The semiclassical zeta function for geodesic flows on negatively curved manifolds

Abstract

We consider the semi-classical (or Gutzwiller–Voros) zeta functions for $C^\infty$ contact Anosov flows. Analyzing the spectra of the generators of some transfer operators associated to the flow, we prove that, for arbitrarily small $\tau >0$, its zeros are contained in the union of the $\tau$-neighborhood of the imaginary axis, $|\mathfrak {R}(s)|<\tau$, and the half-plane $\mathfrak {R}(s)<-\chi _0+\tau$, up to finitely many exceptions, where $\chi _0>0$ is the hyperbolicity exponent of the flow. Further we show that the density of the zeros along the imaginary axis satisfy an analogue of the Weyl law.

Appendices

Appendix A. Proof of Lemma 11.15

As we noted in the text after the statement of Lemma 11.15, the proof is obtained by elementary geometric estimates about diffeomorphisms with some hyperbolicity. We begin with preliminary argument. For definiteness, we assume that

($\bigstar 1$):

the condition (ii) in (7.7) holds, that is, $t\geqslant t_0$, and

($\bigstar 2$):

the condition (10.17) holds.

The cases where these conditions do not hold are treated in a parallel and simpler manner. (See Remark A.2.) Further we may and do assume

($\bigstar 3$):

$\min \{\max \{e^{|m(\mathbf {j})|}, \langle \omega (\mathbf {j})\rangle \}, \max \{e^{|m(\mathbf {j}')|}, \langle \omega (\mathbf {j}')\rangle \}\}$ is large, and

($\bigstar 4$):

$e^{\chi _{\max }\cdot 2t(\omega )}\leqslant \langle \omega \rangle ^{\theta /10}$

by choosing large constant $k_0$ in the definition of the low-frequency part and small constant $\epsilon _0>0$ in the definition of $t(\omega )$ in (7.1) respectively.

For simplicity, we set $\omega =\omega (\mathbf {j})$, $\omega '=\omega (\mathbf {j}')$, $m=m(\mathbf {j})$, $m'=m(\mathbf {j}')$. Below we consider points

$$\begin{aligned}&w''\in \mathrm {supp}\,\rho ^t_{\mathbf {j}\rightarrow \mathbf {j}'}, \end{aligned}$$

(A.1)

$$\begin{aligned}&\mathbf {p}=(w,\xi _w,\xi _z)=(q,p,y,\xi _q,\xi _p,\xi _y,\xi _z)\in \mathrm {supp}\,\Psi ^{\sigma }_{\mathbf {j}}\quad \text{ and }\end{aligned}$$

(A.2)

$$\begin{aligned}&\mathbf {p}'=(w',\xi '_w,\xi '_z)=(q',p',y',\xi '_q,\xi '_p,\xi '_y,\xi '_z)\in \mathrm {supp}\,\Psi ^{\sigma '}_{\mathbf {j}'} \end{aligned}$$

(A.3)

and estimate the quantity (10.25). By changing the coordinates by a transformation in $\mathcal {A} _2$, we may and do assume

($\bigstar 5$):

$\mathfrak {p}_{(x,z)}(w'')=0$ and $\mathfrak {p}_{(x,z)}((\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^{-1}(w''))=0$

without loss of generality. From the assumption that $\mathbf {L}^{t,\sigma \rightarrow \sigma '}_{\mathbf {j}\rightarrow \mathbf {j}'}$ is a component of $\mathbf {L}^{t,\sigma \rightarrow \sigma '}_{\mathrm {hyp},\not \hookrightarrow }$, we have $m\cdot m'\ne 0$ and hence

$$\begin{aligned} e^{\max \{|m|,|m'|\}}\geqslant e^{n_0(\omega )}\geqslant C^{-1}\langle \omega \rangle ^{\theta }. \end{aligned}$$

(A.4)

We can get the conclusion of the lemma for small $\gamma _0$ easily by using (A.4) if either

$$\begin{aligned} \langle \omega '\rangle ^{1/2}|w'\!-\!w''|\geqslant e^{\max \{|m|,|m'|\}/3}\quad \hbox {or}\quad \langle \omega \rangle ^{1/2}|w-(\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^{-1}(w'')|\geqslant e^{\max \{|m|,|m'|\}/3}. \end{aligned}$$

Therefore we will assume

($\bigstar 6$):

$\max \{\;\langle \omega '\rangle ^{1/2}|w'-w''|, \; \langle \omega \rangle ^{1/2}|w-(\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^{-1}(w'')|\;\}< e^{\max \{|m|,|m'|\}/3}$.

We prove the following claim under the additional assumptions above.

Sublemma A.1

There exists a constant $C_0>0$ such that

$$\begin{aligned}&\left| (D^*E_{\omega ,m})^{-1}\left( (\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w), ((D\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^*_{w''})^{-1} \xi _w,\xi _z) -(w',(\langle \xi '_z\rangle /\langle \xi _z\rangle )\xi '_w, \xi _z)\right) \right| \nonumber \\&\qquad > C_0^{-1} e^{(2/3)\max \{|m|,|m'|\}} \langle \omega \rangle ^{-1/2}, \end{aligned}$$

(A.5)

where $D^*E_{\omega ,m}$ is the linear map defined in (6.8).

We defer the proof of this sublemma for a while and finish the proof of Lemma 11.15. We show that w-component of the quantity on the left hand side of (A.5), i.e. $E_{\omega ,m}(\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w)-w')$, is much smaller than the right-hand side of (A.5). In the case $|m|\leqslant n_1(\omega )$, we have $D^*E_{\omega ,m}=\mathrm {Id}$ and $|\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w)-w'|$ is much smaller than the right-hand side of (A.5) from ($\bigstar 4$), ($\bigstar 5$), ($\bigstar 6$) and (A.4). In the case $|m|> n_1(\omega )$, we have $e^{|m|}\geqslant C_0^{-1}\langle \omega \rangle ^{\Theta _1}$ and hence

$$\begin{aligned} |E_{\omega ,m}(\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w)-w')|\leqslant & {} e_{\omega }(m)\cdot |\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w)-w'| \\\leqslant & {} \langle \omega \rangle ^{(1-\beta )(1/2-\theta )+4\theta }\cdot |\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w)-w'| \end{aligned}$$

is again much smaller than the right-hand side of (A.5).

Comparing Sublemma A.1 with what we proved in the last paragraph, we see

$$\begin{aligned} |((D\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^*_{w''})^{-1} (\langle \xi _z\rangle \xi _w) -\langle \xi '_z\rangle \xi '_w| > (2C'_0)^{-1} e^{(2/3)\max \{|m|,|m'|\}} \langle \omega \rangle ^{1/2}. \end{aligned}$$

(A.6)

To finish, we prove

$$\begin{aligned}&\langle \omega \rangle ^{-(1+\theta )/2} \cdot \langle \langle \omega \rangle ^{1/2-4\theta } |\xi _w|\rangle ^{-1/2}\cdot |((D\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^*_{w''})^{-1} (\langle \xi _z\rangle \xi _w) -\langle \xi '_z\rangle \xi '_w|\nonumber \\&\qquad \geqslant (C''_0)^{-1} \max \{\omega , \omega ', e^{|m|}, e^{|m'|}\}^{\gamma _0} \end{aligned}$$

(A.7)

for a small constant $\gamma _0>0$. Clearly the required estimate in Lemma 11.15 follows from this claim. In the case $\langle \omega \rangle ^{1/2-4\theta } |\xi _w|\leqslant 2$, we may neglect the second factor on the left hand side and hence obtain (A.7) immediately using (A.4) and (A.6). Thus we consider the case $\langle \omega \rangle ^{1/2-4\theta } |\xi _w|> 2$ below. In this case, we have

$$\begin{aligned} \langle \omega \rangle ^{1/2}|\xi _w|\leqslant e^{\max \{|m|,|m'|\}+2}\cdot e_{\omega }(|m|) \end{aligned}$$

from (A.2), ($\bigstar 5$) and ($\bigstar 6$). Hence it holds

$$\begin{aligned} \langle \langle \omega \rangle ^{1/2-4\theta } |\xi _w|\rangle ^{-1/2}= & {} \langle \omega \rangle ^{-1/4+2\theta } |\xi _w|^{-1/2} \\\geqslant & {} e^{-\max \{|m|,|m'|\}/2-1}\langle \omega \rangle ^{2\theta }\cdot e_{\omega }(|m|)^{-1/2} \end{aligned}$$

and the left-hand side of (A.7) is bounded from below by

$$\begin{aligned} (e C'_0)^{-1} \langle \omega \rangle ^{(3/2)\theta }\cdot e^{\max \{|m|,|m'|\}/6}\cdot e_{\omega }(|m|)^{-1/2}. \end{aligned}$$

Since $e_\omega (|m|)\leqslant e^{\mu \max \{|m|,|m'|\}}\leqslant e^{\max \{|m|,|m'|\}/20}$, we obtain (A.7) again.

Proof of Sublemma A.1

For the points $\mathbf {p}$ and $\mathbf {p}'$ in (A.2) and (A.3), we set

$$\begin{aligned}&(\hat{x},\hat{y},\hat{\xi }_x,\hat{\xi }_y, \xi _z):=(D^*E_{\omega ,m})^{-1}(\mathbf {p})=(D^*E_{\omega ,m})^{-1}(w,\xi _w, \xi _z),\end{aligned}$$

(A.8)

$$\begin{aligned}&(\hat{x}',\hat{y}',\hat{\xi }'_x,\hat{\xi }'_y, \xi _z):=(D^*E_{\omega ,m'})^{-1}(\mathbf {p}')=(D^*E_{\omega ,m'})^{-1}(w',\xi '_w, \xi _z), \end{aligned}$$

(A.9)

and also set

$$\begin{aligned}&(\tilde{x},\tilde{y},\tilde{\xi }_x,\tilde{\xi }_y, \xi _z):=(D^*E_{\omega ,m})^{-1}(\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}(w), ((D\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^*_{w''})^{-1} \xi _w,\xi _z)\nonumber \\&=(E_{\omega ,m}\circ \breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'}\circ E_{\omega ,m}^{-1}(\hat{w}), (E_{\omega ,m}^*)^{-1}\circ ((D\breve{f}^t_{\mathbf {j}\rightarrow \mathbf {j}'})^*_{w''})^{-1} \circ E_{\omega ,m}^*(\hat{\xi }_w),\xi _z). \end{aligned}$$

(A.10)

The claim of Sublemma A.1 follows if we prove

$$\begin{aligned} |(\tilde{y},\tilde{\xi }_x,\tilde{\xi }_y)-(\hat{y}',(\langle \xi '_z\rangle /\langle \xi _z\rangle )\hat{\xi }'_x,\hat{\xi }'_y)|\geqslant C_0^{-1} e^{(2/3)\max \{|m|,|m'|\}} \langle \omega \rangle ^{-1/2}. \end{aligned}$$

(A.11)

Here we note that $\langle \xi '_z\rangle /\langle \xi _z\rangle$ is bounded because of the assumption ($\bigstar 2$).

For the proof of (A.11), we investigate the conditions on the points (A.8) and (A.9) that come from the choice (A.2) and (A.3) of $\mathbf {p}$ and $\mathbf {p}'$. Then we look into the correspondence from the point (A.8) to (A.10).

From the assumptions ($\bigstar 5$) and ($\bigstar 6$), the points $\mathbf {p}$ and $\mathbf {p}'$ satisfy respectively

$$\begin{aligned} |(q,p)|<e^{\max \{|m|,|m'|\}/3}\langle \omega \rangle ^{-1/2}, \quad |(q',p')|<e^{\max \{|m|,|m'|\}/3}\langle \omega '\rangle ^{-1/2}. \end{aligned}$$

(A.12)

Recall that the function $\Psi ^{\sigma }_{\mathbf {j}}$ is defined in (6.10) using the coordinates (4.24). We let $(\zeta _q,\zeta _p,\tilde{y}, \tilde{\xi }_y)$ and $(\zeta '_q,\zeta '_p,\tilde{y}', \tilde{\xi }'_y)$ be the coordinates (4.24) for the points $\mathbf {p}$ and $\mathbf {p}'$ in (A.2) and (A.3) respectively. Then, from (A.12), we have

$$\begin{aligned} \left| |(\xi _q,\xi _p)|-2^{1/2}\langle \xi _z\rangle ^{-1/2}|(\zeta _q,\zeta _p)|\right| <Ce^{\max \{|m|,|m'|\}/3}\langle \omega \rangle ^{-1/2} \end{aligned}$$

(A.13)

and

$$\begin{aligned} \left| |(\xi '_q,\xi '_p)-2^{1/2}\langle \xi '_z\rangle ^{-1/2}|(\zeta '_q,\zeta '_p)|\right| <Ce^{\max \{|m|,|m'|\}/3}\langle \omega '\rangle ^{-1/2}. \end{aligned}$$

(A.14)

From the former estimate and the condition (A.2), we see that the point (A.8) satisfies the following conditions up to errors bounded by $Ce^{\max \{|m|,|m'|\}/3}\langle \omega \rangle ^{-1/2}$: For the distance from the origin,

$$\begin{aligned} e^{|m|-1}\langle \omega \rangle ^{-1/2}\leqslant |(\hat{\xi }_q,\hat{y},\hat{\xi }_p,\hat{\xi }_y)|\leqslant e^{|m|+1}\langle \omega \rangle ^{-1/2}\quad \text{ if } m\ne 0,\end{aligned}$$

(A.15)

$$\begin{aligned} |(\hat{\xi }_q,\hat{y},\hat{\xi }_p, \hat{\xi }_y)|\leqslant e\cdot \langle \omega \rangle ^{-1/2+\theta }\quad \text{ if } m= 0 \end{aligned}$$

(A.16)

where we write $\hat{\xi }_x=(\hat{\xi }_p,\hat{\xi }_q)$; For the direction from the origin,

$$\begin{aligned}&|(\hat{\xi }_q,\hat{y})|<2 \cdot 2^{-\sigma /2}|(\hat{\xi }_p, \hat{\xi }_y)| \quad \text{ if } m>0,\end{aligned}$$

(A.17)

$$\begin{aligned}&2\cdot 2^{\sigma /2}|(\hat{\xi }_q,\hat{y})|>|(\hat{\xi }_p, \hat{\xi }_y)| \quad \text{ if } m<0. \end{aligned}$$

(A.18)

We have the parallel estimates on the point (A.9) as a consequence of (A.3).

Next we consider the correspondence from the point (A.8) to the point (A.10). By contracting property of $f^t_{\mathbf {j}\rightarrow \mathbf {j}'}$ along the y-axis and Lemma 10.4(1), we have

$$\begin{aligned} |\tilde{y}|\leqslant e^{-\chi _0 t}|\hat{y}|+Ce_{\omega }(m)\cdot \langle \omega \rangle ^{-(1/2+3\theta )}. \end{aligned}$$

(A.19)

Recall the diffeomorphism $h^t_{\mathbf {j}\rightarrow \mathbf {j}'}$ in (10.12), which is roughly the flow $f^t_G$ viewed in the local charts without the factor $E_{\omega ,m}$. By the relation (10.11), we have

$$\begin{aligned} (\tilde{\xi }_x, \tilde{\xi }_y, \xi _z)=(E_{\omega ,m}^*)^{-1}\circ E_{\omega '}^*\circ ((Dh^t_{\mathbf {j}\rightarrow \mathbf {j}'})_{\check{w}}^{*})\circ (E_{\omega }^*)^{-1} \circ E_{\omega ,m}^*(\hat{\xi }_x,\hat{\xi }_y, \xi _z) \end{aligned}$$

(A.20)

where $\check{w}$ is chosen so that $(\check{w},z)=E_{\omega }(w'',z)$.

To proceed, let us first consider the case where $|m|\leqslant n_1(\omega )$ and $|m'|\leqslant n_1(\omega ')$. In this case, we have $e_{\omega }(m)=e_{\omega '}(m')=1$, $E_{\omega ,m}=E_{\omega ',m'}=\mathrm {Id}$ and therefore the correspondence (A.20) is given by nothing but the map $f^t_{\mathbf {j}\rightarrow \mathbf {j}'}$. Then, by Lemma 10.4, we get (A.11) by simple geometric estimates. Indeed, if we ignore

• the difference between $f^t_{\mathbf {j}\rightarrow \mathbf {j}'}$ and its linearization at the origin,

• the errors mentioned about the estimates (A.15)–(A.18) and the corresponding estimates on $(\hat{y}',\hat{\xi }'_x,\hat{\xi }'_y)$, and

• the difference between $\langle \xi '_z\rangle /\langle \xi _z\rangle$ and 1,

then the conclusion (A.11) is an easy consequence of hyperbolicity of $f^t_{\mathbf {j}\rightarrow \mathbf {j}'}$. But we can check easily that the differences above are negligible in fact.

Next we consider the case $|m|\geqslant n_2(\omega )$ and $|m'|\geqslant n_2(\omega ')$, the other extreme. In this case, we have $E_{\omega ,m}=E_{\omega }$ and $E_{\omega ',m'}=E_{\omega '}$ and therefore the correspondence (A.20) is given by the map $h^t_{\mathbf {j}\rightarrow \mathbf {j}'}$. Note that $h^t_{\mathbf {j}\rightarrow \mathbf {j}'}$ is given as iteration of the maps satisfying the conditions in Lemma 10.3 and therefore have nice hyperbolic property. (Note that Lemma 10.3 is valid only for $0\leqslant t\leqslant 2t_0$.) Then we can get the conclusion (A.11) by essentially same manner as in the previous case.

The situations in the middle, i.e. the case where either $n_1(\omega )\leqslant |m|\leqslant n_2(\omega )$ or $n_1(\omega ')\leqslant |m'|\leqslant n_2(\omega ')$ is slightly more complicated. If $||m|-|m'||\geqslant 2\chi _{\max } t$, it is easy to get the conclusion (A.11) because the ratio between $e^{|m|}$ and $e^{|m'|}$ is much larger (or smaller) than the expansion (or contraction) given by the correspondence (A.20). So we may assume $||m|-|m'||\leqslant 2\chi _{\max } t$. If we have $e_\omega (m)=e_{\omega '}(m')$ and $E_{\omega ,m}=E_{\omega ',m'}$, we can see that the correspondence (A.20) has good hyperbolic property in the same manner as in the proof of Lemma 10.4 and hence we can get the conclusion (A.20) again. But, from the slowly varying property (6.7) of $e_{\omega }(m)$, it is clear that the conclusion remains true without this assumption. $\square$

Remark A.2

In the case where the condition (i) in (7.7) holds and $0\leqslant t\leqslant t_0$, we can follow the argument given in the proof above. The only difference is that, instead of the condition $t\geqslant t_0$ that ensure enough hyperbolicity of $f^t_{G}$, we use the fact that $\sigma '< \sigma$ in the estimates. In the case where the condition (10.17) does not hold, the proof is much simpler. We can obtain the conclusion immediately unless both of $|m(\mathbf {j})|$ and $|m(\mathbf {j}')|$ satisfy $|m(\mathbf {j})|>n_2(\omega )$ and $|m(\mathbf {j}')|> n_2(\omega ')$. But, in such case, the conclusion is again easy to obtain as we mentioned in the proof above.

Appendix B. Proof of the main theorem (2): Theorem 2.3

In this section, we prove Theorem 2.3, using the propositions given in Sect. 10. Below we continue to consider the case $\mathcal {L}^t=\mathcal {L}^t_{0,0}$ as in the previous sections. But we can proceed in parallel in the case of vector-valued transfer operators $\mathcal {L}^t_{k,\ell }$ with $(k,\ell )\ne (0,0)$ by regarding them as matrices of scalar valued transfer operators, as we have noted in Remark 2.4 (See also Remark 6.9, Remark 7.15 and Remark 8.11). The argument in this section is essentially parallel to that in [5], where dynamical zeta functions for hyperbolic diffeomorphisms are considered. The main idea is to decompose the operators in consideration into two parts: a trace class part and a “upper-triangular” part whose flat trace is zero. To realize this idea, we will consider the lifted operator $\mathbf {L}^t$ rather than $\mathcal {L}^t$. Below we suppose that the operators are acting on $\mathcal {K}^r(K_0)$ or $\mathbf {K}^r$ if we do not specify otherwise.

1.1 B. 1. Analytic extension of the dynamical Fredholm determinant

The dynamical Fredholm determinant d(s) of the one-parameter group of transfer operators $\mathbb {L}=\{\mathcal {L}^t=\mathcal {L}^t_{0,0}\}$ is well-defined if the real part of s is sufficiently large. In fact, the sum in the definition (2.6) of d(s) converges absolutely if $\mathrm {Re}(s)$ is larger than the topological pressure $P_{top}:=P_{top}(f^t, -(1/2)\log |\det Df^t|_{E_u}|)$ (see [29, Theorem 4.1] for instance). Hence d(s) is a holomorphic function without zeros on $\mathrm {Re}(s)>P_{top}$. We take a constant $P\geqslant P_{top}$ such that

$$\begin{aligned} \Vert \mathcal {L}^t\Vert \leqslant C\Vert {\mathbf {L}}^{t}\Vert \leqslant C e^{P t }\quad \text{ for } \quad t\geqslant t_0 \end{aligned}$$

(B.1)

and consider the function $\log d(s)$ in the disk

$$\begin{aligned} \mathbb {D}(s_0,r_0)=\{z\in \mathbb {C}\mid |z-s_0|<r_0 \} \end{aligned}$$

(B.2)

for $s_0\in \mathbb {C}$ with $\mathrm {Re}(s_0)>P$ and $r_0:=\mathrm {Re}(s_0)+r\chi _0/4$. The n-th coefficient of the Taylor expansion of $\log d(s)$ at the center $s_0$ is

$$\begin{aligned} a_n:=\frac{1}{n!} \left( \frac{d^n}{ds^n}\log d\right) (s_0)=(-1)^{n-1}\frac{1}{n!}\int _{+0}^\infty t^{n-1} e^{-s_0t} \cdot \mathrm {Tr}^\flat \,\mathcal {L}^t dt. \end{aligned}$$

Since we have

$$\begin{aligned} \mathcal {R}(s_0)^n=\frac{1}{(n-1)!}\int _{+0}^\infty t^{n-1}e^{-s_0 t} \mathcal {L}^t dt, \end{aligned}$$

we may write the coefficient $a_n$ as

$$\begin{aligned} a_n=\frac{(-1)^{n-1}}{n}\cdot \mathrm {Tr}^\flat (\mathcal {R}(s_0)^n)\quad \text{ for } \quad n\geqslant 1. \end{aligned}$$

We are going to relate the asymptotic behavior of flat trace $\mathrm {Tr}^\flat (\mathcal {R}(s_0)^n)$ as $n\rightarrow \infty$ with the spectrum of the generator A. Precisely we prove

Proposition B.1

The spectral set of the generator A of $\mathcal {L}^t:{\widetilde{\mathcal {K}}}^{r}(K_0)\rightarrow {\widetilde{\mathcal {K}}}^{r}(K_0)$ in the disk $D(s_0,r_0)$ consists of finitely many eigenvalues $\chi _i\in \mathbb {C}$, $1\leqslant i\leqslant m$, counted with multiplicity. We have the asymptotic formula

$$\begin{aligned} \mathrm {Tr}^\flat \left( \mathcal {R}(s_0)^n\right) =\sum _{i=1}^m \frac{1}{(s_0-\chi _i)^n}+Q_n \end{aligned}$$

where the remainder term $Q_n$ satisfies

$$\begin{aligned} |Q_n|\leqslant C r_0^{-n}\quad \text{ for } n\geqslant 0 \end{aligned}$$

(B.3)

with a constant $C>0$ which may depend on $s_0$.

Theorem 2.3 is an immediate consequence of this proposition. In fact, we have

$$\begin{aligned} \log d(s_0+z)&=\log d(s_0)+\sum _{n=1}^{\infty }a_n z^n\\&=\log d(s_0)+\sum _{i=1}^m \sum _{n=1}^{\infty }\frac{(-1)^{n-1} z^n}{n (s_0-\chi _i)^n}+\sum _{n=1}^{\infty }\frac{(-1)^{n} Q_n}{n} z^n\\&=\log d(s_0)+\sum _{i=1}^m \log \left( 1+\frac{z}{s_0-\chi _i}\right) +\sum _{n=1}^{\infty }\frac{(-1)^{n} Q_n}{n} z^n \end{aligned}$$

and hence

$$\begin{aligned} d(s_0+z)= d(s_0)\cdot \frac{\prod _{i=1}^m ((s_0+z)-\chi _i)}{\prod _{i=1}^m (s_0-\chi _i)}\cdot \exp \left( \sum _{n=1}^{\infty }\frac{(-1)^{n} Q_n}{n} z^n\right) \end{aligned}$$

for $z\in \mathbb {C}$ with sufficiently small absolute value. The right-most factor on the right-hand side extends holomorphically to the disk $D(s_0,r_0)$ and has no zeros on it. So the dynamical Fredholm determinant d(s) extends to the disk $D(s_0,r_0)$ as a holomorphic function and the zeros in $D(s_0,r_0)$ are exactly $\chi _i$, $1\leqslant i\leqslant m$, counted with multiplicity. Note that this conclusion holds for any $s_0\in \mathbb {C}$ with $\mathrm {Re}(s)>P_{top}$ so that the imaginary part of $s_0$ is arbitrary. Therefore, taking $r_*>0$ so large that $r_*\chi _0/4>c$, we obtain the conclusion of Theorem 2.3.

1.2 B. 2. The flat trace of the lifted transfer operators

To proceed, we discuss about the flat trace of the lifted operators and averaging with respect to time. Suppose that $\mathbf {L}:\mathbf {K}^{r}\rightarrow \mathbf {K}^{r}$ is a bounded operator, expressed as

$$\begin{aligned} \mathbf {L}(u_{\mathbf {j}})_{\mathbf {j}\in \mathcal {J}}= \left( \sum _{\mathbf {j}\in \mathcal {J}} \mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}'} u_{\mathbf {j}} \right) _{\mathbf {j}'\in \mathcal {J}}. \end{aligned}$$

(B.4)

If the diagonal components $\mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}}:\mathbf {K}^{r}_{\mathbf {j}}\rightarrow \mathbf {K}^{r}_{\mathbf {j}}$ for $\mathbf {j}\in \mathcal {J}$ are trace class operators and if the sum of their traces converges absolutely, we set

$$\begin{aligned} \mathrm {Tr}^\flat \; \mathbf {L}:=\sum _{\mathbf {j}\in \mathcal {J}} \mathrm {Tr}\; \mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}}=\sum _{\mathbf {j}\in \mathcal {J}} \mathrm {Tr}^\flat \; \mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}} \end{aligned}$$

and call it the flat trace of the operator $\mathbf {L}:\mathbf {K}^{r}\rightarrow \mathbf {K}^{r}$.

Remark B.2

In this definition, we assume that each $\mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}}$ is a trace class operator and hence²⁹ that its trace coincides with the flat trace.

Definition B.3

An operator $\mathbf {L}$ as above is upper triangular (with respect to the index $\tilde{m}(\cdot )$) if the components $\mathbf {L}_{\mathbf {j}\rightarrow \mathbf {j}'}$ vanishes whenever $\tilde{m}(\mathbf {j}') \leqslant \tilde{m}(\mathbf {j})$. (Recall (11.15) for the definition of the index $\tilde{m}(\cdot )$.)

The next lemma is obvious from the definitions.

Lemma B.4

If $\mathbf {L}$ is upper triangular, its flat trace vanishes. If $\mathbf {L}$ and $\mathbf {L}'$ are upper triangular, so are their linear combinations $\alpha \mathbf {L}+\beta \mathbf {L}'$ and their composition $\mathbf {L}\circ \mathbf {L}'$.

Since the flat trace of $\mathcal {L}^t$ is a distribution as a function of t, it takes values against smooth functions $\varphi (t)$ with compact support. Hence, rather than evaluating the flat trace of $\mathcal {L}^t$ itself, it is natural and convenient to consider the flat trace of the integration of $\mathcal {L}^t$,

$$\begin{aligned} \mathcal {L}(\varphi ):=\int \varphi (t) \cdot {\mathcal {L}}^t dt \end{aligned}$$

against a smooth function $\varphi :\mathbb {R}_+\rightarrow \mathbb {R}$ compactly supported on the positive part of the real line. We will also consider the corresponding lifted operator

$$\begin{aligned} \mathbf {L}(\varphi ):=\int _0^\infty \varphi (t) \cdot {\mathbf {L}}^t dt=\mathbf {I}\circ \mathcal {L}(\varphi )\circ \mathbf {I}^*. \end{aligned}$$

(B.5)

Recall the decomposition of the operator $\mathbf {L}^t$,

$$\begin{aligned} \mathbf {L}^t=\mathbf {L}_{\mathrm {low}}^{t}+{\mathbf {L}}^{t}_{\mathrm {hyp}}+\mathbf {L}^{t}_{\mathrm {ctr}}= \mathbf {L}_{\mathrm {low}}^{t}+\mathbf {L}^{t}_{\mathrm {hyp},\hookrightarrow }+\mathbf {L}^{t}_{\mathrm {hyp},\not \hookrightarrow }+\mathbf {L}^{t}_{\mathrm {ctr}}, \end{aligned}$$

that we introduced in Sect. 11.

Lemma B.5

Suppose that $\mathcal {P}$ is a set of $C^\infty$ functions supported on $[0, 2]\subset \mathbb {R}$ and uniformly bounded in the $C^\infty$ sense. Then there exists a constant $C>0$ such that the following holds true: For any $\varphi \in \mathcal {P}$, the operator ${\mathbf {L}}(\varphi )\circ \mathbf {L}^t=\mathbf {L}^t\circ {\mathbf {L}}(\varphi )$ for $t_0\leqslant t\leqslant 2t_0$ is decomposed into two parts

The former $\widehat{\mathbf {L}}$ is upper triangular and satisfies $\Vert \widehat{\mathbf {L}}\Vert \leqslant C$, while the latter is a trace class operator and satisfies ; Further the operator

$$\begin{aligned} (\mathbf {L}^{t}-\mathbf {L}^{t}_{\mathrm {hyp},\hookrightarrow })\circ \mathbf {L}(\varphi )=(\mathbf {L}_{\mathrm {low}}^{t}+\mathbf {L}^{t}_{\mathrm {hyp},\not \hookrightarrow }+\mathbf {L}^{t}_{\mathrm {ctr}})\circ \mathbf {L}(\varphi )\quad \text { for }t_0\leqslant t\leqslant 2t_0 \end{aligned}$$

is a trace class operator and we have $\Vert (\mathbf {L}^{t}-\mathbf {L}^{t}_{\mathrm {hyp},\hookrightarrow })\circ \mathbf {L}(\varphi )\Vert _{\mathrm {Tr}}\leqslant C$.

Proof

The part $\widehat{\mathbf {L}}$ is upper triangular and satisfies $\Vert \widehat{\mathbf {L}}\Vert \leqslant C$ by the definition of the relation $\hookrightarrow ^t$ in Definition 11.11 and Proposition 11.13. From Lemma 11.18 and Proposition 11.13, we know that the operators $\mathbf {L}_{\mathrm {low}}^{s}$ and $\mathbf {L}^{s}_{\mathrm {hyp},\not \hookrightarrow }$ are trace class operators and their trace norms are bounded uniformly for³⁰ $s\in [t_0, 2t_0+2]$. It remains to show that $\int \varphi (s-t) \cdot \mathbf {L}^{s}_{\mathrm {ctr}} ds$ and $\mathbf {L}^{t}_{\mathrm {ctr}}\circ \mathbf {L}(\varphi )$ are trace class operators and that their traces are uniformly bounded for $t_0\leqslant t\leqslant 2t_0$ and $\varphi \in \mathcal {P}$. These claims follow if we show

$$\begin{aligned} \Vert (\mathbf {L}(\varphi )\circ \mathbf {L}^t)_{\mathbf {j}\rightarrow \mathbf {j}'}:\mathbf {K}^{r}_{\mathbf {j}}\rightarrow \mathbf {K}^{r}_{\mathbf {j}'}\Vert _{\mathrm {Tr}}\leqslant C_{\nu } \langle \omega (\mathbf {j})\rangle ^{-\nu } \langle \omega (\mathbf {j})-\omega (\mathbf {j}')\rangle ^{-\nu }\langle |m(\mathbf {j})|\rangle ^{-\nu } \end{aligned}$$

(B.6)

for $\mathbf {j},\mathbf {j}'\in \mathcal {J}$ with $m(\mathbf {j}')=0$ and $t_0\leqslant t\leqslant 2t_0$. Note that, if we apply the argument in the proofs of Lemma 10.11 and Corollary 10.12 to $(\mathbf {L}(\varphi )\circ \mathbf {L}^t)_{\mathbf {j}\rightarrow \mathbf {j}'}$, we obtain the required estimate (B.6) without the term $\langle \omega (\mathbf {j})\rangle ^{-\nu }$ on the right-hand side. To retain the term $\langle \omega (\mathbf {j})\rangle ^{-\nu }$, we make use of the additional integration with respect to time in $\mathbf {L}(\varphi )$. Since $f^{t+t'}_{\mathbf {j}\rightarrow \mathbf {j}'}(w,z)=f^{t}_{\mathbf {j}\rightarrow \mathbf {j}'}(w,z)+(0,t')$ when $|t'|$ is sufficiently small, such integration will reduce the $\mathbf {j}'$-components of the image with large $\omega (\mathbf {j}')$. Indeed, if we (additionally) apply integration by parts to that integral with respect to time using the differential operator $\mathcal {D}=(1-i\xi '_z\partial _{t})(1+|\xi '_z|^2)$ for several times, we obtain the extra factor $\langle \omega (\mathbf {j})\rangle ^{-\nu }$. $\square$

Corollary B.6

If $\varphi :[t_0, \infty )\rightarrow \mathbb {R}$ be a smooth function with compact support, we have $\mathrm {Tr}^\flat \mathbf {L}(\varphi )=\mathrm {Tr}^\flat \mathcal {L}(\varphi )$.

Proof

From the proof of Lemma B.5 above, we see that $\mathrm {Tr}^\flat \mathbf {L}(\varphi )$ is well-defined, that is, the sum over $\mathbf {j}\in \mathcal {J}$ in the definition converges absolutely. Hence we obtain

$$\begin{aligned} \mathrm {Tr}^\flat \mathbf {L}^t= \mathrm {Tr}^\flat (\mathbf {I}^{\sigma }\circ \mathcal {L}^t\circ (\mathbf {I}^{\sigma })^*) = \mathrm {Tr}^\flat (\mathcal {L}^t\circ (\mathbf {I}^{\sigma })^* \circ \mathbf {I}^{\sigma })=\mathrm {Tr}^\flat \mathcal {L}^t \end{aligned}$$

by rotating the order of composition in the middle. $\square$

1.3 B.3. The flat trace of the iteration of the resolvent

Let us put

$$\begin{aligned}&{\mathcal {R}}^{(n)}= \int _0^\infty (1-\chi (t/(2t_0)))\cdot \frac{t^{n-1} e^{-ts_0}}{(n-1)!}\cdot \mathcal {L}^t\, dt \end{aligned}$$

(B.7)

and

$$\begin{aligned}&{\mathbf {R}}^{(n)}= \int _0^\infty (1-\chi (t/(2t_0)))\cdot \frac{t^{n-1} e^{-ts_0}}{(n-1)!}\cdot \mathbf {L}^t\, dt \end{aligned}$$

(B.8)

where the function $\chi (\cdot )$ is that in (5.2).

Remark B.7

The operator ${\mathcal {R}}^{(n)}$ above is defined as an approximation of ${\mathcal {R}}(s_0)^{n}$ and the difference is

$$\begin{aligned} \widetilde{\mathcal {R}}^{(n)}:={\mathcal {R}}(s_0)^{n}-{\mathcal {R}}^{(n)}= \int _0^\infty \chi (t/(2t_0))\cdot \frac{t^{n-1}\cdot e^{-ts_0}}{(n-1)!}\cdot \mathcal {L}^t dt. \end{aligned}$$

(B.9)

We put the part $\widetilde{\mathcal {R}}^{(n)}$ aside because we can not treat the operators $\mathcal {L}^t$ with small $t>0$ in the same way as those with large $t>0$. Since the flat trace and also the operator norm of $\widetilde{\mathcal {R}}^{(n)}$ on ${\widetilde{\mathcal {K}}}^{r}(K_0)$ converges to zero super-exponentially fast as $n\rightarrow \infty$, this does not cause any essential problem, though it introduces some complication in a few places below.

We take constants $r'_0<r''_0$ such that

$$\begin{aligned} r_0=\mathrm {Re}(s_0)+(1/4)r\chi _0<r'_0<r''_0<\mathrm {Re}(s_0)+(1/2)r\chi _0. \end{aligned}$$

Lemma B.8

There exists a constant $C>0$, independent of n, such that the operator ${\mathbf {R}}^{(n)}$ is expressed as a sum and

• (1) is a trace class operator, while

• (2) $\widehat{\mathbf {R}}^{(n)}:\mathbf {K}^{r}\rightarrow \mathbf {K}^{r}$ is upper triangular and satisfies

  $$\begin{aligned} \Vert \widehat{\mathbf {R}}^{(n)}\Vert \leqslant C (r''_0)^{-n}. \end{aligned}$$

Proof

Using the periodic partition of unity $\{q_\omega \}_{\omega \in \mathbb {Z}}$ defined in (6.1), we set³¹

$$\begin{aligned} q_{\omega }^{(n)}(t)=q_{\omega }(t)\cdot (1-\chi (t/(2t_0))\cdot \frac{t^{n-1}e^{-s_0 t}}{(n-1)!}, \end{aligned}$$

so that $\sum _{\omega \geqslant [2t_0]-1} q_{\omega }^{(n)}(t)= (1-\chi (t/(2t_0))\cdot (t^{n-1}e^{-s_0 t})/(n-1)!$ and that

$$\begin{aligned} {\mathbf {R}}^{(n)}=\sum _{\omega =[2t_0]-1}^\infty \mathbf {L}(q_{\omega }^{(n)}) \end{aligned}$$

from the definition (B.5). We deduce the claims of the lemma from $\square$

Claim 1

For arbitrarily small $\tau >0$, there exists a constant $C>0$, independent of n, such that the operators $\mathbf {L}(q_{\omega }^{(n)})$ for $n\geqslant 1$ are decomposed as

where $\widehat{\mathbf {L}}(q_{\omega }^{(n)})$ are upper triangular and satisfy

$$\begin{aligned} \Vert \widehat{\mathbf {L}}(q_{\omega }^{(n)})\Vert \leqslant C \frac{\omega ^{n-1}}{(n-1)!}\cdot e^{-(\mathrm {Re}(s_0) +(1/2)r \chi _0)\omega +\tau n}, \end{aligned}$$

(B.10)

while are trace class operators satisfying

(B.11)

From the claim above, we set

Then the first claim (1) of the lemma follows from (B.11). The second claim (2) also follows because $\widehat{\mathbf {R}}^{(n)}$ is upper triangular from Lemma B.4 and because

$$\begin{aligned} \Vert \widehat{\mathbf {R}}^{(n)}\Vert&\leqslant \sum _{\omega =[2t_0]-1}^\infty \Vert \widehat{\mathbf {L}}(q_{\omega }^{(n)})\Vert \leqslant C \sum _{\omega =[2t_0]-1}^\infty \frac{\omega ^{n-1}}{(n-1)!}\cdot e^{-(\mathrm {Re}(s_0) +(1/2)r \chi _0)\omega +\tau n}\\&\leqslant C\int _0^\infty \frac{t^{n-1}}{(n-1)!} \cdot e^{-(\mathrm {Re}(s_0) +(1/2)r \chi _0)t+\tau n} dt\\&= C e^{\tau n}\cdot \left( \int _0^\infty e^{-(\mathrm {Re}(s_0) +(1/2)r \chi _0)t} dt\right) ^n<C (r''_0)^{-n} \end{aligned}$$

from (B.10). We give the proof of Claim 1 below to complete the proof.

Proof of Claim 1

We first note that the family

$$\begin{aligned} \mathcal {P}=\left\{ (n-1)!\cdot \omega ^{-n+1}\cdot e^{\mathrm {Re}(s_0) \omega -\tau n}\cdot {q}_{\omega }^{(n)}(t+\omega )\;\bigg |\; n\geqslant 1, \omega \geqslant [2t_0]-1 \right\} \end{aligned}$$

(B.12)

satisfies the assumption in Lemma B.5. To proceed, we write an integer $\omega \geqslant [2t_0]-1$ as a sum of real numbers in $[t_0,2t_0]$:

$$\begin{aligned} \omega =\sum _{i=1}^{k(\omega )} t_i,\qquad t_0\leqslant t_i\leqslant 2t_0. \end{aligned}$$

Then we decompose ${\mathbf {L}}(q_{\omega }^{(n)})$ as follows: First we write

$$\begin{aligned} {\mathbf {L}}(q_{\omega }^{(n)})&={\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\circ \mathbf {L}^{\omega }\\&=\mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow }\circ \mathbf {L}(\tilde{q}_{\omega }^{(n)})\circ {\mathbf {L}}^{t_{k(\omega )-1}}\circ \cdots \circ {\mathbf {L}}^{t_{2}}\circ \mathbf {L}^{t_1}\\&\quad + ({\mathbf {L}}^{t_{k(\omega )}}-\mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow })\circ {\mathbf {L}}^{t_{k(\omega )-1}}\circ \cdots \circ {\mathbf {L}}^{t_{2}}\circ \mathbf {L}(\tilde{q}_{\omega }^{(n)})\circ \mathbf {L}^{t_1} \end{aligned}$$

where we set $\tilde{q}_{\omega }^{(n)}(t)=q_{\omega }^{(n)}(t+\omega )$ for brevity. Since the first term on the right-hand side other than the first factor is of the same form as ${\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\circ \mathbf {L}^{\omega }$, we apply the parallel operation to it. If we continue this procedure, we can express $\mathbf {L}(q_{\omega }^{(n)})$ as

$$\begin{aligned}&\mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow }\circ \cdots \circ \mathbf {L}^{t_{2}}_{\mathrm {hyp},\hookrightarrow }\circ \mathbf {L}(\tilde{q}_{\omega }^{(n)})\circ \mathbf {L}^{t_1}\nonumber \\&\quad +\sum _{j=2}^{k(\omega )} \mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow }\circ \cdots \circ \mathbf {L}^{t_{j+1}}_{\mathrm {hyp},\hookrightarrow } \circ ({\mathbf {L}}^{t_{j}}-\mathbf {L}^{t_{j}}_{\mathrm {hyp}, \hookrightarrow })\circ {\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\circ {\mathbf {L}}^{t_{j-1}}\circ \cdots \circ {\mathbf {L}}^{t_{1}}. \end{aligned}$$

(B.13)

Note that, from Lemma B.5 and the estimate noted in the beginning, we see

where $\widehat{\mathbf {L}}$ is upper triangular and $\Vert \widehat{\mathbf {L}}\Vert \leqslant C$ while is in the trace class and , with $C>0$ a constant independent of n and $\omega$. So we may rewrite the first term on the right-hand side of (B.13) as

(B.14)

Let $\widehat{\mathbf {L}}(q_{\omega }^{(n)})$ in Claim 1 be the first term of (B.14) above and be the remainder, that is, the sum of the second term in (B.14) and the sum on the second line of (B.13). By Lemma B.4, $\widehat{\mathbf {L}}(q_{\omega }^{(n)})$ is upper triangular. Since

$$\begin{aligned} \Vert \mathbf {L}^{t}_{\mathrm {hyp},\hookrightarrow }\Vert \leqslant e^{-r\chi _0 t/2 }\quad \text{ for } t_0\leqslant t\leqslant 2t_0 \end{aligned}$$

(B.15)

from Remark 11.16, we obtain the estimate (B.10) immediately. Also, for the second term in (B.14), we have

and this bound is summable with respect to $\omega$. To look into the sum in (B.13), note that the operator $({\mathbf {L}}^{t_{j}}-\mathbf {L}^{t_{j}}_{\mathrm {hyp},\hookrightarrow })\circ {\mathbf {L}}(\tilde{q}_{\omega }^{(n)})$ satisfies

$$\begin{aligned} \Vert ({\mathbf {L}}^{t_{j}}-\mathbf {L}^{t_{j}}_{\mathrm {hyp},\hookrightarrow })\circ {\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\Vert _{\mathrm {Tr}} \leqslant C \cdot \frac{\omega ^{n-1}\cdot e^{-\mathrm {Re}(s_0) \omega +\tau n}}{(n-1)!} \end{aligned}$$

for a constant $C>0$ independent of n and $\omega$, from the latter claim of Lemma B.5 and uniform boundedness of $\mathcal {P}$. Hence we have

$$\begin{aligned}&\sum _{j=2}^{k(\omega )}\Vert \mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow }\circ \cdots \circ \mathbf {L}^{t_{j+1}}_{\mathrm {hyp},\hookrightarrow } \circ ({\mathbf {L}}^{t_{j}}-\mathbf {L}^{t_{j}}_{\mathrm {hyp},\hookrightarrow })\circ {\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\circ {\mathbf {L}}^{t_{j-1}}\circ \cdots \circ {\mathbf {L}}^{t_{1}}\Vert _{\mathrm {Tr}}\\&\quad \leqslant \sum _{j=2}^{k(\omega )}\Vert \mathbf {L}^{t_{k(\omega )}}_{\mathrm {hyp},\hookrightarrow }\Vert \cdots \Vert \mathbf {L}^{t_{j+1}}_{\mathrm {hyp},\hookrightarrow }\Vert \\&\qquad \times \Vert ({\mathbf {L}}^{t_{j}}-\mathbf {L}^{t_{j}}_{\mathrm {hyp},\hookrightarrow }) \circ {\mathbf {L}}(\tilde{q}_{\omega }^{(n)})\Vert _{\mathrm {Tr}}\cdot \Vert {\mathbf {L}}^{t_{j-1}} \circ \cdots \circ {\mathbf {L}}^{t_{1}}\Vert \\&\quad \leqslant C \cdot \frac{\omega ^{n-1}\cdot e^{-(\mathrm {Re}(s_0)-P)\omega +\tau n}}{(n-1)!}\quad \text{ by } \text{(B.1) } \text{ and } \text{(B.15). } \end{aligned}$$

This bound is again summable with respect to $\omega$, provided $\tau$ is sufficiently small. Therefore we obtain the estimate (B.11). $\square$

Corollary B.9

The essential spectral radius of $\mathcal {R}(s_0):\widetilde{\mathcal {K}}^{r}(K_0)\rightarrow \widetilde{\mathcal {K}}^{r}(K_0)$ is bounded by $(r''_0)^{-1}$.

Proof

We consider the decomposition of $\mathcal {R}(s_0)^n:{\widetilde{\mathcal {K}}}^{r}(K_0)\rightarrow {\widetilde{\mathcal {K}}}^{r}(K_0)$ into

where $\widetilde{\mathcal {R}}^{(n)}$ is that in (B.9). From the last lemma, the operator norm of $\widehat{\mathcal {R}}^{(n)}$ on $\mathcal {K}^r(K_0)$ is bounded by $C (r''_0)^{-n}$ with $C>0$ independent of n, and on $\mathcal {K}^r(K_0)$ is a trace class operator. Further it is not difficult to see that these remain true when we regard $\widehat{\mathcal {R}}^{(n)}$ and as operators on ${\widetilde{\mathcal {K}}}^{r}(K_0)$, because $\mathcal {L}^{t_0}:{\widetilde{\mathcal {K}}}^{r}(K_0)\rightarrow \mathcal {K}^r(K_0)$ is bounded from Proposition 7.8. Therefore, recalling Remark B.7 for $\widetilde{\mathcal {R}}^{(n)}$, we obtain that the essential spectral radius of $\mathcal {R}(s_0)^n$ is bounded by $C (r''_0)^{-n}$ and hence by $(r''_0)^{-n}$ from the multiplicative property of essential spectral radius. $\square$

Corollary B.9 implies that the spectral set of $\mathcal {R}(s_0):{\widetilde{\mathcal {K}}}^{r}(K_0)\rightarrow {\widetilde{\mathcal {K}}}^{r}(K_0)$ on the outside of the disk $|z|\leqslant (r'_0)^{-1}$ consists of discrete eigenvalues $\mu _i$, $1\leqslant i\leqslant m$, counted with multiplicity. Since $A \mathcal {R}(s_0)=s_0 \mathcal {R}(s_0)+1$, we have

$$\begin{aligned} \mu -\mathcal {R}(s_0)=\mu \cdot \left( (s_0-\mu ^{-1}) -A\right) \cdot \mathcal {R}(s_0). \end{aligned}$$

This implies that $\mu _i$’s are in one-to-one correspondence to the eigenvalues $\chi _i$, $1\leqslant i\leqslant m$, of the generator A in the disk $D(s_0,r'_0)$ by the relation

$$\begin{aligned} \mu _i=\frac{1}{s_0-\chi _i}=\int _0^\infty e^{-s_0 t}e^{\chi _i t}dt. \end{aligned}$$

Remark B.10

Since the argument above holds for any $s_0$ satisfying $\mathrm {Re}(s_0)>P$, the spectrum of the generator A on the half-plane $\mathrm {Re}(s)>-(1/4)r\chi _0$ consists of discrete eigenvalues with finite multiplicity and the resolvent $\mathcal {R}(s)$ is meromorphic on that half-plane.

Let $\varvec{\pi }:\widetilde{\mathcal {K}}^{r}(K_0)\rightarrow \widetilde{\mathcal {K}}^{r}(K_0)$ be the spectral projector of $\mathcal {R}(s_0)$ for the spectral set $\{\mu _i\}_{i=1}^m$ on the outside of the disk $|z|\leqslant r_0^{-1}$. This is also the spectral projector of the generator A for the spectral set $\{\chi _i\}_{i=1}^m$ and its image is contained in $\mathcal {K}^{r}(K_0)$ from Proposition 7.8. We set $\mathcal {F}(s_0)=\varvec{\pi }\circ \mathcal {R}(s_0)$, so that

$$\begin{aligned} \mathrm {Tr}\,\mathcal {F}(s_0)^n=\mathrm {Tr}^\flat \,\mathcal {F}(s_0)^n=\sum _{i=1}^m \mu _i^n=\sum _{i=1}^m \frac{1}{(s_0-\chi _i)^n}. \end{aligned}$$

Our task is to prove (B.3) in Proposition B.1 for

$$\begin{aligned} Q_n=\mathrm {Tr}^\flat \left( \mathcal {R}(s_0)^n-\mathcal {F}(s_0)^n\right) =\mathrm {Tr}^\flat \left( (1-\varvec{\pi })\circ \mathcal {R}(s_0)^n\right) . \end{aligned}$$

To continue, let $N_0>0$ be a large integer constant which will be specified later in the course of the argument. Consider large integer n and write

$$\begin{aligned} \mathcal {R}(s_0)^n-\mathcal {F}(s_0)^n&=(\mathrm {Id}-{\varvec{\pi }})\circ \mathcal {R}(s_0)^{n(m)}\circ \ \cdots \circ \mathcal {R}(s_0)^{n(1)}. \end{aligned}$$

where $n=n(1)+n(2)+\cdots +n(I)$ with $N_0\leqslant n(i)\leqslant 2N_0$. We decompose each term $\mathcal {R}(s_0)^{n(i)}$ on the right-hand side as

$$\begin{aligned} \mathcal {R}(s_0)^{n(i)}=\mathcal {R}^{(n(i))}+\widetilde{\mathcal {R}}^{(n(i))} \end{aligned}$$

in the same manner as (B.9). From Remark B.7, the part $\widetilde{\mathcal {R}}^{(n(i))}$ is very small in the operator norm if we let the constant $N_0$ be sufficiently large. (And note that the following argument is much simpler if we ignore this part.)

Since the operators $\mathcal {R}^{(n(i))}$ and $\widetilde{\mathcal {R}}^{(n(i))}$ for $1\leqslant i\leqslant I$ commute each other and also with the projection operator $\varvec{\pi }$, we can express $\mathcal {R}(s_0)^n-\mathcal {F}(s_0)^n$ as the sum of the $2^I$ terms of the form

$$\begin{aligned} (1-{\varvec{\pi }})\circ \left( \prod _{i=1}^{I''} \widetilde{\mathcal {R}}^{(n''(i))}\right) \circ \left( \prod _{i=1}^{I'} \mathcal {R}^{(n'(i))}\right) \end{aligned}$$

(B.16)

where $\{n''(1),\ldots , n''(I''), n'(1), \ldots , n'(I')\}$ with $I=I'+I''$ ranges over all the rearrangements of $\{n(1), n(2), \ldots , n(I)\}$. Hence our task is reduced to show

Claim 2

There exists a constant $C>0$ such that

$$\begin{aligned} \left| \mathrm {Tr}^\flat \left( (1-{\varvec{\pi }})\circ \left( \prod _{i=1}^{I''}\widetilde{\mathcal {R}}^{(n''(i))}\right) \circ \left( \prod _{i=1}^{I'} \mathcal {R}^{(n'(i))}\right) \right) \right| \leqslant C r_0^{-n}\cdot 2^{-I}. \end{aligned}$$

(B.17)

Proof of Claim 2

Below we prove the claim in the case $I'\geqslant I''$ because, otherwise, we can get the conclusion by a similar but much easier argument using Remark B.9. We translate the claim to that on the lifted operators. Let us put

$$\begin{aligned} \varvec{\pi }^{\mathrm {lift}}:=\mathbf {I}\circ \varvec{\pi }\circ \mathbf {I}^*:\mathbf {K}^r\rightarrow \mathbf {K}^r. \end{aligned}$$

We write

$$\begin{aligned} \widetilde{\mathbf {R}}^{(n)}= \int _0^\infty \chi (t/(2t_0))\cdot \frac{t^{n-1} e^{-ts_0}}{(n-1)!}\cdot \mathbf {L}^t\, dt, \end{aligned}$$

(B.18)

so that

$$\begin{aligned} {\mathbf {R}}^{(n)}+ \widetilde{\mathbf {R}}^{(n)}=\mathbf {I}\circ \mathcal {R}(s_0)^n\circ \mathbf {I}^* =\int _0^\infty \frac{t^{n-1} e^{-ts_0}}{(n-1)!}\cdot \mathbf {L}^t\, dt. \end{aligned}$$

Then, from Corollary B.6, the inequality in Claim 2 is equivalent to

$$\begin{aligned} \left| \mathrm {Tr}^\flat \,\left( (1-\varvec{\pi }^{\mathrm {lift}})\circ \left( \prod _{i=1}^{I''} \widetilde{\mathbf {R}}^{(n''(i))}\right) \circ \left( \prod _{i=1}^{I'} \mathbf {R}^{(n'(i))}\right) \right) \right| \leqslant Cr_0^{-n}\cdot 2^{-I}. \end{aligned}$$

(B.19)

Since we are assuming $I'\geqslant I''$ and since the operators on the left hand side above commute, we may write the operator on the left-hand side above as

$$\begin{aligned} \prod _{i=1}^{I'} ((1-\varvec{\pi }^{\mathrm {lift}})\circ {\mathbf {R}}_i) \quad \text { setting } {\mathbf {R}}_i= {\left\{ \begin{array}{ll} \widetilde{\mathbf {R}}^{(n''(i))}\circ \mathbf {R}^{(n'(i))},&{}\quad \text{ if } 1\leqslant i\leqslant I'';\\ \mathbf {R}^{(n'(i))},&{}\quad \text{ if } I''<i\leqslant I'. \end{array}\right. } \end{aligned}$$

(B.20)

From the choice of the spectral projector $\varvec{\pi }$ and the fact that $\widetilde{\mathbf {R}}^{(n)}$ has small operator norm by the same reason as noted in Remark B.7, we get the estimate

$$\begin{aligned} \Vert (1-\varvec{\pi }^{\mathrm {lift}})\circ \mathbf {R}_i:\mathbf {K}^{r}\rightarrow \mathbf {K}^{r}\Vert \leqslant (r_0)^{-\tilde{n}(i)}/4\quad \text{ for } 1\leqslant i\leqslant I', \end{aligned}$$

(B.21)

provided that the constant $N_0$ is sufficiently large, where we set

$$\begin{aligned} \tilde{n}(i)={\left\{ \begin{array}{ll} n'(i)+n''(i),&{}\quad \text{ if } 1\leqslant i\leqslant I'';\\ n'(i),&{}\quad \text{ if } I''< i\leqslant I'. \end{array}\right. } \end{aligned}$$

By Lemma B.8, the operators $\mathbf {R}_i$ are decomposed as where

• (1) is a trace class operator and , and

• (2) $\widehat{\mathbf {R}}_i$ is upper triangular and satisfies $\Vert \widehat{\mathbf {R}}_i\Vert \leqslant r_0^{-\tilde{n}(i)}/4$

provided that the constant $N_0$ is sufficiently large. (For the case $1\leqslant i\leqslant I''$, we need a slight modification of Lemma B.8 but the proof goes as well.) In (B.20), we consider the decomposition

and apply the development it in the parallel manner as we used to obtain (B.13). Then, noting that $\mathrm {Tr}^\flat \, (\widehat{\mathbf {R}}_1\circ \cdots \circ \widehat{\mathbf {R}}_{I'})=0$ from Lemma B.4, we obtain

This is bounded in absolute value by

The trace norm is bounded by a constant C independent of j and n. Therefore, using the condition on $\widehat{\mathbf {R}}_i$ and (B.21), we conclude (B.19). This completes the proof of Claim 2 and hence that of Proposition B.1. $\square$