Proof. We mostly follow the strategy of the proof of [Reference Climenhaga10, Theorem 10], while only changing the necessary details. We first prove that the three formulas indeed agree and define the same local measure. Starting from a measurable function $\zeta : \mathcal W^u(q,\delta )\times \mathcal W^{ws}(q,\delta )\to ]0,+\infty [$ , one sees that

$$ \begin{align*} &\int_Z \zeta(\Psi^{-1}_q(z))d((\Psi_q)_*(m^u_{q,V+J^u}\times m^{ws}_{q, J^u}))(z)\\ &\quad=\int_{\Psi_q^{-1}(Z)}\zeta(x,y)\,d((m^u_{q,V+J^u}\times m^{ws}_{q, J^u}))(x,y). \end{align*} $$

Now, this last expression can be seen to be equal to equation (42), using the change of variable by holonomy in equation (35) with the holonomy $\pi :\mathcal W^u(q,\delta )\to R^u_q(y), \pi (x)=[x,y]$ and with $\zeta (z):=e^{w^+_{V+J^u}(z,[z,q])+w^-_{J^u}(z,[q,z])}$ .

$$ \begin{align*} &\int_{R^u_q}\zeta([x,y])\mathbf{1}([x,y])\,d m^u_{q,V+J^u}(x)=\int_{R^u_q(y)\cap Z}\zeta(z)\,d(\pi_*m^u_{q,V+J^u})(z)\\ &\quad =\int_{R^u_q(y)\cap Z}\zeta(z)e^{w^+_{V+J^u}([z,q],z)}\,dm^u_{y,V+J^u}(z)=\int_{R^u_q(y)\cap Z}e^{w^-_{J^u}(z,[q,z])}\,dm^u_{y,V+J^u}(z), \end{align*} $$

where we used the cocycle relation (34). Similarly, we obtain the equivalent equation (43). Next, we show that the set of local measures $\{\nu _q\mid q\in \mathcal M\}$ are compatible in the sense that for any Borel set $ Z\subset R_q\cap R_p$ , one has $ \nu _q(Z)=\nu _p(Z).$ This will then define a global Borel measure. If $p\in \mathcal W^{ws}(q,\delta )$ , then we deduce that the above relation follows from equation (42) and the fact that the system measures $\{m^s_x\mid x\in \mathcal M\}$ is compatible on the intersection. Similarly, the relation for $p\in \mathcal W^{u}(q,\delta )$ follows from the compatibility of $\{m^{ws}_x\mid x\in \mathcal M\}$ and equation (43). The general case then follows from the local product structure. Call $\nu $ the global Borel measure we obtain. The fact that $\nu $ is non-zero and finite follows from the analog fact on the local measures $\nu _q$ and the compactness of the space $\mathcal M$ . What is left to prove is equation (44). It suffices to show that one has

(45) $$ \begin{align} \text{ for all } f\in C^{\infty}(\mathcal M), \ \langle e^{t\mathbf{P}}f,\nu\rangle=e^{tP(V+J^u)}\langle f, \nu\rangle, \end{align} $$

which will clearly imply the first part of equation (44). Because the measure $\nu $ is constructed from its restrictions, it will be enough to prove that

(46) $$ \begin{align} \text{for all } f\in C^{\infty}(\mathcal M),\quad \mathrm{Supp}(f)\cup \mathrm{Supp}(e^{t\mathbf{P}}f) \subset R_q \Rightarrow \nu(e^{t\mathbf{P}}f)=e^{tP(V+J^u)}\nu(f). \end{align} $$

We now use equation (43) to compute explicitly, using the $\varphi _t$ -conformality in equation (30):

$$ \begin{align*} \nu^{V+J^u}(e^{t\mathbf{P}}f)&=\int_{R^{ws}_q}\int_{Z\cap R^{u}_q(y)}e^{S_tV(\varphi_{-t}z)}f(\varphi_{-t}z) e^{w^-_{J^u}(z,y)}\,dm^{u}_{x, V+J^u}(z)\, dm^{ws}_{q,J^u}(y). \end{align*} $$

The cocycle relation (34) gives $e^{w^-_{J^u}(z,y)}=e^{w^-_{J^u}(z,\varphi _{-t}z)}e^{w^-_{J^u}(\varphi _{-t}z,y)}. $ Using equation (33) to get $w^+_{J^u}(z,\varphi _{-t}z)=S_{t}J^u(\varphi _{-t}z)$ , this means that

$$ \begin{align*} &\int_{Z\cap R^{u}_q(y)}e^{S_tV(\varphi_{-t}z)}f(\varphi_{-t}z) e^{w^-_{J^u}(z,y)}\,dm^{u}_{x, V+J^u}(z)\\ &\quad = \int_{Z\cap R^{u}_q(y)}e^{S_t(V+J^u)(\varphi_{-t}z)}f(\varphi_{-t}z) e^{w^-_{J^u}(\varphi_{-t}z,y)}\,dm^{u}_{x, V+J^u}(z)\\ &\quad = \int_{Z\cap R^{u}_q(y)}e^{S_t(V+J^u)(z)}f(z) e^{w^-_{J^u}(z,y)}\,d ((\varphi_{-t})_*m^{u}_{x, V+J^u})(z)\\ &\quad = \int_{Z\cap R^{u}_q(y)}e^{tP(V+J^u)-S_t(V+J^u)(z)}e^{S_t(V+J^u)(z)}f(z) e^{w^-_{J^u}(z,y)}\,dm^{u}_{\varphi_{-t}x, V+J^u}(z)\\ &\quad = e^{tP(V+J^u)} \int_{Z\cap R^{u}_q(y)}f(z) e^{w^-_{J^u}(z,y)}\,dm^{u}_{\varphi_{-t}x, V+J^u}(z), \end{align*} $$

where we used the $\varphi _t$ -conformality, see equation (31). This implies equation (45) and thus

$$ \begin{align*}\mathbf{P}^{*}\nu =P(V+J^u)\nu. \end{align*} $$

The last thing we need to do is to bound the wavefront set of the measure $\nu $ . For this, we will need some regularity result on the conditional measures of the SRB measures and the following adaptation of [Reference Bonthonneau, Guillarmou, Hilgert and Weich5, Lemma 2.9].

Proof. Consider $q\in \mathcal M$ and $\xi \notin E_s^{*}(q)$ , and we shall prove that $(q,\xi )$ is not in the wavefront set. Choose a phase function S such that $d_qS=\xi $ and a cutoff $\chi $ near q, we then use equation (43) to obtain

$$ \begin{align*} &\nu^{V+J^u}_q(\chi e^{i\frac Sh})\\ &\quad :=\!\int_{R^{u}_q}\int_{Z\cap R^{ws}_q(x)}\!\! \chi(z)e^{i\frac Sh} e^{w^+_{V}(z,x)+t(P(V+J^u)-P(V))}e^{w^+_{J^u}(z,x)}\,dm^{ws}_{x,J^u}(z)\,dm^u_{q,V+J^u}(x). \end{align*} $$

Now, we can use [Reference Climenhaga10, Corollary 3.11] to obtain that

$$ \begin{align*}\frac{d\mu_x^{ws}}{d m^{ws}_{x,J^u}}(z)=\frac{e^{w^+_{J^u}(z,x)}}{m^{ws}_x(R_q^{ws}(x))}, \quad h(x):= m^{ws}_x(R_q^{ws}(x)),\end{align*} $$

where $\mu _x^{ws}$ is the conditional measure of the SRB measure on the weak unstable manifold, in other words,

$$ \begin{align*} &\nu^{V+J^u}_q(\chi e^{i\frac Sh})\\ &\quad :=\int_{R^{u}_q}h(x)\bigg(\int_{Z\cap R^{ws}_q(x)}\!\!\chi(z)e^{i(S/h)} e^{w^+_{V}(z,x)+t(P(V+J^u)-P(V))}\mu_x^{ws}(z)\bigg)\,dm^u_{q,V+J^u}(x). \end{align*} $$

Now, we use [Reference De La Llave14, Theorem 3.9] to obtain that the density $\mu _x^{ws}$ is smooth along the leaves $R_q^{ws}(x)$ . By this, we mean that

$$ \begin{align*}\|\mu_x^{ws}\|_{C^k(R_q^{ws}(x))}:=\sup_{z\in R_q^{ws}(x)}\sup_{X_1,\ldots, X_k \in S_z R_q^{ws}(x)}|X_1\cdots X_k(\mu_x^{ws}(z)_{|R_q^{ws}(x)})| \end{align*} $$

is finite for any k and that the norm depends continuously in x.

From the proof of [Reference De La Llave14, Corollary 4.4], one sees that the smoothness of the potential V implies that $w_V^+(z,x)$ is smooth along the leaves $R_q^{ws}(x)$ . More precisely, the proof shows that given a function which is smooth along the leaves, the function defined by the last integral in equation (32) is smooth along the leaves, therefore proving that the holonomy factor is smooth along the leaves. (Note that the fact that the unstable Jacobian is smooth along the leaves in the sense above is non-trivial because it is only Hölder continuous in x.)

Using the fact that each leaf is a smooth submanifold, we can perform integration by parts (in z) to show that the inner integral is $O(h^{\infty })$ as long as $dS_{|R_w^{ws}}(x)$ does not vanish. However, $\xi \notin E_s^{*}$ , so this can be ensured locally near p. This proves that $\xi \notin E_s^{*}$ and we have thus shown that $\mathrm {WF}(\nu )\subset E_s^{*}$ .

We can now finish the proof of Theorem 3.1. For this, we use the characterization in equation (26) to deduce that $\nu $ is indeed a co-resonant state, i.e., $P(V+J^u)$ is a Ruelle resonance.

Proof. The fact that $\nu (|f|)\geq 0$ is a consequence of $\nu $ being a measure (i.e., of $\nu $ being a distribution which is non-negative on non-negative functions). The homogeneity and the triangle inequality are clear. Suppose now $\|f\|_V=0$ , then use equation (41) as well as equation (29) to see that the co-resonant state $\nu $ gives positive measure to any non-empty open set of $\mathcal M$ . Thus, if f is non-zero, then $|f|$ is positive on a small open set, which is a contradiction. The change of variable formula (47) is a direct consequence of equation (45), which was shown in the proof of Theorem 3.1.

Proof. Recall from the proof of Theorem 2.1 (see [Reference Lefeuvre28, Theorem 9.11]) that we can construct an operator $ \tilde Q(\unicode{x3bb} )$ such that, as an equality on operators acting on $C^{\infty }(M)$ ,

(48) $$ \begin{align} (\mathbf{P}+\unicode{x3bb})\tilde Q(\unicode{x3bb})=\mathrm{Id}-\tilde R(\unicode{x3bb}), \end{align} $$

where the remainder is given by

$$ \begin{align*}\tilde R(\unicode{x3bb})=-\int_T^{T+\epsilon}\chi'(t)e^{-t\unicode{x3bb}}e^{t \mathbf{P}}\,dt,\end{align*} $$

where $\chi $ is a cutoff supported in $[0,T+\epsilon [$ and constant equal to $1$ on $[0,T]$ , for some suitable choices of $\epsilon $ and T.

Let $\unicode{x3bb} \in \mathbb C$ such that $\mathrm {Re}(\unicode{x3bb} )>P(V+J^u)$ . Let us show that it is not in the Ruelle spectrum. The projector $\Pi _0(\unicode{x3bb} )$ on the eigenvalue $z=0$ of the Fredholm operator ${F(\unicode{x3bb} ):=\mathrm {Id}-\tilde R(\unicode{x3bb} )}$ is given by the integral

(49) $$ \begin{align} \Pi_0(\unicode{x3bb})=\frac{1}{2\pi i}\int_{|z|=\epsilon}(z\mathrm{Id}-\tilde R(\unicode{x3bb}))^{-1}\,dz \end{align} $$

for a radius $\epsilon>0$ small enough. Note that by the proof of Theorem 2.1, this is the spectral projector for any anisotropic Sobolev space $\mathcal H^s$ for which $\mathrm {Re}(\unicode{x3bb} )\geq -cs+C_0$ and thus equation (49) also holds as a map $C^{\infty }(\mathcal M)\to \mathcal D'(\mathcal M)$ . However, if $f\in L^1(\mathcal M,\nu )$ , then, using Lemma 3.3, one has

$$ \begin{align*} \|\tilde R(\unicode{x3bb})f\|_{V}&\leq \int_T^{T+\epsilon}|\chi'(t)|e^{-t\mathrm{Re}(\unicode{x3bb})}\times \|e^{t\mathbf{P}}f\|_{V}\,dt \\ & \leq \int_T^{T+\epsilon}|\chi'(t)|e^{-t(\mathrm{Re}(\unicode{x3bb})-P(V+J^u))}\times \|f\|_{V}\,dt\leq \frac 1 2\|f\|_{V}, \end{align*} $$

if T is chosen large enough. In particular, this shows that $F(\unicode{x3bb} )$ is invertible on $L^1(\mathcal M,\nu )$ with inverse in $\mathcal L(L^1(\mathcal M,\nu ))$ and thus $\Pi _0(\unicode{x3bb} )=0$ , meaning that $\unicode{x3bb} $ is not a Ruelle resonance.

Proof. Consider $B\in \Psi ^0(\mathcal M)$ such that its principal symbol is equal to $1$ except in a conic neighborhood V of $E_0^{*}$ . We can then write $\tilde R=\tilde RB+(\mathrm {Id}-B)\tilde R(\mathrm {Id}-B)+B\tilde R(\mathrm {Id}-B),$ where $(\mathrm {Id}-B)\tilde R(\mathrm {Id}-B)$ is smoothing thus compact (because $\mathrm {WF}(\tilde R)$ does not intersect $E_0^{*}$ , see [Reference Lefeuvre28, Theorem 9.11] for a detailed computation of the wavefront set). Moreover, we have $\|\tilde RB+B\tilde R(\mathrm {Id}-B)\|_{L^2\to L^2}\leq \tfrac 1 2$ if T is chosen large enough. This shows that the essential spectrum of $\tilde R$ is contained in $B(0,1/2)$ and that the spectrum outside this ball consists of isolated eigenvalues. Using the $L^2\to L^2$ boundedness of $\tilde R(\unicode{x3bb} )$ , we see that for large values of $|z|$ , we get the converging Neumann series (in $\mathcal H^s$ )

(50) $$ \begin{align} (z-\tilde R(\unicode{x3bb}))^{-1}=z^{-1}\sum_{k\geq 0}z^{-k}\tilde R(\unicode{x3bb})^k. \end{align} $$

Now, we use the bound in equation (47) for $f\in C^{\infty }(\mathcal M)$ :

$$ \begin{align*} \|\tilde R(\unicode{x3bb})f\|_{L^{1}(\mathcal M,\nu)}&\leq \int_T^{T+\epsilon}e^{-tP(V+J^u)}|\chi'(t)|\times \|e^{t\mathbf{P}}f\|_{L^{1}(\mathcal M,\nu)}\,dt \\&\leq \int_T^{T+\epsilon}e^{-tP(V+J^u)}|\chi'(t)|e^{tP(V+J^u)}\|f\|_{L^{1}(\mathcal M,\nu)}\,dt \leq \|f\|_{L^{1}(\mathcal M,\nu)}, \end{align*} $$

to get that the above Neumann series actually converges for $|z|>1$ in $\mathcal L(L^{1}(\mathcal M,\nu ))$ and is analytic in $|z|>1$ . By density of smooth functions in $\mathcal H^s $ , this proves that the eigenvalues are contained in the closed unit disk.

Proof. Suppose that $\tau \in \mathbb S^1$ is an eigenvalue of $\tilde R(\unicode{x3bb} _0)$ . We first check that for any smooth function u, one has $\mathbf {P}\tilde R(\unicode{x3bb} _0)u=\tilde R(\unicode{x3bb} _0)\mathbf {P}u.$ In other words, the operator $\tilde R(\unicode{x3bb} _0)$ commutes with $\mathbf {P}$ so that the space $\mathrm {Ran}(\Pi _{\tau }(\unicode{x3bb} _0))$ associated to the eigenvalue $\tau \in \mathbb S^1$ can be decomposed into generalized eigenspaces of $\mathbf {P}$ . Suppose now that $u\in \mathrm {Ran}(\Pi _{\tau }(\unicode{x3bb} _0))$ is an eigenvector for $\mathbf {P}$ for the eigenvalue $\mu $ , then we get

$$ \begin{align*}\tau u=\tilde R(\unicode{x3bb}_0)u=-\int_0^{\infty}\chi'(t)e^{-t\unicode{x3bb}_0+t\mu}u(x)\,dt=\widehat{\chi'}(-i\unicode{x3bb}_0+i\mu)u. \end{align*} $$

This implies that $|\widehat {\chi '}(-i\unicode{x3bb} _0+i\mu )|=1$ . However, using Lemma 3.4, we see that the real part of $\mu $ is smaller than $P(V+J^u)$ , so that

$$ \begin{align*}1=|\tau |= |\widehat{\chi'}(-i\unicode{x3bb}_0-i\mu)|\leq \int_0^{\infty}|\chi'(t)|e^{t\mathrm{Re}(\mu)-tP(V+J^u)}\,dt\leq 1.\end{align*} $$

This implies that $\mathrm {Re}(\mu )=P(V+J^u)$ and that $\cos (t(-i\unicode{x3bb} _0+i\mu ))$ is constant on the support of $\chi '$ . This is possible only if $\unicode{x3bb} _0=\mu $ , which gives $\tau =1$ and shows that u is a resonant state for the Ruelle resonance $\unicode{x3bb} _0$ .

The previous discussion justifies that there exists an integer N such that

$$ \begin{align*}\Pi_0(\unicode{x3bb}_0):=\{u\in \mathcal D^{\prime}_{E_u^{*}}(\mathcal M)\mid (\mathbf P-\unicode{x3bb}_0)^N u=0\}.\end{align*} $$

We will now prove that $\tilde R(\unicode{x3bb} _0)$ has no Jordan block at $1$ and deduce that $N=1$ . Note that the right-hand side of the previous equality does not depend on the choice of cutoff function and so that needs to be the case for the right-hand side. Let f be an eigenfunction of $\tilde R(\unicode{x3bb} _0)$ , then

$$ \begin{align*} R(\unicode{x3bb}_0)f=\mathrm{Id}=-\int_{\mathbb R}\chi'(t)e^{-t\unicode{x3bb}_0+t\mathbf{P}}f\,dt=f. \end{align*} $$

The above relation holds for any cutoff function $\chi $ supported in $[0,T+\epsilon ]$ and equal to $1$ on $[0,T]$ . The only condition being that T must be large enough to ensure the validity of Lemma 3.5. Approximating the Dirac mass at T (for T large enough) yields $e^{-T\unicode{x3bb} _0+T\mathbf {P}}f=f$ and f is thus a resonant state at $\unicode{x3bb} _0$ . The converse is clear and we showed that the eigenfunctions of $\tilde R(\unicode{x3bb} _0)$ at $1$ coincide with the resonant states at $\unicode{x3bb} _0$ . Let $u\in C^{\infty }$ , then near $z=1$ ,

(52) $$ \begin{align} (\tilde R(\unicode{x3bb}_0)-z)^{-1}u=\tilde R_H(z)+\sum_{j=1}^{N}\frac{(\tilde R(\unicode{x3bb}_0) -1)^{j-1}\Pi_0(\unicode{x3bb}_0)u}{(z-1)^j}, \end{align} $$

where $R_+^H(z)$ is the holomorphic part near $z=1$ . We note that $(\tilde R(\unicode{x3bb} _0) -1)^{N-1}\Pi _0(\unicode{x3bb} _0)u$ is an eigenevector of $\tilde R(\unicode{x3bb} )$ for $1$ . By the previous discussion, it is thus a resonant state $\theta $ associated to $\unicode{x3bb} _0$ . By Lemma 3.1, we can consider a co-resonant state $\mu $ such that the pairing $(\theta , \mu )\neq 0$ . On one hand, using equation (52) gives, near $z=1$ ,

(53) $$ \begin{align} \langle (\tilde R(\unicode{x3bb}_0)-z)^{-1}u,\mu\rangle_{\mathcal H^s\times \mathcal H^{-s}}=\frac{(\theta, \mu)}{(z-1)^{N}}+O\bigg(\frac{1}{(1-z)^{N-1}}\bigg). \end{align} $$

However, on the other hand, for any $|z|>1$ , one has, for any $k\geq 0$ ,

$$ \begin{align*}z^{-k}\langle \tilde R(\unicode{x3bb}_0)^ku,\mu\rangle=z^{-k}\langle u,\tilde (R(\unicode{x3bb}_0)^{*})^k\mu\rangle=z^{-k}\langle u,\mu\rangle. \end{align*} $$

Summing the Neumann series then gives

(54) $$ \begin{align} |\langle (\tilde R(\unicode{x3bb}_0)-z)^{-1}u,\mu\rangle_{\mathcal H^s\times \mathcal H^{-s}}|\leq \frac{|\langle u,\mu \rangle |}{|z|^{-1}-1}. \end{align} $$

Going back to equation (41), this is possible only if $N=1$ . In particular, resonances on the critical axis have no Jordan block. Finally, we use that

$$ \begin{align*}\tilde R(\unicode{x3bb}_0)=\Pi_0(\unicode{x3bb}_0)+K(\unicode{x3bb}_0),\quad K(\unicode{x3bb}_0)\Pi_0(\unicode{x3bb}_0)=\Pi_0(\unicode{x3bb}_0)K(\unicode{x3bb}_0)=0,\end{align*} $$

where the spectral radius of $K(\unicode{x3bb} _0)$ on $\mathcal H^s$ is strictly smaller than $1$ . Remember that if $r<1$ is the spectral radius of $K(\unicode{x3bb} _0)$ , then for any $\epsilon>0$ small enough, one has

$$ \begin{align*}\|K(\unicode{x3bb}_0)^n\|_{\mathcal H^s\to \mathcal H^s}\leq (r+\epsilon)^n\to 0. \end{align*} $$

Thus, as bounded operators on $\mathcal H^s$ , we get, as $n\to +\infty $ , $ \tilde R(\unicode{x3bb} _0)^n=\Pi _0(\unicode{x3bb} _0)+K(\unicode{x3bb} _0)^n\to \Pi _0(\unicode{x3bb} _0).$

Proof. Let $v\in C^{\infty }(\mathcal M,[0,+\infty [)$ and first consider $\unicode{x3bb} =P(V+J^u)$ . Then, using Lemma 3.6, we get $\tilde R(\unicode{x3bb} )^n\to \Pi _0(\unicode{x3bb} )$ in $\mathcal L(\mathcal H^s)$ . Now, for any $u\in C^{\infty }(\mathcal M,[0,+\infty [)$ ,

$$ \begin{align*} \langle \tilde R(P(V+J^u))^ku,v\rangle &= -\int_{0}^{+\infty}\chi^{\prime *k}(t)e^{-tP(V+J^u)}\bigg(\int_{\mathcal M}e^{-t\mathbf{P}^{*}}vu\,d\mathrm{vol} \bigg)\,dt\geq 0. \end{align*} $$

This proves that $\langle \Pi _0(P(V+J^u))u,v\rangle \geq 0$ , which gives that $\mu _v$ is a non-negative Radon measure for $v\geq 0$ . Consider a general resonance on the critical axis $\unicode{x3bb} =ib+P(V+J^u)$ with $b\in \mathbb R$ and $v\leq v'\in C^{\infty }(\mathcal M, [0,+\infty [)$ , we get

$$ \begin{align*} \text{for all } u\kern1.3pt{\in}\kern1.3pt C^{\infty}(\mathcal M, \mathbb R),\ |\langle \tilde R^k(\unicode{x3bb})u,v\rangle |\kern1.3pt{\leq}\kern1.3pt \langle \tilde R^k(P(V\kern1.3pt{+}\kern1.3pt J^u))|u|,v\rangle \kern1.3pt{\leq}\kern1.3pt \langle \tilde R^k(P(V\kern1.3pt{+}\kern1.3pt J^u))|u|,v'\rangle. \end{align*} $$

Passing to the limit, we get first that $\mu _v^{\unicode{x3bb} }$ defines an order $0$ distribution and thus defines a (complex) measure. Moreover, the last inequality actually proves that $\mu _v^{\unicode{x3bb} }\ll \mu _{v'}$ with a density bounded by $1$ . In particular, one gets that every measure $ \mu _v^{\unicode{x3bb} }$ is absolutely continuous with respect to the measure $\mu _1$ .

Proof. Consider a co-resonant state $\mu $ for the first Ruelle resonance $P(V+J^u)$ . We know by the previous proposition that $\mu $ is a measure with wavefront set contained in $E_s^{*}$ and we will suppose that it is a non-negative measure. The wavefront set condition, together with [Reference Hormander26, Corollary 8.2.7], justifies that one can restrict the distribution $\mu $ to any unstable manifold $\mathcal W^u(x)$ . We will denote by $\mu _x:=\mu _{|\mathcal W^u(x)}$ . More explicitly, for a smooth function $f\in C^{\infty }(\mathcal M)$ , we can define, for some $s\in \mathbb R$ ,

$$ \begin{align*}(f_{|\mathcal W^u(x)},\mu_x):=(f\delta_{\mathcal W^u(x)}, \mu)=(f\delta_{\mathcal W^u(x)}, \mu)_{\mathcal H^s, \mathcal H^{-s}}, \end{align*} $$

where the bracket denotes the distributional pairing and $\delta _{\mathcal W^u(x)}$ denotes the integration over the unstable manifold $\mathcal W^u(x)$ defined by

(56) $$ \begin{align} (f, \delta_{\mathcal W^u(x)}):=\int_{\mathcal W^u(x)}f(y)\,d\mathrm{vol}_{\mathcal W^u(x)}(y). \end{align} $$

The fact that the distributional pairing coincides with the $\mathcal H^s \times \mathcal H^{-s}$ pairing is a consequence of the wavefront set condition on the distributions. In particular, $\mu _x$ is a non-negative distribution as the product of two non-negative distributions. The restrictions $\{\mu _x \mid x\in \mathcal M\}$ define a system of $\mathcal W^{ws}$ -transversal measures in the sense of [Reference Climenhaga10]. The strategy to show that the first resonance is simple is to use [Reference Climenhaga10, Corollary 3.12], which asserts that the only system (up to constant rescaling) of $\mathcal W^{ws}$ -transversal measures which satisfies $\varphi _t$ -conformality (see equation (30)) and the change of variable by holonomy (see equation (35)) is $m^u_{V+J^u}$ .

The system of measures $\{\mu _x \mid x\in \mathcal M\}$ satisfies $\varphi _t$ -conformality. This is a consequence of the fact that $\mu $ is a co-resonant state. More precisely, for any smooth function ${f\in C^{\infty }(\mathcal M)}$ , we can write

$$ \begin{align*}( f\delta_{\mathcal W^u(x)}, \mu)_{\mathcal H^s \times \mathcal H^{-s}}=e^{-tP(V+J^u)}( e^{t\mathbf P}( f\delta_{\mathcal W^u(x)}),\mu)_{\mathcal H^s \times \mathcal H^{-s}}.\end{align*} $$

Using $e^{t\mathbf P}( f\delta _{\mathcal W^u(x)})=e^{t\mathbf P} fe^{-tX}\delta _{\mathcal W^u(x)}$ , then yields

$$ \begin{align*} \int_{\mathcal W^u(x)}\! f(y)\,d\mu_x(y)=\!\int_{\mathcal W^u(\varphi_t x)}\!\!e^{-tP(V+J^u)}e^{S_tV(\varphi_{-t}y)}f(\varphi_{-t}y) \det (d\varphi_{-t})_{|E_u(y)}\,d\mu_{\varphi_t x}(y).\end{align*} $$

This rewrites

$$ \begin{align*}\int_{\mathcal W^u(x)}f(y)\,d\mu_x(y)=\! \int_{\mathcal W^u(\varphi_t x)}e^{-tP(V+J_u)}e^{S_t(V+J^u)(\varphi_{-t}y)}f(\varphi_{-t}y)\,d\mu_{\varphi_t x}(y),\end{align*} $$

which is exactly $\varphi _t$ -conformality for the potential $V+J^u$ .

The system of measures $\{\mu _x \mid x\in \mathcal M\}$ satisfies the change of variable by holonomy. Investigating the proof of [Reference Climenhaga10, Corollary 3.12], we see that we only need to show the change of variable formula for the standard $\delta _0$ -holonomy given by $\mathcal W^u(x)\ni z\mapsto \pi (z):=[z,x']\in \mathcal W^u(x')$ . We have the following facts.

• • For all $x,x'\kern1.3pt{\in}\kern1.3pt \mathcal M$ , $x\kern1.3pt{\neq}\kern1.3pt x'$ close, one has $L_{x,x'}:\phi \kern1.3pt{\in}\kern1.3pt C^{0}(\mathcal M)\kern1.3pt{\mapsto}\kern1.3pt {(\mu _x\kern1.3pt{-}\kern1.3pt\mu _{x'},\phi )}/{d(x,x')^{\alpha }}$ is linear.

• • For any $\phi \in C^{0}(\mathcal M)$ , one has $\sup _{x\neq x'} |L_{x,x'}(\phi )|<+\infty $ by equation (A.1).

• • The space $C^{0}(\mathcal M)$ is a Banach space.

We can apply the Banach–Steinhauss theorem to get $\sup _{x\neq x'} \|L_{x,x'}\|_{\mathrm {op}}=C<+\infty $ , i.e.,

$$ \begin{align*} \text{for all } \phi\in C^{0}(\mathcal M),\ \text{for all } x,x'\in \mathcal M, \quad |(\mu_x-\mu_{x'},\varphi)|\leq Cd(x,x')^{\alpha}. \end{align*} $$

Now consider $\mathcal W^u(x)\ni z\mapsto \pi (z):=[z,x']\in \mathcal W^u(x')$ . Our goal is to relate $\mu _x$ and $\pi _{x'}^{*}\mu _{x'}$ . Consider $f\in C^0(\mathcal W^u(x,\delta _0))$ . Since the foliation is Hölder, we can extend f locally on a small rectangle by making it constant on weak-stable manifolds, and this defines a continuous function $\bar f$ . We are now left with relating $\mu _x(\bar f)$ and $\mu _{x'}(\bar f)$ .

First, consider the Bowen time $t=t(x,x')$ such that $\varphi _tx\in \mathcal W^s(x')$ . Then, $\varphi _t$ - conformality shows that ${d[(\varphi _t)_*\mu _x]}/{d\mu _{\varphi _tx}}(z)=e^{S_t(V+J^u)(z)-tP(V+J^u)}$ . Now, for any $\tau>0$ , we can use $\varphi _t$ -conformality again to get ${d[(\varphi _\tau )_*\mu _{\varphi _tx}]}/{d \mu _{\varphi _{t+\tau }x}}(z)=e^{S_{\tau }(V+J^u)(z)-\tau P(V+J^u)}$ and ${d[(\varphi _\tau )_*\mu _{x'}]}/{d\mu _{\varphi _\tau x'}}(y)=e^{S_{\tau }(V+J^u)(y)-\tau P(V+J^u)} $ . Now, we have that $d(\varphi _{t+\tau }x, \varphi _\tau x')\leq Ce^{-\eta \tau }$ and the above continuity then shows that $|\mu _{\varphi _{t+\tau }x}( \bar f)-\mu _{\varphi _\tau x'}(\bar f)|\to 0$ when $\tau \to +\infty $ . This means that the overall holonomy factor is given by the following limit:

$$ \begin{align*}\exp(\lim_{\tau \to +\infty} S_t(V+J^u)(z)-tP(V+J^u)+S_\tau(V+J^u)(\varphi_{\tau+t} z)-S_{\tau}(V+J^u)(\varphi_{\tau} y)), \end{align*} $$

which is exactly equal to $e^{w_{V+J^u}^+(z,y)}$ , see the proof of [Reference Climenhaga10, Theorem 3.9] for a more detailed argument.

The first resonance $P(V+J^u)$ is simple. We can use [Reference Climenhaga10, Corollary 3.12] to claim that there is a constant $c>0$ such that for any $x\in \mathcal M$ , one has $\mu _x=c m^u_{x,V+J^u}$ . We now would like to deduce that this implies $\mu =c\nu $ by proving a Fubini-like formula for distributions with wavefront set in $E_s^{*}$ . For this, we first need to disintegrate the volume measure with respect to the unstable foliation and an arbitrary (smooth) transversal complementary foliation G. More precisely, we will use [Reference Bonthonneau, Guillarmou and Weich4, Proposition 1.6], which states that in a product neighborhood U around a point $q\in \mathcal M$ , there is a continuous function $\delta _u: U\to \mathbb R_+$ such that

$$ \begin{align*} \text{for all } f\in C^{\infty}(\mathcal M),\ \int_U f\,d\mathrm{vol}=\int_{G_{\mathrm{loc}}(p)}\bigg(\!\int_{\mathcal W^u(y)}f(z)\delta_u(z)\,d\mathrm{vol}_{\mathcal W^u(y)}(z)\bigg)\,d\mathrm{vol}_p^G(y), \end{align*} $$

where $G_{\mathrm {loc}}(p)$ is the connected component of the element of the partition of G containing p. Actually, the conditional density $\delta _u$ is smooth along the leaves in the following sense:

$$ \begin{align*} \|\delta_u\|_{C^k(R_q^{ws}(x))}:=\sup_{z\in \mathcal W^{u}(x)}\sup_{X_1,\ldots, X_k \in S_z \mathcal W^{u}(x)} |X_1\cdots X_k(\delta_u(z))_{|\mathcal W^{u}(x)}| \end{align*} $$

is finite for any k and that the norm depends continuously in x. This regularity condition allows to integrate by parts in the inner integral and, in particular, an immediate adaptation of [Reference Bonthonneau, Guillarmou and Weich4, Lemma 1.9] yields $\mathrm {WF}(\delta _u(z) \delta _{\mathcal W^u(y)})\subset E_u^{*}\oplus E_0^{*}$ . Here, we have denoted by $\delta _u(z) \delta _{\mathcal W^u(y)}$ the distribution:

$$ \begin{align*}f\mapsto \int_{\mathcal W^u(y)}f(z)\delta_u(z)\,d\mathrm{vol}_{\mathcal W^u(y)}(z).\end{align*} $$

In particular, we consider $f_n\in C^{\infty }(\mathcal M)$ such that $f_n\to \mu $ is $\mathcal D^{\prime }_{\Gamma }(\mathcal M)$ , where $\Gamma $ is a conic neighborhood of $E_s^{*}$ which does not intersect $E_u^{*}\oplus E_0^{*}$ , and we can use the continuity of the distributional product (see [Reference Lefeuvre28, Lemma 4.2.7]) to write for any $f\in C^{\infty }(\mathcal M)$ ,

$$ \begin{align*}(f,\mu)=&\lim_{n\to +\infty}(f,f_n)= \lim_{n\to +\infty}\int_{G_{\mathrm{loc}}(p)} (\delta_u(z) \delta_{\mathcal W^u(y)},f\times f_n)\,d\mathrm{vol}_p^G(y) \\&=\int_{G_{\mathrm{loc}}(p)} (\delta_u(z) \delta_{\mathcal W^u(y)},f\times \mu)\,d\mathrm{vol}_p^G(y) \\&= \int_{G_{\mathrm{loc}}(p)} \mu_y(f\times \delta_u)\,d\mathrm{vol}_p^G(y)=c\int_{G_{\mathrm{loc}}(p)} m^u_{y,V+J^u}(f\times \delta_u)\,d\mathrm{vol}_p^G(y) \\&=c(f,\eta), \end{align*} $$

where we used the continuity of the density and the fact that $\mu _x$ is of order $0$ . Together with the previous lemma, this proves that all co-resonant states are proportional, and thus that $P(V+J^u)$ is simple and concludes the proof.

Proof. The co-resonant state $\nu $ is absolutely continuous with respect to $\mu _{V+J^u}$ . We follow the proof of [Reference Climenhaga10, Theorem 3.10] and prove that there exists a constant $C>0$ independent of $q\in \mathcal M$ and $t\geq 0$ such that

(57) $$ \begin{align} \text{for all } t\geq 0, \text{ for all } q\in \mathcal M, \quad \nu(B_t(q,r))\leq Ce^{S_t(V+J^u)(q)-tP(V+J^u)}, \end{align} $$

which suffices as $\mu _{V+J^u}$ is the the Gibbs state (see [Reference Fisher and Hasselblatt19, Theorem 4.3.26]). From the proof of [Reference Climenhaga10, Theorem 3.10], we get that

$$ \begin{align*} \text{there exists } C>0 \text{ for all } q\in \mathcal M, \text{ for all } t\geq 0, \quad &m^{u}_{V+J^{u}}(B_t(q,\delta))\\ &\qquad\leq C e^{S_t(V+J^u)(q)-tP(V+J^u)}. \end{align*} $$

Now, we can use equation (43), the fact that the integrand $e^{w_{J^u}^-(z,x)}$ is uniformly bounded on $\mathcal M$ , and $m^{ws}_{q,J^u}(R_q^{ws})$ is uniformly bounded in q (see [Reference Climenhaga10, Theorem 3.1]) to get equation (57).

Product formula and averaging formula. Let $\eta $ be the resonant state, which we can obtain by applying Theorem 3.1 to the dual $\mathbf {P}^{*}$ . We have $\mathrm {WF}(\eta ) \subset E_u^{*}$ , $\mathrm {WF}(\nu ) \subset E_s^{*}$ and we have $E_s^{*}\cap E_u^{*} \cap (T^{*}\mathcal M\setminus \{0\}) = \emptyset $ , which shows that the distributional product of $\eta $ and $\nu $ is well defined. Actually, we can use the duality in equation (21) to define the product:

(58) $$ \begin{align} \text{for all } f\in C^{\infty}(\mathcal M),\quad (\eta \times \nu)(f):=(\eta, f \nu)_{\mathcal H^s \times \mathcal H^{-s}}=(f\eta, \nu)_{\mathcal H^s \times \mathcal H^{-s}}, \end{align} $$

where the equality is justified by equation (22). From the fact that $\nu $ and $\eta $ are non-negative measures (see Proposition 3.1), and equation (58), we see that this is also the case of their product $\eta \times \nu $ . It is easily seen to be invariant by the flow:

$$ \begin{align*} (\eta \times \nu)(Xf)&=( X^{*}\eta,f\nu)-(f\eta,X\nu) \\&=P(V+J^u)(\eta,f\nu)-(V\eta,f\nu)+(\mathrm{div}_{\mathrm{vol}}(X)\eta,f\nu) \\&-P(V+J^u)(f\eta,\nu)+(f\eta,V\nu)-(\mathrm{div}_{\mathrm{vol}}(X)f \eta,\nu)=0. \end{align*} $$

We prove that the product is absolutely continuous with respect to $\mu _{V+J^u}$ . For this, we use Proposition 3.1 and the previous proposition to get $\Pi _0(P(V+J^u))1=c\eta $ for some $c>0$ . Now, this means that we can compute $(\eta \times \nu )(f)$ for $f\in C^{\infty }(\mathcal M)$ :

$$ \begin{align*} c(\eta \times \nu)(f)&=( \Pi_0(P(V+J^u))1,f\nu)_{\mathcal H^s \times \mathcal H^{-s}} \\&=-\lim_{k\to +\infty}\int_0^{+\infty}(\chi')^{*k}(t)e^{-tP(V+J^u)}( 1, e^{t\mathbf{P}^{*}}(f\nu))\,dt \\&=-\lim_{k\to +\infty}\int_0^{+\infty}(\chi')^{*k}(t)( e^{tX}f, \nu)\,dt. \end{align*} $$

We use the fact that $\nu $ is absolutely continuous with respect to $\mu _{V+J^u}$ with bounded density to get that $|\nu (e^{tX}f)|\leq C\mu _{V+J^u}(|e^{tX}f|)\leq C\mu _{V+J^u}(|f|)$ because $\mu _{V+J^u}$ is invariant. In particular, we get, by integrating,

$$ \begin{align*}\text{for all } f\in C^{\infty}(\mathcal M),\quad (\eta \times \nu)(|f|)\leq C'\mu_{V+J^u}(|f|) \Rightarrow (\eta \times \nu)\ll \mu_{V+J^u}. \end{align*} $$

This proves that the density of $(\mu \times \nu )$ is invariant by the flow and measurable for the equilibrium state $\mu _{V+J^u}$ . However, this last measure is known to be ergodic so the density is constant, which proves the two formulas.

Weak mixing implies no other resonances on the critical axis. From the averaging formula (8), we see that if $\nu $ gives zero measure to an invariant Borel set, then it is also the case for $\mu _{V+J^u}$ and the two measures are actually equivalent.

Next, we use [Reference Reed and Simon33, Theorem VII.14] to see that the flow is weakly mixing if and only if the only eigenvalue is $1$ and it is a simple eigenvalue. In other words, if

(59) $$ \begin{align} \begin{cases} Xf=i\unicode{x3bb} f, \\ f\in L^2(\mathcal M,\mu_{V+J^u}) \end{cases} \end{align} $$

has no solution except for $\unicode{x3bb} =0$ and f constant. Using Proposition 3.1, the presence of another co-resonant state on the critical axis is equivalent to the existence of an absolutely continuous measure with respect to $\nu $ . Its density with respect to $\nu $ is in $L^{\infty }(\mathcal M,\nu )=L^{\infty }(\mathcal M,\mu _{V+J^u})\subset L^2(\mathcal M,\mu _{V+J^u})$ and is thus a solution of the system above. This shows that weak mixing of the flow with respect to $\mu _{V+J^u}$ implies that there is no other resonance on the critical axis.

Proof. Our first goal is to give meaning to $m^u_V(\varphi )$ for $\varphi \in C^{\infty }(\mathcal M; \Lambda ^{d_s}T^{*}\mathcal M)$ . First, the compatibility statement allows us to only define the duality locally, so let ${\varphi \in C^{\infty }(\mathcal M; \Lambda ^{d_s}T^{*}\mathcal M)}$ be supported in $R_q$ . We can define the duality as follows:

(62) $$ \begin{align} m^u_V(\varphi):=\int_{\mathcal M}m^u_V\wedge \alpha \wedge \varphi=\int_{\mathcal W^u(q,\delta)}\bigg(\!\int_{R_q^{ws}(x)}e^{w_V^+(y,x)}(\varphi\wedge \alpha)(y)\bigg)\,dm^u_V(x), \end{align} $$

where $\ \alpha \in E_0^{*}, \ \alpha (X)=1$ . We see that the previous definition makes sense as $\varphi \wedge \alpha $ is a $d_s+1$ form and as $w_V^+(x,y)$ is smooth in y, in the sense of Lemma 3.2, and continuous in x. The formula clearly defines a current of degree $0$ so $m^u_V\in D' (\mathcal M; \Lambda ^{d_u}(E_u^{*}\oplus E_s^{*}))$ .

Let $\omega _k \in C^{\alpha }(\mathcal M: \Lambda _k^{d_s})$ for some k. Then, the fact that $m^u_V$ is a measure on $\mathcal W^u(q,\delta )$ shows that the pairing $m^u_V(\omega _k)$ is well defined. If $k\geq 1$ , we can use equation (62) and the definition of $E_s^{*}$ to get that $m^u_V(\omega _k)=0$ .

To prove the first part of equation (61), we can consider a $d_s$ form $\varphi $ supported in $R_q$ such that $e^{-t\mathbf {P}}\varphi $ is also supported in $R_q$ . We have to prove that $m^u_V(e^{t\mathbf {P}}\varphi )=e^{tP(V)}m^u_V(\varphi )$ . We first use the Leibniz rule to get $\mathcal L_X(\varphi \wedge \alpha )=\mathcal L_X\varphi \wedge \alpha +\varphi \wedge \mathcal L_X\alpha =\mathcal L_X\varphi \wedge \alpha. $ This allows us to compute explicitly the action of the propagator on the current $m^u_V$ . More precisely, we use the cocycle relation in equation (34), equation (33), as well as $\varphi _t$ conformality in equation (30):

$$ \begin{align*} &m^u_V(e^{t\mathbf{P}}\varphi)=\int_{\mathcal W^u(q,\delta)}\bigg(\!\int_{R_q^{ws}(x)}e^{w_V^+(y,x)}e^{S_tV(\varphi_{-t}y)}e^{t\mathcal L_{{X}}} (\varphi \wedge \alpha)(y)\bigg)\,dm^u_V(x) \\&\quad=\int_{\mathcal W^u(q,\delta)}\bigg(\!\int_{R_q^{ws}(\varphi_tx )}e^{w_V^+(\varphi_t w,x)}e^{S_tV(w)}(\varphi \wedge \alpha)(w)\bigg)\,dm^u_V(x) \\&\quad=\int_{\mathcal W^u(q,\delta)}\!\!e^{w_V^+(\varphi_t x,x)}\bigg(\!\int_{R_q^{ws}(\varphi_t x )}\!\! e^{w_V^+(\varphi_t w,w)}e^{w_V^+(w,\varphi_tx)}e^{S_tV(w)}(\varphi \wedge \alpha)(w)\bigg)\,dm^u_V(x) \\&\quad=\int_{\mathcal W^u(q,\delta)}e^{2tP(V)}e^{-S_tV(x)}\bigg(\!\int_{R_q^{ws}(\varphi_t x )}e^{w_V^+(w,\varphi_tx)}(\varphi \wedge \alpha)(w)\bigg)\,dm^u_V(x) \\&\quad=e^{tP(V)}\int_{\mathcal W^u(q,\delta)}\bigg(\!\int_{R_q^{ws}(z )} e^{w_V^+(z,w)} (\varphi \wedge \alpha)(z)\bigg)\,dm^u_V(z)=e^{tP(V)}m^u_V(\varphi). \end{align*} $$

For the wavefront set condition, we mimic the argument of Lemma 3.2, and consider a smooth $d_s$ -form $\chi $ supported in $R_q$ and S a phase function such that $dS(q)=\xi \notin E_s^{*}$ , and compute

$$ \begin{align*} m^u_V(e^{i(S/h)}\chi)=\int_{\mathcal W^u(q,\delta)}\bigg(\!\int_{R_q^{ws}(x)}e^{w_V^+(y,x)}e^{i({S(y)}/h)}(\chi\wedge \alpha)(y)\bigg)\,dm^u_V(x).\end{align*} $$

Now, the proof is easier than for Lemma 3.2 as the integrand is easily seen to be smooth along the weak-stable leaves (because the potential V is smooth) uniformly in x. We can perform integration by parts and show that the integrand is $O(h^{\infty })$ as long as $dS$ does not vanish on $R_q^{ws}(x)$ , which can be ensured near q by the definition of $E_s^{*}$ . This shows that $\xi \notin \mathrm {WF}(m^u_V)$ and thus $ \mathrm {WF}(m^u_V)\subset E_s^{*}$ . This shows that $P(V)$ is a Ruelle resonance with the associated Ruelle co-resonant state given by $m^u_V$ .

Proof. Suppose $\varphi \in C^{0}(\mathcal M;\Lambda ^{d_s}_0)$ , the bundle is one-dimensional and it thus makes sense to talk of $|\varphi |\wedge \alpha $ as a $d_s+1$ density. If $\varphi (q)\neq 0$ , then by continuity, $\varphi \neq 0$ on a small open set. Then, $\|\varphi \|_{V,0}>0$ because $m^u_V$ gives a positive measure to any open set. The bound in equation (64) follows from the fact that $m^u _V$ is a co-resonant state. The last point follows from the first two points by a direct adaptation of the proof of Lemma 3.4.

Proof. Contracting with a vector in $E_u$ and then wedging with a vector in $E_u^{*}$ sends a section of $C^0(\mathcal M; \Lambda _{k}^{d_s})$ to $C^0(\mathcal M; \Lambda _{k-1}^{d_s})$ , so the formula makes sense by induction. Let us prove that the norm takes finite values. For this, we use the fact that $m^u_V$ is of order zero and Lemma 4.2 to get first that for $f\in C^0(\mathcal M; \Lambda _{0}^{d_s})$ , one has $\|f\|_{V,0}\leq C\|f\|_0$ . This implies, for any $f\in C^0(\mathcal M; \Lambda _{1}^{d_s})$ ,

$$ \begin{align*} \sum_{i=1}^{d_u}\sum_{j=1}^{d_s}\|\iota_{X_u^j}f\wedge Y_u^{i}\|_{V,0}\leq C'\sum_{i,j}\|X_u^j\|_{C^0}\|Y_u^i\|_{C^0}\|f\|_0\leq C"\|f\|_0 \end{align*} $$

with a constant $C"$ independent of the choice of cover or partition of unity. This shows that $\|.\|_{V,1}$ takes finite values and a quick induction shows that it is also the case of $\|.\|_{V,k}$ for any k. The triangle inequality, homogeneity, and non-negativity of $\|.\|_{V,k}$ follow from Lemma 4.2 and an induction. Suppose now that $f \in C^0(\mathcal M; \Lambda _{k}^{d_s})$ and $f\neq 0$ . Then, f is non-zero on a small open set, and since $(X_{s,h}^j)$ and $(Y_u^i)$ are local bases, then there is $(i_0,j_0)$ such that $\iota _{X_{s,h}^{j_0}}f\wedge Y_u^{i_0}$ is continuous and non-zero on that open set. By induction and the fact that $m^u_V$ gives a positive measure to any non-empty open set, we get that $\|f\|_{V,k}>0$ and $\|.\|_{V,k}$ thus defines a norm. We prove equation (66):

$$ \begin{align*} \iota_{X_u^j}e^{t\mathbf{P}}f\wedge Y_u^{i}&=e^{t\mathbf{P}}( \iota_{(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)}f\wedge e^{t\mathcal L_X}Y_u^{i})(y) \\&=e^{t\mathbf{P}}\iota_{(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)(\varphi_t y)}f(y)\wedge e^{t\mathcal L_X}Y_u^{i}(y). \end{align*} $$

Using the invariance of the Anosov decomposition in equation (13), we get

$$ \begin{align*} &\sum_{i=1}^{d_s}\sum_{j=1}^{d_u}e^{t\mathbf{P}}\iota_{(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)(\varphi_t y)}f(y)\wedge e^{t\mathcal L_X}Y_u^{i}(y) \\ &\quad =\sum_{i,j}\|e^{t\mathcal L_X}Y_u^{i}\|\|(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)\| e^{t\mathbf{P}}\iota_{{(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)}/{\|(d\varphi_{-t})_{\varphi_{t}y}(X_u^j)\|}}f(y)\wedge \frac{e^{t\mathcal L_X}Y_u^{i}}{\|e^{t\mathcal L_X}Y_u^{i}\|}. \end{align*} $$

We then have $\|e^{t\mathcal L_X}Y_u^{i}\|,\|(d\varphi _{-t})_{\varphi _{t}y}(X_u^j)\|\leq Ce^{-t\eta }$ for some (uniform) $C,\eta>0$ by the Anosov property. Finally, we obtain

$$ \begin{align*} \sum_{i=1}^{d_s}\sum_{j=1}^{d_u}\|\iota_{X_u^j}e^{t\mathbf{P}}f\wedge Y_u^{i}\|_{V,k-1}\leq Ce^{-2\eta t} \sum_{i=1}^{d_s}\sum_{j=1}^{d_u} \|e^{t\mathbf{P}}(\iota_{\tilde X_u^j}f\wedge \tilde Y_u^{i})\|_{V,k-1} \end{align*} $$

for $\tilde X_u^j:=(d\varphi _{-t})_{\varphi _{t}y}(X_u^j)/\|(d\varphi _{-t})_{\varphi _{t}y}(X_u^j)\|$ and $\tilde Y_u^{i}=e^{t\mathcal L_X}Y_u^{i}/\|e^{t\mathcal L_X}Y_u^{i}\|$ . Passing to the supremum thus gives

$$ \begin{align*}\| e^{t\mathbf{P}}f\|_{V,k-1}\leq Ce^{(P(V)-2(k-1)\eta) t}\|f\|_{V,k-1} \Rightarrow \| e^{t\mathbf{P}}f\|_{V,k}\leq Ce^{(P(V)-2k\eta) t}\|f\|_{V,k}.\end{align*} $$

To conclude the proof of equation (66), we only need to initialize the induction, and this is exactly the statement of Lemma 4.2.

Consider $f\in C^{\infty }(\mathcal M; \mathscr E_{d_s})$ and consider its decomposition

$$ \begin{align*}f= \sum_{k=0}^{d_s} \omega_k, \quad \omega_k \in C^{\alpha}(\mathcal M; \Lambda_k^{d_s}). \end{align*} $$

We define a norm on $C^{\infty }(\mathcal M; \mathscr E_{d_s})$ with the following bound when using the propagator:

$$ \begin{align*}\|f\|_{V}:=\sum_{k=0}^{d_s}\|\omega_k\|_{V,k}, \quad \|e^{t\mathcal L_X}f\|_V\leq Ce^{tP(V)}\|f\|_V.\end{align*} $$

This is the only thing we need to mimic the argument of Lemmas 3.4 and 3.5, and prove that no resonance exists in $\{\mathrm {Re}(\unicode{x3bb} )>P(V)\}$ and that resonances on the critical axis have no Jordan block.

Proof. The argument of Lemma 3.6 goes through and the projector $ \Pi _0(\unicode{x3bb} )$ on the resonant states is obtained as the limit of $\tilde R(\unicode{x3bb} )^n$ as an operator $\mathcal H^s \to \mathcal H^s$ . The formula defining $\tilde R(\unicode{x3bb} )$ is the same. In particular, as in the proof of Proposition 3.1, we obtain every co-resonant state by the following construction:

$$ \begin{align*} &\text{for all } \unicode{x3bb}\kern1.3pt{\in}\kern1.3pt P(V\kern-0.1pt)\kern1.3pt{+}\kern1.3pti\mathbb R, \text{ for all } v\kern1.3pt{\in}\kern1.3pt C^{\infty}(\mathcal M; \Lambda^{\kern-0.1pt d_s}(E_u^{*}\kern1.3pt{\oplus}\kern1.3pt E_s^{*})), \quad \!\Pi_0(\unicode{x3bb})^{\kern-0.1pt*} (v)\kern1.3pt{=}\kern1.3pt\lim_{n\to +\infty}\!\! (\tilde R(\unicode{x3bb})^{\kern-0.1pt*})^{\kern-0.1ptn}(\kern-0.1ptv), \end{align*} $$

where the convergence holds in $\mathcal H^s$ . Consider the pairing between $C^{\infty }(\mathcal M; \Lambda ^{d_s}(E_u^{*}\oplus E_s^{*}))$ and $C^{\infty }(\mathcal M; \Lambda ^{d_u}(E_u^{*}\oplus E_s^{*}))$ defined by $\langle v, u\rangle :=\int _{\mathcal M}u\wedge \alpha \wedge v,$ extended to ${\mathcal H^s\times \mathcal H^{-s}}$ .

Co-resonant states define currents of order 0. Consider two smooth forms ${\omega \in C^{\infty }(\mathcal M; \Lambda ^{d_s}(E_u^{*}\oplus E_s^{*}))}$ and $\theta \in C^{\infty }(\mathcal M; \Lambda ^{d_u}(E_u^{*}\oplus E_s^{*}))$ . We first expand these forms:

(67) $$ \begin{align} \omega=\sum_{k=0}^{d_s}\omega_k,\quad \omega_k\in C^{\alpha}(\mathcal M; \Lambda_k^{d_s}), \quad \theta=\sum_{k=0}^{d_s}\theta_k,\quad \theta_k\in C^{\alpha}(\mathcal M; \Lambda_k^{d_u}). \end{align} $$

Note that the regularity of the component is only Hölder continuous here because of the regularity of the stable and unstable foliation. We see that $\omega _k\wedge \alpha \wedge \theta _l$ is a continuous n-form which is non-zero if and only if $k+l=d_s$ . The previous discussion justifies the following convergence:

$$ \begin{align*} \langle \omega, \Pi_0(\unicode{x3bb})^{*}\theta\rangle&=-\lim_{k\to+\infty}\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\langle e^{t\mathbf P}\omega, \theta\rangle\,dt \\&=- \lim_{k\to+\infty}\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\sum_{k=0}^{d_s}\langle e^{t\mathbf P}\omega_k, \theta_{d_s-k}\rangle\,dt. \end{align*} $$

Here, we have noticed that the integration over all the manifold is an n-current of order zero and thus extends to continuous n-forms.

By Assumption 1, we can find a trivialization basis $(Y^j_{u})_{1\leq j \leq d_s}\in C^0(\mathcal M; E_u^{*})$ such that $ E_u^{*}=\mathrm {Span}\{Y^j_{u}, \ 1\leq j \leq d_s\}$ and $(Y^j_{s})_{1\leq j \leq d_u}\in C^0(\mathcal M; E_s^{*})$ such that $ {E_s^{*}=\mathrm {Span}\{Y^j_{s}, \ 1\leq j \leq d_u\}}$ . Let $Y_s^1\wedge \cdots \wedge Y_s^{d_s}\wedge \alpha \wedge Y_s^1\wedge \cdots \wedge Y_s^{d_u}=: \tilde \omega \wedge \alpha \wedge \tilde \theta $ be a non-vanishing continuous n-form which we suppose to be positive. We notice to start that, using the Anosov property, there is an $\eta>0$ such that

$$ \begin{align*} \int_{\mathcal M}|&e^{t\mathbf{P}}(Y_u^1\wedge \cdots Y_u^{d_s-1}\wedge Y_s^1)\wedge \alpha \wedge Y_u^{d_s}\wedge Y_s^2\wedge \cdots \wedge Y_s^{d_s}| \\&\leq Ce^{-t\eta}\int_{\mathcal M}|e^{t\mathbf{P}}(Y_u^1\wedge \cdots Y_u^{d_s-1})\wedge \alpha \wedge Y_u^{d_s}\wedge Y_s^1\wedge Y_s^2\wedge \cdots \wedge Y_s^{d_s}| \\&\leq Ce^{-2t\eta}\int_{\mathcal M}|e^{t\mathbf{P}}(Y_u^1\wedge \cdots Y_u^{d_s-1} \wedge Y_u^{d_s})\wedge \alpha \wedge Y_s^1\wedge Y_s^2\wedge \cdots \wedge Y_s^{d_s}| \\&\leq Ce^{-2t\eta}\int_{\mathcal M}e^{t\mathbf P}(\tilde \omega) \wedge \alpha \wedge \tilde \theta. \end{align*} $$

Thus, a quick induction yields the following bound:

$$ \begin{align*}\text{ for all } 0\leq k\leq d_s, \quad |\langle e^{t\mathbf P}\omega_k, \theta_{d_s-k}\rangle|\leq C\|\omega\|_{C^0}e^{-2tk\eta} \langle e^{t\mathbf P}\tilde \omega,\tilde \theta\rangle.\end{align*} $$

Now, choose a smooth section $\omega $ and $\theta $ such that $\omega _0$ and $\theta _{d_s}$ are non-vanishing on $\mathcal M$ . Note that such an $\omega $ can be obtained by approximating in $C^0$ norm a non-vanishing continuous section by smooth sections. Then, we get the convergence of

$$ \begin{align*} \ell:=\langle \omega, \Pi_0(\unicode{x3bb})^{*}\theta\rangle =-\lim_{k\to+\infty}\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\langle e^{t\mathbf P}\omega_0, \theta_0\rangle(1+O(e^{-\eta t}))\, dt.\end{align*} $$

Consider thus an arbitrary $\epsilon>0$ and an integer $k_0\geq 0$ such that for any $k\geq k_0$ , one has

$$ \begin{align*}-\epsilon \leq \bigg|\!\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\langle e^{t\mathbf P}\omega_0, \theta_0\rangle(1+O(e^{-\eta t}))\,dt+\ell\bigg| \leq \epsilon. \end{align*} $$

The support of the $(\chi '(t))^{*k}$ goes to infinity with k, so we can suppose that for $k\geq k_0$ , one has $ 1-\epsilon \leq 1+O(e^{-\eta t})\leq 1+\epsilon $ on the support of the integrand. In other words, we have obtained

$$ \begin{align*}\text{ for all } k\geq k_0, \quad \frac{-\epsilon}{1+\epsilon} \leq \bigg|\!\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\langle e^{t\mathbf P}\omega_0, \theta_0\rangle\,dt+\ell\bigg|\leq \frac{\epsilon}{1-\epsilon}. \end{align*} $$

This also proves the boundedness of the above integral with $\omega _0$ and $\theta _0$ replaced by $\tilde \omega $ and $\tilde \theta $ , and this shows that

$$ \begin{align*} \langle\omega, \Pi_0(\unicode{x3bb})^{*}\theta\rangle& =\lim_{k\to+\infty}\int_{\mathbb R}(\chi'(t))^{*k}e^{-t\unicode{x3bb}}\langle e^{t\mathbf P}\omega_0, \theta_0\rangle\,dt \\ &\leq C\|\omega\|_0\limsup\int_{\mathbb R}|(\chi'(t))^{*k}|e^{-tP(V)}\langle e^{t\mathbf P}\tilde \omega, \tilde \theta\rangle\,dt \leq C'\|w\|_{C^0}. \end{align*} $$

The above bound actually holds for any $\theta $ and $\omega $ , and we get that every co-resonant state is an order zero current.

Restrictions of co-resonant state on unstable manifolds are well defined. Following the strategy of the proof of Proposition 3.2, we consider a co-resonant state $\theta $ which satisfies $\mathrm {WF}(\theta )\subset E_s^{*}$ . For any $x\in \mathcal M$ , we can thus define the restriction of $\theta $ to $\mathcal W^u(x)$ . We will denote $\theta _x:=\theta _{|\mathcal W^u(x)}$ . More explicitly, for a smooth function $f\in C^{\infty }(\mathcal M)$ , we can define

$$ \begin{align*}(f_{|\mathcal W^u(x)},\theta_x):=(f[\mathcal W^u(x)], \theta)=(f[{\mathcal W^u(x)}], \theta)_{\mathcal H^s, \mathcal H^{-s}}, \end{align*} $$

where the bracket denotes the distributional pairing and $[\mathcal W^u(x)]$ denotes the $d_u$ -current which consists in integrating over the unstable manifold $\mathcal W^u(x)$ . Then, $\theta _x$ is seen to be of order zero, in other words, $\{\theta _x\mid x\in \mathcal M\}$ defines a system of measures on the unstable leaves.

The system of measures $\{\theta _x\mid x\in \mathcal M\}$ satisfies $\varphi _t$ -conformality. This is again a consequence of the fact that $\theta $ is a co-resonant state. More precisely, for any smooth function $f\in C^{\infty }(\mathcal M)$ , we can write

$$ \begin{align*}( f[\mathcal W^u(x)], \theta)_{\mathcal H^s \times \mathcal H^{-s}}=e^{-tP(V)}( e^{t\mathbf P}( f[\mathcal W^u(x)],\theta)_{\mathcal H^s \times \mathcal H^{-s}}.\end{align*} $$

Using $e^{t\mathbf P}( f[\mathcal W^u(x)])=e^{t\mathbf P} f[\mathcal W^u(\varphi _t x)]$ then yields

$$ \begin{align*}\theta_x(f)=e^{-tP(V)}\theta_{\varphi_t x}(e^{S_tV(\varphi_{-t}y)}f(\varphi_{-t}y)), \end{align*} $$

which is exactly the $\varphi _t$ -conformality for the potential V.

The system of measures $\{\theta _x \mid x\in \mathcal M\}$ satisfies the change of variable by holonomy. This a consequence of the continuity of $x\mapsto \theta _x$ , which is shown as in the proof of Proposition 3.2. Indeed, another way of obtaining $\theta _x$ is to write $\theta $ as a linear combination of smooth $d_u$ -form with distributional coefficients in $\mathcal H^{-s}$ . Then, the restriction $\theta _x$ is obtained by pulling back the smooth forms and the coefficients. The pullback on the coefficients are well defined by the wavefront set condition and we deduce the continuity property by the continuity of the coefficients.

Following the proof of Proposition 3.2 and using [Reference Climenhaga10, Corollary 3.12], we obtain the existence of a constant $c>0$ such that for any $x\in \mathcal M$ , one has $\theta _x=c m^u_{x,V}$ .

The first resonance $P(V)$ is simple. The first part of the proof shows that it suffices to prove that for any $\omega \in C^{\alpha }(\mathcal M; \Lambda _0^{d_s})$ , one has $(\omega , \theta )=c(\omega , m^u_V)$ . Note that such a $\omega $ writes $\omega =f\mathrm {vol}_{\mathcal W^u(x)}$ for some $f\in C^{0}(\mathcal M)$ , we can then write

$$ \begin{align*}\theta_x(f):=( f[\mathcal W^u(x)], \theta)_{\mathcal H^s \times \mathcal H^{-s}}=( f\delta_{\mathcal W^u(x)},\mathrm{vol}_{\mathcal W^u(x)}\wedge \alpha \wedge \theta)_{\mathcal H^s \times \mathcal H^{-s}},\end{align*} $$

where $\delta _{\mathcal W^u(x)}$ is the integration of function on the unstable manifold of x (see equation (56)). Remark that here, the last bracket is the usual distributional pairing. In particular, $\theta _x$ is the restriction to the unstable manifold $\mathcal W^u(x)$ of the measure ${\mathrm {vol}_{\mathcal W^u}\wedge \alpha \wedge \theta }$ . Using the last part of the proof of Proposition 3.2 then yields that

$$ \begin{align*}\text{ for all } f\!\in\! C^{0}(\mathcal M ), \quad (\omega, \theta)=(\mathrm{vol}_{\mathcal W^u}\wedge \alpha \wedge \theta)(f)=c(\mathrm{vol}_{\mathcal W^u}\wedge \alpha \wedge m^u_V)=c(\omega, m^u_V).\end{align*} $$

This concludes the proof.

Proof. The averaging formula is derived from the product formula of Lemma 4.1 and the expression of the projector $ \Pi _0(P(V))$ using the exact same argument as in the proof of Proposition 3.3. Suppose now that there is another co-resonant state $\theta $ associated to a resonance $P(V)+i\unicode{x3bb} $ on the critical axis. We use the fact that $\theta $ is of order $0$ as well as [Reference Simon35, Ch. 6, equation (2.6)] to get that there exists a Borel measure $\mu _{\theta }$ and a $\mu _{\theta }^\unicode{x3bb} $ -measurable function $T_{\theta }$ with values in $\Lambda ^{d_s}(E_u^{*}\oplus E_s^{*})$ normalized $\mu _{\theta }^\unicode{x3bb} $ -almost everywhere such that

$$ \begin{align*}\text{ for all } \omega\in C^{\infty}(\mathcal M; \Lambda^{d_s}(E_u^{*}\oplus E_s^{*})), \quad \langle\omega,\theta\rangle=\int_{\mathcal M}\langle \omega(x), T_{\theta}(x)\rangle\,d\mu^{\unicode{x3bb}}_{\theta}(x), \end{align*} $$

where the bracket here denotes the scalar product on $\Lambda ^{d_s}(E_u^{*}\oplus E_s^{*})$ . Moreover, the measure $\mu _{\theta }^\unicode{x3bb} $ is defined by

(68) $$ \begin{align} \text{ for all } U\subset \mathcal M \text{ open set},\quad \mu^{\unicode{x3bb}}_{\theta}(U):=\sup_{\|w\|_{C^0}\leq 1, \mathrm{supp}(\omega )\subset U}\langle\omega, \theta\rangle. \end{align} $$

The proof of the previous proposition showed that there is a $C>0$ such that

$$ \begin{align*}\text{for all } \omega\in C^{\infty}(\mathcal M; \Lambda^{d_s}(E_u^{*}\oplus E_s^{*})), \quad \langle\omega, \theta\rangle \leq C\langle \theta, m^u_V\rangle \Rightarrow \mu_{\theta}\ll \mu_{m^u_V}=:\mu. \end{align*} $$

Now, using equations (62), (68), as well as the bound

$$ \begin{align*} \text{there exists } C>0\quad \text{for all } q\in\! \mathcal M, \quad\text{for all } t\geq\! 0, \quad m^{u}_{V}(B_t(q,\delta))\leq C e^{S_t(V)(q)-tP(V)}, \end{align*} $$

(which can be found in the proof of [Reference Climenhaga10, Theorem 3.10]), we get that $\mu $ satisfies an upper Gibbs bound. In other words, we have $\mu \ll \mu _V$ . In particular, the Radon–Nikodym density $h:={d\mu _\theta ^\unicode{x3bb} }/{d\mu }$ is an element of $L^{\infty }(\mathcal M, \mu _V)$ .

We use [Reference Reed and Simon33, Theorem VII.14] to see that the flow is weakly mixing if and only if the only eigenvalue is $1$ and it is a simple eigenvalue. In other words, if

(69) $$ \begin{align} \begin{cases} Xf=i\unicode{x3bb} f,\\ f\in L^2(\mathcal M,\mu_{V}) \end{cases} \end{align} $$

has no solution except for $\unicode{x3bb} =0$ and f constant. However, then, we have shown that h is a non-trivial solution of the above system which contradicts weak mixing.