Global Trace Formula for Ultra-Differentiable Anosov Flows

Abstract

Adapting tools that we introduced in Jézéquel (J Spectr Theory 10(1):185–249, 2020) to study Anosov flows, we prove that the trace formula conjectured by Dyatlov and Zworski in (Ann. Sci. Éc. Norm. Supér. (4) 49(3):543–577, 2016) holds for Anosov flows in a certain class of regularity (smaller than ${\mathcal {C}}^\infty$ but larger than the class of Gevrey functions). The main ingredient of the proof is the construction of a family of anisotropic Hilbert spaces of generalized distributions on which the generator of the flow has discrete spectrum.

Appendices

Appendix

Appendix A. Ruelle Resonances are Intrinsic

As pointed out before, the Banach spaces ${\mathcal {B}}$ that appear in Theorem 1.2 are highly non-canonical. To prove that Ruelle resonances do not depend on the choice of these spaces, there is a classical argument based on the investigation of a meromorphic continuation of the resolvent of X as an operator from ${\mathcal {C}}^\infty \left( M\right)$ to ${\mathcal {D}}'\left( M\right)$. To deal with spaces that are not intermediary between ${\mathcal {C}}^\infty \left( M\right)$ and ${\mathcal {D}}'\left( M\right)$, it is easier to use an approach based on the following Lemma A.2, whose proof may be found in [3, Lemma A.3] or [5, Lemma A.1]. Recall first the following definition.

Definition A.1

(Isolated eigenvalue of finite multiplicity and essential spectral radius). If ${\mathcal {B}}$ is a Banach space, X an a priori unbounded operator on ${\mathcal {B}}$ and $\lambda \in {\mathbb {C}}$, we say that $\lambda$ is an isolated eigenvalue of finite multiplicity for X if $\lambda$ is an isolated point of $\sigma \left( X\right)$ and the rank of the spectral projector

$$\begin{aligned} \Pi _{\lambda } = \frac{1}{2 i \pi } \int _{\partial {\mathbb {D}}\left( \lambda ,r\right) } \left( z-X\right) ^{-1} \mathrm {d}z, \end{aligned}$$

where r is any small enough positive real number so that $\sigma \left( X\right) \cap \overline{{\mathbb {D}}}\left( \lambda ,r\right) = \left\{ \lambda \right\}$, is finite (this rank is by definition the algebraic multiplicity of $\lambda$).

Now, if X is bounded we define the essential spectral radius of X as the infimum of $\rho > 0$ such that the intersection of $\sigma \left( X\right)$ with $\left\{ z \in {\mathbb {C}}: \left| z \right| > \rho \right\}$ contains only isolated eigenvalues of finite multiplicity.

Lemma A.2

Let ${\mathcal {B}}$ be a Hausdorff topological vector space. Let ${\mathcal {B}}_1$ and ${\mathcal {B}}_2$ be Banach spaces continuously included in ${\mathcal {B}}$ such that ${\mathcal {B}}_1 \cap {\mathcal {B}}_2$ is dense both in ${\mathcal {B}}_1$ and in ${\mathcal {B}}_2$. Let $L : {\mathcal {B}}\rightarrow {\mathcal {B}}$ be a continuous linear map that preserves ${\mathcal {B}}_1$ and ${\mathcal {B}}_2$. Assume that the maps induced by L on ${\mathcal {B}}_1$ and ${\mathcal {B}}_2$ are bounded operators whose essential spectral radius is smaller than some number $\rho >0$. Then the eigenvalues in $\left\{ z \in {\mathbb {C}}: \left| z \right| > \rho \right\}$ coincide. Furthermore, the corresponding generalized eigenspaces coincide and are contained in ${\mathcal {B}}_1 \cap {\mathcal {B}}_2$.

Applying Lemma A.2 to the resolvent of X, we may prove the two following lemmas. Lemma A.3 asserts that Ruelle resonances are well-defined, while Lemma A.4 ensures that the spectrum of X acting on the space ${\mathcal {H}}$ given by Theorem 1.7 coincides with the Ruelle spectrum (recall Definition 1.3). The proofs of Lemmas A.3 and A.4 are very similar and consequently we will only prove Lemma A.4, in order to show that there are no particular difficulties when working with unusual classes of regularity.

Lemma A.3

Let ${\mathcal {B}}$ and ${\widetilde{{\mathcal {B}}}}$ be two Banach spaces and $A > 0$ be a positive real number. Assume that ${\mathcal {B}}$ and ${\widetilde{{\mathcal {B}}}}$ both satisfy the points (i)–(iv) from Theorem 1.2 for this particular value of A. Then the intersections of $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > - A \right\}$ with the spectrum of X acting on ${\mathcal {B}}$ and ${\widetilde{{\mathcal {B}}}}$ coincide.

Lemma A.4

Assume that M, the flow $\left( \phi ^t\right) _{t \in {\mathbb {R}}}$, and g are ${\mathcal {C}}^{\kappa ,\upsilon }$ for some $\kappa > 0$ and $\upsilon >1$. Let ${\mathcal {B}}$ be a Banach space such that for some ${\tilde{\upsilon }} > \upsilon$ and $A \in {\mathbb {R}}$ we have:

• (i) ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right) \subseteq {\mathcal {B}}\subseteq {\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right) '$, all the inclusions being continuous, the first one having dense image;

• (ii) for all $t \in {\mathbb {R}}_+$, the operator ${\mathcal {L}}_t$ defined by (1.2) is bounded from ${\mathcal {B}}$ to itself;

• (iii) $\left( {\mathcal {L}}_t\right) _{t \ge 0}$ forms a strongly continuous semi-group of operator acting on ${\mathcal {B}}$, whose generator is X;

• (iv) the intersection of the spectrum of X acting on ${\mathcal {B}}$ with $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > -A \right\}$ is made of isolated eigenvalues of finite multiplicity.

Then the intersection of $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > -A \right\}$ with the spectrum of X acting on ${\mathcal {B}}$ is the intersection of $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > -A \right\}$ with the Ruelle spectrum of X from Definition 1.3.

Proof

Apply Theorem 1.2 (with the same value of A) to get a Banach space ${\widetilde{{\mathcal {B}}}}$ such that in particular the intersection of $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > -A \right\}$ with the spectrum of X acting on ${\widetilde{{\mathcal {B}}}}$ coincides with the intersection of $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) > -A \right\}$ with the Ruelle spectrum of X (by definition of the Ruelle spectrum). Now choose a positive real number $z_0$ large enough so that the resolvent $\left( z_0 - X\right) ^{-1}$ is well-defined both on ${\mathcal {B}}$ and ${\widetilde{{\mathcal {B}}}}$. Notice that the resolvents of X acting on ${\mathcal {B}}$ and ${\widetilde{{\mathcal {B}}}}$ coincide on the intersection ${\mathcal {B}}\cap {\widetilde{{\mathcal {B}}}}$. Indeed, from (iii) and [24, Problem 1.15 p.487], it follows that, if $u \in {\mathcal {B}}\cap {\widetilde{{\mathcal {B}}}}$, then $\left( z_0 - X\right) ^{-1}$ is defined as an element of ${\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right) '$ by

$$\begin{aligned} \forall \mu \in {\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right) : \left\langle \left( z_0 - X\right) ^{-1} u,\mu \right\rangle = \int _0^{+ \infty } e^{-z_0 t} \left\langle {\mathcal {L}}_t u,\mu \right\rangle . \end{aligned}$$

Thus we may extend $\left( z_0 - X\right) ^{-1}$ to ${\mathcal {B}}+ {\widetilde{{\mathcal {B}}}}$ by setting that $\left( z_0 - X\right) ^{-1} u$ is equal to $\left( z_0 - X\right) ^{-1} v + \left( z_0 - X\right) ^{-1} w$, if $u = v + w$ with $v \in {\mathcal {B}}$ and $w \in {\widetilde{{\mathcal {B}}}}$ (it does not depend on the choice of v and w). Furthermore, this extension is continuous when ${\mathcal {B}}+ {\widetilde{{\mathcal {B}}}}$ is endowed with the norm $\left\| \cdot \right\| _{{\mathcal {B}}+ {\widetilde{{\mathcal {B}}}}}$ defined by

$$\begin{aligned} \forall u \in {\mathcal {B}}+ {\widetilde{{\mathcal {B}}}} : \left\| u \right\| _{{\mathcal {B}}+ {\widetilde{{\mathcal {B}}}}} = \inf _{\begin{array}{c} u = v + w \\ v \in {\mathcal {B}}, w \in {\widetilde{{\mathcal {B}}}} \end{array}} \left\| v \right\| _{{\mathcal {B}}} + \left\| w \right\| _{{\widetilde{{\mathcal {B}}}}}. \end{aligned}$$

Let $A' <A$ and $R >0$, provided that $z_0$ is large enough we have

$$\begin{aligned} \frac{1}{\sqrt{\left( z_0 + A'\right) ^2 + R^2}} \ge \frac{1}{z_0 + A}. \end{aligned}$$

(A.1)

The map $\lambda \mapsto \left( z_0 - \lambda \right) ^{-1}$ induces a bijection between the spectrum of X acting on ${\mathcal {B}}$ and the spectrum of $\left( z_0 - X\right) ^{-1}$ acting on ${\mathcal {B}}$, but it also sends $\left\{ z \in {\mathbb {C}}: \mathfrak {R}\left( z\right) \le - A \right\}$ into the disc of center 0 and radius $\frac{1}{z_0 + A}$. Consequently, the essential spectral radius of $\left( z_0 - X\right) ^{-1}$ acting on ${\mathcal {B}}$ is less than $\frac{1}{z_0+A}$. The same is true for the same reason replacing ${\mathcal {B}}$ by ${\widetilde{{\mathcal {B}}}}$. Thus we may apply Lemma A.2 (with $\rho = \frac{1}{z_0 + A}$ and ${\mathcal {B}}_1,{\mathcal {B}}_2$ and ${\mathcal {B}}$ being respectively ${\mathcal {B}},{\widetilde{{\mathcal {B}}}}$ and ${\mathcal {B}}+ {\widetilde{{\mathcal {B}}}}$) to see that the spectrum of $\left( z_0 -X\right) ^{-1}$ outside of the disc of center 0 and radius $\frac{1}{z_0 +A}$ is the same on ${\mathcal {B}}$ and on ${\widetilde{{\mathcal {B}}}}$. But the map $\lambda \mapsto \left( z_0 - \lambda \right) ^{-1}$ sends $\left\{ z \in {\mathbb {C}}: -A' \le \mathfrak {R}(z) \le z_0 \text { and } \left| \mathfrak {I}\left( z\right) \right| \le R \right\}$ outside of the disc of center 0 and radius $\frac{1}{z_0+A}$ (see (A.1)). Consequently, the intersection of $\left\{ z \in {\mathbb {C}}: -A' \le \mathfrak {R}(z) \le z_0 \text { and } \left| \mathfrak {I}\left( z\right) \right| \le R \right\}$ with the spectrum of X acting on ${\mathcal {B}}$ coincides with the intersection of $\left\{ z \in {\mathbb {C}}: -A' \le \mathfrak {R}(z) \le z_0 \text { and } \left| \mathfrak {I}\left( z\right) \right| \le R \right\}$ with the set of Ruelle resonances of X. Since $R >0$ and $A' < A$ are arbitrary, and $z_0$ may be chosen arbitrarily large, the lemma is proven. $\quad \square$

Appendix B. Proofs of Lemmas 2.3 and 2.4

Proof of Lemma 2.3

We only need to prove the first point: the same argument with ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$ replaced by ${\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right)$, and ${\mathcal {L}}_t$ and X replaced by their formal adjoints gives the second point.

We start with the case $g = 0$. Using the group property of $\left( {\mathcal {L}}_t\right) _{t \in {\mathbb {R}}}$, we only need to prove differentiability at $t=0$. Then we may cover M by flow boxes, and thus we only need to show that if $u \in {\mathcal {S}}^{{\tilde{\upsilon }}}$ is supported in a compact subset K of ${\mathbb {R}}^{d+1}$ then

$$\begin{aligned} \frac{u\left( \cdot + t e_{d+1}\right) - u}{t} \underset{t \rightarrow 0}{\rightarrow } \partial _{x_{d+1}} u \text { in } {\mathcal {S}}^{{\tilde{\upsilon }}}, \end{aligned}$$

(B.1)

where $e_{d+1}$ denotes the last vector of the canonical basis of ${\mathbb {R}}^{d+1}$. Up to enlarging K we may assume that for all $t \in \left[ -1,1\right]$ the function $u\left( \cdot + t e_{d+1}\right)$ is supported in K. Then if $x \in K$, $\alpha \in {\mathbb {N}}^{d+1}$, and $t \in \left[ -1,1\right]$ we have with Taylor’s formula (for any $\kappa '' >0$):

$$\begin{aligned} \begin{aligned}&\left| \frac{\partial ^\alpha u\left( x+t e_{d+1}\right) - \partial ^{\alpha }u\left( x\right) }{t} - \partial ^{\alpha } \partial _{x_{d+1}} u\left( x\right) \right| \\ {}&\quad \qquad \qquad = \left| \frac{\partial ^\alpha u\left( x+t e_{d+1}\right) - \partial ^{\alpha }u\left( x\right) }{t} - \partial _{x_{d+1}} \partial ^\alpha u\left( x\right) \right| \\ {}&\quad \qquad \qquad \le \frac{\left\| \partial _{x_{d+1}}^2 \partial ^\alpha u \right\| _{\infty }}{2} \left| t \right| \le \frac{\left| t \right| }{2} \left\| u \right\| _{\kappa '',{\tilde{\upsilon }}} \exp \left( \frac{\left( \left| \alpha \right| + 2\right) ^{{\tilde{\upsilon }}}}{\kappa ''}\right) . \end{aligned} \end{aligned}$$

Thus if $\kappa ',\kappa '' > 0$ and for $R > 0$ depending only on K, we have for all $x \in {\mathbb {R}}^{d+1}, \alpha \in {\mathbb {N}}^{d+1}$ and $m \in {\mathbb {N}}$:

$$\begin{aligned} \begin{aligned}&\left( 1 + \left| x \right| \right) ^m \left| \frac{\partial ^\alpha u\left( x+t e_{d+1}\right) - \partial ^{\alpha }u\left( x\right) }{t} - \partial ^{\alpha } \partial _{x_{d+1}} u\left( x\right) \right| \exp \left( - \frac{\left( m + \left| \alpha \right| \right) ^{{\tilde{\upsilon }}}}{\kappa '}\right) \\ {}&\quad \qquad \qquad \le \frac{\left| t \right| }{2} \left\| u \right\| _{\kappa '',{\tilde{\upsilon }}} R^m \exp \left( \frac{\left( \left| \alpha \right| + 2\right) ^{{\tilde{\upsilon }}}}{\kappa ''} - \frac{\left( m + \left| \alpha \right| \right) ^{{\tilde{\upsilon }}}}{\kappa '}\right) . \end{aligned} \end{aligned}$$

Thus if $\kappa ' > 0$ and $\kappa '' > \kappa '$, then there is a constant $C >0$ (that only depends on $K, {\tilde{\upsilon }},\kappa '$, and $\kappa ''$) such that for all $t \in \left[ -1,1\right]$ we have

$$\begin{aligned} \left\| \frac{u\left( \cdot + t e_{d+1}\right) - u}{t} - \partial _{x_{d+1}} u \right\| _{\kappa ',{\tilde{\upsilon }}} \le C \left| t \right| \left\| u \right\| _{\kappa '',{\tilde{\upsilon }}}, \end{aligned}$$

which implies (B.1) and thus ends the proof of the lemma in the case $g =0$.

In order to deduce the result in the case of a general g from the case $g=0$, we only need to prove that the map

$$\begin{aligned} \begin{aligned} t \mapsto \exp \left( \int _0^t g \circ \phi ^\tau \mathrm {d}\tau \right) \end{aligned} \end{aligned}$$

(B.2)

is ${\mathcal {C}}^\infty$ from ${\mathbb {R}}$ to ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$. Indeed, the multiplication is continuous from the product ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right) \times {\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$ to ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$. The map (B.2) is easily seen to be ${\mathcal {C}}^\infty$ from ${\mathbb {R}}$ to ${\mathcal {C}}^0\left( M\right)$, and one may notice that its derivatives are valued in ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}$ (recall that the classes of regularity ${\mathcal {C}}^{\kappa ,{\tilde{\upsilon }}}$, and hence ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}$, are closed by composition) with uniform bounds locally in t. Then, by successive applications of Taylor’s formula at order 1 with integral remainder, one gets that the map (B.2) is ${\mathcal {C}}^\infty$ from ${\mathbb {R}}$ to ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$, ending the proof of the lemma (we use the exact formula for the remainder in order to bound it in ${\mathcal {C}}^{\infty ,{\tilde{\upsilon }}}\left( M\right)$). $\quad \square$

Proof of Lemma 2.4

Denote for now the generator of $\left( {\mathcal {L}}_t\right) _{t \ge 0}$ by ${\widetilde{X}}$. Let $u \in {\mathcal {B}}$ be in the domain of ${\widetilde{X}}$, then the map ${\mathbb {R}}_+ \ni t \mapsto {\mathcal {L}}_t u \in {\mathcal {B}}$ is differentiable at 0 and its derivative at 0 is ${\widetilde{X}}u$ (by definition of ${\widetilde{X}}$). Since ${\mathcal {B}}\subseteq {\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right)$ is continuous, the same is true for the map ${\mathbb {R}}_+ \ni t \mapsto {\mathcal {L}}_t u \in {\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right) '$, whose derivative at 0 is Xu according to Lemma 2.3. Thus ${\widetilde{X}}u = Xu \in {\mathcal {B}}$.

Reciprocally, if $u \in {\mathcal {B}}$ is such that $Xu \in {\mathcal {B}}$, then we may define a ${\mathcal {C}}^1$ map $c: {\mathbb {R}}_+ \rightarrow {\mathcal {B}}$ by $c\left( t\right) = u + \int _0^t {\mathcal {L}}_\tau X u \mathrm {d}\tau$ for all $t \in {\mathbb {R}}_+$. Notice that $c'\left( 0\right) = Xu$. Since ${\mathcal {B}}\subseteq {\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right)$ is continuous, the map c is still ${\mathcal {C}}^1$ when seen as a map from ${\mathbb {R}}_+$ to ${\mathcal {D}}^{{\tilde{\upsilon }}}\left( M\right)$ and we have $c\left( 0\right) = u$ and $c'\left( t\right) = {\mathcal {L}}_t X u$ for all $t \in {\mathbb {R}}_+$, so that $c\left( t\right) = {\mathcal {L}}_t u$ for all $t \in {\mathbb {R}}_+$, using Lemma 2.3. This proves that u belongs to the domain of ${\widetilde{X}}$. $\quad \square$

Appendix C. Factorization of the Dynamical Determinant

We prove here, under the hypotheses of Theorem 1.7, a Hadamard-like factorization (C.3) for the dynamical determinant $d_g$ defined by (1.3). Let $t_0 > 0$ be shorter than any periodic orbit of $\left( \phi ^t\right) _{t \in {\mathbb {R}}}$. Then, working as in the proof of Proposition 1.9, we see that, for $\mathfrak {R}z \gg 1$, the essential spectral radius of

$$\begin{aligned} \begin{aligned} {\mathcal {L}}_{t_0}(z-X)^{-(d+2)} = \frac{1}{(d+1)!}\int _{t_0}^{+ \infty } e^{-z(t-t_0)}(t-t_0)^{d+1} {\mathcal {L}}_t \mathrm {d}t : {\mathcal {H}}\rightarrow {\mathcal {H}}\end{aligned} \end{aligned}$$

(C.1)

is zero. Then, applying holomorphic functional calculus in finite dimension as in the proof of Lemma 6.6, we see that the spectrum of (C.1) is made of the $\frac{e^{\lambda t_0}}{(z-\lambda )^{d+2}}$ for $\lambda$ in the spectrum of X. Then, for $\mathfrak {R}z \gg 1$, Proposition 5.4 implies that the right hand side of (C.1) defines a trace class operator on ${\widetilde{{\mathcal {H}}}}_0$. From Lemma A.2, we see that the spectrum of (C.1) is the same when acting on ${\mathcal {H}}$ or on ${\tilde{h}}_0$. Then, using Lidskii’s Trace Theorem and Proposition 5.4, we see that,¹⁶

$$\begin{aligned} \sum _{\lambda \text { resonance}} \frac{e^{\lambda t_0}}{\left( z - \lambda \right) ^{d+2}} =\frac{1}{\left( d+1\right) !} \sum _{\gamma } T_\gamma ^{\#} \exp \left( \int _\gamma g\right) \left( T_\gamma - t_0\right) ^{d+1} \frac{e^{-z \left( T_\gamma - t_0\right) }}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| }. \end{aligned}$$

For all $\lambda \in {\mathbb {C}}\setminus \left\{ 0 \right\}$ notice that the meromorphic map

$$\begin{aligned} z \mapsto - \sum _{n \ge d+1} \frac{z^n}{\lambda ^{n+1}} e^{-\left( z-\lambda \right) t_0} = \frac{e^{-\left( z-\lambda \right) t_0}}{z- \lambda } + \sum _{n = 0}^{d} \frac{z^n}{\lambda ^{n+1}}e^{-\left( z-\lambda \right) t_0} \end{aligned}$$

has a unique pole in $\lambda$ whose order is 1 and whose residue is 1. Thus there is an entire function $G_{\lambda ,t_0}$ such that for all $z \in {\mathbb {C}}$

$$\begin{aligned} \frac{G_{\lambda ,t_0}'\left( z\right) }{G_{\lambda ,t_0}\left( z\right) } = - \sum _{n \ge d+1} \frac{z^n}{\lambda ^{n+1}} e^{-\left( z-\lambda \right) t_0} = \frac{e^{-\left( z-\lambda \right) t_0}}{z- \lambda } + \sum _{n = 0}^{d} \frac{z^n}{\lambda ^{n+1}}e^{-\left( z-\lambda \right) t_0} \end{aligned}$$

and $G_{\lambda ,t_0}\left( 0\right) = 1$. Choose for $G_{0,t_0}$ any logarithmic primitive of $z \mapsto \frac{e^{-t_0 z}}{z}$.

Now, choose $R > 0$ and if $\left| \lambda \right| \ge 2 R$ notice that for all $z \in {\mathbb {D}}\left( 0,R\right)$ we have

$$\begin{aligned} \left| \frac{G_{\lambda ,t_0}'\left( z\right) }{G_{\lambda ,t_0}\left( z\right) } \right| \le 2 e^{R t_0} R^{d+1} \frac{e^{\mathfrak {R}\left( \lambda \right) t_0}}{\left| \lambda \right| ^{d+2}} \end{aligned}$$

and using the fact that $G_{\lambda ,t_0}$ has a logarithm on ${\mathbb {D}}\left( 0,R\right)$ that vanishes in 0 (since $G_{\lambda ,t_0}$ vanishes only at $\lambda$) we get that, for some constance C depending only on R and all $z \in {\mathbb {D}}\left( 0,R\right)$

$$\begin{aligned} \left| 1 - G_{\lambda ,t_0}\left( z\right) \right| \le C \frac{e^{\mathfrak {R}\left( \lambda \right) t_0}}{\left| \lambda \right| ^{d+2}}. \end{aligned}$$

Using Proposition 1.9, this implies that the infinite product

$$\begin{aligned} {\widetilde{d}}_g\left( z\right) = \prod _{\lambda \text { resonance}} G_{\lambda ,t_0}\left( z\right) \end{aligned}$$

converges uniformly on all compact subset of ${\mathbb {C}}$. Notice that the zeros of ${\widetilde{d}}_g$ are precisely the Ruelle resonances. Now, we find that

$$\begin{aligned} \left( e^{zt_0} \left( \ln G_{\lambda ,t_0}\left( z\right) \right) '\right) ^{\left( d+1\right) } =\left( -1\right) ^{d+1}\left( d+1\right) ! \frac{e^{\lambda t_0}}{\left( z - \lambda \right) ^{d+2}} \end{aligned}$$

and thus, for $\mathfrak {R}z \gg 1$,

$$\begin{aligned} \begin{aligned} \left( e^{zt_0} \left( \ln {\widetilde{d}}_g\left( z\right) \right) '\right) ^{\left( d+1\right) }&= \left( -1\right) ^{d+1}\left( d+1\right) !\sum _{\lambda \text { resonance}} \frac{e^{\lambda t_0}}{\left( z - \lambda \right) ^{d+2}} \\&= \left( -1\right) ^{d+1}\sum _{\gamma } T_\gamma ^{\#} \exp \left( \int _\gamma g\right) \left( T_\gamma - t_0\right) ^{d+1} \frac{e^{-z \left( T_\gamma - t_0\right) }}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| }\\&= \left( e^{z t_0} \sum _{\gamma } T_\gamma ^{\#} \exp \left( \int _\gamma g\right) \frac{e^{- zT_\gamma }}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| }\right) ^{\left( d+1\right) } \\&= \left( e^{z t_0} \left( \ln d_g\left( z\right) \right) '\right) ^{\left( d+1\right) }, \end{aligned} \end{aligned}$$

(C.2)

where $d_g$ is the usual dynamical determinant defined by (1.3). From (C.2) we deduce that there are a polynomial P of degree at most d and $\mu \in {\mathbb {C}}$ such that, for all $z \in {\mathbb {C}}$, we have the Hadamard-like factorization

$$\begin{aligned} d_g\left( z\right) = \mu \exp \left( P\left( z\right) e^{-t_0z}\right) \prod _{\lambda \text { resonance}} G_{\lambda ,t_0}\left( z\right) . \end{aligned}$$

(C.3)

In order to make this factorization more explicit, let us describe the $G_{\lambda ,t_0}$’s. For all $\lambda \in {\mathbb {C}}\setminus \left\{ 0 \right\}$, define the polynomial

$$\begin{aligned} Q_{\lambda ,t_0} = - \sum _{k=0}^{d} \left( \sum _{n=k}^{d} \frac{k!}{n!} \frac{(t_0 - \lambda )^{n-k-1}}{\lambda ^{k+1}} \right) X^k, \end{aligned}$$

and notice that

$$\begin{aligned} \left( Q_{\lambda ,t_0}\left( z\right) e^{-z\left( t_0 - \lambda \right) }\right) ' = \sum _{n = 0}^{d} \frac{z^n}{\lambda ^{n+1}}e^{-\left( z-\lambda \right) t_0}. \end{aligned}$$

Thus we have for all $\lambda \in {\mathbb {C}}\setminus \left\{ 0 \right\}$ and $z \in {\mathbb {C}}$

$$\begin{aligned} \begin{aligned}&G_{\lambda ,t_0}\left( z\right) = \left( 1- \frac{z}{\lambda }\right) \exp \left( Q_{\lambda ,t_0}\left( z\right) e^{-\left( z - \lambda \right) t_0 } - Q_{\lambda ,t_0}(0) e^{\lambda t_0}\right) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \,\, \times \exp \left( z \int _0^1 \frac{e^{-(zu - \lambda )t_0}-1}{zu - \lambda }\mathrm {d}u\right) . \end{aligned} \end{aligned}$$

The last factor is a logarithmic primitive of $z \mapsto \frac{e^{-\left( z-\lambda \right) t_0}-1}{z-\lambda }$.

Appendix D. Applications of the Trace Formula

As applications of the trace formula, we prove here Proposition 1.5 and Corollary 1.6.

Proof of Proposition 1.5

It is folklore to prove the implication (i) $\Rightarrow$ (ii) from residue theorem, see also [30, Theorem 17]. Let us prove the implication (i) $\Rightarrow$ (ii).

Choose $x >0$ large enough so that the series

$$\begin{aligned} \sum _\gamma T_\gamma ^{\#} \frac{e^{-x T_\gamma }}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| } \exp \left( \int _\gamma g\right) \end{aligned}$$

(D.1)

converges absolutely and $x > \mathfrak {R}\left( \lambda \right) +\epsilon$ for all the resonances $\lambda$ and some $\epsilon >0$. Write $k = \lceil \rho \rceil$. Choose $z \in {\mathbb {C}}$ such that $\mathfrak {R}\left( z\right) > x$. Then, we can find a sequence $\left( \varphi _n\right) _{n \in {\mathbb {N}}}$ of ${\mathcal {C}}^\infty$ functions, compactly supported in ${\mathbb {R}}_+^*$ such that

$$\begin{aligned} \lim _{n \rightarrow + \infty } \sup _{t \in {\mathbb {R}}} e^{tx}\left| \varphi _n\left( t\right) - t^{k}e^{-zt} \right| = 0 \end{aligned}$$

(D.2)

and

$$\begin{aligned} \sup _{\begin{array}{c} n \in {\mathbb {N}}\\ t \in {\mathbb {R}}_+^* \end{array}} \left| e^{tx} \varphi ^{\left( k\right) }\left( t\right) \right| < + \infty . \end{aligned}$$

(D.3)

Then, with (D.1) and (D.2), we have

$$\begin{aligned} \sum _\gamma T_\gamma ^{\#} \frac{\varphi _n\left( T_\gamma \right) }{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| } \exp \left( \int _\gamma g\right) \underset{n \rightarrow + \infty }{\rightarrow } \sum _\gamma T_\gamma ^{\#} \frac{e^{-z T_\gamma } T_{\gamma }^{k}}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| } \exp \left( \int _\gamma g\right) . \end{aligned}$$

Now, since the trace formula holds (by assumption), we know that for all $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \sum _\gamma T_\gamma ^{\#} \frac{\varphi _n\left( T_\gamma \right) }{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| } \exp \left( \int _\gamma g\right) = \sum _{\lambda \text { resonances}} L\left( \varphi _n\right) \left( - \lambda \right) . \end{aligned}$$

However, recall that

$$\begin{aligned} L\left( \varphi _n\right) \left( -\lambda \right) = \int _0^\infty e^{\lambda t} \varphi _n\left( t\right) \mathrm {d}t \end{aligned}$$

so that, using (D.2), we have,

$$\begin{aligned} L\left( \varphi _n\right) \left( -\lambda \right) \underset{n \rightarrow + \infty }{\rightarrow } \int _0^\infty t^{k} e^{-\left( z - \lambda \right) t}\mathrm {d}t = \frac{k!}{\left( z-\lambda \right) ^{k+1}}. \end{aligned}$$

Now, if $\lambda$ is non-zero, we have

$$\begin{aligned} L\left( \varphi _n\right) \left( - \lambda \right) = \frac{\left( -1\right) ^{k}}{\lambda ^{k}} \int _0^{+ \infty } e^{\lambda t} \varphi _n^{\left( k\right) } \left( t\right) \mathrm {d}t. \end{aligned}$$

Thus, (D.3), with $x > \mathfrak {R}\left( \lambda \right) +\epsilon$, and the second hypothesis provide a domination of $L\left( \varphi _n\right) \left( - \lambda \right)$, so that we have, using the dominated convergence theorem,

$$\begin{aligned} \sum _{\lambda \text { resonances}} L\left( \varphi _n\right) \left( - \lambda \right) \underset{n \rightarrow + \infty }{\rightarrow } k! \sum _{\lambda \text { resonances}} \frac{1}{\left( z-\lambda \right) ^{k+1}}. \end{aligned}$$

Finally we have (when $\mathfrak {R}\left( z\right) \gg 1$)

$$\begin{aligned} \begin{aligned} k! \sum _{\lambda \text { resonances}} \frac{1}{\left( z-\lambda \right) ^{k+1}}&= \sum _\gamma T_\gamma ^{\#} \frac{e^{-z T_\gamma } T_{\gamma }^{k}}{\left| \det \left( I - {\mathcal {P}}_\gamma \right) \right| } \exp \left( \int _\gamma g\right) \\&= \left( -1\right) ^{k+1} \left( \ln d_g\right) ^{\left( k+1\right) }(z). \end{aligned} \end{aligned}$$

Let P denote the canonical product of genus $k-1$ whose zeros are the Ruelle resonances of X (well-defined by [6, (2.6.4)] thanks to (1.5)). Then we see that, if z is not a Ruelle resonance for X, we have

$$\begin{aligned} \begin{aligned} \left( \ln P\right) ^{(k)}(z) = \left( -1\right) ^k \left( k-1\right) ! \sum _{\lambda \text { resonances}} \frac{1}{(z-\lambda )^k}. \end{aligned} \end{aligned}$$

(D.4)

It follows that $\left( \ln d_g\right) ^{(k+1)} = \left( \ln P\right) ^{(k+1)}$ and consequently there is a complex number a such that for every $z \in {\mathbb {C}}$ that is not a Ruelle resonance for X we have

$$\begin{aligned} \begin{aligned} \left( \ln P\right) ^{(k)}(z) = \left( \ln d_g\right) ^{(k)}(z) + a. \end{aligned} \end{aligned}$$

With (1.5), (D.4) and dominated convergence we see that $\left( \ln P\right) ^{(k)}(r) \underset{\begin{array}{c} r \rightarrow + \infty \\ r \in {\mathbb {R}} \end{array}}{\rightarrow } 0$. By direct inspection, we see that $\left( \ln d_g\right) ^{(k)}(r) \underset{\begin{array}{c} r \rightarrow + \infty \\ r \in {\mathbb {R}} \end{array}}{\rightarrow } 0$, and consequently $a = 0$. Thus, there is a polynomial Q of degree at most $k-1 \le \rho$ such that, for every $z \in {\mathbb {C}}$, we have

$$\begin{aligned} \begin{aligned} d_g(z) = e^{Q(z)} P(z), \end{aligned} \end{aligned}$$

and the result follows since P has order less than $\rho$ by [6, Theorem 2.6.5]. $\quad \square$

Proof of Corollary 1.6

Proposition 1.5 implies that $d_g$ has order less than 1. But notice that $d_g$ is bounded on a line (choose a line parallel to the imaginary axis corresponding to a large positive real part) and thus has to be constant by the Phragmén–Lindelöf Theorem [6, Theorem 1.4.1]. Finally, it has to be constant equal to 1 since $d_g(z) \underset{\begin{array}{c} z \rightarrow + \infty \\ z \in {\mathbb {R}} \end{array}}{\rightarrow } 1$. $\quad \square$

Appendix E. Expanding Maps of the Circle and the Condition

In order to discuss the condition $\upsilon < 2$ in Theorem 1.7, we can consider a very simple example: expanding maps of the circle ${\mathbb {S}}^1 = {\mathbb {R}}/ {\mathbb {Z}}$. An analogue of the space ${\mathcal {H}}$ from Theorem 1.7 would then be an isotropic space of the type (here $\left( {\hat{f}}(n)\right) _{n \in {\mathbb {Z}}}$ denotes the sequence of Fourier coefficient of a function f)

$$\begin{aligned} {\mathcal {H}}_{\alpha ,\beta } = \left\{ f \in {\mathcal {C}}^\infty \left( {\mathbb {S}}^1,{\mathbb {C}}\right) : \sum _{n \in {\mathbb {Z}}} \left| {\hat{f}}\left( n\right) \right| ^2 e^{2 \beta \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha }}} < + \infty \right\} , \end{aligned}$$

where $\beta > 0$ and $\alpha \in \left]\frac{\upsilon - 1}{\upsilon },1 \right[$ (this is the same condition as in Proposition 4.4), endowed with the norm

$$\begin{aligned} \left\| f \right\| _{\alpha ,\beta } = \sqrt{\sum _{n \in {\mathbb {Z}}} \left| {\hat{f}}\left( n\right) \right| ^2 e^{2 \beta \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha }}}}. \end{aligned}$$

Then the transfer operator

$$\begin{aligned} {\mathcal {L}} : f \mapsto \frac{f\left( \frac{\cdot }{2}\right) + f\left( \frac{\cdot + 1}{2}\right) }{2} \end{aligned}$$

associated to the doubling map may be written as

$$\begin{aligned} {\mathcal {L}} = \sum _{n \in {\mathbb {Z}}} \langle e_{2n}, \cdot \rangle _{L^2} e_n, \end{aligned}$$

where $e_n : x \mapsto e^{2 i \pi n x}$ (the sum converges in strong operator topology on the space of continuous endomorphisms of ${\mathcal {H}}_{\alpha ,\beta }$). Thus, the singular values of ${\mathcal {L}}$ acting on ${\mathcal {H}}_{\alpha ,\beta }$ are the $e^{\beta \left( \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha }} - \ln \left( 1 + 2\left| n \right| \right) ^{\frac{1}{\alpha }}\right) }$ for $n \in {\mathbb {Z}}$. Using the fact that

$$\begin{aligned} \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha }} - \ln \left( 1 + 2\left| n \right| \right) ^{\frac{1}{\alpha }} \underset{\left| n \right| \rightarrow + \infty }{=} - \frac{\ln 2}{\alpha } \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha } - 1} + O\left( \ln \left( 1 + \left| n \right| \right) ^{\frac{1}{\alpha } - 2}\right) \end{aligned}$$

we see that ${\mathcal {L}}$ acting on ${\mathcal {H}}_{\alpha ,\beta }$ is trace class when $\alpha < \frac{1}{2}$ and is not trace class when $\alpha > \frac{1}{2}$ (in the case $\alpha = \frac{1}{2}$ it depends on the value of $\beta$). Thus, we need to chose $\alpha < \frac{1}{2}$ if we want ${\mathcal {L}}$ to be nuclear. For general maps, this choice is possible only when $\upsilon < 2$ (see the condition in Proposition 4.4).

Consequently, using our method to prove the trace formula for ${\mathcal {C}}^{\kappa ,\upsilon }$ Anosov flows (or hyperbolic diffeomorphisms as in [21]) would require to construct Hilbert spaces in a totally different way, if $\upsilon \ge 2$.