The Perron–Frobenius theorem

We recall without proof a few facts from [Reference Viana23]; we refer the reader to [Reference Birkhoff4] for the original treatment.

Let E be a real vector space; a cone on E is a subset $C\subset E\setminus \{0 \}$ such that, if $v\in C$, then $tv\in C$ for all t > 0. We define the closure $\bar C$ of C as the set of all $v\in E$ such that there is $w\in C$ and $t_n\searrow 0$ such that $v+t_nw\in C$ for all $n\ge 1$.

In the following, we shall always deal with convex cones C such that

(2.1)\begin{align} \bar C\cap(-\bar C)=\{0 \} . \end{align}

Given $v_1,v_2\in C$, we define

\begin{equation*}\alpha(v_1,v_2)=\sup\{ t \gt 0\colon v_2-tv_1\in C \} , \end{equation*}

\begin{equation*}\frac{1}{\beta(v_1,v_2)}=\sup\{ t \gt 0\colon v_1-tv_2\in C \} \end{equation*}

\begin{equation*}\theta(v_1,v_2)= \log\frac{\beta(v_1,v_2)}{\alpha(v_1,v_2)} . \end{equation*}

We define an equivalence relation on C saying that $v\simeq w$ if v = tw for some t > 0. It turns out that θ is a metric on $\frac{C}{\simeq}$ though it can assume the value $+\infty$. As shown in [Reference Viana23], it separates points thanks to (2.1).

We recall the statement of the Perron–Frobenius theorem.

Theorem 2.1. Let $C\subset E$ be a convex cone satisfying (2.1) and let $L\colon E\rightarrow E$ be a linear map such that $L(C)\subset C$. Let us suppose that

\begin{equation*}D\colon=\sup\{ \theta(L(v_1),L(v_2))\colon v_1,v_2\in C \} \lt +\infty . \end{equation*}

Then, the following three points hold.

1. (1) For all $v_1,v_2\in C$, we have that

   \begin{equation*}\theta(L(v_1),L(v_2))\le(1-{\rm e}^{-D})\theta(v_1,v_2) . \end{equation*}

2. (2) If $(\frac{C}{\simeq},\theta)$ is complete, then L has a unique fixed point $v\in\frac{C}{\simeq}$. Since L is linear, this means that there is $v\in C$, unique up to multiplication by a positive constant, and a unique λ > 0 such that

   \begin{equation*}Lv=\lambda v . \end{equation*}

3. (3) Let $(\frac{C}{\simeq},\theta)$ be complete and let v be the fixed point of point (2). Then, if $v_0\in C$, we have that

   \begin{equation*}\theta(L^n v_0,v)\le \frac{(1-{\rm e}^{-D})^n}{{\rm e}^{-D}}\theta(v_0,v). \end{equation*}

Fractal sets

The fractals we consider are a particular case of what in [Reference Kigami9] (see definitions 1.3.4 and 1.3.13) are called ‘post-critically finite self-similar structures’.

1. (F1) We are given t affine bijections $\psi_i\colon \textbf{R}^d\rightarrow \textbf{R}^d$, $i\in\{1,\dots,t \}$, satisfying

   (2.2)\begin{align} \eta\colon=\sup_{i\in\{1,\dots,t \}}Lip(\psi_i) \lt 1. \end{align}

   By Theorem 1.1.7 of [Reference Kigami9], there is a unique non-empty compact set $G\subset\textbf{R}^d$ such that

   (2.3)\begin{align} G=\bigcup_{i\in\{1,\dots,t \}}\psi_i(G). \end{align}

   In the following, we shall always rescale the norm of R^d in such a way that

   (2.4)\begin{align} {\rm diam}(G)\le 1. \end{align}

   If (F1) holds, then the dynamics of F on G can be coded. Indeed, we define Σ as the space of sequences

   \begin{equation*}\Sigma= \{ ( x_i )_{i\ge 0}\colon x_i\in(1,\dots,n) \} \end{equation*}

   with the product topology. One metric that induces this topology is the following: for $\gamma\in(0,1)$, we set

   (2.5)\begin{align} d_\gamma(( x_i )_{i\ge 0},( y_i )_{i\ge 0})= \inf\{ \gamma^{k+1}\colon x_i=y_i\quad \text{for}\ i\in\{0,\dots,k \} \} \end{align}

   with the convention that the inf on the empty set is 1.

   We define the shift σ as

   \begin{equation*}\sigma\colon \Sigma\rightarrow \Sigma,\qquad \sigma\colon ( x_0,x_1,x_2,\dots )\rightarrow ( x_1,x_2,x_3,\dots ) . \end{equation*}

   If $x=(x_0x_1\cdots)\in\Sigma$, we set $ix=(ix_0x_1\cdots)$.

   For $x=(x_0x_1\cdots)\in\Sigma$, we define the cylinder of Σ

   \begin{equation*}[x_0\cdots x_l]=\{ ( y_i )_{i\ge 0}\in\Sigma\colon y_i=x_i\ \text{for}\ i\in(0,\dots,l) \} . \end{equation*}

   For $x=(x_0x_1\cdots)\in\Sigma$ and $l\in\textbf{N}$, we set

   \begin{equation*}\psi_{x_0\cdots x_l}=\psi_{x_0}\circ\cdots\circ\psi_{x_l} \end{equation*}

   and

   (2.6)\begin{align} [x_0\cdots x_l]_G=\psi_{x_0}\circ\psi_{x_{1}}\circ\cdots\circ\psi_{x_l}(G) . \end{align}

   Formula (2.6) immediately implies that, for all $i\in(1,\dots,t)$,

   (2.7)\begin{align} \psi_i([x_1\cdots x_l]_G)=[i x_1\cdots x_l]_G . \end{align}

   Since the maps ψ_i are continuous and G is compact, the sets $[x_0\cdots x_l]_G\subset G$ are compact. By (2.3), we have that $\psi_i(G)\subset G$ for $i\in(1,\dots,t)$; together with (2.6); this implies that, for all $( x_i )_{i\ge 0}\in\Sigma$ and $l\ge 1$,

   (2.8)\begin{align} [x_0\cdots x_{l-1} x_l]_G\subset [x_0\cdots x_{l-1}]_G . \end{align}

   From (2.2), (2.4) and (2.6), we get that, for all $l\ge 0$,

   (2.9)\begin{align} {\rm diam} ([x_0\cdots x_l]_G)\le\eta^{l+1} . \end{align}

   By (2.8) and (2.9) and the finite intersection property, we get that, if $( x_i )_{i\ge 0}\subset\Sigma$, then

   (2.10)\begin{equation}\bigcap_{l\ge 1}[x_0\cdots x_{l}]_G\end{equation}

   is a single point, which we call $\Phi(( x_i )_{i\ge 0})$; Formula (2.9) implies in a standard way that the map $\Phi\colon \Sigma\rightarrow G$ is continuous. It is not hard to prove, using (2.3), that Φ is surjective. In our choice of the metric d_γ on Σ, we shall always take $\gamma\in(\eta,1)$; we endow G with the Euclidean distance on R^d. With this choice of the metrics, (2.5) and (2.9) easily imply that Φ is 1-Lipschitz. By (2.7) and the definition of Φ, we see that

   (2.11)\begin{align} \psi_i\circ\Phi(x)=\Phi(ix). \end{align}

2. (F2) We ask that, if $i\not=j$, then $\psi_i(G)\cap\psi_j(G)$ is a finite set. We set

   \begin{equation*}{\cal F}\colon=\bigcup_{i\not =j}\psi_i(G)\cap\psi_j(G) . \end{equation*}

3. (F3) We ask that, if $i=(i_0i_1\cdots)\in\Sigma$ and $l\ge 1$, then $\psi_{i_0\cdots i_{l-1}}^{-1}({\cal F})\cap{\cal F}=\emptyset$.

   We briefly prove that (F1)–(F3) imply that the coding Φ is finite-to-one and that the set $N\subset\Sigma$, where Φ is not injective and is countable. More precisely, we assert that the points of G in

   (2.12)\begin{align} H\colon={\cal F}\cup\bigcup_{l\ge 0}\bigcup_{x\in\Sigma} \psi_{x_0\cdots x_l}({\cal F}) \end{align}

   have at most t preimages and that those outside H have only one preimage.

   We begin by showing this latter fact. If $y\in G\setminus{\cal F}$, then $y=\Phi(( x_j )_{j\ge 0})$ can belong by (F2) to at most one $\psi_i(G)$, and thus there is only one choice for x ₀. Using the fact that $\psi_{x_0}$ is a diffeomorphism, we see that there is at most one choice for x ₁ if moreover $y\not\in\psi_{x_0}({\cal F})$; iterating, we see that the points in H ^c have one preimage.

   If $y\in{\cal F}$, then y can belong to at most t sets $\psi_i(G)$, and thus there are at most t choices for x ₀. As for the choice for x ₁, once we have fixed x ₀, then we have that $y\in\psi_{x_0}\circ\psi_{x_1}(G)$, i.e., that $\psi_{x_0}^{-1}(y)\in\psi_{x_1}(G)$. Since $y\in{\cal F}$, (F3) implies that $\psi_{x_0}^{-1}(y)\not\in{\cal F}$ and thus we have at most one choice for x ₁. Iterating, we see that there is at most one choice for $x_2,x_3$, etc. If $y\not\in{\cal F}$ but $y\in\psi_{x_0}({\cal F})$, then we have one choice for x ₀ but t choices for x ₁; arguing as above, we see that there is at most one choice for x ₂, x ₃, etc. Iterating, we get the assertion.

4. (F4) We ask that there are disjoint open sets ${\cal O}_1,\dots,{\cal O}_n\subset\textbf{R}^d$ such that

   \begin{equation*}G\cap{\cal O}_i=\psi_i(G)\setminus{\cal F}\quad \text{for}\ i\in(1,\dots,n) . \end{equation*}

   We define a map $F\colon \bigcup_{i=1}^n{\cal O}_i\rightarrow \textbf{R}^d$ by

   \begin{equation*}F(x)=\psi_i^{-1}(x)\quad \text{if}\ x\in{\cal O}_i . \end{equation*}

   This implies the second equality below; the first one holds by the formula above if we ask that $\psi({\cal O}_i)\subset{\cal O}_i$.

   (2.13)\begin{align} F\circ\psi_i(x)=x\qquad\forall x\in{\cal O}_i\quad \text{and}\quad \psi_i\circ F(x)=x\qquad\forall x\in{\cal O}_i . \end{align}

   We call a_i the unique fixed point of ψ_i; note that, by (2.3), $a_{i}\in G$. If $x\in{\cal F}$, we define $F(x)=a_{i}$ for some arbitrary a_i. This defines F as a Borel map on all of G, which satisfies (2.13).

From (2.6) and the definition of Φ, we see that $\Phi([x_0\cdots x_l])=[x_0\cdots x_l]_G$. We saw earlier that Φ is finite-to-one and that there are at most countably many points with more that one preimage; an immediate consequence is that the set

\begin{equation*}\Phi^{-1} ([x_0\cdots x_l]_G)\setminus [x_0\cdots x_l] \end{equation*}

is at most countable. Let us suppose that $\Phi(x_0x_1\cdots)\in{\cal O}_i$; the discussion after (F3) implies that there is only one choice for x ₀ and it is $x_0=i$; by the first formula of (2.13), this implies the second equality below; the first and last ones are the definition of Φ.

\begin{equation*}F\circ\Phi(x_0x_1\cdots)=F\left( \bigcap_{l\ge 0}\psi_{x_0x_1\cdots x_l}(G) \right) = \bigcap_{l\ge 0}\psi_{x_1x_2\cdots x_l}(G)=\Phi\circ\sigma(x_0x_1\cdots) . \end{equation*}

This yields the first equality below, while the second one is (2.11).

(2.14)\begin{align} \left\{ \begin{matrix} \Phi\circ\sigma(x)=F\circ\Phi(x)\ \text{save possibly when } \Phi(x)\in {\cal F}\text{,}\\ \Phi(ix) =\psi_i(\Phi(x)) \qquad\forall x\in\Sigma, \quad\forall i\in (1,\dots,n) . \end{matrix} \right. \end{align}

In other words, up to a change of coordinates, shifting the coding one place to the left is the same as applying F.

A particular case we have in mind is the harmonic Sierpinski gasket on R² (see for instance [Reference Kajino7]). We set

\begin{equation*}T_1=\left( \begin{matrix} \frac{3}{5},&0\cr 0,&\frac{1}{5} \end{matrix} \right) , \quad T_2=\left( \begin{matrix} \frac{3}{10},&\frac{\sqrt 3}{10}\cr \frac{\sqrt 3}{10},&\frac{1}{2} \end{matrix} \right) ,\quad T_3=\left( \begin{matrix} \frac{3}{10},&-\frac{\sqrt 3}{10}\cr -\frac{\sqrt 3}{10},&\frac{1}{2} \end{matrix} \right) , \end{equation*}

\begin{equation*}A=\left( \begin{matrix} 0\cr 0 \end{matrix} \right),\qquad B=\left( \begin{matrix} 1\cr \frac{1}{\sqrt 3} \end{matrix} \right) ,\qquad C=\left( \begin{matrix} 1\cr -\frac{1}{\sqrt 3} \end{matrix} \right) \end{equation*}

and

\begin{equation*}\psi_1(x)=T_1(x),\quad \psi_2(x)= B +T_2\left( x-B \right),\quad \psi_3(x)= C +T_3\left( x-C \right) . \end{equation*}

Referring to Figure 1, ψ ₁ brings the triangle $G_0\colon\!=ABC$ into Abc; ψ ₂ brings G ₀ into Bac and ψ ₃ brings G ₀ into Cba. We set ${\cal F}=\{a,b,c \}$ and take ${\cal O}_1$, ${\cal O}_2$, ${\cal O}_3$ as three disjoint open sets which contain, respectively, the triangle Abc minus $b,c$, Bca minus $c,a$ and Cba minus $a,b$; in this way, hypotheses (F1)–(F4) hold. It is easy to check that also hypothesis (ND)₁ is satisfied.

Figure 1. The first two prefractals of the harmonic Gasket.

We define the map F as

\begin{equation*}F(x)=\psi_i^{-1}(x)\quad \text{if}\ x\in{\cal O}_i \end{equation*}

and we extend it arbitrarily on ${\cal F}=\{a,b,c \}$, say $F(a)=A$, $F(b)=B$ and $F(c)=C$.

For $q\in(1,\dots,d)$, we consider ${\Lambda^q}(\textbf{R}^d)$, the space of q-forms on R^d; if $(\cdot,\cdot)$ is an inner product on R^d, it induces an inner product on the monomials of $\Lambda^q(\textbf{R}^d)$ by

\begin{equation*}((v_1\wedge\cdots\wedge v_q),(w_1\wedge\cdots\wedge w_q))= (v_1\wedge\cdots\wedge v_q) (i(w_1),\cdots, i(w_q)), \end{equation*}

where $i\colon \Lambda^1(\textbf{R}^d)\rightarrow \textbf{R}^d$ is the Riesz map, i.e., the natural identification of a Hilbert space with its dual. This product extends by linearity to all $\Lambda^q(\textbf{R}^d)$, which thus becomes a Hilbert space.

As a consequence, if A is a matrix bringing $\Lambda^q(\textbf{R}^d)$ into itself, we can define its adjoint ${{}^{t}} A$. We shall denote by M ^q the space of all self-adjoint operators (a.k.a. symmetric matrices) on ${\Lambda^q}(\textbf{R}^d)$. Again since ${\Lambda^q}(\textbf{R}^d)$ is a Hilbert space, we can define the trace ${\rm tr}(A)$ of $A\in M^q$ and thus the Hilbert–Schmidt product

\begin{equation*}(\cdot,\cdot)_{\rm HS}\colon M^q\times M^q\rightarrow \textbf{R}\end{equation*}

\begin{equation*}(A,B)_{\rm HS}\colon={\rm tr}(A{{}^{t}} B)={\rm tr}(AB), \end{equation*}

where the second equality comes from the fact that B is symmetric.

Thus, M ^q is a Hilbert space whose norm is

\begin{equation*}||A||_{\rm HS}=\sqrt{(A,A)_{\rm HS}}.\end{equation*}

Since $\Lambda^k(\textbf{R}^d)$ is a Hilbert space, we shall see its elements b as column vectors, on which the elements $M\in M^q$ act to the left: Mb.

Matrix-valued measures

Let $(E,\hat d)$ be a compact metric space; in the following, $(E,\hat d)$ will be either one of the spaces $(\Sigma,d_\gamma)$ or $(G,||\cdot||)$ defined above. We denote by $C(E,M^q)$ the space of continuous functions from E to M ^q and by ${\cal M}(E,M^q)$ the space of the Borel measures on E valued in M ^q. If $\mu\in{\cal M}(E,M^q)$, we denote by $||\mu||$ its total variation measure; it is standard [Reference Rudin20] that $||\mu||(E) \lt +\infty$ and that

\begin{equation*}\mu=M\cdot||\mu||\end{equation*}

where $M\colon E\rightarrow M^q$ is Borel and

(2.15)\begin{align} ||M(x)||_{\rm HS}=1\quad \text{for}\ ||\mu||\text{-a.e.}\ x\in E. \end{align}

Let now $\mu\in{\cal M}(E,M^q)$ and let $f\colon E\rightarrow M^q$ be a Borel function such that $||f||_{\rm HS}\in L^1(E,||\mu||)$. We define

(2.16)\begin{align} \int_E(f, {\rm d} \mu)_{\rm HS}\colon= \int_E(f(x),M(x))_{\rm HS} {\rm d} ||\mu||(x) . \end{align}

Note that the integral on the right converges: indeed, the function $x\rightarrow(f(x),M(x))$ is in $L^1(G,||\mu||)$ by our hypotheses on f, (2.15) and Cauchy–Schwarz. A particular case we shall use often is when $f\in C(E,M^q)$.

If $\mu\in{\cal M}(E,M^q)$ and $a,b\colon E\rightarrow {\Lambda^q}(\textbf{R}^d)$ are Borel functions such that $||a||\cdot||b||\in L^1(E,||\mu||)$, we can define

\begin{equation*}\int_E(a, {\rm d} \mu\cdot b)\colon= \int_E( a(x),M(x)b(x) )\, {\rm d} ||\mu||(x) . \end{equation*}

Note again that the function $x\rightarrow(a(x),M(x)b(x))$ belongs to $L^1(G,||\mu||)$ by our hypotheses on a, b, (2.15) and Cauchy–Schwarz. We have written $M(x)b(x)$ because, as we said above, we consider the elements of $\Lambda^q(\textbf{R}^d)$ as column vectors. Since

\begin{equation*}(a\otimes b,M)_{\rm HS}=(a,Mb), \end{equation*}

we have that

(2.17)\begin{align} \int_E(a, {\rm d} \mu\cdot b)= \int_E(a\otimes b, {\rm d} \mu)_{\rm HS} . \end{align}

By Riesz’s representation theorem, the duality coupling ${\langle \cdot ,\cdot\rangle}$ between $C(E,M^q)$ and ${\cal M}(E,M^q)$ is given by

\begin{equation*}{\langle f ,\mu\rangle}\colon= \int_E(f, {\rm d} \mu)_{\rm HS}, \end{equation*}

where the integral on the right is defined by (2.16).

Let $\mu\in{\cal M}(E,M^q)$; we say that $\mu\in{\cal M}_+(E,M^q)$ if $\mu(A)$ is a semi-positive-definite matrix for all Borel sets $A\subset E$. By Lusin’s theorem (see [Reference Bessi2] for more details), this is equivalent to saying that

(2.18)\begin{align} {\langle f ,\mu\rangle}\ge 0 \end{align}

for all $f\in C(E,M^q)$ such that $f(x)\ge 0$ (i.e., it is semi-positive definite) for all $x\in E$. An equivalent characterization is that, in the polar decomposition (2.15), $M(x)\ge 0$ for $||\mu||$-a.e. $x\in E$.

As a consequence of the characterization (2.18), ${\cal M}_+(E,M^q)$ is a convex set closed for the weak$\ast$ topology of ${\cal M}(E,M^q)$.

If $\mu\in{\cal M}(E,M^q)$ and $A\colon E\rightarrow M^q$ belongs to $L^1(E,||\mu||)$, we define the scalar measure $(A,\mu)_{\rm HS}$ by

\begin{equation*}\int_E f(x)\, {\rm d} (A,\mu)_{\rm HS}(x)= \int_E(f(x)A(x), {\rm d} \mu(x))_{\rm HS}\end{equation*}

for all $f\in C(E,\textbf{R})$. Recalling the definition (2.16) of the integral on the right, we have that

\begin{equation*}(A,\mu)_{\rm HS}=(A,M)_{\rm HS}\cdot||\mu||, \end{equation*}

where $\mu=M\cdot||\mu||$ is the polar decomposition of µ as in (2.15).

We recall (a proof is, for instance, in [Reference Bessi2]) that, if $A,B,C\in M^q$, if $A\ge B$ and $C\ge 0$, then

(2.19)\begin{align} (A,C)_{\rm HS}\ge(B,C)_{\rm HS} . \end{align}

As a consequence, we have that, if $f,g\in C(E,M^q)$ satisfy $f(x)\ge g(x)$ for all $x\in E$ and if $\mu\in{\cal M}_+(E,M^q)$, then

(2.20)\begin{align} \int_E(f(x), {\rm d} \mu(x))_{\rm HS}\ge \int_E(g(x), {\rm d} \mu(x))_{\rm HS}. \end{align}

Let $Q\in C(E,M^q)$ be such that $Q(x) \gt 0$ for all $x\in E$; by compactness, we can find ϵ > 0 such that

(2.21)\begin{align} \epsilon Id\le Q(x)\le\frac{1}{\epsilon}Id\qquad\forall x\in E . \end{align}

It is shown in [Reference Bessi2] that there is $C_1=C_1(\epsilon) \gt 0$ such that, if Q satisfies (2.21) and $\mu\in{\cal M}_+(E,M^q)$, then

(2.22)\begin{align} \frac{1}{C_1(\epsilon)}||\mu||\le (Q,\mu)_{\rm HS}\le C_1(\epsilon)||\mu||. \end{align}

Let $Q\in C(E,M^q)$ satisfy (2.21); we say that $\mu\in{\cal P}_Q$ if $\mu\in{\cal M}_+(E,M^q)$ and

(2.23)\begin{align} {\langle Q ,\mu\rangle}=1 . \end{align}

We saw after Formula (2.18) that ${\cal M}_+(E,M^q)$ is convex and closed for the weak$\ast$ topology; Formula (2.23) implies that ${\cal P}_Q$ is also closed and convex; (2.22) and (2.23) imply that it is compact.

Cones in function spaces

We say that $f\in C_+(E,M^q)$ if $f\in C(E,M^q)$ and f(x) is positive-definite for all $x\in E$. It is easy to see that $C_+(E,M^q)$ is a convex cone in $C(E,M^q)$ satisfying (2.1). We let $\theta^+$ denote the hyperbolic distance on $\frac{C_+(E,M^q)}{\simeq}$.

Let a > 0 and $\alpha\in(0,1]$; we say that $A\in Log C^{a,\alpha}_+(E)$ if $A\in C_+(E,M^q)$ and for all $x,y\in E$, we have

\begin{equation*}A(y)\,{\rm e}^{-a\hat d(x,y)^\alpha}\le A(x)\le A(y)\,{\rm e}^{a\hat d(x,y)^\alpha}, \end{equation*}

where the inequality is the standard one between symmetric matrices.

It is easy to check that $LogC^{a,\alpha}_+(E)$ is a convex cone satisfying (2.1). We let θ ^a denote the hyperbolic distance on $\frac{LogC^{a,\alpha}_+(E)}{\simeq}$.

Next, we introduce the subcone of the elements $A\in LogC^{a,\alpha}_+(E)$ whose eigenvalues are bounded away from zero. In other words, we define

\begin{equation*}||A||_{\infty}=\sup_{x\in E}||A(x)||_{\rm HS},\end{equation*}

and for ϵ > 0, we set

\begin{equation*}LogC^{a,\alpha}_\epsilon(E)=\{ A\in LogC_+^{a,\alpha}(E)\colon A(x)\ge\epsilon||A||_\infty Id\qquad\forall x\in E \}. \end{equation*}

The Ruelle operator

Recall that the maps ψ_i of § (2.2) are affine and injective; in particular, $D\psi_i$ is a constant, non-degenerate matrix. We define the pull-back $(D\psi_i)_\ast\colon \Lambda^q(\textbf{R}^d)\rightarrow \Lambda^q(\textbf{R}^d)$ as in (1) and the push-forward $\Psi_i\colon M^q\rightarrow M^q$ as in (2). Let $\alpha\in(0,1]$ and let $V\in C^{0,\alpha}(G,\textbf{R})$.

The Ruelle operator ${\cal L}_{G,V}$ on $C(G,M^q)$ is the one defined by (3).

We also define a Ruelle operator ${\cal L}_{\Sigma,V}$ on $C(\Sigma,M^q)$: defining (ix) as in § 2, we set

\begin{equation*}{\cal L}_{\Sigma,V}\colon C(\Sigma,M^q)\rightarrow C(\Sigma,M^q),\end{equation*}

(2.24)\begin{align} ({\cal L}_{\Sigma,V} A)(x)=\sum_{i=1}^t {\rm e}^{V\circ\Phi(ix)}\Psi_i(A(ix)) . \end{align}

Note that $V\circ\Phi$ is α-Hölder: indeed, V is α-Hölder and we saw in § 2 that Φ is 1-Lipschitz.

In order to apply the Perron–Frobenius theorem, we need a non-degeneracy hypothesis.

(ND)${}_q$

We suppose that there is γ > 0 such that, if $c,e\in{\Lambda^q}(\textbf{R}^d)$, we can find $i\in(1,\dots,t)$ such that

\begin{equation*}|((D\psi_{i})_\ast c, e)|\ge\gamma ||c||\cdot||e|| . \end{equation*}

In other words, we are asking that, for all $c\in\Lambda^q(\textbf{R}^d)\setminus\{0 \}$, $(D\psi_{i})_\ast c$ is a base of $\Lambda^q(\textbf{R}^d)$; clearly, this implies that $t \ge \left( {\matrix{\matrix{d \hfill \cr q \hfill \cr} \cr } } \right)$.

We state the standard theorem on the existence of Gibbs measures, whose proof [Reference Parry and Pollicott15, Reference Viana23] adapts easily to our situation; the details are in [Reference Bessi2] and [Reference Bessi3].

We recall Lemma 4.6 of [Reference Bessi3], which shows that there is a natural relationship between Gibbs measures on Σ and G; the notation is that of Proposition 2.2.

Just a word on the relationship with [Reference Feng and Käenmäki5] and [Reference Morris13]. It is standard [Reference Bessi2] that, if $V\equiv 0$, then the eigenfunction Q is constant and $D_1=1$ in point (5) of Proposition 2.2. On constant matrices, our operator ${\cal L}_{\Sigma,V}$ and the operator L_A of Proposition 15 of [Reference Morris13] coincide; in particular, they have the same spectral radius and the same maximal eigenvector, Q. Looking at the eigenvalues, this means that our $\beta_{\Sigma,V}$ coincides with their ${\rm e}^{P(A,2)}$. Moreover, when $V\equiv 0$, $\kappa_{\Sigma,V}$ coincides with the measure m on the shift defined in [Reference Feng and Käenmäki5] and [Reference Morris13]; we briefly prove why this is true.

A theorem of [Reference Feng and Käenmäki5], recalled as Theorem 2 in [Reference Morris13], says that the probability measure m of [Reference Morris13] satisfies the following: there are C > 0 and $P(A,2)\in\textbf{R}$ such that, for all $x_1,\dots,x_n\in\{1,\dots,t \}$, we have

\begin{equation*}\frac{1}{C}m([x_1\cdots x_n])\le \frac{||A_{x_n}\cdot\cdots\cdot A_{x_1}||_{\rm HS}^2}{{\rm e}^{nP(A,2)}}\le Cm([x_1\cdots x_n]) . \end{equation*}

We have already seen that $P(A,2)=\beta_V$. Together with the Gibbs property of Proposition 2.2, the formula above implies that m is absolutely continuous with respect to $\kappa_{\Sigma,V}$. Since m is invariant and $\kappa_{G,V}$ is ergodic, they must coincide.

Before stating the last lemma of this section, we need to define the Lyapounov exponents of the fractal. We consider the map

\begin{equation*}H\colon G\rightarrow M^d\end{equation*}

\begin{equation*}H\colon x\rightarrow (D\psi_i)_\ast\quad \text{if}\ x\in{\cal O}_i, \end{equation*}

which is defined $\kappa_{G,V}$-a.e. on G. For the expanding map F defined in (F4), we set

\begin{equation*}H^l(x)=H(x)\cdot H(F(x))\cdot\cdots\cdot H(F^{l-1}(x)). \end{equation*}

Since Kusuoka’s measure $\kappa_{G,V}$ is ergodic for the map F, we can apply the results of [Reference Ruelle21] (see [Reference Kelliher8] for a more elementary presentation) and get that for $\kappa_{G,V}$-a.e. $x\in G$, there is $\Lambda_x\in M^d$ such that

(2.25)\begin{align} \lim_{l\rightarrow+\infty} \left[H^l(x)\cdot\mu_{G,V}(G)\cdot{{}^{t}} H^l(x)\right]^\frac{1}{2l}=\Lambda_x. \end{align}

The fact that $\kappa_{G,V}$ is ergodic for F implies by [Reference Ruelle21] that neither the eigenvalues of $\Lambda_x$ nor their multiplicities depend on x.

Though we would not need it in the following, we recall Lemma 6.2 of [Reference Bessi3].

Lemma 2.4. Let the maps ψ_i, the Gibbs measure $\mu_{G,V}$ and Kusuoka’s measure $\kappa_{G,V}$ be as in Theorem 2.1. Let us suppose that the maximal eigenvalue of $\Lambda_x$ is simple and let us call $\hat E_x$ its eigenspace; let $P_{\hat E_x}$ be the orthogonal projection on $\hat E_x$. Then, we have that

\begin{equation*}\mu_{G,V}=P_{\hat E_x}||\mu_{G,V}|| . \end{equation*}

Moreover, we have that, for $\kappa_{G,V}$-a.e. $x\in G$,

(2.26)\begin{align} P_{\hat E_x}=\lim_{l\rightarrow+\infty} \frac{H^l(x)\cdot\mu_{G,V}(G)\cdot{{}^{t}} H^l(x)}{|| H^l(x)\cdot\mu_{G,V}(G)\cdot{{}^{t}} H^l(x) ||_{\rm HS}} . \end{align}

Proof of Theorem 1

As we saw at the beginning of this section, it suffices to show Formula (3.3), and this is what we shall do.

For $\{m,M \}\in{\cal K}_V(\Sigma)$ and $x\in\Sigma$, we define the following two functions of x.

(3.19)\begin{align} q_i(x)=m[i|\sigma^{-1}(x)],\qquad p_i(x)={\rm e}^{V(i\sigma(x))}(Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M\circ\sigma(x))_{\rm HS}. \end{align}

By (3.7), $\{q_i \}_i$ satisfies (3.16) for m-a.e. $x\in\Sigma$; we show that, setting $\lambda=\beta_V$, $\{p_i \}_i$ too satisfies (3.16) m-a.e.. The first equality below comes from the definition of p_i in the formula above, and for the second one, we transpose; the third one follows by the definition of ${\cal L}_V={\cal L}_{\Sigma,V}$ in (2.24); the fourth one comes from the fact that ${\cal L}_{V}Q_V=\beta_V Q_V$ and the last equality follows from the fact that $(m,M)\in{\cal K}_V(\Sigma)$.

\begin{equation*}\sum_{i=1}^t p_i=\sum_{i=1}^t{\rm e}^{V(i\sigma(x))}\cdot (Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M\circ\sigma(x))_{\rm HS} \end{equation*}

\begin{equation*}=\sum_{i=1}^t {\rm e}^{V(i\sigma(x))}(\Psi_i\cdot Q_V(i\sigma(x)),M\circ\sigma(x))_{\rm HS} \end{equation*}

\begin{equation*}=(({\cal L}_{V}Q_V)(\sigma(x)),M\circ\sigma(x))_{\rm HS}= \beta_V\cdot (Q_V\circ\sigma(x),M\circ\sigma(x))_{\rm HS}= \beta_V. \end{equation*}

Since (3.16) is satisfied, Formula (3.17) holds, implying that for m-a.e. $x\in G$, we have

\begin{equation*}-\sum_{i=1}^tq_i(x)\,\log q_i(x)+ \sum_{i=1}^t q_i(x)[V(i\sigma(x))+\log(Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M\circ\sigma(x))_{\rm HS}]\le \log\beta_V. \end{equation*}

Integrating against m, we get that

\begin{equation*}-\sum_{i=1}^t\int_\Sigma q_i(x)\,\log q_i(x)\, {\rm d} m(x)\end{equation*}

\begin{equation*}+\sum_{i=1}^t\int_\Sigma q_i(x)[V(i\sigma(x))+\log( Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M\circ\sigma(x))_{\rm HS}] {\rm d} m(x)\le \log\beta_V. \end{equation*}

By the first one of (3.19) and (3.8), we get that

\begin{equation*}-\sum_{i=1}^t\int_\Sigma q_i(x)\,\log q_i(x)\, {\rm d} m(x)=h(m). \end{equation*}

The first one of (3.19) and (3.18) imply that

\begin{equation*}\sum_{i=1}^t\int_\Sigma q_i(x) V(i\sigma(x))\, {\rm d} m(x)= \int_\Sigma V(x)\, {\rm d} m(x). \end{equation*}

The last three formulas imply that, if $(m,M)\in{\cal K}_V(\Sigma)$ and $I_\Sigma$ is defined as in (3.4), then

\begin{equation*}I_\Sigma(m,M)\le\log\beta_V. \end{equation*}

To end the proof of (3.3), it suffices to show that, in the formula above, equality is attained when $\{m,M\}=\{\kappa_V,M_V \}$. The only inequality in the derivation of the formula above is that of (3.17), which becomes an equality if

(3.20)\begin{align} q_i(x)=\frac{p_i(x)}{\beta_V}\quad \text{for}\ i\in(1,\dots,t)\ \text{and}\ \kappa_V\text{-a.e.}\ x\in\Sigma. \end{align}

This is what we are going to prove. The first equality below is the definition of q_i in (3.19) for $m=\kappa_V$, and the second one is (3.14).

(3.21)\begin{align} q_i(x)=\kappa_V[i|\sigma^{-1}(x)]= \frac{ {\rm e}^{V(i\sigma(x))} }{ \beta_V(Q_V\circ\sigma(x),({{}^{t}}\Psi_i)^{-1}\cdot M_V(i\sigma(x)))_{\rm HS} }. \end{align}

We assert that (3.20) would follow if there were some positive Borel function $\lambda\colon \Sigma\rightarrow \textbf{R}$ such that

(3.22)\begin{align} {{}^{t}}\Psi_i\cdot M_V\circ\sigma(x)=\lambda(x) M_V(i\sigma(x)) \end{align}

for m-a.e. $x\in\Sigma$.

Indeed, this formula implies the two outermost equalities below; the two innermost ones come from the fact that $\{\kappa_V,M_V \} \in{\cal K}_V(\Sigma)$.

\begin{equation*}\frac{ 1 }{ (Q_V\circ\sigma(x),{{}^{t}}\Psi_i^{-1}\cdot M_V(i\sigma(x)))_{\rm HS} } = \frac{ \lambda(x) }{ (Q_V\circ\sigma(x),M_V\circ\sigma(x))_{\rm HS} } =\lambda(x)\end{equation*}

\begin{equation*}=\lambda(x)(Q_V(i\sigma(x)),M_V(i\sigma(x)))_{\rm HS}= (Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M_V\circ\sigma(x))_{\rm HS}. \end{equation*}

The first equality below is (3.21); the second one is the formula above; the last one is (3.19).

\begin{equation*}q_i(x)= \frac{ {\rm e}^{V(i\sigma(x))} }{ \beta_V(Q_V\circ\sigma(x),({{}^{t}}\Psi_i)^{-1}\cdot M_V(i\sigma(x)))_{\rm HS} }\end{equation*}

\begin{equation*}=\frac{1}{\beta_V}\,{\rm e}^{V(i\sigma(x))}\cdot (Q_V(i\sigma(x)),{{}^{t}}\Psi_i\cdot M_V\circ\sigma(x))_{\rm HS}= \frac{p_i(x)}{\beta_V}, \end{equation*}

proving (3.20).

In order to show (3.22), it suffices to recall that, for κ_V-a.e. $x\in\Sigma$,

\begin{equation*}M_V(x)\cdot\frac{{\rm d} \kappa_V}{{\rm d} ||\mu||}(x) =\lim_{l\rightarrow+\infty} \frac{ {{}^{t}}\Psi_{x_0\cdots x_{l-1}}\cdot\mu_V(E) }{ ||{{}^{t}}\Psi_{x_0\cdots x_{l-1}}\cdot\mu_V(E)||_{\rm HS} } . \end{equation*}

The proof of this equality follows from point (5) of Proposition 2.2 and Rokhlin’s theorem; the details are in [Reference Bessi3]. By (10), the last formula easily implies (3.22).

Proof of point (1) of Theorem 2

The first equality below comes from (4.18) and requires (4.4); for the second one, we use (4.10).

\begin{equation*}\zeta(z,V)= \frac{ 1 }{ {\rm det}(Id-zL_V) } \end{equation*}

\begin{equation*}=\frac{1}{1-\beta z}\cdot \prod_{j=2}^{\hat d}\frac{1}{1-\lambda_{j,V}z}. \end{equation*}

In other words,

(4.22)\begin{align} \zeta(z,V)\cdot (1-\beta z)=\tilde\alpha(z) \end{align}

and, by the inequality of (4.10), there is ϵ > 0 such that $\tilde\alpha$ is non-zero and analytic in

(4.23)\begin{align} B_\epsilon\colon=\left\{ z\in\textbf{C}\colon |z| \lt \frac{{\rm e}^\epsilon}{\beta} \right\} . \end{align}

The first equality below follows taking derivatives in the last expression of (4.21); recall that $\lambda(\tau)$ is defined in (4.19). Differentiating the product in (4.22) and recalling that $\tilde\alpha\not=0$ in B_ϵ, we see that the second equality below holds for a function α analytic in the ball B_ϵ of (4.23).

\begin{equation*}\sum_\tau\sum_{j=1}^{m}\sum_{k\ge 1}\lambda(\tau)\cdot\alpha^k_{\tau,j}\cdot {\rm e}^{kV(\tau)}\cdot z^{k\cdot\lambda(\tau)-1}=\frac{ \zeta^{\prime}(z,V) }{ \zeta(z,V) } \end{equation*}

(4.24)\begin{align} =\frac{\beta}{1-\beta z}+\alpha(z). \end{align}

From now on we shall suppose that $V\equiv 0$. We rewrite (4.24) under this assumption; for the term on the left, we sum on the periodic orbits $\tau^{\prime}=(\tau,k)$ and use (4.20); for the term on the right, we use the geometric series.

\begin{equation*}\frac{1}{z}\sum_{\tau^{\prime}}\Lambda(\tau^{\prime}) {\rm tr}{{}^{t}}\Psi_{\tau^{\prime}} \cdot z^{\lambda(\tau^{\prime})}= \frac{ \zeta^{\prime}(z,V) }{ \zeta(z,V) } \end{equation*}

\begin{equation*}=\frac{1}{z}\sum_{n\ge 1}(\beta z)^n+\alpha(z). \end{equation*}

Subtracting the sum on the right, we get that

\begin{equation*}\sum_{n\ge 1}z^{n-1}\left[ \sum_{\lambda(\tau^{\prime})=n}\Lambda(\tau^{\prime}){\rm tr}{{}^{t}}\Psi_{\tau^{\prime}} -\beta^n \right] =\alpha(z) . \end{equation*}

Since α is analytic in the disc B_ϵ of (4.23), we get that the radius of convergence of the series on the left is $\frac{{\rm e}^\epsilon}{\beta} \gt \frac{1}{\beta}$. Since the general term of a converging series tends to zero, we deduce that, for any smaller ϵ > 0,

(4.25)\begin{align} \frac{{\rm e}^{n\epsilon}}{\beta^n}\left[ \sum_{\lambda(\tau^{\prime})=n}\Lambda(\tau^{\prime}){\rm tr} {{}^{t}}\Psi_{\tau^{\prime}}-\beta^n \right] \rightarrow 0\quad \text{as}\ n\rightarrow+\infty. \end{align}

In particular, the sequence above is bounded by some $D_1 \gt 0$.

We consider the case β > 1; we define $\eta(r)$ in the first equality below, where the sum is over the periodic orbits $\tau^{\prime}=(\tau,k)$, not just the prime ones; λ and Λ are defined in (4.20). The second equality comes (for r integer) by adding and subtracting.

\begin{equation*}\eta(r)\colon=\sum_{\lambda(\tau^{\prime})\le r} \Lambda(\tau^{\prime}){\rm tr}{{}^{t}}\Psi_{\tau^{\prime}}\end{equation*}

(4.26)\begin{align} =\sum_{n\in[1,r]}\left[ \sum_{\lambda(\tau^{\prime})=n}\left( \Lambda(\tau^{\prime}){\rm tr}{{}^{t}}\Psi_{\tau^{\prime}} \right) -\beta^n \right] + \beta\frac{\beta^r-1}{\beta-1}. \end{align}

By (4.26) and the remark after (4.25), there is $D_1 \gt 0$, independent of r, such that the first inequality below holds; the second one holds for some $D_2=D_2(\beta) \gt 0$ independent of r by the properties of the geometric series; note that $\frac{\beta}{e^\epsilon} \gt 1$ if ϵ is small enough, since we are supposing that β > 1.

(4.27)\begin{align} \left\vert \eta(r)-\frac{\beta^{r+1}-\beta}{\beta-1} \right\vert \le D_1\sum_{n\in[1,r]}\left(\frac{\beta}{e^{\epsilon }}\right)^n\le D_2\frac{\beta^r}{e^{\epsilon r}} . \end{align}

For ${\rm tr}{{}^{t}}\Psi_\tau$ defined as in (4.19) and $r\ge 1$, we set

(4.28)\begin{align} \pi^{\prime}(r)=\sum_{\lambda(\tau)\le r } {\rm tr}({{}^{t}}\Psi_\tau), \end{align}

where τ varies among the prime orbits.

For γ > 1, we set $r=\gamma y$; (4.28) implies the equality below; the inequality follows since $\frac{\lambda(\tau)}{y} \gt 1$ on the set of the first sum, while $\frac{\lambda(\tau)}{y}\ge 0$ for all τ; recall that ${\rm tr}{{}^{t}}\Psi_\tau^k\ge 0$ by Lemma 4.2. The second equality comes from the definition of η in (4.26) and the notation (4.20).

\begin{equation*}\pi^{\prime}(r)=\pi^{\prime}(y)+ \sum_{\{\tau\colon y \lt \lambda(\tau)\le r \}}{\rm tr}({{}^{t}}\Psi_{\tau})\end{equation*}

\begin{equation*}\le\pi^{\prime}(y)+\sum_{\{(k,\tau)\colon k\cdot\lambda(\tau)\le r \}}\frac{\lambda(\tau)}{y} {\rm tr}({{}^{t}}\Psi_{\tau})^k= \pi^{\prime}(y)+\frac{1}{y}\eta(r) . \end{equation*}

If we multiply this inequality by $\frac{r}{\beta^r}$ and recall that $r=\gamma y$, we get

(4.29)\begin{align} \frac{r\cdot\pi^{\prime}(r)}{\beta^r} \le \frac{(\gamma y)\cdot\pi^{\prime}(y)}{\beta^{\gamma y}}+ \frac{\gamma\cdot\eta(r)}{\beta^r}. \end{align}

In Lemma 4.4, we are going to show that, if $\gamma^{\prime} \gt 1$, then

(4.30)\begin{align} \frac{\pi^{\prime}(y)}{\beta^{\gamma^{\prime} y}}\rightarrow 0\quad \text{as}\ y\rightarrow+\infty. \end{align}

We show that the last two formulas imply

(4.31)\begin{align} \limsup_{r\rightarrow+\infty} \frac{r\cdot\pi^{\prime}(r)}{\beta^r}\le \gamma\cdot\frac{\beta}{\beta-1}. \end{align}

For the proof, we fix $\gamma^{\prime}\in(1,\gamma)$; (4.29) implies the first inequality below, while (4.30) and the fact that $\beta^{\gamma-\gamma^{\prime}} \gt 1$ imply the second one; the last one comes from (4.27).

\begin{equation*}\limsup_{r\rightarrow+\infty}\frac{r\cdot\pi^{\prime}(r)}{\beta^r}\le \gamma\limsup_{y\rightarrow+\infty}\frac{y}{\beta^{(\gamma-\gamma^{\prime})y}}\cdot \frac{\pi^{\prime}(y)}{\beta^{\gamma^{\prime} y}} + \gamma\limsup_{r\rightarrow+\infty}\frac{\eta(r)}{\beta^r}\end{equation*}

\begin{equation*}\le\gamma\limsup_{r\rightarrow+\infty}\frac{\eta(r)}{\beta^r}\le \gamma\cdot\frac{\beta}{\beta-1} . \end{equation*}

This shows (4.31).

Since γ > 1 is arbitrary, (4.31) implies that

\begin{equation*}\limsup_{r\rightarrow+\infty}\frac{\pi^{\prime}(r)r}{\beta^r}\le\frac{\beta}{\beta-1} . \end{equation*}

Consider now the function π of (11) with $\hat V\equiv 1$; in this case, it is immediate that the constant c of Theorem 2 is $\log\beta \gt 0$, where β is the eigenvalue for the zero potential. By the definition of $\pi^{\prime}$ in (4.28), we get that

\begin{equation*}\pi(r)=\sum_{{\rm e}^{c\lambda(\tau)}\le r}{\rm tr}\Psi_\tau= \pi^{\prime}\left( \frac{\log r}{c} \right) . \end{equation*}

Since $c=\log\beta$, the last two formulas imply (12).

The case β < 1 is simpler: (4.26) and (4.28) imply the first inequality below, the second one comes from (4.25) and the last one from the fact that β < 1.

\begin{equation*}\pi^{\prime}(r)\le\eta(r)\end{equation*}

\begin{equation*}\le\sum_{n\in[1,r]}\left[ \beta^n+D_1\frac{\beta^n}{{\rm e}^{n\epsilon}} \right] \le D_2. \end{equation*}

Now (12) follows.

Now there is only one only missing ingredient, the proof of (4.30); we begin with some preliminaries.

We note that, by point (5) of Lemma 4.2,

\begin{equation*}\sum_{i=1}^m\alpha_{\tau,i}^k\ge 0. \end{equation*}

If we assume that x > 0, this implies the first inequality below; the second inequality comes from the fact that $x\ge\log(1+x)$ and the formula above again. The two equalities that bookend the formula come from the properties of the logarithm.

\begin{equation*}\prod_{i=1}^m\frac{1}{1-\alpha_{\tau,i}\cdot x}= \exp\left[ \sum_{k\ge 1}\frac{1}{k}x^k\left( \sum_{i=1}^m\alpha_{\tau,i}^k \right) \right] \end{equation*}

\begin{equation*}\ge\exp\left[ \left( \sum_{i=1}^m\alpha_{\tau,i} \right)x \right] \end{equation*}

(4.32)\begin{align} \ge\exp\left[ \left( \sum_{i=1}^m\alpha_{\tau,i} \right)\log(1+x) \right] = \left( 1+x \right)^{\sum_{i=1}^m\alpha_{\tau,i}}. \end{align}

Proof. Since $\gamma^{\prime} \gt 1$ and we are supposing β > 1,

\begin{equation*}\frac{1}{\beta^{\gamma^{\prime}}} \lt \frac{1}{\beta} . \end{equation*}

Together with Proposition 4.3, this implies that $\frac{1}{\beta^{\gamma^{\prime}}}$ belongs to the disk where $\zeta(z,0)$ converges. Now we take y > 1; since $\frac{1}{\beta^{\gamma^{\prime}}} \gt 0$ and ${\rm tr}(\Psi_\tau) \gt 0$ by Lemma 4.2, (4.21) implies the first inequality below, while the first equality comes from the properties of the logarithm; recall that $V\equiv 0$. For the second inequality, we apply (4.32) to the innermost product. For the third inequality, we note that $1+\frac{1}{\beta^{\gamma^{\prime}\cdot y}}$ is smaller than all the factors in the product. The second equality comes from the definition of $\pi^{\prime}$ in (4.28). The second last inequality follows from the binomial theorem, while the last one is obvious.

\begin{equation*}\zeta\left(\frac{1}{\beta^{\gamma^{\prime}}},0\right)\ge \exp\sum_{\lambda(\tau)\le y}\sum_{j=1}^m\sum_{k\ge 1} \frac{\alpha^k_{\tau,j}\,{\rm e}^{kV(\tau)}\frac{1}{\beta^{\gamma^{\prime} k\lambda(\tau)}}}{k}\end{equation*}

\begin{equation*}=\prod_{\lambda(\tau)\le y}\prod_{i=1}^m \frac{1}{1-\frac{\alpha_{\tau,i}}{\beta^{\gamma^{\prime}\cdot\lambda(\tau)}}} \end{equation*}

\begin{equation*}\ge\prod_{\lambda(\tau)\le y}\left( 1+\frac{1}{\beta^{\gamma^{\prime}\cdot\lambda(\tau)}} \right)^{\sum_{i=1}^m\alpha_{\tau,i}} \ge \prod_{\lambda(\tau)\le y } \left( 1+\frac{1}{\beta^{\gamma^{\prime}\cdot y}} \right)^{\sum_{i=1}^m\alpha_{\tau,i}}\end{equation*}

\begin{equation*}=\left( 1+\frac{1}{\beta^{\gamma^{\prime}\cdot y}} \right)^{\pi^{\prime}(y) }\ge 1+\frac{\pi^{\prime}(y)}{\beta^{\gamma^{\prime}\cdot y}}\ge \frac{\pi^{\prime}(y) }{\beta^{\gamma^{\prime}\cdot y}} . \end{equation*}

Thus, $\frac{\pi^{\prime}(y)}{\beta^{\gamma^{\prime}\cdot y}}$ is bounded for all $\gamma^{\prime},y \gt 1$, which implies that

\begin{equation*}\frac{\pi^{\prime}(y)}{\beta^{\gamma^{\prime}\cdot y}}\rightarrow 0\quad \text{for}\ y\rightarrow+\infty \end{equation*}

for all $\gamma^{\prime} \gt 1$.

Standing hypothesis

Our standing hypothesis from now on is that, when c > 0, the function $\zeta_{-V}(s)$ can be extended to a continuous function on $\{{\rm Re}(s)\ge 1 \}\setminus \{1 \}$, with a simple pole at s = 1. When c < 0, we do not make any hypothesis: it will suffice the fact, which we saw above, that ζ _−V extends to a holomorphic function in ${\rm Re}(s) \gt -1$. In other words, we are supposing that

(5.4)\begin{align} \zeta_{-V}(s)=\left\{ \begin{aligned} \frac{\tilde\alpha_+(s)}{(s-1)}&\quad \text{if}\ c \gt 0\cr \tilde\alpha_-(s)&\quad \text{if}\ c \lt 0, \end{aligned} \right. \end{align}

with $\tilde\alpha_+$ analytic and non-zero in ${\rm Re}(s)\ge 1$ if c > 0, and $\tilde\alpha_-$ analytic and non-zero in ${\rm Re}(s) \gt -1$ if c < 0.

Remark. In the scalar case of [Reference Parry and Pollicott14], the fact that the zeta function has no zeroes on the line ${\rm Re}(s)=1$ has a connection with the dynamics, being equivalent to the fact that the suspension flow by $\hat V$ is topologically mixing. One of the ways to express the topologically mixing property is the following. We define the operator

\begin{equation*}V\colon C(\Sigma,\textbf{C})\rightarrow C(\Sigma,\textbf{C}), \end{equation*}

\begin{equation*}(Vw)(x)={\rm e}^{-ia\hat V(x)}w\circ\sigma(x) . \end{equation*}

We say that the suspension flow (which we leave undefined) is weakly topologically mixing if Vw = w implies that a = 0 and that w is constant. As we said, we shall not study the connection between this property and the behaviour of the zeta function on the critical line.

Next, we set

(5.5)\begin{align} N(\tau)={\rm e}^{V(\tau)}={\rm e}^{|c|\hat V(\tau)}. \end{align}

With this notation, we can rewrite one equality of (5.3) as

\begin{equation*}\zeta_{-V}(s)=\exp\sum_{k\ge 1}\sum_\tau\sum_{i=1}^{m} \frac{N(\tau)^{-sk}}{k}\alpha_{\tau,i}^k . \end{equation*}

As we saw after (5.2), this holds in $Re(s) \gt 1$ if c > 0, and in $Re(s) \gt -1$ if c < 0.

Now we take derivatives in two different expressions for $\log\zeta_{-V}(s)$; for the equality on the left, we differentiate the formula above, for that on the right we differentiate (5.4) and recall that $\tilde\alpha_\pm\not=0$; the function $\alpha_+$ is analytic in an open half-plane containing $Re(s)\ge 1$ and $\alpha_-$ is analytic in the open half-plane $Re(s) \gt -1$.

\begin{equation*}-\sum_{k\ge 1}\sum_\tau\sum_{i=1}^{m} [\log N(\tau)]\cdot N(\tau)^{-sk}\cdot\alpha_{\tau,i}^k\end{equation*}

(5.6)\begin{align} =\frac{ \zeta_{-V}^{\prime}(s) }{\zeta_{-V}(s) }= \left\{ \begin{aligned} -\frac{1}{s-1}+\alpha_+(s)&\quad \text{if}\ c \gt 0\cr \alpha_-(s)& \quad \text{if}\ c \lt 0 . \end{aligned} \right. \end{align}

We denote by $\lfloor a\rfloor$ the integer part of $a\ge 0$; recall that we saw at the beginning of this section that $V=|c|\hat V \gt 0$; by (5.5), this implies that $\log N(\tau)$ is always positive. As a consequence, for $r\ge 1$, we can define $n(r,\tau)\ge 0$ by

\begin{equation*}n(r,\tau)\colon= \left\lfloor \frac{ \log r }{ \log N(\tau) } \right\rfloor . \end{equation*}

We define the function $S\colon (1,+\infty)\rightarrow \textbf{R}$ in the first equality below; the second one comes from the definition of $n(r,\tau)$ above and the fact that $N(\tau)\ge 1$.

\begin{equation*}S(r)\colon=\sum_{\{(\tau,n)\colon N(\tau)^n\le r\}}\sum_{i=1}^{m} ( \log N(\tau) )\cdot\alpha_{\tau,i}^n\end{equation*}

(5.7)\begin{align} =\sum_{\{\tau\colon N(\tau)\le r \}}( \log N(\tau) )\cdot \sum_{n=1}^{n(r,\tau)} \left( \sum_{i=1}^{m} \alpha_{\tau,i}^n \right) . \end{align}

Note that S is monotone increasing and thus its derivative $ {\rm d} S$ is a positive measure; from the first sum above, we gather that each orbit $\tau^{\prime}=(\tau,k)$ contributes to $ {\rm d} S$ the measure

\begin{equation*}\sum_{i=1}^{m} \left(\log N(\tau)\right) \alpha_{\tau,i}^k \cdot\delta_{N(\tau)^k}, \end{equation*}

where δ_x denotes the Dirac delta centred at x. Let now c > 0; the second equality below is the right hand side of (5.6); the first equality follows from the formula above, the left hand side of (5.6) and the fact that $N(\tau)\ge 1$ by (5.5). The integral is Stieltjes and $Re(s) \gt 1$.

(5.8)₊\begin{align} -\int_1^{+\infty}r^{-s}\, {\rm d} S(r)= \frac{ \zeta^{\prime}_{-V}(s) }{ \zeta_{-V}(s) }=\frac{-1}{s-1}+\alpha_+(s). \end{align}

When c < 0, we do the same thing, but using the second expression on the right of (5.6); the equalities hold for $s \gt -1$.

(5.8)₋\begin{align} -\int_{1}^{+\infty}r^{-s}\, {\rm d} S(r)= \frac{ \zeta^{\prime}_{-V}(s) }{ \zeta_{-V}(s) }=\alpha_-(s). \end{align}

We recall the Wiener–Ikehara Tauberian theorem. A proof is in Appendix I of [Reference Parry and Pollicott15]; for the whole theory, we refer the reader to [Reference Wiener25].

Theorem 5.2. Assume that the formula below holds, where $A\in\textbf{R}$, the integral on the left is Stieltjes, α is analytic in ${\rm Re}(s) \gt 1$ and has a continuous extension to ${\rm Re}(s)\ge 1$; the function S is monotone increasing.

\begin{equation*}\int_1^{+\infty}r^{-s}\, {\rm d} S(x)=\frac{A}{s-1}+\alpha(s). \end{equation*}

Then,

(5.9)\begin{align} \lim_{s\rightarrow+\infty}\frac{S(s)}{s}=A. \end{align}

For $N(\tau)$ defined as in (5.5), we set, as in (11),

(5.10)\begin{align} \pi(r)=\sum_{N(\tau)\le r }\sum_{i=1}^{m}\alpha_{\tau,i}. \end{align}

Lemma 5.3. Let the potentials $V,\hat V \gt 0$ be as at the beginning of this section; let π be defined as in (5.10) and let

(5.11)\begin{align} \hat\pi(r)= \sum_{{\rm e}^{\hat V(\tau)}\le r }{\rm tr}{{}^{t}}\Psi_\tau. \end{align}

Then,

(5.12)\begin{align} \hat\pi(r)= \pi(r^{|c|}). \end{align}

In particular, if

(5.13)\begin{align} \lim_{r\rightarrow+\infty}\frac{\pi(r)\,\log r}{r}\le A , \end{align}

then

(5.14)\begin{align} \lim_{r\rightarrow+\infty}\frac{\hat\pi(r)\,\log r}{r^{|c|}}\le\frac{A}{|c|} . \end{align}

Lastly, if we had ≥ in (5.13), then we would have ≥ also in (5.14).

Proof. We define $N(\tau)$ as in (5.5); the first equality below comes from the fact that $V=|c|\hat V$ and the second one is (5.5).

\begin{equation*}{\rm e}^{\hat V(\tau)}={\rm e}^{\frac{1}{|c|}V(\tau)}=N(\tau)^\frac{1}{|c|} . \end{equation*}

The first equality below is the definition of $\hat\pi$ in (5.11); since $|c| \gt 0$, the second one comes from the formula above and the third one is the definition of π in (5.10).

\begin{equation*}\hat\pi(r)= \sum_{{\rm e}^{\hat V(\tau) }\le r }\sum_{i=1}^m\alpha_{\tau,i}= \sum_{{\rm e}^{V(\tau) }\le r^{|c|} }\sum_{i=1}^m\alpha_{\tau,i}= \pi(r^{|c|}) . \end{equation*}

This shows (5.12); now a simple calculation shows that (5.13) implies (5.14).

Proof of point (2) of Theorem 2

By Lemma 5.3, formula (13) of Theorem 2 follows if we show (5.13) with A = 1 when c > 0 and with A = 0 when c < 0; this is what we shall do.

We distinguish two cases: when c > 0, (5.8)₊ implies that S satisfies the hypotheses of Theorem 5.2 for A = 1 and thus (5.9) holds, always for A = 1; a consequence is the asymptotic sign in the formula below, which holds for $r\rightarrow+\infty$. The equality comes from (5.7).

(5.15)\begin{align} r\simeq S(r)= \sum_{N(\tau)\le r} (\log N(\tau)) \sum_{n=1}^{n(t,\tau)} \left( \sum_{i=1}^{m}\alpha_{\tau,i}^n \right). \end{align}

Now we consider γ > 1 and, for $r\ge 1$, we define $y=y(r)$ by the formula below.

(5.16)\begin{align} 1\le y=r^\frac{1}{\gamma} \lt r. \end{align}

The first equality below comes from the definition of $\pi(r)$ in (5.10); for the first inequality, we use point (5) of Lemma 4.2 and the fact that $\frac{\log N(\tau)}{\log y}\ge 1$ if $N(\tau)\ge y$ and $\frac{\log N(\tau)}{\log y}\ge 0$ in all cases; this follows from (5.5) and the fact that V > 0. The second equality comes from (5.7); the equality at the end comes by (5.16).

\begin{equation*}\pi(r)=\pi(y)+ \sum_{y \lt N(\tau)\le r }{\rm tr}({{}^{t}}\Psi_{\tau})\end{equation*}

\begin{equation*}\le\pi(y)+\sum_{\{(\tau,k)\colon N(\tau)^k\le r \}}\frac{\log N(\tau)}{\log y} {\rm tr}({{}^{t}}\Psi_{\tau}^k)\end{equation*}

\begin{equation*}=\pi(y)+\frac{1}{\log y}S(r)=\pi(y)+\frac{\gamma}{\log r}S(r). \end{equation*}

If we multiply the last formula by $\frac{\log r}{r}$ and recall (5.16), we get

(5.17)\begin{align} (\log r)\cdot\frac{\pi(r)}{r}\le \frac{\pi(y)}{y^\gamma}\cdot\gamma\cdot\log y+\gamma\cdot\frac{S(r)}{r} . \end{align}

In Lemma 4.2, we are going to see that, if $\gamma^{\prime} \gt 1$, then

(5.18)\begin{align} \limsup_{y\rightarrow+\infty}\frac{\pi(y)}{y^{\gamma^{\prime}}} \lt +\infty. \end{align}

Let now $\gamma^{\prime}\in(1,\gamma)$; Formula (5.17) implies the inequality below, (5.18) and (5.15) imply the equality.

\begin{equation*}\limsup_{r\rightarrow+\infty}\frac{\pi(r)\,\log r}{r}\end{equation*}

\begin{equation*}\le\gamma\cdot\limsup_{y\rightarrow+\infty} \frac{\pi(y)}{y^{\gamma^{\prime}}}\cdot\frac{\log y}{y^{\gamma-\gamma^{\prime}}}+ \gamma\cdot\limsup_{r\rightarrow+\infty}\frac{S(r)}{r}=\gamma. \end{equation*}

Recalling that γ > 1 is arbitrary, we get the inequality of (5.13) with A = 1, i.e., Formula (13) of Theorem 2.

The case c < 0 is similar. Formula (5.8)₊ implies that S satisfies the hypotheses of Theorem 5.2 for A = 0; in turn, this implies that

\begin{equation*}\frac{S(r)}{r}\rightarrow 0\ \text{as}\ r\rightarrow+\infty . \end{equation*}

Now the same argument as above implies that

\begin{equation*}\frac{\pi(r)\log(r)}{r}\rightarrow 0\quad \text{as}\ r\rightarrow+\infty . \end{equation*}

Now (5.18) is the only missing ingredient of the proof of Theorem 2.

Proof. We show the case c > 0, since the other one is analogous. We take γ > 1 and y > 1; recall that we saw after (5.2) that $\zeta_{-V}(s)$ is analytic in $Re(s) \gt 1$; as a consequence, (5.3) holds for $s=\gamma$; by point (5) of Lemma 4.2, this yields the first inequality below. As for the equality, it comes from the properties of the logarithm as in (5.3). For the second inequality, we use (4.32). For the third inequality, we note that $(1+y^{-\gamma})$ is the smallest argument; the last inequality follows by the binomial theorem.

\begin{equation*}\zeta_{-V}(\gamma)\ge \exp\sum_{N(\tau)\le y}\sum_{k\ge 1}\sum_{i=1}^m \frac{{\rm e}^{-k V(\tau)\cdot\gamma}\alpha^k_{\tau,i}}{k}\end{equation*}

\begin{equation*}=\prod_{N(\tau)\le y}\prod_{i=1}^m\frac{1}{1-{\rm e}^{-V(\tau)\cdot \gamma}\alpha_{\tau,i}}\ge \prod_{N(\tau)\le y}\prod_{i=1}^m[ 1+N(\tau)^{-\gamma} ]^{\alpha_{i,\tau}} \end{equation*}

\begin{equation*}\ge(1+y^{-\gamma})^{\pi(y)}\ge\frac{\pi(y)}{y^\gamma}.\end{equation*}

Thus, for y sufficiently large, $\frac{\pi(y)}{y^\gamma}$ is smaller than $\zeta_{-V}(\gamma)$.