2. Preliminaries of Riemannian geometry

Let us introduce some basic notions of Riemannian geometry. Given an infinite-dimensional $C^{\infty}$ manifold (M, g) equipped with a smooth Riemannian metric g, let $T M$ be the tangent bundle and $T_{1} M$ be the set of unit norm tangent vectors of (M, g), the unit tangent bundle. Let $\chi(M)$ be the set of $C^{\infty}$ vector fields of M.

In [Reference Biliotti and Mercuri3], several results for Riemannian metrics on infinite-dimensional manifolds are presented. We will not use any of the results of that paper.

The only infinite-dimensional manifold we will be interested in here is $\mathcal{N}$ which is the set of Hölder equilibrium probabilities (which was initially defined in [Reference Giulietti, Kloeckner, Lopes and Marcon10]). Tangent vectors, differentiability, analyticity, etc., should be always considered in the sense of the setting described in Sections 2.3 and 5.1 in [Reference Giulietti, Kloeckner, Lopes and Marcon10] (see also [Reference Bomfim, Castro and Varandas4] and [Reference da Silva, da Silva and Souza8]). We will elaborate on this later.

So in our case, $M= \mathcal{N}$, and g is the L ² metric, $g_{A}(X,Y) = \int X Y\,\mathrm{d}\mu_{A}$.

For practical purposes, we shall call Energy the function $E(v) = g(v,v)$, $v \in T \mathcal{N}$, although in mechanics the energy is rather defined by $\frac{1}{2}g(v,v)$.

Given a smooth function $f :\mathcal{N} \longrightarrow \mathbb{R}$, the derivative of f with respect to a vector field $X \in \chi (\mathcal{N} )$ will be denoted by X(f). The Lie bracket of two vector fields $X, Y \in \chi(\mathcal{N} )$ is the vector field whose action on the set of functions $f: \mathcal{N} \longrightarrow \mathbb{R}$ is given by $[X,Y](f) = X(Y(f)) - Y(X(f))$.

The Levi-Civita connection of $(\mathcal{N} ,g)$, $\nabla : \chi(\mathcal{N} )\times \chi(\mathcal{N} ) \longrightarrow \chi(\mathcal{N} )$, with notation $\nabla(X,Y) = \nabla_{X}Y$, is the affine operator characterized by the following properties:

1. (1) Compatibility with the metric g:

   \begin{equation*} Xg(Y,Z) = g(\nabla_{X}Y, Z) + g(Y, \nabla_{X}Z) \end{equation*}

   for every triple of vector fields $X, Y, Z$.

2. (2) Absence of torsion:

   \begin{equation*} \nabla_{X}Y - \nabla_{Y}X = [X,Y].\end{equation*}

3. (3) For every smooth scalar function f and vector fields $X,Y \in \chi(\mathcal{N} )$, we have

   • • $ \nabla_{fX}Y = f\nabla_{X}Y$,

   • • Leibniz rule: $ \nabla_{X}(fY) = X(f)Y + f\nabla_{X}Y$.

The expression of $\nabla_{X}Y$ can be obtained explicitly from the expression of the Riemannian metric, in dual form. Namely, given two vector fields $X, Y \in \chi(\mathcal{N} )$ and $Z \in \chi(\mathcal{N} )$, we have

\begin{eqnarray*} g(\nabla_{X}Y, Z) & = & \frac{1}{2}(Xg(Y,Z) + Yg(Z, X) -Zg(X,Y) \\ & - & g([X,Z], Y) -g([Y,Z],X) -g([X,Y], Z)) , \end{eqnarray*}

2.1. Curvature tensor and sectional curvatures

We follow [Reference do Carmo9] for the definitions in the subsection. To simplify the notation, from now on, we shall adopt the convention $g(X,Y) = \langle X, Y \rangle$. The curvature tensor

\begin{equation*} \mathcal{R} : \chi(\mathcal{N}) \times \chi(\mathcal{N}) \times \chi(\mathcal{N}) \longrightarrow \chi(\mathcal{N}) \end{equation*}

is defined in terms of the Levi-Civita connection as follows

(7)\begin{equation} \mathcal{R} (X, Y)Z = \nabla_{Y}\nabla_{X}Z - \nabla_{X}\nabla_{Y}Z + \nabla_{[X,Y]}Z . \end{equation}

The sectional curvature of the plane generated by two vector fields $X, Y$ at the point $A \in \mathcal{N}$, which are orthonormal at A, is given by

(8)\begin{equation} K(X,Y) = \langle \nabla_{Y}\nabla_{X}X - \nabla_{X}\nabla_{Y}X + \nabla_{[X,Y]}X, Y \rangle = \langle \mathcal{R}(X,Y)X,Y\rangle.\end{equation}

Let A be a normalized Hölder potential. Let us consider a local smooth surface $S(t,s)$, for $\mid t \mid, \mid s \mid \leq \epsilon$ small, tangent to the plane $ \{A + tX + sY\} $ generated by $X, Y$ at the point $A = S(0,0)$. Let $\bar{X}$, $\bar{Y}$ be the coordinate vector fields of the surface and suppose that $\bar{X}_{A} =X$, $\bar{Y}_{A} =Y$. In § 4.2, we shall exhibit such local surfaces.

Lemma 2.1. The expression of the sectional curvature of the plane generated by the two orthonormal vectors $X, Y$ is

(9)\begin{align} K(X,Y) &= -\frac{1}{2}(\bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2}) )+ \bar{Y}(\bar{Y}(\parallel \bar{X} \parallel^{2}) )) + \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2} + \bar{Y}(\bar{X}\langle \bar{X}, \bar{Y} \rangle )\nonumber\\ &\quad\, - \langle \nabla_{\bar{X}}\bar{X} , \nabla_{\bar{Y}}\bar{Y} \rangle . \end{align}

Proof. The fact that $\bar{X}$ and $\bar{Y}$ commute implies that $\nabla_{\bar{X}}\bar{Y} = \nabla_{\bar{Y}}\bar{X}$ and

\begin{equation*} \langle \mathcal{R}(\bar{X},\bar{Y})\bar{X},\bar{Y} \rangle = \langle \nabla_{\bar{Y}}\nabla_{\bar{X}}\bar{X} - \nabla_{\bar{X}}\nabla_{\bar{Y}}\bar{X} , \bar{Y} \rangle .\end{equation*}

The first term of $\langle \mathcal{R}(\bar{X},\bar{Y})\bar{X},\bar{Y} \rangle$ gives

\begin{eqnarray*} \langle \nabla_{\bar{Y}}\nabla_{\bar{X}}\bar{X} , \bar{Y} \rangle & = & \bar{Y}\langle \nabla_{\bar{X}}\bar{X}, \bar{Y} \rangle - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle \\ & = & \bar{Y}(\bar{X} \langle \bar{X}, \bar{Y} \rangle - \langle \bar{X}, \nabla_{\bar{X}}\bar{Y}\rangle ) - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle \\ & = & \bar{Y}(\bar{X} \langle \bar{X}, \bar{Y} \rangle - \langle \bar{X}, \nabla_{\bar{Y}}\bar{X}\rangle ) - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle \\ & = & \bar{Y}(\bar{X} \langle \bar{X}, \bar{Y} \rangle - \frac{1}{2}\bar{Y} (\parallel \bar{X} \parallel^{2}))- \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle \\ & = & - \frac{1}{2}\bar{Y}(\bar{Y} (\parallel \bar{X} \parallel^{2})) + \bar{Y}(\bar{X} \langle \bar{X}, \bar{Y} \rangle ) - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle. \end{eqnarray*}

The second term of the formula gives

\begin{eqnarray*} \langle \nabla_{\bar{X}}\nabla_{\bar{Y}}\bar{X} , \bar{Y} \rangle & = & \bar{X}\langle \nabla_{\bar{Y}}\bar{X},\bar{Y} \rangle - \langle \nabla_{\bar{Y}}\bar{X}, \nabla_{\bar{X}}\bar{Y}\rangle \\ & = & \bar{X}\langle \nabla_{\bar{X}}\bar{Y}, \bar{Y}\rangle - \langle \nabla_{\bar{Y}}\bar{X}, \nabla_{\bar{Y}}\bar{X}\rangle \\ & = & \frac{1}{2} \bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2})) - \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2}. \end{eqnarray*}

Subtracting the second term from the first one we obtain the lemma.

3. The analytic structure of the set of normalized potentials

Definition 3.1. Let $ (X, | .|)$ and $(Y, |.|)$ Banach spaces and V an open subset of $ X.$ Given $k\in \mathbb{N}$, a function $F : V\to Y$ is called k-differentiable in x, if for each $j=1, \ldots, k$, there exists a j-linear bounded transformation

\begin{equation*}D^j F(x) : \underbrace{X \times X \times ... \times X}_j \to Y,\end{equation*}

such that,

\begin{equation*}D^{j -1}F(x + v_j )(v_1, \ldots, v_{j-1}) - D^{j-1}F(x)(v_1, \ldots, v_{j-1}) = D^jF(x)(v_1, \ldots, v_j ) + o_ j (v_j ), \end{equation*}

where

\begin{equation*} o_j : X \to Y, \text{satisfies,} \lim_{v\to 0} \frac{|o_j (v)|_Y}{ |v|_X }= 0. \end{equation*}

By definition F has derivatives of all orders in V, if for any $x\in V$ and any $k\in \mathbb{N}$, the function F is k-differentiable in x.

Definition 3.2. Let $X, Y$ be Banach spaces and V an open subset of X. A function $F : V \to X$ is called analytic on V, when F has derivatives of all orders in V, and for each $x \in V$ there exists an open neighbourhood V_x of x in V, such that, for all $v\in V_x$, we have that

\begin{equation*} F(x + v) - F(x) = \sum_{j=1}^\infty \frac{1}{n !} D^j F(x)v^j,\end{equation*}

where $D^j F(x)v^j = D^j F(x)(v, \ldots, v) $ and $ D_j F(x) $ is the j-th derivative of F in x.

Above we use the notation of Section 3.2 in [Reference da Silva, da Silva and Souza8].

$\mathcal{N}$ can be expressed locally in coordinates via analytic charts (see [Reference Giulietti, Kloeckner, Lopes and Marcon10]).

3.1. Some more estimates from thermodynamic formalism

Given a potential $B \in \text{Hol} $, we consider the associated Ruelle operator $\mathscr{L}_B$ and the corresponding main eigenvalue λ_B and eigenfunction h_B.

The function

(10)\begin{equation} \Pi (B) = B + \log(h_{B}) - \log(h_{B}(T)) -\log(\lambda_{B}) \end{equation}

describes the projection of the space of potentials B on Hol onto the analytic manifold of normalized potentials $\mathcal{N}$.

We identify below $T_A \mathcal{N}$ with the affine subspace $\{A + X : X \in T_A \mathcal{N}\}.$

The function Π is analytic (see [Reference Giulietti, Kloeckner, Lopes and Marcon10]) and therefore has first and second derivatives. Given the potential B, then the map $D_B \Pi : T_{B}\mathcal{N} \longrightarrow T_{\Pi(B)}\mathcal{N} $ given by

\begin{equation*} D_B \Pi (X) = \frac{\partial}{\partial t}(\Pi(B+ tX)_{t=0} \end{equation*}

should be considered as a linear map from Hol to itself (with the Hölder norm on Hol). Moreover, the second derivative $D^2_B \Pi$ should be interpreted as a bilinear form from Hol × Hol to Hol and is given by

\begin{equation*} D^2_B \Pi(X,Y) = \frac{\partial^2}{\partial t \partial s}(\Pi(B+ tX +sY)_{t=s=0}. \end{equation*}

We denote by $||A||_\alpha$ the α-Hölder norm of an α-Hölder function A.

When B is normalized the eigenvalue is 1 and the eigenfunction is equal to 1. We would like to study the geometry of the projection Π restricted to the tangent space $T_{A}\mathcal{N}$ into the manifold $\mathcal{N}$ (namely, to get bounds for its first and second derivatives with respect to the potential viewed as a variable) for a given normalized potential A.

The space $T_{A}\mathcal{N}$ is a linear subspace of functions and the derivative map $D \Pi$ is analytic when restricted to it.

We denote by $E_0 =E_0^A$ the set of Hölder functions g, such that, $\int g \,\mathrm{d} \mu_A =0,$ where µ_A is the equilibrium probability for the normalized potential A. Note that $E_0^A$ is contained in $T_{A} (\mathcal{N}).$

Most of the claims of the next Lemma are based mainly on results of [Reference Giulietti, Kloeckner, Lopes and Marcon10] (see also [Reference da Silva, da Silva and Souza8], [Reference Bomfim, Castro and Varandas4]).

Lemma 3.3. Let $\Lambda : \text{Hol} \longrightarrow \mathbb{R}$, $H : \text{Hol} \longrightarrow \text{Hol}$ be given, respectively, by $ \Lambda (B) = \lambda_{B},$ $ H(B) = h_{B}$. Then we have

1. (1) The maps Λ, H and $A \longrightarrow \mu_{A}$ are analytic.

2. (2) For a normalized B, we get that $D_{B}\log(\Lambda) (\psi) = \int \psi \,\mathrm{d}\mu_{B}.$

3. (3) $ D^{2}_{B}\log(\Lambda) (\eta, \psi) = \int \eta \psi \,\mathrm{d}\mu_{B},$ where $\psi , \eta $ are at $T_{B}\mathcal{N}$, and B is normalized.

4. (4) If A is a normalized potential, then for every function $X \in T_{A}\mathcal{N}$, we have

   • • $\int X \,\mathrm{d}\mu_{A} =0$.

   • • $D_{A}\Pi(X) = X$.

In order to simplify the notation, from now on, unless is necessary for the understanding, we will denote $(I - \mathscr{L}_{T,A}|_{E_0^A})^{-1}$ by $ (I - \mathscr{L}_{T,A})^{-1}.$

Items (2) and (3) are taken from Theorem D in [Reference Giulietti, Kloeckner, Lopes and Marcon10]. Item $\int X \,\mathrm{d}\mu_{A} =0$ in (3) follows from Theorem A and Corollary B in [Reference Giulietti, Kloeckner, Lopes and Marcon10], and the other item in (4) is trivial.

The analyticity of Λ and H of the item (1) are well-known facts (see Chapter 4 in [Reference Parry and Pollicott18] or Corollary B in [Reference Giulietti, Kloeckner, Lopes and Marcon10]) which was also proved in [Reference Bomfim, Castro and Varandas4].

The law that takes a Hölder potential B to its normalization A is differentiable according to Section 2.2 in [Reference Giulietti, Kloeckner, Lopes and Marcon10].

Note that the derivative linear operator $X \to D_{A}H(X)$ is zero when A is normalized.

Remark 1: Item (1) above means that for a fixed Hölder function f the map $A \to \int f \,\mathrm{d} \mu_A$ is differentiable on A (see Theorem B in [Reference Bomfim, Castro and Varandas4]).

Questions related to second derivatives on thermodynamic formalism are considered in [Reference Ma and Pollicott16], [Reference Petkov and Stoyanov19] and [Reference Pollicott and Sharp21].

4. Evaluating the sectional curvatures of the Riemannian metric

The goal of the section is to calculate the sectional curvature $K(X,Y)$ of the plane generated by two orthogonal vector fields tangent to $A \in \mathcal{N}$ applying the calculus of thermodynamic formalism. We start with a technical result that is a consequence of formula 6. This lemma will be extensively used in the article.

4.1. Leibniz rule of differentiation

Lemma 4.1. Let $A \in \mathcal{N}$ and let $\gamma: (-\epsilon, \epsilon) \longrightarrow \mathcal{N}$ be a smooth curve such that $\gamma(0) =A$. Let $X(t) = \gamma'(t)$, and let Y be a smooth vector field tangent to $\mathcal{N}$ defined in an open neighbourhood of A. Denote by $Y(t)= Y(\gamma(t))$. Then the derivative of $\int Y(t) \,\mathrm{d}\mu_{\gamma(t)}$ with respect to the parameter t is

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} \int Y(t)\mathrm{d}\mu_{\gamma(t)} = \int \frac{\mathrm{d}Y(t)}{\mathrm{d}t}\mathrm{d}\mu_{\gamma(t)} + \int Y(t)X(t)\mathrm{d}\mu_{\gamma(t)} \end{equation*}

for every $ t\in (-\epsilon, \epsilon)$.

Proof. The idea of the proof is very simple and based on the fact that the function $Q: \chi(\mathcal{N})\times m_{T} \longrightarrow \mathbb{R}$ given by

\begin{equation*}Q(X,\mu) = \int X\mathrm{d}\mu\end{equation*}

is a bilinear form, where $\chi(\mathcal{N})$ is the set of C ¹ vector fields tangent to $\mathcal{N}$ and m_T is the set of invariant measures of the map T. So the derivative of a function of the type $Q(X(t), \mu(t))$ satisfies a sort of Leibniz rule. Let us check.

Let us calculate the derivative at t = 0, for every other $t \in (-\epsilon, \epsilon)$, the calculation is analogous. We have

\begin{eqnarray*} \frac{\mathrm{d}}{\mathrm{d}t}\int Y(t) \,\mathrm{d}\mu_{\gamma(t)} \mid_{t=0} & = & \lim_{t \rightarrow 0}\frac{1}{t}(\int Y(t)\,\mathrm{d}\mu_{\gamma(t)} - \int Y(0) \,\mathrm{d}\mu_{A} )\\ & = & \int \lim_{t\rightarrow 0}\frac{1}{t}(Y(t) -Y(0)) \,\mathrm{d}\mu_{\gamma(t)} \\ & + & \lim_{t \rightarrow 0} \frac{1}{t}(\int Y(0) \,\mathrm{d}\mu_{\gamma(t)} - \int Y(0)\,\mathrm{d}\mu_{A}) \\ &= & \int \frac{\mathrm{d}Y(t)}{\mathrm{d}t}\,\mathrm{d}\mu_{A}+ \lim_{t \rightarrow 0} \frac{1}{t}(\int Y(0) \,\mathrm{d}\mu_{A + tX(0)} - \int Y(0)\,\mathrm{d}\mu_{A}) \end{eqnarray*}

where in the last step we use the fact that the derivative with respect to t only depends on the vector X(0) and not on the curve through A tangent to X(0). By Equation (6), the second term in the above equality is just $\frac{\mathrm{d}}{\mathrm{d}t} \int Y(0) \,\mathrm{d}\mu_{A+tX(0)} \mid_{t=0}$, which equals $\int X(0) Y(0) \,\mathrm{d}\mu_{A }$. This finishes the proof of the lemma.

From now on, we shall adopt the notation $\frac{\partial Y}{\partial t}= Y' = Y_{t}$; the second one applies when there is only one parameter involved in the calculations, and the third one will be used otherwise.

4.2. Auxiliary local surfaces in $\mathcal{N}$

Next, given a normalized potential A and $X, Y$ orthonormal vector in the tangent space of A, we proceed to construct a local surface $S(t,s)$, $\mid t \mid, \mid s \mid \lt \epsilon$ small, such that $S(0,0) =A$, and the tangent space of $S(t,s)$ at A is the plane generated by $X, Y$. Let us consider the plane

\begin{equation*}P(t,s) = A + tX + sY \end{equation*}

where $t, s, \in \mathbb{R} $, that is a subset of $T_{A} \mathcal{N}$, and let Π be the projection into $\mathcal{N}$ defined in Equation (10). The vector fields $X_{P(t,s) } = \frac{\partial}{\partial t} P(t,s) = X$, $Y_{P(t,s)} = \frac{\partial}{\partial s} P(t,s) = Y$ are tangent to the plane P of course.

Let $S(t,s) = \Pi(P(t,s))$. By Lemma 3.3 item (5), the restriction of the map Π to the plane $P(t,s)$ is a local diffeomorphism onto its image, so there exists ϵ > 0 small such that $S(t,s)$ is an analytic embedding of the rectangle $\{\mid t \mid \lt \epsilon\} \times \{\mid s \mid \lt \epsilon \}$.

The coordinate vector fields of $S(t,s)$ are $\bar{X}_{S(t,s)} = \frac{\partial }{\partial t}( \Pi(P(t,s))) = D_{P(t,s)} \Pi(X)$, $\bar{Y}_{S(t,s)} = \frac{\partial }{\partial s}( \Pi (P(t,s)) = D_{P(t,s)} \Pi(Y)$, so $\bar{X}, \bar{Y}$ are extensions of $X, Y$.

Moreover, we have the following result from thermodynamic formalism (for derivatives of high order see (3.4) in [Reference Ma and Pollicott16]):

Lemma 4.2. Suppose $\psi:\{1,2, \ldots,d\}^{\mathbb{N}} \to \mathbb{R}$ is Hölder, normalized and µ denotes the associated equilibrium probability. Assume also that the Hölder function ϕ satisfies $\mathscr{L}_\psi (\phi)=0$. Denote by λ_t and w_t, $t\in \mathbb{R}$, respectively, the eigenvalue and the eigenfunction for the Ruelle operator $\mathscr{L}_{\psi + t \phi}$. Then, we have

1. (1) The derivative of w_t satisfies

   (11)\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} w_t(x)|_{t=0}=c, \text{for all}\, x, \end{equation}

   for some constant c.

2. (2) Moreover, as ψ is normalized

   (12)\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} \log (w_t(x)|_{t=0})=c, \text{for all}\, x. \end{equation}

3. (3) Suppose $\overline{X}$ is an analytic vector field, extending the tangent vector X, defined in a neighbourhood of ϕ. Let $\gamma : (-\epsilon, \epsilon)\to \mathcal{N}$ be an integral curve of $\overline{X}$, with $\gamma(0)=\phi$, and let w_t be the curve of eigen functions for the Ruelle operator of $\gamma(t)$. Then,

   (13)\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} w_t(x)=c_t, \text{for all}\, x, \end{equation}

   is a curve of constant functions which is analytic on t.

4. (4) For any tangent vector X (in the kernel of the Ruelle operator), the directional derivative

   (14)\begin{equation} D_{\psi}H (X) =c_X = D_{\psi}\log H (X), \end{equation}

   where c_X depends on X and ψ.

5. (5) From Equation (11), we get

   (15)\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} w_t(T(x))|_{t=0}=c, \text{for all}\, x, \end{equation}

   and for the same constant c of Equation (11).

Proof. We are going to take derivative on the Hölder direction ϕ. Assume that ϕ satisfies $\mathscr{L}_\psi (\phi)=0,$ which implies that $\int \phi \,\mathrm{d} \mu=0.$ This is so because iterates of a function under the Ruelle–Perron–Frobenius operator converge to the integral of that function against the eigenmeasure.

Denote by $w(t,x)=w_t(x)$, the normalized eigenfunction for $\mathscr{L}_{\psi + t \phi}$ associated with the eigenvalue λ_t. That is

(16)\begin{equation} \mathscr{L}_{\psi + t \phi} (w_t)= \lambda_t w_t . \end{equation}

Taking derivative on t:

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} \mathscr{L}_{\psi + t \phi} (w(t,.) ) (x) = \mathscr{L}_{\psi + t \phi } ( \phi(.) w(t,.)) (x)+\mathscr{L}_{\psi +t \phi} (\frac{\mathrm{d}}{\mathrm{d}t} w(t,.)) (x). \end{equation*}

Therefore, for all x, when t = 0, we get

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} \mathscr{L}_{\psi + t \phi} (w(t,.) ) (x)|_{t=0} = \mathscr{L}_{\psi } ( \phi(.) ) (x)+\mathscr{L}_{\psi } (\frac{\mathrm{d}}{\mathrm{d}t} w(t,.)|_{t=0} ) (x)=\end{equation*}

\begin{equation*}0 +\mathscr{L}_{\psi } (\frac{\mathrm{d}}{\mathrm{d}t} w(t,.)|_{t=0} ) (x) . \end{equation*}

On the other hand, for all x and t

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t}[ \lambda_t w(t,x) ] = w(t,x) \frac{\mathrm{d}}{\mathrm{d}t}\lambda_t + \lambda_t \frac{\mathrm{d}}{\mathrm{d}t} w(t,x). \end{equation*}

Then, taking t = 0,

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t}[ \lambda_t w(t,x) ]|_{t=0} = w(0,x) \frac{\mathrm{d}}{\mathrm{d}t}\lambda_t|_{t=0} + \lambda_t \frac{\mathrm{d}}{\mathrm{d}t}|_{t=0} w(t,x)= \end{equation*}

\begin{equation*}\int \phi \,\mathrm{d} \mu +\frac{\mathrm{d}}{\mathrm{d}t}|_{t=0} w(t,x)= \frac{\mathrm{d}}{\mathrm{d}t}|_{t=0} w(t,x). \end{equation*}

Denote $g(x) = \frac{\mathrm{d}}{\mathrm{d}t} w(t,.)|_{t=0} (x) .$

Then, $\forall x$, we get from the above and Equation (16)

\begin{equation*}\mathscr{L}_{\psi } (g) (x) = g(x),\end{equation*}

for the normalized potential ψ. But, the only continuous eigenfunctions for $\mathscr{L}_{\psi }$, which are associated to the eigenvalue 1 are the constant functions.

Therefore, there exists c such that $ \frac{\mathrm{d}}{\mathrm{d}t} w_t(x)|_{t=0}=c$, for all x.

As, for all t and x

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} \log (w_t(x)|_{t=0})= \frac{\frac{\mathrm{d}}{\mathrm{d}t} w_t(x)|_{t=0}}{w_0(x)}=\frac{\mathrm{d}}{\mathrm{d}t} w_t(x)|_{t=0} ,\end{equation*}

we get Equation (12).

Equation (14) follows at once from the above.

Expression (13) is obtained in the same way as it was derived Equation (11) (applying the argument for each value t), and, finally, Equation (15) follows trivially from Equation (11).

We will use the above result on the next lemma.

Lemma 4.3. The derivatives with respect to $t , s$ of the coordinate vector fields $\bar{X}$, $\bar{Y}$ at the point A (a normalized potential) are

1. (1) $\frac{\partial }{\partial t} \bar{X} =\frac{\partial }{\partial s} \bar{Y} = - 1 $

2. (2) $\frac{\partial }{\partial s} \bar{X} = \frac{\partial }{\partial t}\bar{Y} = 0$.

Proof. We assume that the tangent vector is Hölder and in the kernel of the Ruelle operator $\mathscr{L}_A$. The proof of the lemma will be a direct consequence of Lemma 4.2 taking $\psi=A$ and $\phi=X$. We will prove first the item (1) above.

The local surface $S(t,s)$ is contained in the manifold of normalized potentials, and we denote, respectively, the corresponding eigenvalue by $\lambda _{S(t,s)}$ and the associated eigenfunction by $h_{S(t,s)}$ (of the Ruelle operator associated with $S(t,s)$).

Let I be the identity map. The expression of the projection Π (Equation (10)) is

\begin{equation*} \Pi (B) = I(B) + \log(h_{B}) - \log(h_{B}(T)) -\log(\lambda_{B}). \end{equation*}

By definition, we have

(17)\begin{equation} \frac{\partial }{\partial t}( \bar{X}_{S(t,0)})_{t=0} = \frac{\partial }{\partial t}( D_{P(t,0)}\Pi(X_{P(t,0)}))_{t=0}. \end{equation}

Lemma 3.3 grants that all the functions involved in the expression of Π are differentiable, so we get at the point t = 0,

(18)\begin{eqnarray} \begin{aligned} \frac{\partial }{\partial t}( D_{P(t,0)}\Pi(X_{P(t,0)}))_{t=0} & = \frac{\partial}{\partial t}(D_{P(t,0)}I(X_{P(t,0)}) )_{t=0}\\ & + \frac{\partial}{\partial t} ((D_{P(t,0)}\log (H) )(X_{P(t,0)}))_{t=0} \\ & - \frac{\partial}{\partial t}((D_{P(t,0)}\log(H\circ T))(X_{P(t,0)}))_{t=0} \\ & - \frac{\partial}{\partial t} ((D_{P(t,0)} \log (\Lambda)) (X_{P(t,0)}))_{t=0}. \end{aligned} \end{eqnarray}

The first term gives at t = 0,

\begin{equation*} \frac{\partial}{\partial t}(D_{P(t,0)}I(X_{P(t,0)}))_{t=0} = \frac{\partial}{\partial t}(X)_{t=0}=0,\end{equation*}

since X does not depend on t.

Claim 1:

The second and third term cancel due to Equations (11) and (15).

Indeed, the curves

\begin{equation*}\alpha(t) = D_{P(t,0)}\log (H) (X_{P(t,0)}), \beta(t) = D_{P(t,0)} \log (H \circ T) (X_{P(t,0)})\end{equation*}

coincide by Equations (11) and (15) with the expression

\begin{equation*} \alpha (t) = \beta (t) = \frac{\frac{\mathrm{d}}{\mathrm{d}t}c_{X_{P(t,0)}}}{c_{X_{P(t,0)}}}\end{equation*}

for each t, where c_X is given in Lemma 4.2. These curves are analytic and therefore differentiable, so their derivatives with respect to t coincide. Since derivatives $\alpha'(t)$, $\beta '(t)$ appear with opposite signs in Equation (18), they add up to zero in this formula. This proves the Claim.

Finally, the fourth line of Equation (18) gives by Lemma 3.3 item (3),

\begin{equation*}- \frac{\partial}{\partial t} ((D_{P(t,0)} \log (\Lambda)) (X_{P(t,0)}))_{t=0} = -\int X^{2} \,\mathrm{d}\mu = -1\end{equation*}

since X has L ² norm equal to 1. The same argument applies replacing X by Y in the above proof, so this finishes the proof of item (1).

Item (2) follows the same type of reasoning and using Equation (13). By definition, we have

\begin{eqnarray*} \frac{\partial }{\partial s} (\bar{X}_{S(0,s)})_{s=0} &= &\frac{\partial }{\partial s}( D_{P(t,s)}\Pi(X_{P(t,s)})_{t=0})_{s=0} . \end{eqnarray*}

This expression, according to Equation (18), is

\begin{eqnarray*} \frac{\partial }{\partial s}( D_{P(t,s)}\Pi(X_{P(t,s)}))_{t=s=0} & = & \frac{\partial}{\partial s}(D_{P(t,s)}I(X_{P(t,s)}) )_{t=s=0}\\ & + & \frac{\partial}{\partial s} ((D_{P(t,s)}\log (H) )(X_{P(t,s)}))_{t=s=0} \\ & - & \frac{\partial}{\partial s}((D_{P(t,s)}\log(H\circ T))(X_{P(t,s)}))_{t=s=0} \\ & - & \frac{\partial}{\partial s} ((D_{P(t,s)} \log (\Lambda)) (X_{P(t,s)}))_{t=s=0}. \end{eqnarray*}

The first term gives at t = 0,

\begin{equation*} \frac{\partial}{\partial s}(D_{P(t,0)}I(X_{P(t,0)}))_{t=0} = \frac{\partial}{\partial s}(X)_{t=0}=0\end{equation*}

since X does not depend on $t,s$. The fourth term is, by Lemma 3.3 item (3),

\begin{equation*}- \frac{\partial}{\partial s} (D_{P(t,s)} \log (\Lambda) (X_{P(t,s)}))_{t=s=0} = - D^{2}_{A}\log (\Lambda)(Y,X) = -\int XY \,\mathrm{d}\mu_{A} =0. \end{equation*}

As for the second and third terms, we have

Claim 2:

\begin{equation*}\frac{\partial}{\partial s} ((D_{P(t,s)}\log (H) )(X_{P(t,s)}))_{t=s=0}= \frac{\partial}{\partial s}((D_{P(t,s)}\log(H\circ T))(X_{P(t,s)}))_{t=s=0}.\end{equation*}

The proof goes as in Claim 1, letting

\begin{equation*}\alpha_{s}(t) = D_{P(t,s)}\log (H) )(X_{P(t,s)}), \beta_{s}(t) = D_{P(t,s)} \log (H \circ T) (X_{P(t,s)})\end{equation*}

we have by Lemma 4.2 items (3) and (5) that $\alpha_{s}(t) = \beta_{s}(t)$ is an analytic curve of constant functions for each given s. Therefore, the function

\begin{equation*} w(t,s) = \alpha_{s}(t) = \beta_{s}(t) \end{equation*}

is an analytic function of the parameters $t,s$ and therefore, the derivatives of $\alpha_{s}(t)$ and $\beta_{s}(t)$ with respect to s coincide and give a family of constant functions in the local surface $S(t,s)$. This finishes the proof of Claim 2.

Claim 2 yields that the sum of the second and third terms of the expression of $\frac{\partial }{\partial s}( D_{P(t,s)}\Pi(X_{P(t,s)}))_{t=s=0}$ vanishes, just finishing the proof of item (2).

4.3. The expression of $K(X,Y)$ in terms of the calculus of thermodynamic formalism

Let us first state some notation. Let $\bar{X}_{t}$ be the derivative of the vector field $\bar{X}$ with respect to the parameter t and $\bar{X}_{s}$ be the derivative of the vector field $\bar{X}$ with respect to the parameter s. The same convention applies to $\bar{Y}_{t}$, $\bar{Y}_{s}$. The notation $\bar{X}(Y) = \frac{\partial}{\partial t}\bar{Y} = \bar{Y}_{t}$ will always represent derivatives with respect to the vector field $\bar{X}$, while $\bar{X}\bar{Y}$ or $\bar{X}\times \bar{Y}$ will represent the product of the functions $\bar{X}$ and $\bar{Y}$. Through the section, this double character of the vectors tangent to the manifold $\mathcal{N}$ which are also functions will show up in all statements and proofs.

Theorem 4.4. Let A be a normalized potential, let $X, Y \in T_{A} \mathcal{N}$ be a pair of orthonormal vector fields, and let $S: (-\epsilon, \epsilon)\times (-\delta, \delta) \longrightarrow \mathcal{N}$ be the local surface defined in the previous subsection with $S(0,0)= A$, $\bar{X} $, whose coordinate vector fields are $\bar{X}$, $\bar{Y}$, with $\bar{X}(A) =X$, $\bar{Y}(A) = Y$. Then the sectional curvature $K(X,Y)$ at A of the plane generated by $X, Y$ is given by the expression

\begin{equation*} K(X,Y)= \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2} - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle. \end{equation*}

We shall subdivide the proof into several steps.

Lemma 4.5. We have that $\bar{X}_{s} = \bar{Y}_{t}$ in the local surface S.

This is a straightforward consequence of the fact that the vector fields $\bar{X}, \bar{Y}$ commute.

Next, let us evaluate the terms of the sectional curvature in Lemma 2.1,

\begin{eqnarray*} K(X,Y) & = & -\frac{1}{2}(\bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2})) + \bar{Y}(\bar{Y}(\parallel \bar{X} \parallel^{2})) ) + \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2} \\ & + &\bar{Y}(\bar{X}\langle \bar{X}, \bar{Y} \rangle ) - \langle \nabla_{\bar{X}}\bar{X} , \nabla_{\bar{Y}}\bar{Y} \rangle . \end{eqnarray*}

Lemma 4.6. At every point $p \in S(t,s)$, we have

1. (1) $ \bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2}) )= 2\int \bar{Y}\bar{Y}_{tt}\,\mathrm{d}\mu_{p} - \int \bar{Y}^{2} \,\mathrm{d}\mu_{p} +\int \bar{X}^{2}\bar{Y}^{2} \,\mathrm{d}\mu_{p} .$

2. (2) $ \bar{Y}(\bar{Y}(\parallel \bar{X} \parallel^{2} ))= 2\int \bar{X}\bar{X}_{ss}\,\mathrm{d}\mu_{p} - \int \bar{X}^{2} \,\mathrm{d}\mu_{p} +\int \bar{X}^{2}\bar{Y}^{2} \,\mathrm{d}\mu_{p} .$

In particular, if p = A, we have

1. (1) $ \bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2}) )= 2\int \bar{Y}\bar{Y}_{tt}\,\mathrm{d}\mu_{A} -1 +\int \bar{X}^{2}\bar{Y}^{2} \,\mathrm{d}\mu_{A} .$

2. (2) $ \bar{Y}(\bar{Y}(\parallel \bar{X} \parallel^{2} ))= 2\int \bar{X}\bar{X}_{ss}\,\mathrm{d}\mu_{A} - 1 +\int \bar{X}^{2}\bar{Y}^{2} \,\mathrm{d}\mu_{A} .$

Proof. The expression follows from the application of the Leibniz rule to differentiate $\parallel \bar{Y} \parallel^{2}= \int \bar{Y}^{2} \,\mathrm{d}\mu_{p}$ (we shall omit for convenience the p in the notation of the measure $\mathrm{d}\mu_{p}$ ):

\begin{eqnarray*} \bar{X}(\bar{X}\int \bar{Y}^{2} \,\mathrm{d}\mu) & = & \bar{X} ( 2\int \bar{Y} \bar{Y}_{t} \,\mathrm{d}\mu + \int \bar{X} \bar{Y}^{2}\,\mathrm{d}\mu) \\ & =& 2 \int (\bar{Y}_{t})^{2} \,\mathrm{d}\mu + 2 \int \bar{Y}\bar{Y}_{tt} \,\mathrm{d}\mu + 2\int \bar{Y} \bar{X} \bar{Y}_{t} \,\mathrm{d}\mu \\ & + & \int \bar{X}_{t} \bar{Y}^{2} \,\mathrm{d}\mu + 2 \int \bar{X}\bar{Y} \bar{Y}_{t} \,\mathrm{d}\mu + \int \bar{X}^{2} \bar{Y}^{2} \,\mathrm{d}\mu \\ & = & 2\int (\bar{Y}_{t})^{2} \,\mathrm{d}\mu + 2\int \bar{Y}\bar{Y}_{tt}\,\mathrm{d}\mu + 4\int \bar{X}\bar{Y}\bar{Y}_{t} \,\mathrm{d}\mu \\ & + & \int \bar{X}_{t}\bar{Y}^{2} \,\mathrm{d}\mu + \int \bar{X}^{2}\bar{Y}^{2} \,\mathrm{d}\mu . \end{eqnarray*}

Since by Lemma 4.3 we have that $\bar{X}_{s} = \bar{Y}_{t}=0$, $\bar{X}_{t}= \bar{Y}_{s} = -1$, we get item (1) just by replacing this values in the integral expressions above.

Interchanging $\bar{X}$ and $\bar{Y}$, t and s, in the above formula, we get item (2). At the point p = A, we have that $\int \bar{X}^{2} \,\mathrm{d}\mu_{A} = \int \bar{T}^{2} \,\mathrm{d}\mu_{A} = 1$, so replacing these values in the formula we finish the proof of the lemma.

Lemma 4.7. The expression of $\bar{Y}(\bar{X}\langle \bar{X}, \bar{Y} \rangle ) = \bar{Y}(\bar{X}\int \bar{X}\bar{Y} \,\mathrm{d}\mu_{p}) $ is

\begin{equation*} \bar{Y}(\bar{X}\int \bar{X}\bar{Y}\,\mathrm{d}\mu_{p}) = \int \bar{Y}\bar{X}_{ts}\,\mathrm{d}\mu_{p} + 1 - \int \bar{Y}^{2}\,\mathrm{d}\mu_{p} + \int \bar{X} \bar{Y}_{ts}\,\mathrm{d}\mu_{p} - \int \bar{X}^{2}\,\mathrm{d}\mu_{p} + \int \bar{X}^{2} \bar{Y}^{2}\,\mathrm{d}\mu_{p} \end{equation*}

at every point $p \in S(t,s)$. In particular, at p = A, we have

\begin{equation*} \bar{Y}(\bar{X}\int \bar{X}\bar{Y}\,\mathrm{d}\mu_{A}) = \int \bar{Y}\bar{X}_{ts}\,\mathrm{d}\mu_{A} + \int \bar{X} \bar{Y}_{ts}\,\mathrm{d}\mu_{A} - 1 + \int \bar{X}^{2} \bar{Y}^{2}\,\mathrm{d}\mu_{A}. \end{equation*}

Proof. We apply the Leibniz rule,

\begin{eqnarray*} \bar{Y}(\bar{X}\int \bar{X}\bar{Y}\,\mathrm{d}\mu) & = & \bar{Y}( \int \bar{X}_{t}\bar{Y}\,\mathrm{d}\mu + \int \bar{X}\bar{Y}_{t}\,\mathrm{d}\mu + \int \bar{X}^{2} \bar{Y}\,\mathrm{d}\mu )\\ & = & \int \bar{X}_{ts} \bar{Y}\,\mathrm{d}\mu + \int \bar{X}_{t} \bar{Y}_{s}\,\mathrm{d}\mu + \int \bar{X}_{t} \bar{Y}^{2}\,\mathrm{d}\mu \\ & + & \int \bar{X}_{s} \bar{Y}_{t}\,\mathrm{d}\mu + \int \bar{X} \bar{Y}_{ts}\,\mathrm{d}\mu + \int \bar{X} \bar{Y}_{t} \bar{Y}\,\mathrm{d}\mu \\ & + & \int \bar{Y}_{s}\bar{X}^{2}\,\mathrm{d}\mu + 2\int \bar{Y} \bar{X} \bar{X}_{s}\,\mathrm{d}\mu + \int \bar{X}^{2}\bar{Y}^{2}\,\mathrm{d}\mu. \end{eqnarray*}

Since by Lemma 4.5 we have that $\bar{X}_{s} = \bar{Y}_{t}$ we get the following formula just adding the terms in the above formula:

\begin{eqnarray*} \bar{Y}(\bar{X}\int \bar{X}\bar{Y}\,\mathrm{d}\mu) & = & \int \bar{Y}\bar{X}_{ts}\,\mathrm{d}\mu + \int \bar{X}_{t}\bar{Y}_{s}\,\mathrm{d}\mu + \int \bar{X}_{t} \bar{Y}^{2}\,\mathrm{d}\mu + \int (\bar{X}_{s})^{2}\,\mathrm{d}\mu \\ & + & \int \bar{X} \bar{Y}_{ts}\,\mathrm{d}\mu + 3\int \bar{X}\bar{X}_{s}\bar{Y}\,\mathrm{d}\mu + \int \bar{Y}_{s}\bar{X}^{2}\,\mathrm{d}\mu + \int \bar{X}^{2} \bar{Y}^{2}\,\mathrm{d}\mu. \end{eqnarray*}

By Lemma 4.3, $\bar{X}_{s}= \bar{Y}_{t}=0$, $\bar{X}_{t}= \bar{Y}_{s}=-1$, and replacing these values in the integral expression above we obtain the formula in the statement. Moreover, if p = A, we know that $\int \bar{X}^{2}\,\mathrm{d}\mu_{A} = \int ^{2}\,\mathrm{d}\mu_{A} = 1$, as well as $\int \bar{Y}^{2}\,\mathrm{d}\mu_{A} = \int Y^{2}\,\mathrm{d}\mu_{A} = 1$, thus concluding the proof of the Lemma.

Corollary 4.8. The term $-\frac{1}{2}(\bar{X}(\bar{X}(\parallel \bar{Y} \parallel^{2}) )+ \bar{Y}(\bar{Y}(\parallel \bar{X} \parallel^{2}))) + \bar{Y}(\bar{X}\langle \bar{X}, \bar{Y} \rangle )$ in the expression of $K(X,Y)$ at the point A vanishes.

Proof. To shorten notation, we shall omit the dependence of A in the expressions. According to Lemma 4.5, we have that

1. (1) $ \int \bar{X}\bar{X}_{ss}\,\mathrm{d}\mu = \int \bar{X} \bar{Y}_{ts}\,\mathrm{d}\mu$.

2. (2) $\int \bar{Y} \bar{X}_{st}\,\mathrm{d}\mu = \int \bar{Y} \bar{Y}_{tt}\,\mathrm{d}\mu$.

Replacing the above equalities in the expressions of Lemmas 4.6 and 4.7, and adding the resulting formulae we get Corollary 4.8.

Theorem 4.4 follows at once from Corollary 4.8.

5. Cristoffel coefficients at the expression of $K(X,Y)$

We denote by $\{X_{i} \}$, $i \in \mathbb{N}$, a complete orthonormal base of the vector space $T_{A} \mathcal{N} \subset L^2(\mu)$ (for the Gibbs probability µ associated with the normalized potential A).

The main goal of the section is to obtain the expression for the sectional curvature in Theorem 1.1.

Namely, let $A \in \mathcal{N}$ be a point in the manifold of normalized potentials, let $X,Y \in T_{A} \mathcal{N}$ be two orthonormal tangent vectors. Then the expression of the curvature of the plane generated by $X, Y$ is

(19)\begin{equation} K(X,Y) = \frac{1}{4}[ \sum_{i=1}^\infty ( \int X Y X_i \,\mathrm{d} \mu)^2 - \sum_{i=1}^\infty \int X^2 X_i \,\mathrm{d} \mu \int Y^2 X_i\,\mathrm{d} \mu ]. \end{equation}

In Proposition 5.2, we will show that the above sum is well-defined.

The proof is a direct calculation of the terms $ \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2}, \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle$ that appear in the expression of the curvature in Theorem 4.4. We shall subdivide the calculation in several lemmas.

We follow the notation of the previous section. Let $S(t,s)$ be the local surface given in § 4 tangent to the plane generated by the vectors $X, Y$, satisfying $S(0,0) = A$, let $\bar{X}, \bar{Y}$ be the local extensions of the vectors $X, Y$ obtained by projecting by the map Π the plane generated by $X, Y$ at $T_{A} \mathcal{N}$ into the tangent space of $\mathcal{N}$.

Let us define local extensions $\bar{X}_{i}$ of the vector fields X_i in an analogous way we defined the extensions of $X, Y$: let S_k be the plane generated by $X_{1}, X_{2}, \ldots,X_{k}$ and let us project by Π the tangent space of S_k into $T\mathcal{N}$ by the differential of the projection into $\mathcal{N}$.

The terms $ \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2}, \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y}\rangle$ involve the Cristoffel symbols of the vector fields $\bar{X}, \bar{Y}$, at the point A we have:

\begin{equation*} \nabla_{\bar{X}_{k}}\bar{X}_{l} = \sum_{i=1}^{\infty} \Gamma_{kl}^{i}X_{i} \end{equation*}

where $\Gamma_{kl}^{i} = \langle \nabla_{\bar{X}_{k}}\bar{X}_{l}, \bar{X}_{i} \rangle $ is the Cristoffel coefficient. We follow [Reference do Carmo9] for the definitions and basic properties of Cristoffel coefficients.

The coefficient $\Gamma_{ij}^{k}$ can be calculated in terms of the coefficients of the first fundamental form of the metric at A, the inner products $g_{ij} = \langle X_{i}, X_{j} \rangle $ by the following formula:

\begin{equation*} \Gamma_{kl}^{i} = \frac{1}{2}g^{im}(g_{mk,l} + g_{ml,k} - g_{kl,m}) \end{equation*}

where g ^im is the coefficient of the inverse of the first fundamental form of index im, $g_{mk,l}$ is the derivative with respect to $\bar{X}_{l}$ of the coefficient g_mk and the above notation is Einstein’s convention for the sum on the index m.

The expression ‘inverse of the first fundamental form’ requires some explanation since we are dealing with an infinite-dimensional Riemannian manifold. One natural rigorous approach is to evaluate the series $\sum_{i=1}^{\infty} \Gamma_{kl}^{i}X_{i}$ as the limit of its partial sums $\sum_{i=1}^{n} \Gamma_{kl}^{i}X_{i}$ that includes the Cristoffel coefficients in the subspace of $T_{A}\mathcal{N}$ generated by $\{X_{1},X_{2}, \ldots,X_{n}\}$. The first fundamental form restricted to this subspace is a n × n matrix that, under our assumptions, is the identity. Its inverse is of course the identity. This allows us to define all the terms in the partial sum, then we take the limit as $n \rightarrow \infty$ to get the series. We shall prove that the series converges absolutely, so the above procedure provides the expression of $\nabla_{\bar{X}_{k}}\bar{X}_{l}$ as an infinite series.

In particular, since the basis $\{X_{1},X_{2}, \ldots,X_{n}, \ldots\}$ is orthonormal, the indices in the sum of the expression of $\nabla_{\bar{X}_{k}}\bar{X}_{l}$ according to Einstein’s convention just reduce to ii,kk, ll, depending on the case, and $g_{kl} = g^{kl} = \delta_{kl}$. So at the point A we get the formula

\begin{equation*} \Gamma_{kl}^{i} = \frac{1}{2}(g_{ik,l} + g_{il,k} - g_{kl,i}). \end{equation*}

Lemma 5.1. The term $g_{ik,l}$ at A, for any permutation of the indices, is

\begin{equation*} g_{ik,l} = \int X_{i}X_{k}X_{l}\,\mathrm{d}\mu. \end{equation*}

Then,

\begin{equation*} \nabla_{\bar{X}_{k}}\bar{X}_{l} = \frac{1}{2} \sum_{i=1}^{\infty} (\int X_{i}X_{k}X_{l}\,\mathrm{d}\mu_{A}) X_{i}. \end{equation*}

Proof. We have that $g_{ik,l} = \bar{X}_{l} \langle \bar{X}_{i}, \bar{X}_{k} \rangle = \bar{X}_{l} \int \bar{X}_{i}\bar{X}_{k}\,\mathrm{d}\mu$. By the Leibniz rule, we have

\begin{equation*} \bar{X}_{l} \int \bar{X}_{i}\bar{X}_{k}\,\mathrm{d}\mu = \int \frac{\partial}{\partial \bar{X}_{l}} (\bar{X}_{i})\bar{X}_{k}\,\mathrm{d}\mu + \int \bar{X}_{i}\frac{\partial}{\partial \bar{X}_{l}}(\bar{X}_{k})\,\mathrm{d}\mu + \int \bar{X}_{i}\bar{X}_{k} \bar{X}_{l}\,\mathrm{d}\mu \end{equation*}

where $\frac{\partial}{\partial \bar{X}_{l}} (\bar{X}_{i})$ is the derivative of the vector field $\bar{X}_{i}$ in the direction of $\bar{X}_{l}$.

Notice that Lemma 4.3 extends to the submanifolds S_k for every $k \in \mathbb{N}$. So we have

1. (1) $\frac{\partial}{\partial \bar{X}_{l}} (\bar{X}_{i}) = 0 $ if l ≠ i,

2. (2) $\frac{\partial}{\partial \bar{X}_{l}} (\bar{X}_{i}) = -1$ if l = i.

In both cases, since $\int \bar{X}_{i}\,\mathrm{d}\mu =0$ for every i, we get $g_{ik,l} = \int X_{i}X_{k}X_{l}\,\mathrm{d}\mu$ as claimed.

The expression for $\nabla_{\bar{X}_{k}}\bar{X}_{l}$ is straightforward from this formula.

Corollary 5.2. Let us assume that $X=X_{1}$ and $ Y= X_{2}$ are the first two vectors of the orthonormal base $\{X_{i} \}$. For the normalized potential $A= S(0,0)$, we get the following expressions

\begin{equation*} \nabla_{\bar{X}_{1}} \bar{X}_{1} = \frac{1}{2} \sum_{i=1}^{\infty} (\int X_{1}^{2}X_{i}\,\mathrm{d}\mu_{A}) X_{i}\end{equation*}

\begin{equation*} \nabla_{\bar{X}_{2}} \bar{X}_{2} = \frac{1}{2} \sum_{i=1}^{\infty}(\int X_{2}^{2}X_{i}\,\mathrm{d}\mu_{A}) X_{i}\end{equation*}

\begin{equation*}\nabla_{\bar{X}_{1}} \bar{X}_{2} = \frac{1}{2}\sum_{i=1}^{\infty} (\int X_{1}X_{2}X_{i}\,\mathrm{d}\mu_{A}) X_{i}.\end{equation*}

Moreover, for any pair $X,Y \in T_{A} \mathcal{N}$ the sums

\begin{equation*}\sum_{i=1}^\infty ( \int X Y X_i \,\mathrm{d} \mu)^2 \text{and} \sum_{i=1}^\infty \int X^2 X_i \,\mathrm{d} \mu \int Y^2 X_i \,\mathrm{d} \mu \end{equation*}

are both finite.

Proof. We consider an extension of the family X_r, $r \in \mathbb{N}$, to all $ L^2(\mu)$ and we get a complete orthonormal base of the vector space $ L^2(\mu)$, given by $X_r, Y_s$, $r,s \in \mathbb{N}$. The first three expressions in the statement are straightforward from Lemma 5.1.

Given two elements $X,Y \in T_{A} \mathcal{N}$ consider $f=X Y = \sum_r a_r^f X_r + \sum_s b_s^f Y_s \in L^2(\mu)$, then,

\begin{equation*} (\int X Y X_i \,\mathrm{d} \mu)^2 = |a_i^f|^2. \end{equation*}

It follows that $\sum_{i=1}^\infty ( \int X Y X_i \,\mathrm{d} \mu)^2 = \sum_{i=1}^\infty |a_i^f|^2\leq \parallel f\parallel^2$ is finite.

Denote $g= X^2 =\sum_r a_r^g X_r + \sum_s b_s^g Y_s $ and $h=Y^2= \sum_r a_r^h X_r + \sum_s b_s^h Y_s$. Therefore,

\begin{equation*}\int g h \,\mathrm{d} \mu= \sum_{i=1}^\infty a_i^g a_i^h + \sum_{j=1}^\infty b_j^g b_j^h . \end{equation*}

Form this follows that $\sum_{i=1}^\infty a_i^g a_i^h $ converges. Note that $ \int X^2 X_i \,\mathrm{d} \mu = a_i^g $ and $ \int Y^2 X_i \,\mathrm{d} \mu = a_i^h.$ Then,

\begin{equation*} \sum_{i=1}^\infty \int X^2 X_i \,\mathrm{d} \mu \int Y^2 X_i \,\mathrm{d} \mu = \sum_{i=1}^\infty a_i^g a_i^h \end{equation*}

converges.

Theorem 1.1 follows from direct calculation applying Corollary 5.2 to the expression of $K(X,Y)$.

6. A worked example in the Markov case: an orthonormal basis for the kernel of the Ruelle operator

From now on $M=\{0,1\}^{\mathbb{N}}$ and we denote by $\mathcal{K}$ the set of stationary Markov probabilities taking values in $\{0,1\}$.

In this section, given a probability $\mu_A\in K$, we will exhibit an orthonormal basis for the tangent space to $\mathcal{N}$ (the kernel of the Ruelle operator) at µ_A.

Given a finite word $x =(x_1,x_2, \ldots,x_k)\in \{0,1\}^k$, $k \in \mathbb{N}$, we denote by $[x]$ the associated cylinder set in $M=\{0,1\}^{\mathbb{N}}$.

Consider an invariant Markov probability µ obtained from a row stochastic matrix $(P_{i,j})_{i,j=0,1}$ and an initial left invariant vector of probability $\pi=(\pi_0,\pi_1)\in \mathbb{R}^2$.

Given $r \in (0,1)$ and $s\in (0,1)$, we denote

(20)\begin{equation} P= \left( \begin{array}{cc} P_{0,0} & P_{0,1}\\ P_{1,0} & P_{1,1} \end{array}\right)= \left( \begin{array}{cc} r & 1-r\\ 1-s & s \end{array}\right) . \end{equation}

In this way $(r,s)\in (0,1) \times (0,1)$ parameterize all row stochastic matrices.

The explicit expression is

(21)\begin{equation} \mu [x_1,x_2, \ldots,x_n] = \pi_{x_1} P_{x_1,x_2} P_{x_2,x_3} \cdots P_{x_{n-1},x_n}. \end{equation}

Definition 6.1. Denote by $J:\{0,1\}^{\mathbb{N}} \to \mathbb{R}$ the Jacobian associated to P. This function J is such that is constant equal

\begin{equation*}J_{i,j}=\frac{\pi_i P_{i,j}}{\pi_j}\end{equation*}

on the cylinder $[i,j]$, $i,j=0,1$.

According to our previous notation $\mu_A=\mu_{\log J}$ (which in this section will be called just µ).

Definition 6.2. The Ruelle operator for $\log J$ acts on continuous functions φ and is given by: for each $\varphi:M \to \mathbb{R}$, we get that

(22)\begin{equation} \mathscr{L}_{\log J} (\varphi) (x_1,x_2,x_3, \ldots)= \frac{\pi_0 P_{0,x_1}}{\pi_{x_1}} \varphi(0,x_1,x_2, \ldots)+ \frac{\pi_1 P_{1,x_1}}{\pi_{x_1}} \varphi(1,x_1,x_2, \ldots). \end{equation}

It is known that $\mathscr{L}_{\log J}^* (\mu)=\mu$ (see [Reference Parry and Pollicott18]).

We also consider the action of $\mathscr{L}_{\log J}$ on $L^2(\mu)$ and we are interested in the kernel of this operator when acting on Holder functions.

Given a finite word $x=(x_1,x_2, \ldots,x_n)$, depending of the context $[x]$ will either denote the word or the corresponding cylinder set in $ \{0,1\}^{\mathbb{N}}.$ The empty word is also considered a finite word.

We start by recalling that, given a Markov probability µ on $\{0,1\}^{\mathbb{N}}$, the family of Holder functions

(23)\begin{align} e_{[x]} =\frac{1}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x0]} - \frac{1}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x1]}, \end{align}

where $x=(x_1,x_2, \ldots,x_n)$ is a finite word on the symbols $\{0,1\}$, is an orthonormal set for $\mathscr{L}^2 (\mu)$ (see [Reference Kessebohmer and Samuel12] for a general expression and [Reference Cioletti, Hataishi, Lopes and Stadlbauer7] for the specific expression we are using here). In order to get a (Haar) basis, we should add $e_{[\emptyset]}^0=\frac{1}{\sqrt{\mu ([0])}} \mathfrak{1}_{[0]}$ and $ e_{[\emptyset]}^1=\frac{1}{\sqrt{\mu ([1])}} \mathfrak{1}_{[1]}$ to this family.

Definition 6.3. Given a finite word $x=(x_1,x_2, \ldots,x_n)$, we denote

(24)\begin{equation} a_x=\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_n]}. \end{equation}

It will follow from Equations (33) and (34) that the terms $| a_x |$ are uniformly bounded away from zero (the minimum value is 2). Moreover, they depend just on the first letter of the word $[x]$.

Definition 6.4. We denote by

(25)\begin{equation} \hat{a}_x=\frac{1}{|a_x|} a_x, \end{equation}

the normalization of a_x.

In order to get a complete orthonormal set for the kernel of the Ruelle operator, we will have to add to the functions of the form (25) two more functions: $\hat{a}_{[\emptyset]}^0$ and $ \hat{a}_{[\emptyset]}^0$ to be set in Definition 6.8. To show this result is our main goal in this section. This family will be later denoted by $\mathcal{F}$ according to Definition 6.9.

In this direction, we first consider the problem of exhibiting an orthogonal family which is a basis for the kernel of the Ruelle operator, and later via normalization, we will get a complete orthonormal family which is a basis for the kernel of the Ruelle operator.

Following this line of reasoning, one of our main tasks in this section is to show the following:

We will address first the issue related to the functions a_x, and later to questions regarding the functions $e_{[\emptyset]}^0$ and $e_{[\emptyset]}^1$.

First note that as the family $e_{[x]}$, where x is a finite word, is orthonormal, then, a_x, where x is a finite word with size bigger or equal to 1, is an orthogonal family.

Indeed, it follows from the fact that the family $e_{[x]}$ defined by Equation (23) is orthogonal, and the bilinearity of the inner product, that

\begin{equation*} \langle a_x,a_z \rangle= \end{equation*}

\begin{equation*} \langle \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}P_{0,x_1} }} e_{[0,x]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}P_{1,x_1} }}e_{[1,x]},\frac{\sqrt{\pi_{z_1} }} {\sqrt{\pi_{0}P_{0,z_1} }} e_{[0,z]} - \frac{\sqrt{\pi_{z_1} }} {\sqrt{\pi_{1}P_{1,z_1} }}e_{[1,z]}\rangle =0,\end{equation*}

for all $x=(x_1,x_2, \ldots,x_n)\neq z= (z_1,z_2, \ldots,z_k)$.

We shall subdivide the proof of Theorem 6.5 into several steps. First of all, we have that:

Proposition 6.6. Given $x=[x_1,x_2, \ldots,x_n]$ with a size larger or equal to 1,

(26)\begin{equation} \mathscr{L}_{\log J} ( e_{[x_1,x_2, \ldots,x_n]} ) =\frac{\sqrt{\pi_{x_1}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}} e_{[x_2,x_3, \ldots,x_n]}. \end{equation}

From this follows that all elements in the orthogonal family

(27)\begin{equation} a_x=\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2,..,x_n]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2,..,x_n]}, \end{equation}

indexed by words $x=(x_1,x_2, \ldots,x_n)$, are in the kernel of the Ruelle operator $\mathscr{L}_{\log J}$.

Proof. We consider finite words x with size larger or equal to 1.

Indeed, given the word $x=(x_1,x_2, \ldots,x_n)$, let $L = \mathscr{L}_{\log J} ( e_{[x_1,x_2, \ldots,x_n]} )$, then we get

\begin{eqnarray*} L & =& \frac{\pi_{x_1}}{\pi_{x_2}} P_{x_1,x_2} [ \frac{1}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]} \\ & -& \frac{1}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} ] \\ & = & \frac{\pi_{x_1}}{\pi_{x_2}}\sqrt{P_{x_1,x_2}} [ \frac{\sqrt{P_{x_1,x_2}}}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]} \\ & - & \frac{\sqrt{P_{x_1,x_2}}}{\sqrt{\mu([x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} ]. \end{eqnarray*}

This is equal to

\begin{equation*} \frac{\pi_{x_1}}{\pi_{x_2}}\frac{\sqrt{\pi_{x_2}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}} \frac{\sqrt{P_{x_1,x_2}}}{\sqrt{\pi_{x_1} P_{x_1,x_2} P_{x_2,x_3} \cdots P_{x_{n-1},x_n}}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]}\end{equation*}

\begin{equation*}-\frac{\pi_{x_1}}{\pi_{x_2}}\frac{\sqrt{\pi_{x_2}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}}\frac{\sqrt{P_{x_1,x_2}}}{\sqrt{\pi_{x_1} P_{x_1,x_2} P_{x_2,x_3} \cdots P_{x_{n-1},x_n}}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} \end{equation*}

which is equivalent to

\begin{equation*}\frac{\pi_{x_1}}{\pi_{x_2}} \frac{\sqrt{\pi_{x_2}}}{\sqrt{\pi_{x_1} }} \sqrt{P_{x_1,x_2}} \frac{1}{\sqrt{\pi_{x_2} P_{x_2,x_3} \cdots P_{x_{n-1},x_n}}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]}\end{equation*}

\begin{equation*}-\frac{\pi_{x_1}}{\pi_{x_2}}\frac{\sqrt{\pi_{x_2}}}{\sqrt{\pi_{x_1} }} \sqrt{P_{x_1,x_2}} \frac{1}{\sqrt{\pi_{x_2} P_{x_2,x_3} \cdots P_{x_{n-1},x_n}}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} \end{equation*}

that yields

\begin{eqnarray*} L& = & \frac{\sqrt{\pi_{x_1}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}} \frac{1}{\sqrt{\mu [x_2,x_3, \ldots,x_n]}} [ \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]} - \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} ]\\ & = & \frac{\sqrt{\pi_{x_1}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}} e_{x_2,x_3, \ldots,x_n}. \end{eqnarray*}

Then,

\begin{equation*} \mathscr{L}_{\log J} ( \frac{\sqrt{\pi_{x_2} }} {\sqrt{\pi_{x_1}}\sqrt{P_{x_1,x_2} }} e_{[x_1,x_2, \ldots,x_n]} ) \end{equation*}

\begin{equation*}=\frac{1}{\sqrt{\mu [x_2,x_3, \ldots,x_n]}}[ \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_2, \ldots,x_n,0]} - \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_2, \ldots,x_n,1]} ]= e_{[x_2,x_3, \ldots,x_n]}\end{equation*}

and therefore,

\begin{equation*} \mathscr{L}_{\log J} ( \frac{\sqrt{\pi_{x_2} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_2} }} e_{[0,x_2, \ldots,x_n]} ) =\mathscr{L}_{\log J} ( \frac{\sqrt{\pi_{x_2} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_2} }} e_{[1,x_2, \ldots,x_n]} ) .\end{equation*}

For each finite word $(x_1,x_2, \ldots,x_n)$ denote

\begin{eqnarray*} a_x & =& \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_n]}\\ & = & \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} \frac{1}{\sqrt{\mu([0 x])}} [ \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[0x0]} - \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[0x1]} ] \end{eqnarray*}

(28)\begin{equation} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} \frac{1}{\sqrt{\mu([1 x])}} [ \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[1x0]} - \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[1x1]} ]. \end{equation}

From the above reasoning, it follows that the family a_x is in the kernel of the Ruelle operator.

For words, x of size greater or equal to 1 the function a_x is constant in cylinder sets of size equal to the length of x plus 2.

As an example, we get that

\begin{eqnarray*} a_0 & =&\frac{\sqrt{\pi_{0} }} {\sqrt{\pi_{0}}\sqrt{P_{0,0} }} \frac{1}{\sqrt{\mu([0 0])}} [ \sqrt{\frac{P_{0,1}}{P_{0,0} }} \mathfrak{1}_{[000]} - \sqrt{\frac{P_{0,0}}{P_{0,1} }} \mathfrak{1}_{[001]} ] \end{eqnarray*}

(29)\begin{equation} - \frac{\sqrt{\pi_{0} }} {\sqrt{\pi_{1}}\sqrt{P_{1,0} }} \frac{1}{\sqrt{\mu([1 0])}} [ \sqrt{\frac{P_{0,1}}{P_{0,0} }} \mathfrak{1}_{[100]} - \sqrt{\frac{P_{0,0}}{P_{0,1} }} \mathfrak{1}_{[101]} ] \end{equation}

is constant on cylinders of size 3.

Note that if x and z are different words, then, 1x, 0x, 0z and 1z are four different words.

Note that

(30)\begin{align} e_{[x]}^2 =\frac{1}{\mu([x])} \frac{P_{x_n,1}}{P_{x_n,0} } \mathfrak{1}_{[x0]} + \frac{1}{\mu([x])} \frac{P_{x_n,0}}{P_{x_n,1} } \mathfrak{1}_{[x1]}. \end{align}

Therefore,

\begin{eqnarray*} a_x^2=a_x a_x & = & \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_n]}^2 \\ & = & [ \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } \frac{1}{\mu([0 x])} \frac{P_{x_n,1}}{P_{x_n,0} } \mathfrak{1}_{[0x0]} + \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } \frac{1}{\mu([0 x])} \frac{P_{x_n,0}}{P_{x_n,1} } \mathfrak{1}_{[0x1]} ] \end{eqnarray*}

(31)\begin{equation} +[ \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } \frac{1}{\mu([1 x])} \frac{P_{x_n,1}}{P_{x_n,0} } \mathfrak{1}_{[1x0]} + \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } \frac{1}{\mu([1 x])} \frac{P_{x_n,0}}{P_{x_n,1} } \mathfrak{1}_{[1x1]} ] . \end{equation}

From the above, it follows that

(32)\begin{equation} | a_x |=\sqrt{\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } }. \end{equation}

Using the notation in the variables $r,s$ for the matrix P, when $x_1=0$ we get

(33)\begin{equation}|a_x|= \sqrt{\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } }=(\sqrt{r (1-r)})^{-1} \end{equation}

and when $x_1=1$ we get

(34)\begin{equation}|a_x|= \sqrt{\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } }=(\sqrt{s (1-s)})^{-1}. \end{equation}

As we mentioned before, we will have to add two more functions in order to get a basis (a completely orthogonal set in the Hilbert space) for the kernel of the Ruelle operator $\mathscr{L}_{\log J}$.

We claim that the orthogonal pair (constant in cylinders of size 2)

\begin{equation*} V_1 = \pi_1 P_{1,0} \mathfrak{1}_{[00]} - \pi_0 P_{0,0} \mathfrak{1}_{[10]}\end{equation*}

(35)\begin{equation} V_2 = \pi_0 P_{0,1} \mathfrak{1}_{[11]} - \pi_1 P_{1,1} \mathfrak{1}_{[01]}\end{equation}

is in the kernel of the Ruelle operator (see Proposition 6.11). The functions V ₁ and V ₂ are orthogonal to all $\hat{a}_x \in \tilde{\mathcal{F}}$ and they depend on the first two coordinates $x_1,x_2$ of x.

The vectors $\hat{V}_1 = \frac{V_1}{|V_1| }$ and $\hat{V}_2 = \frac{V_2}{|V_2| }$ are normalized and orthogonal to all $\hat{a}_x$. This claim will be proved in Proposition 6.11.

One can show that

(36)\begin{equation}|V_1| = \sqrt{ \pi_1^2 P_{1,0}^2 \pi_0 P_{0,0} + \pi_0^2 P_{0,0}^2 \pi_1 P_{1,0} }= \sqrt{\frac{(1-r) r (s-1)^3}{(-2 + r + s)^3} } \end{equation}

and

(37)\begin{equation}|V_2|= \sqrt{ \pi_0^2 P_{0,1}^2 \pi_1 P_{1,1} + \pi_1^2 P_{1,1}^2 \pi_0 P_{0,1} } = \sqrt{\frac{(1-s) s (r-1)^3}{(-2 + r + s)^3} }. \end{equation}

These two functions are constant in cylinders of size 2.

Proposition 6.11. The orthogonal pair

\begin{equation*} V_1 = \pi_1 P_{1,0} \mathfrak{1}_{[00]} - \pi_0 P_{0,0} \mathfrak{1}_{[10]}\end{equation*}

(38)\begin{equation} V_2 = \pi_0 P_{0,1} \mathfrak{1}_{[11]} - \pi_1 P_{1,1} \mathfrak{1}_{[01]}\end{equation}

is such that, each one of them is orthogonal to the other elements $\hat{a}_x$, where x ranges in all finite words with size bigger or equal to 1. V₁ and V₂ are on the kernel of the Ruelle operator $\mathscr{L}_{\log J}$.

Proof. Note first that $\mathfrak{1}_{[00]} $ is orthogonal to all a_x, where $x=(x_1,x_2, \ldots,x_n)$ is a word with size equal or greater than 1. This claim follows from (28). Indeed, if $x_1=0$, we get that

\begin{equation*}\langle \mathfrak{1}_{[00]} , \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[0x0]} - \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[0x1]}\rangle= \end{equation*}

\begin{equation*} \sqrt{P_{x_n,1}} \pi_0 P_{0,x_1 } P_{x_1,x_2 } \cdots P_{x_{n-1},x_n } \sqrt{P_{x_n,0 }} -\end{equation*}

\begin{equation*} \sqrt{P_{x_n,0}} \pi_0 P_{0,x_1 } P_{x_1,x_2 } \cdots P_{x_{n-1},x_n } \sqrt{P_{x_n,1 }}=0. \end{equation*}

If $x_1=1$ the claim follows at once.

Using the same reasoning one can show that $\mathfrak{1}_{[01]},\mathfrak{1}_{[10]},\mathfrak{1}_{[11]} $ are orthogonal to all a_x, where length of x is bigger than zero. It follows that linear combinations of this functions are also orthogonal to all a_x. It follows that V ₁ and V ₂ are orthogonal to all a_x, where the length of x is bigger than zero.

We will show that V ₁ is in the kernel of the Ruelle operator (for V ₂ the proof is similar). Given $y=(y_1,y_2, \ldots,y_n, \ldots)\in M$, suppose first that $y_1=0$, then, we get

\begin{equation*}\mathscr{L}_{\log J} (V_1)= \mathscr{L}_{\log J} ( \pi_1 P_{1,0} \mathfrak{1}_{[00]} - \pi_0 P_{0,0} \mathfrak{1}_{[10]}) (y)= \end{equation*}

\begin{equation*} \pi_1 P_{1,0} ( J_{0,y_1} \mathfrak{1}_{[00]}(0,y_1,y_2, \ldots) + J_{1,y_1} \mathfrak{1}_{[00]} (1,y_1,y_2, \ldots) ) - \end{equation*}

\begin{equation*} \pi_0 P_{0,0} ( J_{0,y_1} \mathfrak{1}_{[10]}(0,y_1,y_2, \ldots) + J_{1,y_1} \mathfrak{1}_{[10]} (1,y_1,y_2, \ldots) ) = \end{equation*}

\begin{equation*} \pi_1 P_{1,0} J_{0,0} - \pi_0 P_{0,0} J_{1,0}= \pi_1 P_{1,0} \frac{\pi_0 P_{0,0} }{\pi_0} - \pi_0 P_{0,0}\frac{\pi_1 P_{1,0} }{\pi_0} =0.\end{equation*}

In the case $y_1=1$, we get

\begin{equation*}\mathscr{L}_{\log J} (V_1)= \pi_1 P_{1,0} ( J_{0,y_1} \mathfrak{1}_{[00]}(0,y_1,y_2, \ldots) + J_{1,y_1} \mathfrak{1}_{[00]} (1,y_1,y_2, \ldots) ) - \end{equation*}

\begin{equation*} \pi_0 P_{0,0} ( J_{0,y_1} \mathfrak{1}_{[10]}(0,y_1,y_2, \ldots) + J_{1,y_1} \mathfrak{1}_{[10]} (1,y_1,y_2, \ldots) ) = 0. \end{equation*}

Remark 6.12. A function of the form $w=r_1 \mathfrak{1}_{[0]} + r_2 \mathfrak{1}_{[1]}$ is in the kernel of $\mathscr{L}_{\log J}$ only in the case where $P_{01}=(1-r)= s=P_{11}$. In this case

(39)\begin{equation} w =(1-r) \mathfrak{1}_{[0]} - (1-s) \mathfrak{1}_{[1]} \end{equation}

is such that $\mathscr{L}_{\log J} (w)=0.$

We do not have to take into account in our future reasoning this function because

\begin{equation*}w = \frac{1}{r} V_1 + \frac{1}{r-1} V_2.\end{equation*}

Proof. From Proposition 6.6, we know that given $x=[x_1,x_2, \ldots,x_n]$

(40)\begin{equation} \mathscr{L}_{\log J} ( e_{[x_1,x_2, \ldots,x_n]} ) =\frac{\sqrt{\pi_{x_1}}}{\sqrt{\pi_{x_2} }} \sqrt{P_{x_1,x_2}} e_{[x_2,x_3, \ldots,x_n]}. \end{equation}

Suppose φ is in the kernel of the Ruelle operator. We will show that φ can be expressed as an infinite linear combination of the normalized functions $\hat{a}_x \in \mathcal{F}$.

We can express φ as

\begin{equation*}\varphi = \sum_{\text{words} \,y} c_y e_{[y]} . \end{equation*}

When applying $\mathscr{L}_{\log J}$ on φ, we separate the infinite sum in subsums of the form

\begin{equation*} c_{0,\alpha_2, \ldots,\alpha_n } e_{[0,\alpha_2, \ldots,\alpha_n]} + c_{1,\alpha_2, \ldots,\alpha_n } e_{[1,\alpha_2, \ldots,\alpha_n]}.\end{equation*}

Assuming that φ is in the kernel of $\mathscr{L}_{\log J}$, we get from Equation (40) that

\begin{eqnarray*} 0 & = & \mathscr{L}_{\log J}( \sum_n \sum_{\alpha_2, \ldots,\alpha_n} [ c_{0,\alpha_2, \ldots,\alpha_n } e_{[0,\alpha_2, \ldots,\alpha_n]} + c_{1,\alpha_2, \ldots,\alpha_n } e_{[1,\alpha_2, \ldots,\alpha_n]} ] ) \\ & = & \sum_n \sum_{\alpha_2, \ldots,\alpha_n} [ \frac{\sqrt{\pi_{0}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{0,\alpha_2}} c_{0,\alpha_2, \ldots,\alpha_n } e_{[\alpha_2, \ldots,\alpha_n]} + \frac{\sqrt{\pi_{1}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{1,\alpha_2}} c_{1,\alpha_2, \ldots,\alpha_n } e_{[\alpha_2, \ldots,\alpha_n]} ] \\ & = & \sum_n \sum_{\alpha_2, \ldots,\alpha_n} [ \frac{\sqrt{\pi_{0}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{0,\alpha_2}} c_{0,\alpha_2, \ldots,\alpha_n } + \frac{\sqrt{\pi_{1}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{1,\alpha_2}} c_{1,\alpha_2, \ldots,\alpha_n } ] e_{[\alpha_2, \ldots,\alpha_n]} . \end{eqnarray*}

Then, for fixed n and $(\alpha_2,\alpha_3, \ldots,\alpha_n)$

\begin{equation*} \frac{\sqrt{\pi_{0}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{0,\alpha_2}} c_{0,\alpha_2, \ldots,\alpha_n } = - \frac{\sqrt{\pi_{1}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{1,\alpha_2}} c_{1,\alpha_2, \ldots,\alpha_n } ,\end{equation*}

which means

\begin{equation*} c_{0,\alpha_2, \ldots,\alpha_n } = - \frac{\sqrt{\pi_{1}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{1,\alpha_2}} \frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{0} }} \frac{1}{\sqrt{P_{0,\alpha_2}}} c_{1,\alpha_2, \ldots,\alpha_n } .\end{equation*}

Then, the sum

\begin{equation*}c_{0,\alpha_2, \ldots,\alpha_n } e_{[0,\alpha_2, \ldots,\alpha_n]} + c_{1,\alpha_2, \ldots,\alpha_n } e_{[1,\alpha_2, \ldots,\alpha_n]} \end{equation*}

is equal to

\begin{equation*} - c_{1,\alpha_2, \ldots,\alpha_n } [ \frac{\sqrt{\pi_{1}}}{\sqrt{\pi_{\alpha_2} }} \sqrt{P_{1,\alpha_2}} \frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{0} }} \frac{1}{\sqrt{P_{0,\alpha_2}}} e_{[0,\alpha_2, \ldots,\alpha_n]} - e_{[1,\alpha_2, \ldots,\alpha_n]}]. \end{equation*}

Multiplying the above expression by $\frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{1} }} \frac{1}{\sqrt{P_{1,\alpha_2}}} $, we get

\begin{equation*} \frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{1} }} \frac{1}{\sqrt{P_{1,\alpha_2}}} [ c_{0,\alpha_2, \ldots,\alpha_n } e_{[0,\alpha_2, \ldots,\alpha_n]} + c_{1,\alpha_2, \ldots,\alpha_n } e_{[1,\alpha_2, \ldots,\alpha_n]}]\end{equation*}

which is equal to

\begin{eqnarray*} & - & c_{1,\alpha_2, \ldots,\alpha_n } [ \frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{0} }} \frac{1}{\sqrt{P_{0,\alpha_2}}} e_{[0,\alpha_2, \ldots,\alpha_n]} - \frac{\sqrt{\pi_{\alpha_2}}}{\sqrt{\pi_{1} }} \frac{1}{\sqrt{P_{1,\alpha_2}}} e_{[1,\alpha_2, \ldots,\alpha_n]}]\\ & = & - c_{1,\alpha_2, \ldots,\alpha_n } a_{[a_2, \ldots,a_n]}. \end{eqnarray*}

Then, $( c_{0,\alpha_2, \ldots,\alpha_n } e_{[0,\alpha_2, \ldots,\alpha_n]} + c_{1,\alpha_2, \ldots,\alpha_n } e_{[1,\alpha_2, \ldots,\alpha_n]} )$ is a multiple of the function $\hat{a}_{[\alpha_2, \ldots,\alpha_n]}.$ Since the above reasoning was done for a generic choice of $(\alpha_2,\alpha_3, \ldots,\alpha_n),$ we conclude that for each n the sum $\sum_{\text{words} \,y \,\text{of length} \,n} c_y e_{[y]}$ can be expressed as a linear combination of elements $\hat{a}_x$, using words of length n − 1, n > 1.

From this follows that each element in the kernel of $\mathscr{L}_{\log J} $ can be expressed as an infinite linear combination of the functions $\hat{a}_x$.

Theorem 6.5 follows from the combination of Propositions 6.6 and 6.13.

The above shows that the set $\mathcal{F}$ is a complete orthonormal set for the kernel of the Ruelle operator acting on $\mathscr{L}^2(\mu).$

7. A worked example in the Markov case: preliminary calculations of the terms in $K(X,Y)$

In this section, we shall devote ourselves to the calculation of the sectional curvatures in the case of Markov stationary probabilities on $M=\{0,1\}^{\mathbb{N}}$.

We denote by $K\subset \mathcal{N},$ the set of Markov invariant probabilities. We will consider this section the sectional curvature for points in $\mathcal{K}$ for general orthogonal pairs of tangent vectors to $\mathcal{N}.$

We can also consider $\mathcal{K}$ as a two-dimensional manifold carrying the Riemannian structure induced by $\mathcal{N}.$ From this point of view, there exists just one orthonormal pair to be considered. One of our main results (see Theorem 7.14) claims that for the two-dimensional manifold $K,$ for any point in $\mathcal{K}$, the sectional curvature for the pair of tangent vectors to $\mathcal{K}$ is always zero.

We will consider in our reasoning the empty word as a regular word. $\hat{a}_\emptyset^0 $ and $\hat{a}_\emptyset^1 $ are two elements in $\mathcal{F}$ associated with the empty word.

Definition 7.1. We say that z is a subprefix of x, if x and z satisfy

\begin{equation*} [x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [z]=[x_1,x_2, \ldots,x_k],\end{equation*}

where $n \geq k$.

Note that, even when z is not a subprefix of x and x is not a subprefix of z, they can share some common subprefix. Note also that if x and z do not share a common subprefix, then z is not a subprefix of x and x is not a subprefix of z.

If $[x]=[z]$, then, x is a subprefix of $z.$

Definition 7.2. We say that z is a strict subprefix of x, if x and z satisfy

\begin{equation*} [x]=[x_1,x_2, \ldots, x_k,x_{k+1}, \ldots, x_n] \subset [z]=[x_1,x_2, \ldots,x_k],\end{equation*}

where n > k.

Two different words with the same length cannot be subprefix of each other. If the length of z is strictly larger than the length of x, then, z cannot be a subprefix of x.

If for example $ x=(0,0,0)$ and $z=(0,0,0,1)$, then

\begin{equation*}D[x,z]=\{\hat{a}_\emptyset^0,(0), (0,0),(0,0,0)\}.\end{equation*}

In the case $x=(0,1,0,0,1) $ and $z=(0,1,1)$ we get that $D[x,z]= \{\hat{a}_\emptyset^0(0),(0,1),\}.$

Another example: $D[a_{0,0}, \hat{a}_\emptyset^0]=\{\hat{a}_\emptyset^0 \}$ and $D[a_{0,0}, \hat{a}_\emptyset^1]=\emptyset.$

Note that in the case $z=(z_1,z_2, \ldots,z_k)$ is a subprefix of $x=(x_1,x_2, \ldots,x_n)$, n > k, then, $z_1=x_1$. Then, it follows from (32) that $|a_x|=|a_z|$.

Proposition 7.4. Assume that x is not a subprefix of z and z is not a subprefix of x. Then,

\begin{equation*}a_x a_z=0.\end{equation*}

Note that the hypothesis of the last proposition is equivalent to saying that the cylinders $[x]$ and $[z]$ are disjoint.

Corollary 7.5. Given a word x assume that x is not a subprefix of y and y is not a subprefix of x. Then,

\begin{equation*}\hat{a}_x^2 \hat{a}_y=0.\end{equation*}

Note that if x and y have the same length, but they are different, then $\int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu=0.$

From Proposition 7.4, it follows:

Corollary 7.6. Assume that x is not a subprefix of z and z is not a subprefix of x. Then, we get that the products (part of the first sum contribution in Equation (44)) satisfy

(41)\begin{equation} \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu=0, \end{equation}

for all word y.

Remember that $\mathcal{F}$ (defined in last section) is the set of all functions of the form

(42)\begin{equation} \hat{a}_x = \frac{1}{\sqrt{\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } }} a_x,\end{equation}

where $x= (x_1,x_2, \ldots,x_k)$ is a general finite word, plus the functions $\hat{a}_{[\emptyset]}^0 $ and $\hat{a}_{[\emptyset]}^1$.

Remember that Proposition 6.13 of last section claims that the family of functions $\mathcal{F}$ determines an orthonormal basis for the Kernel of the Ruelle operator.

We want to estimate for $X=\hat{a}_x,$ $Y= \hat{a}_z \in \mathcal{F}$ and the orthogonal basis $X_i= \hat{a}_y\in \mathcal{F}$ the explicit expression of the curvature which was described in Theorem 1.1

(43)\begin{equation} K(X,Y) = \frac{1}{4}[ \sum_{i=1}^\infty ( \int X Y X_i \,\mathrm{d} \mu)^2 - \sum_{i=1}^\infty \int X^2 X_i \,\mathrm{d} \mu \int Y^2 X_i \,\mathrm{d} \mu ]. \end{equation}

We will not present the explicit expression of the sectional curvature $K(X,Y)$ for any pair of vectors $X,Y$ in the kernel, but just for the case where the functions $X,Y$ are part of the family $\hat{a}_x\in \mathcal{F}$.

An important issue is: $0=\langle \hat{a}_z^2, \hat{a}_y\rangle = \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu$, when the length of y is strictly larger than the length of z (as will be proved in § 8 and 9). We mention this point to stress the point that the last sum in expression (44) is a sum of a finite number of terms.

Our main result in this section concerns the Markov case:

Theorem 7.7. For a fixed pair $\hat{a}_x, \hat{a}_z\in \mathcal{F}$ (with z different from x) the value

(44)\begin{equation} K(\hat{a}_z,\hat{a}_x) = \frac{1}{4}[ \sum_{\text{word}\, y}( \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2 - \sum_{\text{word}\, y} \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu ].\end{equation}

In the case the length of x is strictly larger than the length of z we get that Equation (44) can be expressed in a more simplified form as:

(45)\begin{equation} \frac{1}{4} [ (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2 - \sum_{y \in D[x,z]} \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu]. \end{equation}

In this case, the above expression is a sum of a finite number of terms.

In the general case, the value $\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu$ is zero if z is not a subprefix of x. If z is a strict subprefix of x and y is a strict subprefix of z, then the term

(46)\begin{equation}- \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu \end{equation}

is non-positive. Moreover, by Equation (75), we get $\int \hat{a}_z^2 \hat{a}_x \,\mathrm{d} \mu=0$. Then, it follows that Equation (44) is a sum of a finite number of terms, for any given x and z, with z ≠ x.

The proof of this result will take several sections and subsections. Proposition 7.12 will summarize several explicit computations that are necessary in our reasoning.

We will also provide an explicit expression for the curvature (45) in terms of the words $x,z$ and the probability µ (which is indexed by (r, s) of expression (20)). This will follow from explicit expressions for $(\int a_x a_z a_y \,\mathrm{d} \mu)^2 $, $ \int a_z^2 a_y \,\mathrm{d} \mu $ and $ \int a_x^2 a_y \,\mathrm{d} \mu$, for all finite words $x,z,y$, that will be presented the Propositions 7.9 and 7.12 (which will be proved in § 8 and 9).

It will also follow that when x and z do not share a common subprefix y, then the curvature $K(\hat{a}_z,\hat{a}_x)$ is equal to 0 (see Proposition 7.10).

There are examples (for instance, the case $ x=(0,1,0)$ and $z=(0,1,0,0)$) where the curvature $K(\hat{a}_z,\hat{a}_x) $ is positive for some values of the parameters (r, s) and negative for others (see Example 7.19). We can show from the explicit expressions we obtain that for fixed values of the parameters (r, s) the curvature $K(\hat{a}_z,\hat{a}_x)$ can be very negative if both words $x,z$ have large lengths and share common subprefix with large length (see Remark 7.17). In Example 7.20, we show that $K( \hat{a}_{(0)}, \hat{a}_{(0,0)} )= -0.205714 \cdots $, when $r=0.1,s=0.3$. In Proposition 7.18, we show the curvature $K(\hat{a}_{[\emptyset]}^0 ,\hat{a}_0)$ can be positive for some pairs $r,s\in(0,1)$. It follows from the expressions of Proposition 7.12 that all sectional curvatures $K( \hat{a}_{z}, \hat{a}_{x} )$ are equal to $-1/2$, when $r=1/2=s$, the size of z is bigger than 1 and z is a strict subprefix of x. See also Proposition 7.18, when $r=1/2=s$, for the computation of $ K(\hat{a}_{[\emptyset]}^0,\hat{a}_0)=1/2$.

Proposition 7.9. Assume that the length of x is larger than the length of z. The first sum on expression (44) is given by

(47)\begin{equation} \sum_{\text{word}\, y }( \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2 = (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2 + (\int \hat{a}_z^2 \hat{a}_x \,\mathrm{d} \mu)^2= (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2. \end{equation}

For a proof of this claim see expression (78) in § 9. This term in the sum (44) is the part that contributes to the curvature to be more positive. The second term in the sum (44) will contribute to the curvature becoming more negative (see Proposition 7.12).

Note that Equation (47) does not depend on y. Note also that from expression (21) one can get explicitly the values (47) as a function of (r, s).

In Proposition 7.4, we show that if x is not a subprefix of z and z is not a subprefix of x, we get that $\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu=0$. In this case, the contribution of Equation (47) for the curvature will be null.

Proposition 7.10. When z and x do not share common subprefix the curvature

\begin{equation*}K(\hat{a}_{z}, \hat{a}_{x})=0.\end{equation*}

Proposition 7.12. Assume that the length of x is larger or equal to the length of y. Then we have:

(a) $\int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu=0$, if y is not a subprefix of x. This also includes the case where x ≠ y and length of x is equal to the length of y.

(b.0) Assume that $[x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [y]=[x_1,x_2, \ldots,x_k]$, where n > k, and $x_{k+1}=0$. Note that from Equation (32) we get that $|a_x| =|a_y|$. Then,

\begin{equation*} \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu =\end{equation*}

\begin{equation*} \frac{1}{|a_y|^3} \frac{\sqrt{P_{x_k,1}}}{\sqrt{P_{x_k,0}} }\{( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y])}} - ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y])}} \} =\end{equation*}

(48)\begin{equation} \frac{1}{|a_x|^3} \sqrt{P_{x_k,1}}\{( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y 0])}} - ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y 0])}} \} . \end{equation}

(b.1) Assume that $[x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [y]=[x_1,x_2, \ldots,x_k]$, where n > k, and $x_{k+1}=1$. Then,

\begin{equation*} \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu = \end{equation*}

\begin{equation*}\frac{1}{|a_y|^3} \frac{\sqrt{P_{x_k,0}}}{\sqrt{P_{x_k,1}} }\{ - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y])}} +( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y ])}} \}= \end{equation*}

(49)\begin{equation}\frac{1}{|a_x|^3} \sqrt{P_{x_k,0}}\{ - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y 1])}} +( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y 1])}}\}. \end{equation}

(b.2) Assume, $[x]=[x_1,x_2, \ldots,x_n] = [y].$

Then,

\begin{equation*} \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu= \int \hat{a}_y^3 \,\mathrm{d} \mu=\end{equation*}

(50)\begin{equation} \frac{1}{|a_x|^3} \{( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([0 y 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([0 y 1])}} ] -(\frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([1 y 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([1 y 1])}} ]\} . \end{equation}

(b.3) If $x_1=0$, then

(51)\begin{equation} \int \hat{a}_x^2\hat{a}_{[\emptyset]}^0 \,\mathrm{d}\mu= \frac{1}{|a_x|^2 | V_1| }( \frac{\pi_1 P_{1,0} } {P_{0,0} } + \frac{\pi_0^2 P_{0,0} } {\pi_1 P_{1,0} })= \end{equation}

\begin{equation*} \frac{(s-1) (1- 2 r + 2 r^2) }{\sqrt{\frac{(1-r) r (s-1)^3}{(-2 + r + s)^3} } (-2 + r +s)} \gt 0,\end{equation*}

and

\begin{equation*}\int \hat{a}_x^2\hat{a}_{[\emptyset]}^1 \,\mathrm{d}\mu=0.\end{equation*}

When $r=1/2=s$, we get that for any word x (with size bigger or equal to 1), such that, $x_1=0$

(52)\begin{equation} \int \hat{a}_x^2\hat{a}_{[\emptyset]}^0 \,\mathrm{d}\mu = \sqrt{2}. \end{equation}

For the proof of this proposition, see § 8.1 and 8.2.

We previously denoted by $\mathcal{K}$ the two-dimensional manifold of Markov invariant probabilities (the set of equilibrium probabilities for potentials depending on two coordinates and parametrized by $r,s$).

Given the Markov invariant probability µ associated with the parameters $r,s$, the set of vectors which are tangent to $\mathcal{K}$ at this point is the set of functions that depend on two coordinates $(x_1,x_2)$. The ones that are on the kernel of $\mathscr{L}_{\log J}$ are $\hat{a}_{[\emptyset]}^0 $ and $\hat{a}_{[\emptyset]}^1.$

Proof. Remember that $ V_1 = \pi_1 P_{1,0} \mathfrak{1}_{[00]} - \pi_0 P_{0,0} \mathfrak{1}_{[10]}$ and $ V_2 = \pi_0 P_{0,1} \mathfrak{1}_{[11]} - \pi_1 P_{1,1} \mathfrak{1}_{[01]}$ determine an orthogonal basis for the tangent space to $\mathcal{K}$ at µ.

We claim that the curvature $K(\hat{a}_\emptyset^0,\hat{a}_\emptyset^1)= 0$.

Indeed, take $X_i= \hat{a}_z$, for some finite word $z=(x_1,x_2, \ldots,x_k)$. If we assume that $x_1=1$, then

\begin{eqnarray*} V_1^2 a_z & =& [ \pi_1^2 P_{1,0}^2 (\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } \frac{1}{\mu([0 z])} \frac{P_{x_k,1}}{P_{x_k,0} })^{1/2} \mathfrak{1}_{[0z0]} \mathfrak{1}_{[00]} \\ & - & \pi_1^2 P_{1,0}^2 (\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } \frac{1}{\mu([0 z])} \frac{P_{x_k,0}}{P_{x_k,1} })^{1/2} \mathfrak{1}_{[0z1]} \mathfrak{1}_{[00]}] \\ & - & [\pi_0^2 P_{0,0}^2 (\frac{\pi_{x_1}} {\pi_{1} P_{1,x_1} } \frac{1}{\mu([1 z])} \frac{P_{x_k,1}}{P_{x_k,0} })^{1/2} \mathfrak{1}_{[1z0]} \mathfrak{1}_{[10]}\\ & -& \pi_0^2 P_{0,0}^2 (\frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } \frac{1}{\mu([1 z])} \frac{P_{x_k,0}}{P_{x_k,1} } )^{1/2} \mathfrak{1}_{[1z1]}\mathfrak{1}_{[00]} ] =0. \end{eqnarray*}

Above we use the fact that $\mathfrak{1}_{[0 1 x_2...x_k 0]} \mathfrak{1}_{[00]}=0$, etc.

Therefore, it follows that:

\begin{equation*}\int \hat{V}_1^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{V}_2^2 \hat{a}_y \,\mathrm{d} \mu =0.\end{equation*}

If we assume that $x_1=0$, then in a similar way $\int \hat{a}_z V_2^2 \,\mathrm{d} \mu =0$, and therefore, $\int \hat{a}_z \hat{V}_1^2 \,\mathrm{d} \mu \int \hat{a}_z \hat{V}_2^2 \,\mathrm{d} \mu =0.$

Note that $V_1 V_2=0.$

Then, for any word y, we get $\int \hat{V}_1 \hat{V}_2 \hat{a}_y \,\mathrm{d} \mu=0.$ In the same way, $\int \hat{V}_1^2 \hat{V}_2 \,\mathrm{d} \mu=0$ and $\int \hat{V}_2^2 \hat{V}_1 \,\mathrm{d} \mu=0$.

Finally, we get

(53)\begin{equation} \frac{1}{4} [ (\int (\hat{a}_\emptyset^0)^2 \hat{a}_\emptyset^1 \,\mathrm{d} \mu)^2 - \sum_{y} (\int (\hat{a}_\emptyset^0)^2 \hat{a}_y \,\mathrm{d} \mu ) (\int (\hat{a}_\emptyset^1)^2 \hat{a}_y \,\mathrm{d} \mu)] =0. \end{equation}

Remark 7.15. Recall that the expression of the Gauss sectional curvature $K_{M}(X,Y)$ of an isometric immersion $(M,g_{M})$, submanifold of the Riemannian manifold (N, g), at the plane generated by two orthogonal vector fields $X, Y$ tangent to $\mathcal{K}$, is given by

\begin{equation*} K_{M} (X,Y) = K(X,Y) + \langle \nabla^{\perp}_{X}X, \nabla^{\perp}_{Y}Y \rangle - \parallel \nabla^{\perp}_{X}Y \parallel^{2}\end{equation*}

according to Gauss formula (see for instance [Reference do Carmo9]). Here, the operator $\nabla^{\perp}_{X}Y$ is the component of the covariant derivative $\nabla_{X}Y$ of the Riemannian manifold (N, g) that is normal to $(M,g_{M})$.

Notice that the sectional curvature

\begin{equation*} K(X,Y)= \parallel \nabla_{\bar{Y}}\bar{X} \parallel^{2} - \langle \nabla_{\bar{X}}\bar{X}, \nabla_{\bar{Y}}\bar{Y} \rangle \end{equation*}

includes all the terms of the normal component of the covariant derivative of $X,Y$. By Theorem 7.14, all the components of the covariant derivative of a certain pair of orthogonal vector fields tangent to the surface of Markov probabilities vanish. In particular, all the terms of the normal covariant derivative of $X,Y$ vanish. Therefore, Theorem 7.14 yields that the Gaussian curvature of the surface of Markov probabilities vanishes, its intrinsic curvature as an isometric immersion of the manifold of normalized potentials is zero. This is a remarkable fact, which implies for instance that the surface would be totally geodesic in the manifold of normalized potentials provided that geodesics exist. We won’t consider the problem of the existence of geodesics in this article, we shall study this problem in further papers.

Proposition 7.16. When r = 0.5 and s = 0.5, we get that

(54)\begin{equation} K(\hat{a}_y,\hat{a}_x) =-1/2, \end{equation}

for words $x,y$ with size bigger or equal to 1

Proof. It follows from the above proposition that due to symmetry, when r = 0.5 and s = 0.5, we get $\int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu=0$, for words with size bigger or equal to 1. Moreover, $\int \hat{a}_x^2 w \,\mathrm{d} \mu=0$, for any a_x. In this case, if $x_1=0$ and x is a subprefix of y, we get that for words with size bigger or equal to 1 (see Equation (52)),

(55)\begin{equation} K(\hat{a}_y,\hat{a}_x)= - 1/4 \int \hat{a}_x^2\hat{a}_{[\emptyset]}^0\,\mathrm{d}\mu \int \hat{a}_y^2\hat{a}_{[\emptyset]}^0\,\mathrm{d}\mu =-1/2. \end{equation}

Proof. Note that Equation (45) can be expressed as

\begin{equation*} K(\hat{a}_{[\emptyset]}^0,\hat{a}_0)=\frac{1}{4} [ (\int \hat{a}_0^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu)^2 - \int \hat{a}_0^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu \int (\hat{a}_{[\emptyset]}^0)^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu) ].\end{equation*}

For any $r,s$, it is known from Equation (51) that

(56)\begin{equation} \int \hat{a}_0^2\hat{a}_{[\emptyset]}^0 \,\mathrm{d}\mu= \frac{1}{|a_0|^2 | V_1| }( \frac{\pi_1 P_{1,0} } {P_{0,0} } + \frac{\pi_0^2 P_{0,0} } {\pi_1 P_{1,0} }) \gt 0. \end{equation}

Note that

\begin{equation*} V_1^2 V_1=\end{equation*}

\begin{equation*} (\pi_1^2 P_{1,0}^2 \mathfrak{1}_{[00]} + \pi_0^2 P_{0,0}^2 \mathfrak{1}_{[10]})\times ( \pi_1 P_{1,0} \mathfrak{1}_{[00]} - \pi_0 P_{0,0} \mathfrak{1}_{[10]})=\end{equation*}

\begin{equation*} \pi_1^3 P_{1,0}^3 \mathfrak{1}_{[00]} - \pi_0^3 P_{0,0}^3 \mathfrak{1}_{[10]}.\end{equation*}

Then,

\begin{equation*} \int V_1^2 V_1 = \pi_1^3 P_{1,0}^3 \mu([00]) - \pi_0^3 P_{0,0}^3 \mu([10]), \end{equation*}

which is equal to $ \frac{1}{2}^6 \frac{1}{2}^2 - \frac{1}{2}^6 \frac{1}{2}^2 =0$, in the case $r=1/2=s$.

Therefore,

\begin{equation*} K(\hat{a}_{[\emptyset]}^0,\hat{a}_0)=\frac{1}{4} ( \int \hat{a}_0^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu)^2 = \frac{1}{4} \sqrt{2}^2=1/2 \gt 0.\end{equation*}

In other examples, we used the software Mathematica for getting explicit computations.

Example 7.19. Consider the case where $ z=(0,1,0)$ and $x=(0,1,0,0).$

$\int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu =0$, unless $\hat{a}_y$ is such that

\begin{equation*}y\in D[(0,1,0),(0,1,0,0)]=\{(0),(0,1),(0,1,0)\}, \text{or}\, \hat{a}_y=\hat{a}_{[\emptyset]}^0.\end{equation*}

Note that $\int \hat{a}_z^2 \hat{a}_{[\emptyset]}^1 \,\mathrm{d} \mu=\int \hat{a}_x^2 \hat{a}_{[\emptyset]}^1 \,\mathrm{d} \mu=0.$

Using Mathematica and the formulas of Proposition 7.12, we made computations when r = 0.1 and s = 0.3. In this case, $\pi_0=0.4375$ and $\pi_1=0.5625$ and from Equation (25) we get $|a_{(0,1,0)}|=|a_{(0,1,0,0)}|=3.33 \cdots $ and $|V_1|= 0.086 \cdots.$ Finally, $\frac{1}{|a_x|^2 |a_z|}= \frac{1}{|a_z|^3}=\frac{1}{|a_x|^2 |a_y|}=\frac{1}{|a_z|^2 |a_y|}=0.027 \cdots. $

We will show that $K( \hat{a}_{(0,1,0)}, \hat{a}_{(0,1,0,0)} )=35.9142 \cdots.$

We get the following values:

\begin{equation*}\text{using } (50) \int \hat{a}_z^2 \hat{a}_z \,\mathrm{d} \mu= \frac{1}{|a_z|^3 } \int a_{(0,1,0)}^2 a_{(0,1,0)} \,\mathrm{d} \mu = 107,51 \cdots,\end{equation*}

\begin{equation*}\text{using } (48) \int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu = \frac{1}{|a_x|^2 |a_z|} \int a_{(0,1,0,0)}^2 a_{(0,1,0)} \,\mathrm{d} \mu =120.949 \cdots,\end{equation*}

\begin{equation*} (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2 = (\int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0,1,0)} \,\mathrm{d} \mu)^2= (16.93 \cdots)^2 = 14628.7 \cdots,\end{equation*}

\begin{equation*} \text{using } (48) \int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0,1)} \,\mathrm{d} \mu = 38.2473 \cdots,\end{equation*}

\begin{equation*}\text{using } (48) \int \hat{a}_{(0,1,0)} ^2 \hat{a}_{(0,1)} \,\mathrm{d} \mu =38.2473 \cdots, \end{equation*}

\begin{equation*} \text{using } (49) \int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu= -1.34387 \cdots,\end{equation*}

\begin{equation*}\text{using } (49) \int \hat{a}_{(0,1,0)} ^2 \hat{a}_{(0)} \,\mathrm{d} \mu = -1.34387 \cdots,\end{equation*}

and finally, using Equation (51)

\begin{equation*}\int \hat{a}_{(0,1,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu=\int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu=\frac{1}{|a_{(0,1,0)}|^2 |V_1|}\int a_{(0,1,0)}^2 V_1 \,\mathrm{d} \mu= 4.13241 \cdots.\end{equation*}

Using Equations (51) and (36) (note that $x_1=0$), we get that the expression (45) can be written in this case as

\begin{eqnarray*} K( \hat{a}_{(0,1,0)}, \hat{a}_{(0,1,0,0)} ) & = &\frac{1}{4} (\int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0,1,0)} \,\mathrm{d} \mu)^2 \\ & - & \frac{1}{4}[ \int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0,1,0)} \,\mathrm{d} \mu \int \hat{a}_{(0,1,0)}^2 \hat{a}_{(0,1,0)} \,\mathrm{d} \mu]\\ & - & \frac{1}{4}[\int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0,1)} \,\mathrm{d} \mu \int \hat{a}_{(0,1,0)}^2 \hat{a}_{(0,1)} \,\mathrm{d} \mu ] \\ & - & \frac{1}{4}[\int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu \int \hat{a}_{(0,1,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu ] \\ &-&\frac{1}{4} \int \hat{a}_{(0,1,0,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu \int \hat{a}_{(0,1,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu = 35.9142 \cdots.\\ \end{eqnarray*}

Taking $r=0.8,s=0.5$, we get $K( \hat{a}_{(0,1,0)}, \hat{a}_{(0,1,0,0)} )=-3.17713 \cdots.$ When $r=1/2=s$, we get $K( \hat{a}_{(0,1,0)}, \hat{a}_{(0,1,0,0)} )=-1/2$.

$ \diamondsuit$

Example 7.20. Consider the case where $ z=(0)$ and $x=(0,0).$ Then, $D[(0),(0,0)] =\{\hat{a}_0, \hat{a}_{[\emptyset]}^0\}.$ Therefore,

\begin{eqnarray*} K( \hat{a}_{(0)}, \hat{a}_{(0,0)} ) & = &\frac{1}{4} (\int \hat{a}_{(0,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu)^2 \\ & - & \frac{1}{4}[ \int \hat{a}_{(0,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu \int \hat{a}_{(0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu]\\ &-&\frac{1}{4} \int \hat{a}_{(0,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu \int \hat{a}_{(0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu. \end{eqnarray*}

In this case, using Mathematica, one can show that $K( \hat{a}_{(0)}, \hat{a}_{(0,0)} )\leq 0$, for all values $r,s\in(0,1)$. For r = 0.1, s = 0.3, we will show that $K( \hat{a}_{(0)}, \hat{a}_{(0,0)} )= -0.205714 \cdots.$

When, $r=0.1,s=0.3$, we get

\begin{equation*} |a_0|=3.333 \cdots, \end{equation*}

\begin{equation*}|V_1|= 0.086 \cdots,\end{equation*}

\begin{equation*}\int \hat{a}_{(0,0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu=\frac{1}{|a_{(0)}|^3} \end{equation*}

\begin{equation*}\int \hat{a}_{(0)}^2 \hat{a}_{(0)} \,\mathrm{d} \mu= \frac{1}{|a_{(0)}|^3 }\end{equation*}

\begin{equation*}\int \hat{a}_{(0,0)}^2 \hat{a}_{[\emptyset]}^0 \,\mathrm{d} \mu = \frac{1}{|a_{(0)}|^2 |V_1|} \int a_{(0)}^2 V_1 \,\mathrm{d} \mu = \frac{1}{|a_{(0)}|^2 |V_1|} 3.96.\end{equation*}

Finally, when r = 0.1, s = 0.3, we get $K( \hat{a}_{(0)}, \hat{a}_{(0,0)} )= -0.205714 \cdots.$

$ \diamondsuit$

8. Computations for the integral $\int X^2 Y $

Our purpose in this section is to evaluate the integral

(57)\begin{equation} \sum_{\text{word}\, y } \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu, \end{equation}

for any given pair of words $x,z$. This corresponds to the second term in the sum given by expression (44).

We assume that x is different from z.

From Proposition 7.5, if x is not a subprefix of y and y is not a subprefix of x, and x ≠ y, then:

\begin{equation*}\hat{a}_x^2 \hat{a}_y=0.\end{equation*}

In the same way, if z is not a subprefix of y and y is not a subprefix of z, and z ≠ y, then:

\begin{equation*}\hat{a}_z^2 \hat{a}_y=0.\end{equation*}

If y has the same length as x but y ≠ x, then $\hat{a}_x^2 \hat{a}_y=0.$

In this way, for a fixed pair of words $x,z$, several words y do not contribute to the sum (74).

8.1. The value of $\langle \hat{a}_x^2, \hat{a}_y\rangle $ when length of x is larger or equal than the length of y

We want to compute $\langle \hat{a}_x^2, \hat{a}_y\rangle =\int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu$ in the case where the length of x is larger or equal to the length of y.

Our computation is in fact for $\langle a_x^2, a_y\rangle $ and after that, of course, to get $\langle \hat{a}_x^2, \hat{a}_y\rangle $ it will be necessary to divide by $|a_x|^2 |a_y|.$

We assume that $[x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [y]=[x_1,x_2, \ldots,x_k]$, where $n \geq k$ (otherwise we get zero).

Note that these assumptions include the integral $\int \hat{a}_x^3 \,\mathrm{d} \mu$, that is, the case x = y (see (iii) below).

(i) Case n > k – We will assume first that $x_{k+1}=0$ in the word $[x]$.

Given the words $z=(v_1, \ldots,v_t)$ and $v=(v_1,v_2, \ldots,v_t, v_{t+1}, \ldots,v_m)$, assume $v_{t+1}=0$, then, from Equations (23) and (58)

\begin{equation*} e_{[v]}^2 e_{[z]}=[\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} } \mathfrak{1}_{[v_1, \ldots,v_t,0,v_{t+2}, \ldots,v_m, 0]} + \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} } \mathfrak{1}_{[ v_1, \ldots,v_t,0,v_{t+2}, \ldots,v_m ,1]}] \end{equation*}

\begin{equation*}\times [\frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,1}}{P_{v_t,0} }} \mathfrak{1}_{[v_1, \ldots,v_t, 0]} - \frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,0}}{P_{v_t,1} }} \mathfrak{1}_{[v_1, \ldots,v_t , 1]} ] \end{equation*}

\begin{equation*}=(\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} }) ( \frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,1}}{P_{v_t,0} }} ) \mathfrak{1}_{[v_1, \ldots,0, v_{t+2}, \ldots,v_m,0]} \end{equation*}

(58)\begin{align} + ( \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} }) (\frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,1}}{P_{v_t,0} }} ) \mathfrak{1}_{[v_1,...,0,v_{t+2},...,v_m, 1]}. \end{align}

Note that in the above reasoning when going from the second to the third line the term multiplying $\mathfrak{1}_{[v_1,...,v_t , 1]} $ disappear because we assume that $v_{t+1}=0.$

We are going to apply the above when $z=[0 y], z=[1 y], v=[0 x], v=[1 x], m=n$ and $t+1=k$.

Then, from Equations (27), (31) and (58) and using the fact that

\begin{equation*}e_{[0,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}^2 e_{[1,x_1, x_2, \ldots,x_k]}=0,\end{equation*}

\begin{equation*}e_{[1,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}^2 e_{[0,x_1, x_2, \ldots,x_k]}=0,\end{equation*}

we get

\begin{equation*} a_x^2 a_y=[\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_k, 0,x_{k+2}, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}^2] \end{equation*}

\begin{equation*} \times [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_k]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_k]}] \end{equation*}

\begin{equation*}= ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{1}{\mu([0 x])} \frac{P_{x_n,1}}{P_{x_n,0} }) (\frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }} ) \mathfrak{1}_{[0x0]} \end{equation*}

\begin{equation*}+ (\frac{1}{\mu([0 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) ( \frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }}) \mathfrak{1}_{[0x1]} ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ (\frac{1}{\mu([1 x])} \frac{P_{x_n,1}}{P_{x_n,0} } ) ( \frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }) } \mathfrak{1}_{[1x0]} \end{equation*}

\begin{equation*} + (\frac{1}{\mu([1 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) (\frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }}) \mathfrak{1}_{[1x1]} ] . \end{equation*}

Finally, as the matrix P is row stochastic

\begin{equation*} \int a_x^2 a_y \,\mathrm{d} \mu= ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{P_{x_n,1}}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }} ) + (\frac{P_{x_n,0}}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }}) ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ ( \frac{P_{x_n,1}}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }}) + (\frac{P_{x_n,0}}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,1}}{P_{x_k,0} }}) ] \end{equation*}

\begin{equation*}= (P_{x_n,1} +P_{x_n,0} ) \sqrt{P_{x_k,1}}\{ ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{1}{\sqrt{\mu([0 y 0])}} ] \end{equation*}

\begin{equation*} -( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ \frac{1}{\sqrt{\mu([1 y 0])}} ] \}= \end{equation*}

(59)\begin{equation} \sqrt{P_{x_k,1}}\{ ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y 0])}} -( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y 0])}} \} . \end{equation}

(ii) Case n > k – If we assume $x_{k+1}=1$ in the word $[x]$, then we get in a similar way as before

\begin{equation*} \int a_x^2 a_y \,\mathrm{d} \mu= \end{equation*}

(60)\begin{equation}\sqrt{P_{x_k,0}}\{ - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y 1])}} +( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y 1])}} \}. \end{equation}

Indeed, given the words $z=(v_1, \ldots,v_t)$ and $v=(v_1,v_2, \ldots,v_t, v_{t+1}, \ldots,v_m)$, assume $v_{t+1}=1$, then, from Equations (23) and (58)

\begin{equation*} e_{[v]}^2 e_{[z]}=[\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} } \mathfrak{1}_{[v_1, \ldots,v_t,1,v_{t+2}, \ldots,v_m, 0]} + \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} } \mathfrak{1}_{[ v_1, \ldots,v_t,1,v_{t+2}, \ldots,v_m ,1]}] \end{equation*}

\begin{equation*}\times [\frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,1}}{P_{v_t,0} }} \mathfrak{1}_{[v_1, \ldots,v_t, 0]} - \frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,0}}{P_{v_t,1} }} \mathfrak{1}_{[v_1, \ldots,v_t , 1]} ] \end{equation*}

\begin{equation*}= - [(\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} }) ( \frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,0}}{P_{v_t,1} }} ) \mathfrak{1}_{[v_1, \ldots,1, v_{t+2}, \ldots,v_m,0]} ]\end{equation*}

(61)\begin{align} + ( \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} }) (\frac{1}{\sqrt{\mu([z])}} \sqrt{\frac{P_{v_t,0}}{P_{v_t,1} }} ) \mathfrak{1}_{[v_1, \ldots,1,v_{t+2}, \ldots,v_m, 1]}. \end{align}

We are going to apply the above when $z=[0 y], z=[1 y], v=[0 x], v=[1 x], m=n$ and $t=k+1$.

Then, from Equations (27), (31) and (61) and using the fact that

\begin{equation*}e_{[1,x_1, x_2, \ldots,x_k,1,x_{k+2}, \ldots,x_n]}^2 e_{[0,x_1, x_2, \ldots,x_k]}=0,\end{equation*}

\begin{equation*}e_{[0,x_1, x_2, \ldots,x_k,1,x_{k+2}, \ldots,x_n]}^2 e_{[1,x_1, x_2, \ldots,x_k]}=0,\end{equation*}

we get

\begin{equation*} a_x^2 a_y=[\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2,..,x_k, 1,x_{k+2}, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_k,1,x_{k+2}, \ldots,x_n]}^2] \end{equation*}

\begin{equation*} \times [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_k]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_k]}] \end{equation*}

\begin{equation*}= - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{1}{\mu([0 x])} \frac{P_{x_n,1}}{P_{x_n,0} }) (\frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }} ) \mathfrak{1}_{[0x0]} \end{equation*}

\begin{equation*}+ (\frac{1}{\mu([0 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) ( \frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }}) \mathfrak{1}_{[0x1]} ] \end{equation*}

\begin{equation*}+( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ (\frac{1}{\mu([1 x])} \frac{P_{x_n,1}}{P_{x_n,0} } ) ( \frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }) } \mathfrak{1}_{[1x0]} \end{equation*}

\begin{equation*}+ (\frac{1}{\mu([1 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) (\frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }}) \mathfrak{1}_{[1x1]} ] . \end{equation*}

Finally, as the matrix P is row stochastic

\begin{equation*} \int a_x^2 a_y \,\mathrm{d} \mu= - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{P_{x_n,1}}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }} ) + (\frac{P_{x_n,0}}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }}) ] \end{equation*}

\begin{equation*}+( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ ( \frac{P_{x_n,1}}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }}) + (\frac{P_{x_n,0}}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_k,0}}{P_{x_k,1} }}) ]= \end{equation*}

\begin{equation*}\sqrt{P_{x_k,0}}\{ - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_n,1}+ P_{x_n,0}}{\sqrt{\mu([0 y 1])}} ]+( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ \frac{P_{x_n,1}+ P_{x_n,0}}{\sqrt{\mu([1 y 1])}} ] \}= \end{equation*}

(62)\begin{equation}\sqrt{P_{x_k,0}}\{ - ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0 y 1])}} +( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1 y 1])}} \}.\end{equation}

(iii) Case n = k – We assume $[x]=[x_1,x_2, \ldots,x_n] = [y]$, otherwise $ \int \hat{a}_x^2 \hat{a}_y \,\mathrm{d} \mu=0. $ Then, one can show that

(63)\begin{equation} \int a_x^2 a_y \,\mathrm{d} \mu= \int a_x^3 \,\mathrm{d} \mu= \end{equation}

\begin{equation*} ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([0 x 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([0 x 1])}} ] -( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([1 x 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([1 x 1])}} ] .\end{equation*}

Indeed, note first that from Equation (58), $v=[v_1,v_2, \ldots,v_m]$

\begin{equation*} e_{[v]}^2 e_{[v]}=[\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} } \mathfrak{1}_{[v_1, \ldots,v_m, 0]} + \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} } \mathfrak{1}_{[ v_1, \ldots,v_m ,1]}] \end{equation*}

\begin{equation*}\times [\frac{1}{\sqrt{\mu([v])}} \sqrt{\frac{P_{v_m,1}}{P_{v_m,0} }} \mathfrak{1}_{[v_1, \ldots,v_m, 0]} - \frac{1}{\sqrt{\mu([v])}} \sqrt{\frac{P_{v_m,0}}{P_{v_m,1} }} \mathfrak{1}_{[v_1, \ldots,v_m , 1]} ] \end{equation*}

\begin{equation*}=(\frac{1}{\mu([v])} \frac{P_{v_m,1}}{P_{v_m,0} }) ( \frac{1}{\sqrt{\mu([v])}} \sqrt{\frac{P_{v_m,1}}{P_{v_m,0} }} ) \mathfrak{1}_{[v_1, \ldots,v_m,0]} \end{equation*}

(64)\begin{align} - ( \frac{1}{\mu([v])} \frac{P_{v_m,0}}{P_{v_m,1} }) (\frac{1}{\sqrt{\mu([v])}} \sqrt{\frac{P_{v_m,0}}{P_{v_m,1} }} ) \mathfrak{1}_{[v_1, \ldots,v_m, 1]}. \end{align}

Then, from Equations (27), (31) and (61)

\begin{equation*} a_x^2 a_x=[\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_n]}^2] \end{equation*}

\begin{equation*} \times [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_n]}] \end{equation*}

\begin{equation*}= ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{1}{\mu([0 x])} \frac{P_{x_n,1}}{P_{x_n,0} }) (\frac{1}{\sqrt{\mu([0 x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) \mathfrak{1}_{[0x0]} \end{equation*}

\begin{equation*}- (\frac{1}{\mu([0 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) ( \frac{1}{\sqrt{\mu([0 x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[0x1]} ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ (\frac{1}{\mu([1 x])} \frac{P_{x_n,1}}{P_{x_n,0} } ) ( \frac{1}{\sqrt{\mu([1 x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }) } \mathfrak{1}_{[1x0]} \end{equation*}

\begin{equation*} - (\frac{1}{\mu([1 x])} \frac{P_{x_n,0}}{P_{x_n,1} }) (\frac{1}{\sqrt{\mu([1 x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[1x1]} ] . \end{equation*}

Therefore,

\begin{equation*} \int a_x^2 a_x \,\mathrm{d} \mu= ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{P_{x_n,1}}{\sqrt{\mu([0 x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) - (\frac{P_{x_n,0}}{\sqrt{\mu([0 x])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ ( \frac{P_{x_n,1}}{\sqrt{\mu([1 x])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }}) - (\frac{P_{x_n,0}}{\sqrt{\mu([1 x)}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) ]= \end{equation*}

\begin{equation*} ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([0 x 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([0 x 1])}} ] -( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ ( \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([1 x 0])}} ) - (\frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([1 x 1])}} ] =\end{equation*}

(65)\begin{equation} ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([0 x 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([0 x 1])}} ] -( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ \frac{P_{x_n,1}^{3/2}}{\sqrt{\mu([1 x 0])}} - \frac{P_{x_n,0}^{3/2}}{\sqrt{\mu([1 x 1])}} ] . \end{equation}

The above reasoning shows (iii).

Given the word $[x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] $, we get n words y, such that the cylinder $[x]\subset [y]=[x_1,x_2, \ldots,x_k]$, where $n\geq k$.

Given x and z, with length larger than y, then $\int \hat{a}_{x}^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_{z}^2 \hat{a}_y \,\mathrm{d} \mu$ will be non-zero only for the subprefixes y which are common to both x and z (see Proposition 7.5). If there are no common subprefixes for x and z, then the contribution $\int \hat{a}_{x}^2 \hat{a}_y \,\mathrm{d} \mu \int \hat{a}_{z}^2 \hat{a}_y \,\mathrm{d} \mu$, for words y of length strictly smaller than the length of x and z, in the sum (45) is null.

8.2. The values of $\langle \hat{a}_x^2, \hat{a}_{[\emptyset]}^0\rangle $ and $\langle \hat{a}_x^2, \hat{a}_{[\emptyset]}^1\rangle $ when x is a finite word

Denote $[x]=[x_1,x_2, \ldots, x_n]$. We assume that $n\geq 2.$

In fact, we will compute $\langle a_x^2, V_1\rangle $ and $\langle a_x^2, V_2\rangle $. In order to compute $\langle \hat{a}_x^2, \hat{a}_{[\emptyset]}^0\rangle $ and $\langle \hat{a}_x^2, \hat{a}_{[\emptyset]}^1\rangle $, it will be necessary to normalize.

(i) Case $\langle a_x^2, V_1\rangle $

We will consider first the case $x_1=0$.

Denote $y=(y_1,y_2, \ldots,y_k)$. If we assume $y_1=0,y_2=0$, then, from Equation (58)

\begin{equation*} e_{[y]}^2 V_1=[\frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \mathfrak{1}_{[ y_1,y_2, \ldots,y_k ,1]}] \end{equation*}

\begin{equation*}\times [ \pi_1 P_{1,0} \mathfrak{1}_{[0,0]} - \pi_0 P_{0,0} \mathfrak{1}_{[1,0]} ]= \end{equation*}

(66)\begin{align} \frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \pi_1 P_{1,0} \mathfrak{1}_{[y_1, y_2, \ldots,y_k,0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \pi_1 P_{1,0} \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 1]}. \end{align}

If we assume $y_1=1,y_2=0$, then, from Equation (58)

\begin{equation*} e_{[y]}^2 V_1=[\frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \mathfrak{1}_{[ y_1,y_2, \ldots,y_k ,1]}] \end{equation*}

\begin{equation*}\times [ \pi_1 P_{1,0} \mathfrak{1}_{[0,0]} - \pi_0 P_{0,0} \mathfrak{1}_{[1,0]} ]= \end{equation*}

(67)\begin{align} - [\frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \pi_0 P_{0,0} \mathfrak{1}_{[y_1, y_2, \ldots,y_k,0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \pi_0 P_{0,0} \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 1]}]. \end{align}

As we assume that $x_1=0$, then, from Equations (27), (31) and (61), we get

\begin{equation*} a_x^2 V_1=[\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_k, 0,x_{k+2}, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}^2] \end{equation*}

\begin{equation*}\times [ \pi_1 P_{1,0} \mathfrak{1}_{[0,0]} - \pi_0 P_{0,0} \mathfrak{1}_{[1,0]} ]= \end{equation*}

\begin{equation*}= \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } [ \frac{1}{\mu([0x])} \frac{P_{x_n,1}}{P_{x_n,0} } \pi_1 P_{1,0} \mathfrak{1}_{[0,x_1, x_2, \ldots,x_n,0]} + \frac{1}{\mu([0x])} \frac{P_{x_n,0}}{P_{x_n,1} } \pi_1 P_{1,0} \mathfrak{1}_{[0,x_1,x_2, \ldots,x_n, 1]}]+ \end{equation*}

\begin{equation*} \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } [ \frac{1}{\mu([1x])} \frac{P_{x_n,1}}{P_{x_n,0} } \pi_0 P_{0,0} \mathfrak{1}_{[1,x_1, x_2, \ldots,x_n,0]} + \frac{1}{\mu([1x])} \frac{P_{x_n,0}}{P_{x_n,1} } \pi_0 P_{0,0} \mathfrak{1}_{[1,x_1,x_2, \ldots,x_n, 1]}] .\end{equation*}

Therefore,

\begin{equation*}\int a_x^2 V_1\,\mathrm{d}\mu=\end{equation*}

\begin{equation*} \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } [ \frac{1}{\mu([0x])} \frac{P_{x_n,1}}{P_{x_n,0} } \pi_1 P_{1,0} \mu[0,x,0] + \frac{1}{\mu([0x])} \frac{P_{x_n,0}}{P_{x_n,1} } \pi_1 P_{1,0} \mu[0,x, 1] ] + \end{equation*}

\begin{equation*} \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } [ \frac{1}{\mu([1x])} \frac{P_{x_n,1}}{P_{x_n,0} } \pi_0 P_{0,0} \mu[1,x,0] + \frac{1}{\mu([1x])} \frac{P_{x_n,0}}{P_{x_n,1} } \pi_0 P_{0,0} \mu[1,x, 1]] =\end{equation*}

\begin{equation*} \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } [ P_{x_n,1} \pi_1 P_{1,0} + P_{x_n,0} \pi_1 P_{1,0} ] + \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } [ P_{x_n,1} \pi_0 P_{0,0} + P_{x_n,0} \pi_0 P_{0,0} ] =\end{equation*}

\begin{equation*} \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } \pi_1 P_{1,0} + \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } \pi_0 P_{0,0}.\end{equation*}

As we assumed that $x_1=0$ we get

(68)\begin{equation} \int \hat{a}_x^2\hat{a}_{[\emptyset]}^0\,\mathrm{d}\mu= \frac{1}{|a_x|^2 | V_1| }( \frac{\pi_1 P_{1,0} } {P_{0,0} } + \frac{\pi_0^2 P_{0,0} } {\pi_1 P_{1,0} }) . \end{equation}

(ii) $\langle a_x^2, V_2\rangle $

Now we will compute $ \int a_x^2 V_2\,\mathrm{d}\mu.$

Denote $y=(y_1,y_2, \ldots,y_k)$. If we assume $y_1=0,y_2=0$, then, from Equation (58)

\begin{equation*} e_{[y]}^2 V_2=[\frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \mathfrak{1}_{[ y_1,y_2, \ldots,y_k ,1]}] \end{equation*}

\begin{equation*}\times [ \pi_0 P_{0,1} \mathfrak{1}_{[1,1]} - \pi_1 P_{1,1} \mathfrak{1}_{[0,1]} ]=0. \end{equation*}

If we assume $y_1=1,y_2=0$, then, from Equation (58)

\begin{equation*} e_{[y]}^2 V_2=[\frac{1}{\mu([y])} \frac{P_{y_k,1}}{P_{y_k,0} } \mathfrak{1}_{[y_1,y_2, \ldots,y_k, 0]} + \frac{1}{\mu([y])} \frac{P_{y_k,0}}{P_{y_k,1} } \mathfrak{1}_{[ y_1,y_2, \ldots,y_k ,1]}] \end{equation*}

\begin{equation*} \times [ \pi_0 P_{0,1} \mathfrak{1}_{[1,1]} - \pi_1 P_{1,1} \mathfrak{1}_{[0,1]} ]=0. \end{equation*}

As we assumed that $x_1=0$, then, from Equations (27), (31) and (61), we get

\begin{equation*} a_x^2 V_2=[\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_n]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_n]}^2] \end{equation*}

\begin{equation*}\times [ \pi_0 P_{0,1} \mathfrak{1}_{[1,1]} - \pi_1 P_{1,1} \mathfrak{1}_{[0,1]} ]=0. \end{equation*}

Therefore, if $x_1=0$, we get

(69)\begin{equation} \int \hat{a}_x^2 \hat{a}_{[\emptyset]}^1 \,\mathrm{d}\mu=0. \end{equation}

The case $x_1=1$ is left for the reader.

8.3. The value of $\langle \hat{a}_z^2, \hat{a}_y\rangle $ when length of y is strictly larger than the length of z

Now we want to estimate $\langle \hat{a}_z^2, \hat{a}_y\rangle =\int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu$ in the case that the length of y is strictly larger than the length of z. We will show that $\int \hat{a}_z^2 \hat{a}_y \,\mathrm{d} \mu=0.$

We assume that $[y]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [z]=[x_1,x_2, \ldots,x_k]$, where n > k (otherwise we get that$\int \hat{a}_z^2 \hat{a}_y d \mu$ is zero from Proposition 7.5).

In fact, we will show that $\int a_z^2 a_y \,\mathrm{d} \mu=0.$

(i) If we assume $x_{k+1}=0$ in the word $[y]$, then, from Equation (58)

\begin{equation*} e_z^2 e_y =[\frac{1}{\mu([z])} \frac{P_{x_k,1}}{P_{x_k,0} } \mathfrak{1}_{[x_1, \ldots,x_k, 0]} + \frac{1}{\mu([z])} \frac{P_{x_k,0}}{P_{x_k,1} } \mathfrak{1}_{[ x_1, \ldots,x_k ,1]}] \end{equation*}

\begin{equation*}\times [\frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_1, \ldots,x_k,0,x_{k+2}, \ldots,x_n, 0]} - \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_1, \ldots,x_k,0,x_{k+2}, \ldots,x_n, 1]} ] \end{equation*}

\begin{equation*}= (\frac{1}{\mu([z])} \frac{P_{x_k,1}}{P_{x_k,0} }) ( \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) \mathfrak{1}_{[x_1, \ldots,x_k,0, x_{k+2}, \ldots,x_n,0]} \end{equation*}

(70)\begin{align} - (\frac{1}{\mu([z])} \frac{P_{x_k,1}}{P_{x_k,0} }) ( \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} ) \mathfrak{1}_{[x_1, \ldots,x_k,0, x_{k+2}, \ldots,x_n,1]}. \end{align}

Note that above, from the second to the third line, we use the fact that

\begin{equation*} \mathfrak{1}_{[ x_1, \ldots,x_k ,1]} \mathfrak{1}_{[x_1, \ldots,x_k,0,x_{k+2}, \ldots,x_n, 0]} =0\end{equation*}

and

\begin{equation*} \mathfrak{1}_{[ x_1, \ldots,x_k ,1]} \mathfrak{1}_{[x_1, \ldots,x_k,0,x_{k+2}, \ldots,x_n, 1]} =0.\end{equation*}

Then, from Equations (27), (70) and (31)

\begin{equation*} a_z^2 a_y= [\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_k]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_k]}^2] \end{equation*}

\begin{equation*}\times [\frac{\sqrt{\pi_{x_1}} } {\sqrt{\pi_{0} P_{0,x_1}} } e_{[0,x_1, x_2, \ldots,x_k, 0,x_{k+2}, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} } } {\sqrt{\pi_{1}P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}] \end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{1}{\mu([0 z])} \frac{P_{x_k,1}}{P_{x_k,0} }) (\frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) \mathfrak{1}_{[0y0]} \end{equation*}

\begin{equation*}- (\frac{1}{\mu([0 z])} \frac{P_{x_k,1}}{P_{x_k,0} }) ( \frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[0y1]} ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ (\frac{1}{\mu([1 z])} \frac{P_{x_k,1}}{P_{x_k,0} } ) ( \frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }) } \mathfrak{1}_{[1y0]} \end{equation*}

\begin{equation*}- (\frac{1}{\mu([1 z])} \frac{P_{x_k,1}}{P_{x_k,0} }) (\frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[1y1]} ] . \end{equation*}

Finally,

\begin{equation*} \int a_z^2 a_y\,\mathrm{d} \mu \end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_k,1}}{\mu([0 z 0])} \sqrt{\mu([0 y 0])} \sqrt{P_{x_n,1}} - \frac{P_{x_k,1}}{\mu([0 z 0]} \sqrt{\mu([0 y 1])} \sqrt{P_{x_n,0} } ] \end{equation*}

\begin{equation*}+ ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [- \frac{P_{x_k,1}}{\mu([1 z 0])} \sqrt{\mu([1 y 0])} \sqrt{P_{x_n,1} } + \frac{P_{x_k,1}}{\mu([1 z 0])} \sqrt{\mu([1 y 1])} \sqrt{P_{x_n,0}} ] \end{equation*}

\begin{equation*}=P_{x_k,1}\sqrt{\mu([y])} \{ ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{\sqrt{P_{x_n,1} P_{0,x_1} P_{x_n,0}} }{\mu([0 z 0])} - \frac{\sqrt{P_{x_n,0} P_{0,x_1} P_{x_n,1} }}{\mu([0 z 0]} ] \end{equation*}

(71)\begin{equation} + P_{x_k,1} ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [- \frac{\sqrt{P_{x_n,1} P_{1,x_1} P_{x_n,0} }}{\mu([1 z 0])} + \frac{\sqrt{P_{x_n,0} P_{1,x_1} P_{x_n,1}}}{\mu([1 z 0])} ] \} =0. \end{equation}

(ii) If we assume $x_{k+1}=1$ in the word $[y]$, then, from Equation (58)

\begin{equation*} e_z^2 e_y =[\frac{1}{\mu([z])} \frac{P_{x_k,1}}{P_{x_k,0} } \mathfrak{1}_{[x_1, \ldots,x_k, 0]} + \frac{1}{\mu([z])} \frac{P_{x_k,0}}{P_{x_k,1} } \mathfrak{1}_{[ x_1, \ldots,x_k ,1]}] \end{equation*}

\begin{equation*} \times [\frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} \mathfrak{1}_{[x_1, \ldots,x_k,1,x_{k+2}, \ldots,x_n, 0]} - \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} \mathfrak{1}_{[x_1, \ldots,x_k,1,x_{k+2}, \ldots,x_n, 1]} ] \end{equation*}

\begin{equation*}=(\frac{1}{\mu([z])} \frac{P_{x_k,0}}{P_{x_k,1} }) ( \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) \mathfrak{1}_{[x_1, \ldots,x_k,1, x_{k+2}, \ldots,x_n,0]} \end{equation*}

(72)\begin{align} - (\frac{1}{\mu([z])} \frac{P_{x_k,0}}{P_{x_k,1} }) ( \frac{1}{\sqrt{\mu([y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }} ) \mathfrak{1}_{[x_1, \ldots,x_k,1, x_{k+2}, \ldots,x_n,1]} \end{align}

Then, from Equations (27), (72) and (31)

\begin{equation*} a_z^2 a_y= [\frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} } e_{[0,x_1, x_2, \ldots,x_k]}^2 + \frac{\pi_{x_1} } {\pi_{1}P_{1,x_1} } e_{[1,x_1, x_2, \ldots,x_k]}^2] \end{equation*}

\begin{equation*}\times [\frac{\sqrt{\pi_{x_1}} } {\sqrt{\pi_{0} P_{0,x_1}} } e_{[0,x_1, x_2, \ldots,x_k, 0,x_{k+2}, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} } } {\sqrt{\pi_{1}P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_k,0,x_{k+2}, \ldots,x_n]}] \end{equation*}

\begin{equation*}= ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ (\frac{1}{\mu([0 z])} \frac{P_{x_k,0}}{P_{x_k,1} }) (\frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }} ) \mathfrak{1}_{[0y0]} \end{equation*}

\begin{equation*} - (\frac{1}{\mu([0 z])} \frac{P_{x_k,0}}{P_{x_k,1} }) ( \frac{1}{\sqrt{\mu([0 y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[0y1]} ] \end{equation*}

\begin{equation*}-( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [ (\frac{1}{\mu([1 z])} \frac{P_{x_k,0}}{P_{x_k,1} } ) ( \frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_n,1}}{P_{x_n,0} }) } \mathfrak{1}_{[1y0]} \end{equation*}

\begin{equation*}- (\frac{1}{\mu([1 z])} \frac{P_{x_k,0}}{P_{x_k,1} }) (\frac{1}{\sqrt{\mu([1 y])}} \sqrt{\frac{P_{x_n,0}}{P_{x_n,1} }}) \mathfrak{1}_{[1y1]} ] . \end{equation*}

Finally,

\begin{equation*} \int a_z^2 a_y \,\mathrm{d} \mu\end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{P_{x_k,0}}{\mu([0 z 1])} \sqrt{\mu([0 y 0])} \sqrt{P_{x_n,1}} - \frac{P_{x_k,0}}{\mu([0 z 1]} \sqrt{\mu([0 y 1])} \sqrt{P_{x_n,0} } ] \end{equation*}

\begin{equation*}+( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [- \frac{P_{x_k,1}}{\mu([1 z 0])} \sqrt{\mu([1 y 0])} \sqrt{P_{x_n,1} } + \frac{P_{x_k,1}}{\mu([1 z 0])} \sqrt{\mu([1 y 1])} \sqrt{P_{x_n,0}} ] \end{equation*}

\begin{equation*}=P_{x_k,0}\sqrt{\mu([y])} \{ ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} [ \frac{\sqrt{P_{x_n,1} P_{0,x_1} P_{x_n,0}} }{\mu([0 z 1])} - \frac{\sqrt{P_{x_n,0} P_{0,x_1} P_{x_n,1}} }{\mu([0 z 1]} ] \end{equation*}

(73)\begin{equation}+ P_{x_k,1}( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} [- \frac{\sqrt{P_{x_n,1} P_{1,x_1} P_{x_n,0} }}{\mu([1 z 1])} + \frac{\sqrt{P_{x_n,0} P_{1,x_1} P_{x_n,1}} }{\mu([1 z 1])} ] \}=0 . \end{equation}

9. Computations for the integral $\int X Y Z $

Our purpose on this section is: given x and z, we want to compute for all y

(74)\begin{equation} \sum_{\text{word}\, y }( \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2, \end{equation}

which corresponds to the first term in the sum given by expression (44).

Remember that from Corollary 7.12 if x is not a subprefix of z and z is not a subprefix of x, we get that for any y

\begin{equation*}\int \hat{a}_x \hat{a}_z \hat{a}_y\,\mathrm{d} \mu=0.\end{equation*}

Without loss of generality, we assume that z is a subprefix of x (see Proposition 7.4). The only possible non-zero value for Equation (74) is $\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu.$ This justify the first term in the sum (45).

We assume first that:

$[y]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n,x_{n+1}, \ldots,x_j] \subset [x]=[x_1,x_2, \ldots,x_k,x_{k+1}, \ldots, x_n] \subset [z]=[x_1,x_2, \ldots,x_k]$, where $j \gt n\geq k$.

We will show in all cases that $ \int a_x a_y a_z \,\mathrm{d} \mu=0.$ This includes the case

(75)\begin{equation} \int a_z^2 a_x \,\mathrm{d} \mu=0. \end{equation}

(i) First we assume that $x_{k+1} =0= x_{n+1}.$

Then,

\begin{equation*} a_x a_y a_z =\end{equation*}

\begin{equation*} [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_k]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_k]}] \end{equation*}

\begin{equation*} \times [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_j]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_j]}] \end{equation*}

\begin{equation*} \times [\frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{0}}\sqrt{P_{0,x_1} }} e_{[0,x_1, x_2, \ldots,x_n]} - \frac{\sqrt{\pi_{x_1} }} {\sqrt{\pi_{1}}\sqrt{P_{1,x_1} }} e_{[1,x_1, x_2, \ldots,x_n]}] \end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0x])\mu([0z])\mu([0y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} P_{x_j,1}}{P_{x_n,0}P_{x_k,0}P_{x_j,0}}} \mathfrak{1}_{[0y0]} \end{equation*}

\begin{equation*}- \sqrt{\frac{P_{x_n,1}P_{x_k,1}P_{x_j,0}}{P_{x_n,0}P_{x_k,0} P_{x_j,1}}} \mathfrak{1}_{[0y 1]} ] \end{equation*}

\begin{equation*} - ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1x])\mu([1z])\mu([1y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} P_{x_j,1}}{P_{x_n,0}P_{x_k,0}P_{x_j,0}}} \mathfrak{1}_{[1y0]} \end{equation*}

\begin{equation*}- \sqrt{\frac{P_{x_n,1}P_{x_k,1}P_{x_j,0}}{P_{x_n,0}P_{x_k,0}P_{x_j,1} }} \mathfrak{1}_{[1y 1]} ] .\end{equation*}

Note that for all j

(76)\begin{equation} \sqrt{\frac{P_{x_j,1}}{P_{x_j,0} }} \mu([0y0])= \sqrt{P_{x_j,1 } P_{x_j,0} } \mu([0y])= \sqrt{\frac{P_{x_j,0}}{P_{x_j,1} }} \mu([0y1]) \end{equation}

and

(77)\begin{equation}\sqrt{\frac{P_{x_j,1}}{P_{x_j,0}} } \mu([1y0]) = \sqrt{P_{x_j,1 } P_{x_j,0} } \mu([1y]) = \sqrt{\frac{P_{x_j,0}}{P_{x_j,1}} } \mu([1y1]). \end{equation}

Finally, from Equations (76) and (77)

\begin{equation*} \int a_x a_y a_z \,\mathrm{d} \mu=\end{equation*}

\begin{equation*} ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0x])\mu([0z])\mu([0y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} P_{x_j,1}}{P_{x_n,0}P_{x_k,0}P_{x_j,0}}} \mu([0y0]) \end{equation*}

\begin{equation*}- \sqrt{\frac{P_{x_n,1}P_{x_k,1}P_{x_j,0}}{P_{x_n,0}P_{x_k,0} P_{x_j,1}}} \mu([0y1]) ] \end{equation*}

\begin{equation*}- ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1x])\mu([1z])\mu([1y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} P_{x_j,1}}{P_{x_n,0}P_{x_k,0}P_{x_j,0}}} \mu([1y0]) \end{equation*}

\begin{equation*} -\sqrt{\frac{P_{x_n,1}P_{x_k,1}P_{x_j,0}}{P_{x_n,0}P_{x_k,0}P_{x_j,1} }} \mu([1y1]) ] =\end{equation*}

\begin{equation*} ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0x])\mu([0z])\mu([0y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} }{P_{x_n,0}P_{x_k,0}}} \sqrt{\frac{P_{x_j,1}}{P_{x_j,0}} } \mu([0y0]) \end{equation*}

\begin{equation*}- \sqrt{\frac{P_{x_n,1}P_{x_k,1}}{P_{x_n,0}P_{x_k,0} }} \sqrt{\frac{P_{x_j,0}}{P_{x_j,1}} } \mu([0y1]) ] \end{equation*}

\begin{equation*}- ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1x])\mu([1z])\mu([1y])}} [ \sqrt{\frac{P_{x_n,1}P_{x_k,1} }{P_{x_n,0}P_{x_k,0}}} \sqrt{\frac{P_{x_j,1}}{P_{x_j,0}} } \mu([1y0]) \end{equation*}

\begin{equation*} -\sqrt{\frac{P_{x_n,1}P_{x_k,1}}{P_{x_n,0}P_{x_k,0}}} \sqrt{\frac{P_{x_j,0}}{P_{x_j,1}} } \mu([1y1]) ] = 0-0=0 .\end{equation*}

(ii) Now we assume that $x_{k+1} =1= x_{n+1}.$ In a similar way as before

\begin{equation*} \int a_x a_y a_z \,\mathrm{d} \mu\end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{1}{\sqrt{\mu([0x])\mu([0z])\mu([0y])}} [ \sqrt{\frac{P_{x_n,0}P_{x_k,0} P_{x_j,1}}{P_{x_n,1}P_{x_k,1}P_{x_j,0}}} \mu([0y0]) \end{equation*}

\begin{equation*} -\sqrt{\frac{P_{x_n,0}P_{x_k,0}P_{x_j,0}}{P_{x_n,1}P_{x_k,1} P_{x_j,1}}} \mu([0y1]) ] \end{equation*}

\begin{equation*} - ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{1}{\sqrt{\mu([1x])\mu([1z])\mu([1y])}} [ \sqrt{\frac{P_{x_n,0}P_{x_k,0} P_{x_j,1}}{P_{x_n,1}P_{x_k,1}P_{x_j,0}}} \mu([1y0]) \end{equation*}

\begin{equation*}- \sqrt{\frac{P_{x_n,0}P_{x_k,0}P_{x_j,0}}{P_{x_n,1}P_{x_k,1}P_{x_j,1} }} \mu([1y1]) ] \end{equation*}

\begin{equation*} = ( \frac{\pi_{x_1} } {\pi_{0} P_{0,x_1} })^{3/2} \frac{\sqrt{\mu([0y0]) P_{x_j,1}}}{\sqrt{\mu([0x])\mu([0z])}} [ \sqrt{\frac{P_{x_n,0}P_{x_k,0} }{P_{x_n,1}P_{x_k,1}}} - \sqrt{\frac{P_{x_n,0}P_{x_k,0}}{P_{x_n,1}P_{x_k,1} }} ] \end{equation*}

\begin{equation*} - ( \frac{\pi_{x_1} } {\pi_{1} P_{1,x_1} } )^{3/2} \frac{\sqrt{\mu([1y0]) P_{x_j,1}}}{\sqrt{\mu([1 x])\mu([1 z])}} [ \sqrt{\frac{P_{x_n,0}P_{x_k,0} }{P_{x_n,1}P_{x_k,1}}} - \sqrt{\frac{P_{x_n,0}P_{x_k,0}}{P_{x_n,1}P_{x_k,1}}} ]= 0-0=0 .\end{equation*}

(iii) Now if we assume that $x_{k+1} =0 $ and $ x_{n+1}=1$ or that $x_{k+1} =1 $ and $ x_{n+1}=0$, we get that in a similar way that

\begin{equation*} \int \hat{a}_x \hat{a}_y \hat{a}_z \,\mathrm{d} \mu=0.\end{equation*}

After all these computations, for fixed $\hat{a}_x$ and $\hat{a}_z$, we want to compute $K(\hat{a}_z,\hat{a}_x)$. In this direction, we have to consider Equation (74) which is the first sum in expression (44).

We wonder for which y we have that $ (\int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2\neq 0.$ We assumed without loss of generality that z is a subprefix of x. In this case, the length of x is strictly larger than the length of z.

Considering first the case where the length of y is larger than z and x, it follows from the above that

\begin{equation*}\sum_{\text{word}\, y\,\text{with length larger than}\ x\ \text{and}\ z}( \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2=0.\end{equation*}

Now we consider the case where the length of y is strictly smaller than the length of z and x.

For the case where the length of y is strictly smaller than z and x, we need to assume that y is a subprefix of z (otherwise $\hat{a}_y \hat{a}_z=0$ and we get $ (\int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2=0$). If y is a strict subprefix of z and z is a strict subprefix of s we get from the above that $ (\int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2=0$.

Finally, we assume that the length of y is strictly smaller than x and strictly larger than z. In this case, we have to assume that x is a subprefix of y and y is a subprefix of z (otherwise by Proposition 7.4 we have $ \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu^2=0$). It follows from the above that also in this case $ (\int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2=0$.

Therefore, in the estimation of expression (74), it follows from our reasoning that all elements in this sum are zero up to expressions $(\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2$ and $(\int \hat{a}_z^2 \hat{a}_x \,\mathrm{d} \mu)^2$, that is, the cases where y = x or y = z. From Proposition 7.5, we have to assume that x is a subprefix of z or vice versa. The explicit expressions for these two cases were analysed in § 8.1 and 8.3.

If the length of x is larger than the length of z, then, from Equation (73) we get $(\int \hat{a}_z^2 \hat{a}_x d \mu)^2 =0.$

The final conclusion is that

(78)\begin{equation} \sum_{\text{word}\, y }( \int \hat{a}_x \hat{a}_z \hat{a}_y \,\mathrm{d} \mu)^2 = (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2 + (\int \hat{a}_z^2 \hat{a}_x \,\mathrm{d} \mu)^2= (\int \hat{a}_x^2 \hat{a}_z \,\mathrm{d} \mu)^2. \end{equation}