\section*{1. Introduction}
 In this short note we present Veech's proof of Sarnak's theorem on the  M\"{o}bius flow \cite{VeechNotes}, \cite{VeechNotes2}. Of-course, this proof is connected to Sarnak and Chowla conjectures. Moreover, let us stress that our exposition is self-contained as much as possible. 
 \medskip

 
Roughly speaking, Chowla conjecture assert that the Liouville function is normal, and Sarnak conjecture assert that the M\"{o}bius randomness law holds for any dynamical sequence with zero topological entropy. For more details on the M\"{o}bius randomness law  we refer to \cite{Kowalski}.
\medskip

 It is turn out that Veech's proof in combine with the recent result of Tao \cite{BlogTao} yields that Sarnak conjecture implies Chowla conjecture. Indeed, Tao's result assert that if the even logarithmic Chowla conjecture holds then there exists a subsequence $\mathcal{N}$ with logarithmic density $1$ along which the Chowla conjecture holds, and from Veech's proof we will see that this is enough to conclude that Chowla conjecture holds. We remind that T. Tao obtained as a corollary the recent result of Gomilko-Kwietniak-Lema\'{n}czyk  \cite{lem}.
\medskip

Let us further point out that the proof of Gomilko-Kwietniak-Lema\'{n}czyk is based essentially on Tao's theorem on logarithmic Sarnak and Chowla conjectures.
\medskip

We further notice, as T. Tao pointed out, that the proof of Gomilko-Kwietniak-Lema\'{n}czyk  use only that the M\"{o}bius function is bounded.
\medskip

Here, as mentioned before, combining Tao's result with Sarnak's theorem as established by W. Veech, we deduce that Sarnak conjecture holds if and only if Chowla conjecture holds.
\medskip

The more striking result that follows from Veech's proof is the connection between Sarnak conjecture and Hadamard matrix. 
\medskip

We remind that the matrix $H$ of order $n$ is a Hadamard matrix if 
$H$ is a $n \times n$ matrix with entries $\pm 1$ such that 
$HH^{\textrm{T}}=nI_n$, where $I_n$ is the identity matrix. The Hadamard matrix are named after Hadamard since the equality in the famous Hadamard determinant inequality  holds if and only if the matrix is a Hadamard matrix.
\medskip

It is well known that Hadamard matrix exist when $n=1,2$ or $n$ is a multiple of $4$.
\medskip

The Hadamard conjecture states that there is a Hadamard matrix for every any multiple of 4. In the opposite direction, the circulant Hadamard matrix conjecture state that the only circulant Hadamard matrix are matrix  of order $1$ and $4$. We recall that a circulant matrix of order $m$ is an $m \times m$ matrix for which
each row except the first is a cyclic permutation of the previous row by one position to the right. 
\medskip

The conjectures of Hadamard are two of the most outstanding unsolved problems in mathematics nowadays.
\medskip

It is well known that the Hadamard matrix is related to the so-called Barker sequences. The Barker sequence is a sequence of $\pm 1$ for which the autocorrelation coefficients are bounded by $1$. We remind that the autocorrelation of a sequence $(x_j)_{j=0}^{N-1}$ are given by
$$c_k =\sum_{j=0}^{N-k-1} x_j x_{j+k},~~~~~~~k \geq 1$$
with
$$c_k =\overline{c_{-k}},~~~~\textrm{if} k < 0.$$
For the special real case we have $c_k =c_{-k}$. To be more precise, it is well known that if a Barker sequence of
even length $n$ exists, then so does a circulant Hadamard matrix of order $n$. But, very recently, the author established that there are only finitely many Barker sequences, that is, Turyn-Golay's conjecture is true \cite{elabdal-Erdos}. For more details on the Hadamard matrix, we refer to \cite{Hor}.\\


\setcounter{section}{2}
\section*{2. Setup and the main result}
The M\"{o}bius function $\bm \mu$ is related intimately to the Liouville function $\bm \lambda$ which is defined by $\bm \lambda(n)= 1$ if the number of prime factor of $n$ is even and
 $-1$ otherwise. Precisely, the M\"{o}bius function $\bm \mu$ is given by 
\[
\bm \mu(n)  = \begin{cases} 
1, \ &\textrm{if } n = 1 \\
 \bm \lambda(n), \ &\textrm{if } n   \textrm{~is square-free, } \\
 0, \ &\textrm{otherwise.}
\end{cases}
\]
We remind that $n$ is square-free if $n$ has no factor in the subset
$\mathcal{P}_2\stackrel{\textrm {def}}{=}\big\{p^2/ p \in \mathcal{P}\big\}$, where
as customary,  $\mathcal{P}$ denote the subset of prime numbers.\\

In his seminal paper  \cite{Sar}, P. Sarnak makes the following conjecture.
\begin{sarnak}\label{conj-sarnak} For any dynamical flow $(X,T)$ with topological entropy zero, for any continuous function $f \in C(X)$, for any point $x \in X$, 
	\begin{eqnarray}\label{sarnak-conj}
	\frac{1}{N}\sum_{n=1}^{N}\bm \mu(n)f(T^nx) \tend{N}{+\infty}0.
	\end{eqnarray}
\end{sarnak}
The popular Chowla conjecture on the correlation of the M\"{o}bius function state that 
\begin{Chowla}\label{conj-chowla}
	For any $r\geq 0$, $1\leq a_1<\dots<a_r$, $i_s\in \{1,2\}$ not all equal to $2$, we have
	\begin{equation}\label{cza}
	\sum_{n\leq N}\bm \mu^{i_0}(n)\bm \mu^{i_1}(n+a_1)\cdot\ldots\cdot\bm \mu^{i_r}(n+a_r)={\rm o}(N).
	\end{equation} 
\end{Chowla}
This conjecture is related to the weaker conjecture stated in \cite{chowla}. We refer to \cite{chowla} for more details.
\medskip

In his breakthrough paper \cite{Taol}, T. Tao proposed the following logarithmic version of Sarnak and Chowla conjectures. 

\begin{Taos}\label{Tconj-sarnak} For any dynamical flow $(X,T)$ with topological entropy zero, for any continuous function $f \in C(X)$, for any point $x \in X$, 
	\begin{eqnarray}\label{Tsarnak-conj}
	\frac{1}{\log(N)}\sum_{n=1}^{N}\frac{\bm \mu(n)f(T^nx)}{n} \tend{N}{+\infty}0.
	\end{eqnarray}
\end{Taos}

The logarithmic Chowla conjecture can be stated as follows:

\begin{Taoc}
	For any $r\geq 0$, $1\leq a_1<\dots<a_r$, $i_s\in \{1,2\}$ not all equal to $2$, we have
	\begin{equation}\label{Tcza}
	\sum_{1 \leq n\leq N}\frac{\bm \mu^{i_0}(n)\bm \mu^{i_1}(n+a_1)\cdot\ldots\cdot\bm \mu^{i_r}(n+a_r)}{n}={\rm o}(\log(N)).
	\end{equation} 
\end{Taoc}
We remind that the logarithmic density of a subset $E \subset \mathbb{N}$ is given by
the following limit (if it exists)

$$\lim_{N \rightarrow \infty} \frac{1}{\log(N)}\sum_{1}^{N} \frac{{\mathbb{I}}_{E}(n)}{n}.$$

Let us further notice that one can replace $\log(N)$ by $\ell_N=\sum_{n=1}^{N}\frac{1}{n}.$ Thanks to Euler estimation.
\medskip
  
 Following L. Mirsky \cite{Mir} and P. Sarnak \cite{Sar}, the subset $A \subset \mathbb{N}$ is admissible if the cardinality $t(p,A)$ of classes modulo $p^2$ in $A$ given by
\[ 
t(p,A) \stackrel{\textrm {def}}{=} \left|\bigl\{z\in\mathbb{Z}/p^2\mathbb{Z}: \exists n\in A, n=z\ [p^2]\bigr\}\right| \] satisfy

	\begin{equation}
	\label{eq:def-admissible}
	\forall p \in \mathcal{P} ,\ t(p,A)<p^2.
	\end{equation}
In other words, for every prime $p$ the image of $A$ under reduction mod $p^2$ is proper in $\mathbb{Z}/p^2\mathbb{Z}$.
\medskip

	Let $X_3$ be the set $\{0,\pm 1\}^\mathbb{N}$ and $X_2\stackrel{\textrm {def}}{=}\{0, 1\}^\mathbb{N}$, and for each $i=1,2$, let $X_i$ be equipped with the product topology. Therefore, $X_3$ and $X_2$ are a compact set.  We denote by $\mathcal{M}_1(X_i)$,  $i=1,2$, the set of the probability measures on $X_i$. it is turn out that $\mathcal{M}_1(X_i)$, $i=1,2$ is a compact set for the weak-star topology by Banach-Alaoglu-Bourbaki  theorem. Let $x \in X_i$, $i=1,2$, and for each $N \in \mathbb{N}$, put 
	$$m_N(x)=\frac{1}{N}\sum_{n=0}^{N-1}\delta_{S^nx},$$
	where $\delta_y$ is the Dirac measure on $y$ and $S$ is the canonical shift map $(Sx)_n=x_{n+1}$, for each $n \in \mathbb{N}$. Therefore $m_N(x) \in \mathcal{M}_1(X_{A_i}).$ 
\medskip
	 
	We thus get that the weak-star closure  $\mathcal{I}_S(x)$ of the set $\Big\{m_N(x)\Big\} $ is not empty.  We further define the square map $s$ on $X_3$ by
	$s(x)=(x_n^2)$ for any $x \in X_3$. 
	 
\begin{defn}
	 An infinite sequence $x=(x_n)_{n\in\mathbb{N}^*}\in X_3$ is said to be \emph{admissible} if its \emph{support}
	$\{n\in\mathbb{N}^*: x_n \neq 0\}$ is admissible. In the same way, a finite block $x_1\ldots x_N\in\{0,\pm 1\}^N$ is \emph{admissible} if 
	$\{n\in\{1,\ldots,N\}: x_n \neq 0\}$ is admissible. In the same manner, we define the admissible sets in $X_2$.
\end{defn}
\medskip

For each $i=1,2$, we denote by $X_{\mathcal{A}_i}$ the set of all admissible sequences in  $X_i$. Since a set is admissible if and only if each of its finite subsets is admissible, and a translation of a admissible set is admissible,   $X_{\mathcal{A}_i}$ is a closed and shift-invariant subset of $X_i$, \textit{i.e.} a subshift.
We further have that $\bm \mu^2$ is admissible, and 
$X_{\mathcal{A}_3}=s^{-1}(X_{\mathcal{A}_2}).$
\medskip

Let us notice that the previous notions has been extended to the so-called $\mathcal{B}$-free setting by el Abdalaoui-Lema\'{n}czyk-de-la-Rue in \cite{elabd-bfree}. Therein , the authors produced a dynamical proof of the Mirsky theorem on the pattern of $\bm \mu^2$ which assert that the indicator function of the square-free integers is generic for the Mirsky measure $\nu_{M}$, that is, $\bm \mu^2$ is generic for the push-forward measure of the Haar measure $\mu_h$ of the group $G=\prod_{p}\mathbb{Z}/p^2\mathbb{Z}$ under the map $\varphi~:~G \longrightarrow X_2$ defined by 
\begin{equation}
\label{eq:def_phi} \forall g \in G,~~~
\varphi(g) \stackrel{\textrm {def}}{=} \Bigl( f(T^ng)\Bigr)_{n\in\mathbb{N}^*},
\end{equation}
where $T$ is the translation by $\bm1 \stackrel{\textrm {def}}{=}=(1,1,\cdots)$ and 
$f$ is defined by 
\[
f(g)\stackrel{\textrm {def}}{=} \begin{cases}
0 \text{ if there exists $k\ge1$ such that $g_k=0$,}\\
1 \text{ otherwise.}
\end{cases}
\]
We thus get that $\bm \mu^2=(f(T^n\bm O)),$ $\bm O=(0,0,\cdots)$.
\medskip

We further have that for each measurable subset $C\subset X_2$,
$\nu_{M}(C)\stackrel{\textrm {def}}{=} \nu_h(\varphi^{-1}C)$. Then $\nu_{M}=\varphi(\mu_h)$ is shift-invariant, and it can be shown that $\nu_{M}$ is concentrated on $X_{\mathcal{A}_2}$. Moreover, the measurable dynamical system $(X_{\mathcal{A}_2},\nu_{M},S)$ is a factor of $(G,\nu_h,T)$. In particular, it is ergodic, and for any $\eta \in \mathcal{I}_S(\bm \mu)$, we have
$s\eta=\nu_M$, that is, $\eta(s^{-1}A)=\nu_M(A),$ for any Borel set of $X_{\mathcal{A}_2}$. For more details, we refer to \cite{elabd-bfree}.
\medskip

For any finite sets $A,B \subset \mathbb{N}$, we denote  by $F_{A,B}$ the function
$$F_{A,B}(x)=\Big(\prod_{a \in A}\pi_a(x)\Big) 
\Big(\prod_{b \in B}\pi_b(x)^2\Big),$$
where $\pi_n$ is the $n^{\rm{th}}$ canonical projection given by 
$\pi_n(x)=x_n$, $n \in \mathbb{N}$. Obviously, 
$F_{A,B} \in C(X_{A_3})$, where $C(X_{A_3})$ is the space of continuous
function on $X_{A_3}$. We further have 
$F_{A,B}=F_{A,B\setminus (A \cap B)}$, so we can assume always that 
$A$ and $B$ are disjoint.\\

Following W. Veech \cite{VeechNotes}, \cite{VeechNotes2}, we introduce also the notion of admissible measure. 
\begin{defn}A measure $m \in \mathcal{M}_1(X_{\mathcal{A}_3})$ is admissible if 
\begin{enumerate}[(i)]
	\item $Sm=m$, that is, $m(S^{-1}A)=m(A)$, for each Borel set $A \subset X_{\mathcal{A}_3}$.
	\item $s(m)=\nu_M$, and
	 $$\int_{X_{\mathcal{A}_3}}F_{A,B}(x) dm(x)=0,$$ for any $A \neq \emptyset.$ and $B$ finite sets of $\mathbb{N}$.
\end{enumerate}
\end{defn}

We are now able to state the main result.
\begin{thm}[Sarnak's Theorem on M\"{o}bius flow \cite{Sar}]\label{main1} There exists a unique admissible measure $\mu_M$ on $X_{\mathcal{A}_3}$ which is ergodic with the Pinsker algebra
	$$\mathcal{P}i_{\mu_M}=s^{-1}\Big(\mathcal{B}(\mathcal{A}_2)\Big).$$
Moreover, $\mathbb{E}(\pi_1|\mathcal{P}i_{\mu_M})=0.$	
\end{thm} 
Following W. Veech \cite{VeechNotes2}, the measure $\mu_M$ is called Chowla measure.
\medskip

\begin{rem}Furthermore, as pointed out by Veech, the existence of the putative "Chowla measure" does not depend on the Chowla conjecture.
\end{rem}
 
For the proof of our second main result, we need the following result due to T. Tao \cite{BlogTao}.
\begin{thm}[Tao's theorem on logarithmic and non-logarithmic Chowla conjectures\cite{BlogTao}]\label{Taoquasigeneric} 
Let $k$ be a natural number. Assume that the logarithmically averaged Chowla conjecture is true for $2k$. Then there exists a set ${{\mathcal N}}$ of natural numbers of logarithmic density $1$ such that
\[
\lim_{\overset{N \rightarrow \infty} 
	{N \in {\mathcal N}}} \frac{1}{N}\sum_{1}^{N} \lambda(n+h_1) \dots \lambda(n+h_k) = 0,
\]
for any distinct ${h_1,\dots,h_k}.$ 	
\end{thm} 
As a corollary, T. Tao obtain the following
\begin{cor}[Gomilko-Kwietniak-Lema\'{n}czyk's theorem \cite{lem}]
If Sarnak's conjecture holds then there exists a set ${{\mathcal N}}$ of natural numbers such that
for any $r\geq 0$, $1\leq a_1<\dots<a_r$, $i_s\in \{1,2\}$ not all equal to $2$, we have
\begin{equation}\label{gkl}
\frac{1}{N}\sum_{n\leq N}\bm \mu^{i_0}(n)\bm \mu^{i_1}(n+a_1)\cdot\ldots\cdot\bm \mu^{i_r}(n+a_r)
\tend[{N \in{{\mathcal N}} }]{N}{+\infty}0.
\end{equation}  
\end{cor}
Combining Sarnak's Theorem \ref{main1} with Tao's Theorem \ref{Taoquasigeneric}, we get the following
\begin{cor} Sarnak conjecture \ref{conj-sarnak} is equivalent to Chowla conjecture \ref{conj-chowla}.
\end{cor}
\begin{proof}The proof of the implication follows from Tao's Theorem \ref{Taoquasigeneric} and since the admissible measure is unique. Indeed, Sarnak's conjecture implies Chowla logarithmic conjecture along a subsequence of full density logarithmic. Therefore, by Tao's theorem \ref{Taoquasigeneric} combined with the admissibility of the measure, we get that Chowla conjecture holds. For the converse, there are several proofs by Sarnak \cite{lett}, Tao \cite{Taol}, Veech \cite{VeechNotes2},  and \linebreak el Abdalaoui-Kua\l ga-Przymus-Lema\'{n}czyk-de la Rue\cite{elabdal-chowla}.
\end{proof}
\setcounter{section}{3}
\section*{3. Proof of the main result.}
We start by proving the following proposition related to Hadamard matrix. For that,
let $E$ be a finite nonempty set, and $\P(E)$ be the set of subset of $E$. For any $A,B \in \P(E)$, put 
$$C(A,B)=(-1)^{|A \cap B|},$$
where $|.|$ is the cardinality function.  Therefore $C$ is a matrix of order $2^{|E|}$, we further have
\begin{prop}\label{Hadamard}With the notations above, 
	$$\det(C)=\begin{cases} 
	2^{|E|2^{|E|-1}}, \ &\textrm{if $|E|>1$}\\
	 -2, \ &\textrm{otherwise}.
	\end{cases}
	$$
Moreover, if the vector $(\nu(B))_{B \in \P(E)}$ satisfy 
$$\sum_{B \in \P(E)}C(A,B)\nu(B)= 
\begin{cases} a, \ &\textrm{if $A=\emptyset$}\\
 0, \ & \textrm{otherwise}.
\end{cases}
$$
Then $$\nu(B)=\frac{a}{2^{|E|}}.$$	
\end{prop}
\begin{proof} The proof of the first part of the proposition can be found in \cite[p.42]{Book}, but for the sake of completeness we include an alternative proof of it.
\medskip
	
 We start by recalling the Hadamard determinant inequality. Let $M$ be a matrix of order $n$ with real entries and columns $m_1,\cdots,m_n$, then 
	$$\big|\det(M)\big| \leq \prod_{j=1}^{n}\big\|m_j\big\|_2,$$
where $\|.\|_2$ is the usual Euclidean norm. Therefore, if all the entries are in the interval $[-1,1]$, we get 
$$\big|\det(M)\big| \leq n^{\frac{n}{2}},$$
with equality if and only if $M$ is a Hadamard matrix . For short and elementary proof of the Hadamard determinant inequality we refer to \cite [pp.40-41]{Book}, \cite{lang}.\\

We thus need to check that $C$ is a Hadamard matrix. For that, we proceed by induction. For $n=1$, the matrix is given by
$$C=\begin{pmatrix}
C(\emptyset,\emptyset) & C(\emptyset,\{1\}) \\
C(\{1\},\emptyset) & C(\{1\},\{1\})
\end{pmatrix}=
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
$$
Assume that the property is true for $n \geq 1$, and let 
$E_{n+1}=\{1,2,\cdots,n+1\}=E_n \cup \{n+1\}$. We assume that the subsets of 
$E_{n+1}$ are ordered as those of $E_n$. Notice that this does not affect our proof since the determinant does not depend upon any ordering of the elements of $2^{E}$. It follows that the resulting $2^{n+1}\times 2^{n+1}$ matrix has block form 
$$C_{n+1}=\begin{pmatrix}
C_n & C_n \\
C_n & -C_n
\end{pmatrix}.$$ We thus get, by Sylvester observation, that $C_{n+1}$ is a Hadamard matrix. For the second part, let $$
\left(
\begin{array}{c}
p\\
q
\end{array} \right)\in \mathbb{R}^{2^n} \times \mathbb{R}^{2^n}$$ such that
$$ C_{n+1} \left(
\begin{array}{c}
p\\
q\\
\end{array}
\right)=a. \delta_{\emptyset}(A).$$
Then, for $n=1$, we have
$$\begin{cases}
p+q=a,\ &\textrm{if $A=\emptyset$}\\
p-q=0,\ & \textrm{if not}.
\end{cases}$$
Obviously, we get $p=q=\frac{a}{2}.$ Assume that the property is true for $n$. Then 
$$p+q=\Big(\frac{a}{2^n},\cdots,\frac{a}{2^n}\Big).$$
Moreover, since $\det(C_n) \neq 0$, we get $p=q$, that is,
$$p=q=\Big(\frac{a}{2^{n+1}},\cdots,\frac{a}{2^{n+1}}\Big).$$
The proof of the lemma is complete.
\end{proof}
 
For the proof of the Sarnak's theorem \ref{main1}, we need also to characterize the Chowla measure. For that, let us put
$${{\mathcal{Q}}r}_n=\Big\{x \in X_{2}|  \textrm{~~supp($x$)~is~~admissible}  \Big\},\quad \quad n >0$$  
and 
$$C(x)=\bigcap_{j=1}^{n}\Big\{y \in X_{\mathcal{A}_2}| \pi_j(x)=x_j\Big\}.$$
Define the partition ${{\mathcal{P}}r}_n$ by
$${{\mathcal{P}}r}_n=\Big\{C(x)| x \in {\mathcal{Q}}r_n \Big\}.$$
If follows that any $C(x) \in {\mathcal{P}}r_n$ admits a partition into $2^{|\textrm{supp}(x)|}$ "cylinder" since 
$$s^{-1}(C(x)) \subset s^{-1}\big(X_{\mathcal{A}_2}\big)=X_{\mathcal{A}_3}.$$ 
More precisely, if $A \subset \textrm{supp}(x)$, then $A$ can be seen as a element $y(A) \in X_{\mathcal{A}_2}$. We thus denote by $C(x,y(A))$ 
the subset of $X_{\mathcal{A}_3}$ such that $z \in C(x,y(A))$ if and only if 
the first $n$ coordinates of $z$ are $-1$ on $A$, $1$ on $\textrm{supp}(x)\setminus A$ and $0$ on $[1,n]\setminus \textrm{supp}(x)$. This allows us to see that
$$s^{-1}(C(x))=\bigcup_{A \subseteq \textrm{supp}(x) }C(x,y(A)).$$
Now, for any $A \subset \textrm{supp}(x),$ put
$$G_{A,\textrm{supp}(x)\setminus A}=
F_{A,\textrm{supp}(x)\setminus A} \prod_{c \in [1,n] \setminus \textrm{supp}(x)}(1-\pi_c(y)^2).$$
It is straightforward that $G_{A,\textrm{supp}(x)\setminus A} \in C(X_{\mathcal{A}_3})$. Moreover, 
\begin{eqnarray}\label{expand}
G_{A,\textrm{supp}(x)\setminus A}|s^{-1}\big(C(x)\big)=
F_{A,\textrm{supp}(x)\setminus A}|s^{-1}\big(C(x)\big)
\end{eqnarray}
and  $G_{A,\textrm{supp}(x)\setminus A}$ is identically null on $X_{A_3}\setminus s^{-1}\big(C(x)\big).$\\

Expand the product in the definition of $G_{A,\textrm{supp}(x)\setminus A}$, we get 
$$ G_{A,\textrm{supp}(x)\setminus A} =
\sum_{B \subset [1,n]\setminus \textrm{supp}(x)}
(-1)^{|B|} F_{A,\textrm{supp}(x)\setminus A \cup B}.$$
Now, let $m$ be an admissible measure and assume that $A \neq \emptyset$. Then 
\begin{eqnarray}\label{ve:1}
\int_{X_{\mathcal{A}_3}} G_{A,\textrm{supp}(x)\setminus A} m(dz)
&=&
\int_{s^{-1}C(x)} F_{A,\textrm{supp}(x)\setminus A}(z) m(dz)\\
&=& 0.
\end{eqnarray}

This gives, for $A \subset \textrm{supp}(x)$ and $A \neq \emptyset$,
$$\sum_{B \subset \textrm{supp}(x)}
(-1)^{|A \cap B|} m(C(x,y(B))=0,$$
since $F_{A,\textrm{supp}(x)\setminus A}$ is constant on each "cylinder" set
$C(x,y(B))$ with the constant value equal to 
$(-1)^{|A \cap B|}.$ \\

We proceed now to evaluate the expression when $A=\emptyset.$ Since 
$sm=\nu_M$, we obtain
$$\nu_M(C(x))=\sum_{B \subset \textrm{supp}(x)} m(C(x,y(B))=0.$$
This combined with Proposition \ref{Hadamard} yields that 
for any $C(x) \in {{\mathcal{P}}r}_n$, for any $B \subset \textrm{supp}(x)$, we have
$$m(C(x,y(B)))=\frac{\nu_M(C(x))}{2^{|\textrm{supp}(x)|}}.$$
Summarizing, we conclude that $m$ is completely determined on the partition
${{\mathcal{P}}r}_n$, i.e., if an admissible measure exists, then it is unique.\\

We proceed now to the proof  of Sarnak 's theorem \ref{main1}.
\medskip

Consider the canonical dynamical system $\big(X_{\mathcal{A}_2} \times \{\pm 1\}^\mathbb{N}, S \times S, \nu_M \otimes m_B(\frac12,\frac12)\big),$ where $m_B(\frac12,\frac12)$ is 
the Bernoulli measure. Therefore, by Furstenberg theorem (Proposition I.3 in \cite{Fur}), the dynamical system $\big(X_{\mathcal{A}_2} \times \{\pm 1\}^\mathbb{N}, S \times S, \nu_M \otimes m_B(\frac12,\frac12)\big),$ is ergodic. We further have, by Theorem 18.13 from \cite[p.325]{Elie}, that the Pinsker algebra satisfies
$$\mathcal{P}i_{\nu_M \times m_B(\frac12,\frac12)}=
\mathcal{P}i_{\nu_M} \otimes \mathcal{P}i_{m_B(\frac12,\frac12)}=
\mathcal{B}(X_{\mathcal{A}_2}) \times \big\{\emptyset, \{\pm 1 \}^\mathbb{N}\big\}$$
up to the null set with respect to $\nu_M \otimes m_B(\frac12,\frac12)$.
\medskip

Now we define a coordinate-wise multiplicative map $\Pi : X_{\mathcal{A}_2} \times  \{\pm 1 \}^\mathbb{N} \longrightarrow X_{\mathcal{A}_3}$ by 
$$\Pi(x,\omega)=x.\omega,$$
that is, 
$$\pi_n(\Pi(x,\omega))=x_n \omega_n,\quad \quad n>0.$$
Therefore the dynamical system $(X_{\mathcal{A}_3},S,\mu_M=\Pi(\nu_M \otimes m_B(\frac12,\frac12)))$, where $\mu_M$ is the push-forward measure under $\Pi$, satisfies
\begin{itemize}
	\item $s\mu_M=\nu_M,$ and 
	\item For any $A \neq \emptyset$, we have
	$$\int_{X_{\mathcal{A}_3}}F_{A,B}(z) \mu_M(dz)=0.$$
\end{itemize}
Whence, $\mu_M$ is admissible, ergodic and $\mathcal{P}i_{\mu_M}=s^{-1}\mathcal{B}(X_{\mathcal{A}_2})$ up to $\mu_M$ null set. This last fact follows from the following  
$$\Pi^{-1}(\bigcap_{n}S^{-n}\mathcal{B}(X_{A_3}))\subseteq 
\bigcap_{n}(S \times S)^{-n}\mathcal{B}(X_{A_2} \times \{\pm 1\}^\mathbb{N}).$$
To finish the proof, we need only to notice that 
$\pi_1=F_{\{1\}, \emptyset}$, and for any finite set $B \subset \mathbb{N}$,
$$\int_{X_{A_3}}\pi_1(y) F_{\emptyset, B}(y) \mu_M(dy)=\int_{X_{A_3}}F_{\{1\},B}(y) \mu_M(dy).$$
We thus conclude that 
$$\mathbb{E}(\pi_1|\mathcal{P}i_{\mu_M})=0,$$ 
up to $\mu_M$ null sets, since the family $\{F_{A,B}\}$ are dense in $\mathbb{C}(X_{\mathcal{A}_3}),$ by the classical Stone-Weierstrass theorem.
\begin{que} Let $O(\bm \mu) \subset X_{\mathcal{A}_3}$ be the orbit closure of $\bm \mu$ 
under the left shift $S$. Do we have that $O(\bm \mu)=X_{\mathcal{A}_3}$?
\end{que}  
\begin{Thanks}The author would like to thanks Jean-Paul Thouvenot for the simulating discussion on Sarnak and Chowla conjectures. He is indebted to W. Veech for sending him his notes\footnote{W. Veech in his letter indicated to me that there is only four persons in the world who has a copy of his notes including me.} and for many e-simulating discussions on the subject. The author Would like also to express his thanks to Mahendra Nadkarni, Mahesh Nerurka and Forni Giovanni for the discussion  or e-discussion on the subject durant the prepartion of this revised version. 
\end{Thanks}
