\rho} \rangle &= \langle \delta_{\rho'}, |r_0 BT_H^{(k)} B^* \chi_{\sigma} \widetilde{\exp}^{\circ} \delta_{\rho} \rangle \\
&= \langle \varphi_0 | \hat{T}_{-\rho'} T_k \hat{T}_{\rho'} \varphi_{\rho-\rho'} \rangle e^{i\xi'(x'-x)} \chi_{\sigma}(\rho-\rho') \\
&= \underset{(C.3)}{\langle \varphi_0 | \hat{T}_{\rho'}^{\dagger} e^{-i\xi'x'} T_k \hat{T}_{\rho'} \varphi_{\rho-\rho'} \rangle} e^{i\xi'(x'-x)} \chi_{\sigma}(\rho-\rho') \\
&= \langle \varphi_{\rho'} | T_k \varphi_{\rho} \rangle \chi_{\sigma}(\rho-\rho') = \langle \delta_{\rho'}, |BT_k B^{\dagger} \delta_{\rho} \rangle \chi_{\sigma}(\rho-\rho').
\end{align*}
$$

□

APPENDIX D. LINEAR CONTRACTING MAPS

Let $(E, g)$ a finite dimensional vector space with Euclidean metric $g$. In this section
we consider a **linear** invertible and **contracting map** $\phi : (E, g) \to (E, g)$, i.e.

$$
(D.1) \qquad \lambda_{\pm} := \lim_{t \to \pm\infty} \log \left\| \phi^t \right\|^{1/t}
$$

satisfy<fnref n="15" /> $\lambda_- \le \lambda_+ < 0$. Recall that $\mathrm{Op}(\Phi) := \frac{(Y(\Phi))^{1/2} \mathcal{B}^\dagger \Phi^{-\circ} \mathcal{B}}{\mathcal{B} : S(E) \to S(E \oplus E^*)}$ with $\Phi := \phi^{-1} \oplus \phi^* : E \oplus E^* \to E \oplus E^*$ being the induced map on cotangent space and $\mathcal{B} : S(E) \to S(E \oplus E^*)$ defined in (C.8). The pull back operator $\phi^\circ : S(E) \to S(E)$ has been defined in (A.4). The purpose of this section is to study the spectrum of the following operator

$$
(D.2) \qquad \mathrm{Op}(\Phi) \underset{(C.45)}{=} |\det\phi|^{1/2} \phi^\circ \quad : S(E) \to S(E)
$$

on an adequate Hilbert space that contains $S(E)$ (notice that $\mathrm{Op}(\Phi)$ is unitary in $L^2(E)$ and has essential spectrum on the unit circle).

For $k \in \mathbb{N}$, consider $T_k$ the finite rank projector defined in (C.53). The vector space $\mathrm{Im}(T_k) \subset S'(E)$ is finite dimensional. Since $\phi$ is a linear map, we have for any $k \in \mathbb{N}$,

$$
(D.3) \qquad [\phi^\circ, T_k] = 0,
$$

hence $\mathrm{Op}(\Phi) : \mathrm{Im}(T_k) \to \mathrm{Im}(T_k)$ is invariant and has finite rank (we can compute
explicitly its spectrum from the spectrum of $\phi$). Let

$$
(D.4) \qquad \gamma_k^{\pm} := \lim_{t \to \pm\infty} \log \left\| \mathrm{Op}(\Phi^t) \bigg|_{\mathrm{Im}(T_k)} \right\|^{1/t}.
$$

We can<fnref n="16" /> compute $\gamma_k^{\pm}$ from the eigenvalues of $\phi$, see Remark [18, rem. 3.4.7]. As in (1.7), we have

$$
(D.5) \qquad \left(\frac{d}{2} + k\right) \lambda_{-} \le \gamma_{-} \le \gamma_{+} \le \left(\frac{d}{2} + k\right) \lambda_{+},
$$

with $d = \dim E$. For every $k \in \mathbb{N}$, we have $\gamma_k^- \le \gamma_k^+ \le \gamma_{k+1}^+ \le \gamma_k^+$. The spectrum of $\mathrm{Op}(\Phi^t) : \mathrm{Im}(T_k) \to \mathrm{Im}(T_k)$ is discrete and contained in the annulus $\{z \in \mathbb{C}, e^{t\gamma_k^-} \le$

<footnote n="15">¹⁵Equivalently $\lambda_- \le \lambda_+ < 0$ are the log of the minimal/maximal modulus of eigenvalues of $\phi$ given by its Jordan decomposition.</footnote>

<footnote n="16">¹⁶In the simple $1 - \dim$ case $\phi(x) = e^{\lambda} x$ on $\mathbb{R}$, with $\lambda_{\pm} = \lambda < 0$, we get $\gamma_k^+ = \gamma_k^- = (\frac{1}{2} + k)\lambda$.</footnote>
$|z| \le e^{t\gamma_k^+}$. However we want to understand the action of $\text{Op}(\Phi)$ on every function in $S(E)$. For $K \in \mathbb{N}$, let

$$T_{\ge (K+1)} := \text{Id}_{S(E)} - \left( \sum_{k=0}^{K} T_k \right).$$

We will show in Proposition D.3 that $\|\text{Op}(\Phi^t) T_{\ge (K+1)}\|_{\mathcal{H}_W(E)} \le C_\epsilon e^{(\gamma_{K+1}^+ + \epsilon)t}$ for some adequate norm $\|\cdot\|_{\mathcal{H}_W(E)}$ that we first define.

D.1. **Anisotropic Sobolev space $\mathcal{H}_W(E)$**. Let $0 < \gamma < 1$, $R \in \mathbb{R}$. We define a weight function $\mathcal{W}: E \oplus E^* \to \mathbb{R}^+$ similar to (4.49) as follows. For $\rho = (x, \xi) \in E \oplus E^*$, let

$$h_{\gamma}(\rho) = \langle \|\rho\|_{g \oplus g^{-1}} \rangle^{-\gamma}$$

and

$$(D.6) \qquad \mathcal{W}(\rho) := \frac{\langle h_{\gamma}(\rho) \|\xi\|_{g^{-1}} \rangle^R}{\langle h_{\gamma}(\rho) \|x\|_g \rangle^R}.$$

**Definition D.1.** For $u \in S(E)$, we define the norm

$$(D.7) \qquad \|u\|_{\mathcal{H}_W(E)} := \|\mathcal{W}Bu\|_{L^2(E \oplus E^*)}$$

(where $\mathcal{W}$ denotes multiplication operator by $\mathcal{W}$) and the **Sobolev space**

$$(D.8) \qquad \mathcal{H}_W(E) := \overline{\{u \in S(E)\}},$$

where the completion is with the norm $\|u\|_{\mathcal{H}_W(E)}$.

**Remark D.2.**

* For $R = 0$ we have $\mathcal{W} \equiv 1$ hence $\mathcal{H}_W(E) = L^2(E)$.

* We have that $\mathcal{W}(0, \xi) \le \langle |\xi| \rangle^r$ and $\mathcal{W}(x, 0) \le \langle |x| \rangle^{-r}$ with the order $r = R(1 - \gamma)$ as in (4.20) (except for the factor $\frac{1}{2}$ that came from the metric).

For a non-linear map (Axiom A diffeomorphisms), it has been explained in [20, section 4] that it is necessary to use a metric on phase space as (2.25) with $\delta^{\perp}(\eta) = \langle |\eta| \rangle^{-\alpha^{\perp}}$ and exponent $\alpha^{\perp} \ge 1/2$ to perform the micro-local analysis and get Ruelle spectrum. In this section, we consider the special case of a linear map $\phi: E \to E$ is linear and for that reason we can use the constant Euclidean metric $g \oplus g^{-1}$ on $T^*E = E \oplus E^*$ (i.e. exponent $\alpha^{\perp} = 0$) to get the Ruelle spectrum of the operator $\text{Op}(\Phi)$ in (D.2).

D.2. **Result.** The result that we will show is

**Proposition D.3.** For any $\epsilon > 0$, $K \in \mathbb{N}$, $R$ in (D.6) large enough so that

$$\lambda_+ R (1 - \gamma) < \gamma_{K+1}^+,$$

$$\exists C_{\epsilon} > 0, \forall t \ge 0,$$

$$(D.9) \qquad \| \text{Op}(\Phi^t) T_{\ge (K+1)} \|_{\mathcal{H}_W(E)} \le C_{\epsilon} e^{(\gamma_{K+1}^+ + \epsilon)t},$$

with $\gamma_{K+1}^+$ defined in (D.4).
D.2.1. *Proof of Proposition D.3.* We have $\tilde{\text{Op}}(\Phi) = (Y(\Phi))^{1/2} \mathcal{P}\Phi^{-\circ} \mathcal{P}$ and let

$$ \tilde{\text{Op}}_{\mathcal{W}}(\Phi) := \mathcal{W}\tilde{\text{Op}}(\Phi) \mathcal{W}^{-1}. $$

From (C.38) and (D.7) we have the commutative diagram (with the notation of weighted norm $\|u\|_{L^2(E \oplus E^*; W^2)} := \int \mathcal{W}^2(x, \xi) u(x, \xi) \frac{dxd\xi}{(2\pi)^{\dim E}})$

$$ \begin{array}{ccc} \text{H}_{\mathcal{W}}(E) & \xrightarrow{\mathcal{B}} & L^2(E \oplus E^*; \mathcal{W}^2) \xrightarrow{\mathcal{W}} L^2(E \oplus E^*) \\ \downarrow \text{Op}(\Phi) & & \downarrow \tilde{\text{Op}}(\Phi) \downarrow \tilde{\text{Op}}_{\mathcal{W}}(\Phi) \\ \text{H}_{\mathcal{W}}(E) & \xrightarrow{\mathcal{B}} & L^2(E \oplus E^*; \mathcal{W}^2) \xrightarrow{\mathcal{W}} L^2(E \oplus E^*) \end{array} \quad \text{(D.10)} $$

where horizontal arrows are isometries by definition. Hence, the study of $\text{Op}(\Phi^t) : \text{H}_{\mathcal{W}}(E) \to \text{H}_{\mathcal{W}}(E)$ is equivalent to study of $\tilde{\text{Op}}_{\mathcal{W}}(\Phi^t) : L^2(E \oplus E^*) \to L^2(E \oplus E^*)$.

**Lemma D.4** ("Decay property of $\mathcal{W}$ with respect to $\Phi$"). There exists $C > 0$, such that for any $t \ge 0$, there exists $C_t$ such that for any $\rho \in E \oplus E^*$

$$ \begin{array}{l} \displaystyle \frac{\mathcal{W}(\Phi^t(\rho))}{\mathcal{W}(\rho)} \le C \\ \text{(D.11)} \qquad \qquad \qquad \qquad \le C e^{-\Lambda t} \quad \text{if } \|\rho\|_g \ge C_t \end{array} $$

with

$$ \text{(D.12)} \qquad \Lambda = -\lambda_+ R(1-\gamma) > 0. $$

*Proof.* Similar to the proof in [20, Thm 5.9]. Write $\rho = (x, \xi) \in T^*E = E \oplus E^*$. We consider different zones of $T^*E$:

(1) If $|x|, |\xi| \le 1$ then $W(\rho) \asymp 1$.

(2) If $|x| > 1$ and $|\xi| \le |x|^\gamma$ then $W(\rho) \asymp |x|^{-R(1-\gamma)}$.

(3) If $|\xi| > 1$ and $|x| \le |\xi|^\gamma$ then $W(\rho) \asymp |\xi|^{R(1-\gamma)}$.

(4) Otherwise $W(\rho) \asymp (\frac{|\xi|}{|x|})^R$.

We observe that for $t \ge 0$, $|\phi^{-t}(x)| \ge e^{-\lambda_+ t} |x|$ and $|\phi^{*t}(\xi)| \le e^{\lambda_+ t} |\xi|$ from which we deduce Lemma D.4. $\square$

**Truncation in phase space near the trapped set.** Let $\chi \in C_c^\infty(\mathbb{R}^+; [0, 1])$ such that

$$ \chi(x) = 1 \text{ if } x \le 1, $$

$$ \chi(x) = 0 \text{ if } x \ge 2. $$

Let $\sigma > 0$. For $\rho \in E \oplus E^*$ let

$$ \chi_{\sigma}(\rho) := \chi\left(\frac{\|\rho\|_{g}}{\sigma}\right). $$

**Lemma D.5.** We have $\forall R > 0, \forall \varepsilon > 0, \exists C > 0, \forall t \ge 0, \exists \sigma_t > 0, \forall \sigma > \sigma_t$,

$$ \text{(D.13)} \qquad \left\| \tilde{\text{Op}}_{\mathcal{W}}(\Phi^t) (1 - \chi_{\sigma}) \right\|_{L^2} \le C e^{(-\Lambda + \varepsilon)t}, $$

where $\Lambda$ is given in (D.12). The operator $\tilde{\text{Op}}_{\mathcal{W}}(\Phi^t) \chi_{\sigma}$ is Trace class in $L^2$. Consequently for any $t > 0$, the essential spectral radius is

$$ \text{(D.14)} \qquad r_{\text{ess.}}(\tilde{\text{Op}}_{\mathcal{W}}(\Phi^t)_{L^2 \to L^2}) \le e^{-\Lambda t}. $$
**Proof.** For (D.13), see [20, thm 5.13]. □

**Lemma D.6.** Let $k \in \mathbb{N}$. If $k + 1 < r$ in (4.20), then $T_k : \mathcal{H}_W(E) \to \mathcal{H}_W(E)$ is a bounded operator.

**Proof.** We check that in (C.53), $\|x^\alpha\|_{\mathcal{H}_W(E)} < \infty$ and $\|\delta_0^{(\alpha)}\|_{\mathcal{H}_W(E)^*} < \infty$. □

**Lemma D.7.** Let $K \in \mathbb{N}$ and $R > 0$ large enough such that $-\Lambda < \gamma_{K+1}^+$ in (D.12). For $t > 0$, the spectral radius is

$$r_{\text{spec}}((\text{Op}(\Phi^t) T_{\ge (K+1)})_{\mathcal{H}_W(E)}) = e^{t\gamma_{K+1}^+}.$$

**Remark D.8.** Lemma D.7 is similar to [18, Claim (2) in Prop. 3.4.6].

**Proof.** First, from (D.14) and because $\text{Op}(\Phi^t) T_k$ is finite rank, we have that

$$r_{\text{ess}}((\text{Op}(\Phi^t) T_{\ge (K+1)})_{\mathcal{H}_W(E)}) \le e^{-\Lambda t},$$

i.e. $\text{Op}(\Phi^t) T_{\ge (K+1)} : \mathcal{H}_W(E) \to \mathcal{H}_W(E)$ has discrete spectrum outside the disk of radius $e^{-\Lambda t}$. From Taylor-Lagrange remainder formula, for any $u, v \in S(E)$, we have

$$(D.15) \qquad |\langle v | \text{Op}(\Phi^t) T_{\ge (K+1)} u \rangle_{L^2}| \le C e^{t\gamma_{K+1}^+} \|u\|_{C^{K+1}} \|x^{K+1}v\|_{L^1},$$

i.e. correlation functions decay faster than $e^{t\gamma_{K+1}^+}$. We deduce that $\text{Op}(\Phi^t) T_{\ge (K+1)}$ has no spectrum on $|z| > e^{t\gamma_{K+1}^+}$. □

Finally Lemma D.7 is equivalent to Proposition D.3.

**D.3. Discrete Ruelle spectrum in a simple toy model.** We give here a very simple example that illustrates Appendix D and some mechanisms that play a role in this paper. (See the lecture notes [14] for further details.) On $\mathbb{R}_x$ let us consider the vector field

$$(D.16) \qquad X = -x \frac{\partial}{\partial x}$$

that gives a flow $\Phi^t(x) = e^{-t}x$ that is contracting for $t > 0$, hence we think $E_s = \mathbb{R}_x$ as a stable direction. We observe that for any $k \in \mathbb{N}$, the monomial $x^k$ is an eigenfunction of $X$ (hence of $e^{tX}$) with eigenvalue $(-k)$:

$$X x^k = (-k) x^k$$

and the spectral projector is

$$T_k = x^k \langle \frac{1}{k!} \delta^{(k)} | \cdot \rangle,$$

where $\delta^{(k)}$ is the $k$-th derivative of the Dirac distribution (notice indeed from $\langle \frac{1}{k!} \delta^{(k)} | x^{k'} \rangle = \delta_{k=k'}$ that $(\delta^{(k)})_k$ forms a dual basis to $(x^{k'})_{k'}$).

However $x^k, \delta^{(k)}$ do not belong to $L^2(\mathbb{R})$. In the Hilbert space $L^2(\mathbb{R})$ one has $X^\dagger = -X + 1 \Leftrightarrow (X - \frac{1}{2})^\dagger = -(X - \frac{1}{2})$ that implies that the spectrum of $X$ is on the vertical axis $\frac{1}{2} + i\mathbb{R}$ with some essential spectrum. Some better Hilbert space $\mathcal{H}_W(\mathbb{R})$ is constructed as in (D.8) in order to study the pull back operator $(e^{tX}u)(x) = u(\Phi^t(x)) =$
$u(e^{-t}x)$ on functions $u \in S(\mathbb{R})$. We consider the induced (pull-back) flow on the cotangent space $T^*E_s = T^*\mathbb{R}$ given by

$$ \tilde{\phi}^t(x, \xi) = (e^t x, e^{-t} \xi). $$

For $0 < h \ll 1$, $R \ge 0$ we define the “escape function” or “Lyapunov function” on $T^*\mathbb{R}$ for $\tilde{\phi}^t$ that is

$$ W(x, \xi) = \frac{\langle \sqrt{h\xi} \rangle^R}{\langle \sqrt{hx} \rangle^R}. $$

Indeed, it satisfies $\frac{W \circ \tilde{\phi}^t}{W} \le C$ everywhere and $\frac{W \circ \tilde{\phi}^t}{W} \le e^{-Rt}$ far from the “trapped set” or “non-wandering set” (0, 0). In the Hilbert space

$$ \mathcal{H}_W(\mathbb{R}) := \text{Op}(W^{-1}) L^2(\mathbb{R}), $$

for $-R + \frac{1}{2} < -k$, we have that $x^k \in \mathcal{H}_W(\mathbb{R})$ and $\|T_k\|_{\mathcal{H}_W} \le C$ is a bounded operator. See Figure D.1. For large time $t \gg 1$, the emerging behavior of $e^{tX}u$ is given by

$$ \begin{aligned} e^{tX}u &= \Pi_0 u + O_{\mathcal{H}_W}(e^{-t}) \\ &= 1 u(0) + O_{\mathcal{H}_W}(e^{-t}), \end{aligned} $$

i.e. projection onto the constant function.

<figure><img src="image_29.png" /><figcaption>FIGURE D.1. In blue, discrete spectrum of the operator (D.16), $X = -x \frac{\partial}{\partial x}$ in $\mathcal{H}_W(\mathbb{R})$. In brown, the essential spectrum moves far away if $R \to +\infty$.</figcaption></figure>

## ACKNOWLEDGMENTS

F. Faure acknowledges Claude Gignoux, Colin Guillarmou, Victor Maucout, Malik Mezzadri, Stéphane Nonnenmacher for their support, for interesting and motivating discussions. F. Faure acknowledges M.S.R.I. and organizers of the micro-local semester 2020 where a part of this work has been developed.

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