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Proof. Since $\lvert \overline\varphi \rvert < 1$, we get $\pm\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$. If $t = a+n\varphi$, we necessarily have $n \neq 0$, for otherwise $t$ and therefore $x$ is $0$. Wihout loss of generality, we may assume $n > 0$. By the previous lemma (and the discussion after), we get $t \in T$, so $x = \pm\varphi^{-n} \cdot t$ as desired. $\square$

Proof. We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we can produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick a generator in $T$ for each prime ideal. Since all elements of $T$ and all chosen representatives are positive, we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$

Proof. If $t = \pm\varphi^n \cdot x$, then $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$ since $\lvert \overline\varphi \rvert < 1$. If $t = a+n\varphi$, we necessarily have $n \neq 0$, for otherwise $t$ and therefore $x$ is $0$. Wihout loss of generality, we may assume $n > 0$. By the previous lemma (and the discussion after), we get $t \in T$, so $x = \pm\varphi^{-n} \cdot t$ as desired. $\square$

Proof. We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick the generator for each prime ideal to be in $T$. Since all elements of $T$ and all chosen representatives are positive, we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$

The main reference is the 2-page paper [Arnoux, 1989], explaining the relation to $\mathbf Z[\varphi]$. In addition, [Zhuravlev, 2007] notes that every nonzero element of $\mathbf Z[\varphi]$ can be written (non-uniquely) as $\varphi^{-n}t$ for $t \in T$ and $n \in \mathbf N$, giving the promised computation of $T^{\text{gp}}$ of the corollary below.

Definition. Define the subset $Z \subseteq t^2\mathbf Z[t]$ of elements of the form $P(t)=\sum_{i=2}^r a_it^i$ with all $a_i \in \{0,1\}$ and $a_ia_{i+1} = 0$ for all $i$. We remove $0$ from this set and add $1$, since this will be our multiplicative unit. Let $T \subseteq \mathbf Z[\varphi]$ be the image of $Z$ under $f$. Then Zeckendorf's theorem shows that $f$ and $g$ give bijections $$Z \stackrel\sim\to T \stackrel\sim\to \mathbf N.$$ By definition, the map $h$ has the property $h(xy) = h(x)h(y)$ if $x,y \in Z$. But $Z$ is not closed under multiplication, so this doesn't produce a monoid isomorphism $h \colon Z \stackrel\sim\to \mathbf N$. The key point is:

Lemma [Arnoux, 1989]. The set $T$ is given by the elements $t=a+n\varphi$ with $a,n \in \mathbf N$ such that $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$. The map $g^{-1}$ is given by $n \mapsto a_n + n\varphi$, where $a_n = \lfloor (n+1)\tfrac{-1+\sqrt{5}}{2} \rfloor$ is Hofstadter's $G$-sequence (OEIS A005206).

See Lemmas 2 and 3 in Arnoux. The final statement is not there, but can easily be deduced from the first. This description shows immediately that $T$ is closed under multiplication as $(\varphi-2,\varphi-1) \subseteq (-1,1)$.

Lemma. Every positive element $x \in \mathbf Z[\varphi] \cap \mathbf R_{>0}$ can be written as $\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.

The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include an argument here.

Proof. Let $M \colon \mathbf Z[\varphi] \to \mathbf Z[\varphi]$ be multiplication by $\varphi$, so $M$ has eigenvalues $\lambda_1 = \varphi$ and $\lambda_2 = \overline\varphi$. If $x = a + b\varphi$ with $a,b \in \mathbf Z$, then we can write the vector $\big(\begin{smallmatrix}a \\ b\end{smallmatrix}\big) \in \mathbf R^2$ as $\mathbf v_1 + \mathbf v_2$ for real eigenvectors $\mathbf v_1,\mathbf v_2$ of $M$ with eigenvalues $\lambda_1$ and $\lambda_2$ respectively. Since $\lvert \lambda_1 \rvert > 1 > \lvert \lambda_2 \rvert$, we have $M^nx - \varphi^n \mathbf v_1 \to 0$ as $n \to \infty$. But $\mathbf v_1$ is given by $$\mathbf v_1 = c\begin{pmatrix}\varphi-1 \\ 1\end{pmatrix}$$ for some $c \in \mathbf R$, which is nonzero since $\mathbf v_2 \not\in \mathbf Z^2$. Since $x>0$, we conclude that $t:=\varphi^n x$ has positive coordinates for all $n \gg 0$. On the other hand, we have $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$ since $\lvert \overline{\varphi}\rvert < 1$. Taking $n$ large enough therefore gives $t \in T$ by the previous lemma, so $x = \varphi^{-n} \cdot t$ as desired. $\square$

Proof. We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we can produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick a generator in $T$ for each prime ideal. Since all elements of $T$ and all chosen representatives are positive (for the usual real embedding), we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$

The main reference is the 2-page paper [Arnoux, 1989], explaining the relation to $\mathbf Z[\varphi]$. In addition, [Zhuravlev, 2007] notes that every nonzero element of $\mathbf Z[\varphi]$ can be written (non-uniquely) as $\pm\varphi^{-n}t$ for $t \in T$ and $n \in \mathbf N$, giving the promised computation of $T^{\text{gp}}$ of the corollary below.

Definition. Define the subset $Z \subseteq t^2\mathbf Z[t]$ of elements of the form $P(t)=\sum_{i=2}^r a_it^i$ with all $a_i \in \{0,1\}$ and $a_ia_{i+1} = 0$ for all $i$. We remove $0$ from this set and add $1$, since this will be our multiplicative unit. Let $T \subseteq \mathbf Z[\varphi]$ be the image of $Z$ under $f$. Then Zeckendorf's theorem shows that $f$ and $g$ give bijections $$Z \stackrel\sim\to T \stackrel\sim\to \mathbf N.$$ By definition, the map $h$ has the property $h(xy) = h(x)h(y)$ for all $x,y \in Z$. But $Z$ is not closed under multiplication, so this doesn't produce a monoid isomorphism $h \colon Z \stackrel\sim\to \mathbf N$. The key point is:

Lemma [Arnoux, 1989]. The set $T \setminus \{1\}$ is given by the elements $t=a+n\varphi$ with $a,n \in \mathbf Z_{>0}$ such that $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$. The map $g^{-1}$ is given by $n \mapsto a_n + n\varphi$, where $$a_n = \left\lfloor (n+1)\tfrac{-1+\sqrt{5}}{2} \right\rfloor$$ is Hofstadter's $G$-sequence (OEIS A005206).

See Lemmas 2 and 3 in Arnoux. The final statement is not there, but can easily be deduced from the first. This shows that $T$ is closed under multiplication as $(\varphi-2,\varphi-1) \subseteq (-1,1)$. In addition, it gives the clean formula $$n \circ m = nm + na_m + ma_n.$$ Note that in the first statement, we don't need the assumption $a > 0$. Indeed, if $a \leq 0$ we get $\overline{\!\ t\ \!} = a+n\overline\varphi \leq \overline\varphi < \varphi-2$ since $\overline\varphi < 0$ and $n \geq 1$.

Lemma. Every element $x \in \mathbf Z[\varphi]$ can be written as $\pm\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.

Since $T$ and $\varphi$ are positive, the sign agrees with the sign of $x$. The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include an argument here.

Proof. Since $\lvert \overline\varphi \rvert < 1$, we get $\pm\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$. If $t = a+n\varphi$, we necessarily have $n \neq 0$, for otherwise $t$ and therefore $x$ is $0$. Wihout loss of generality, we may assume $n > 0$. By the previous lemma (and the discussion after), we get $t \in T$, so $x = \pm\varphi^{-n} \cdot t$ as desired. $\square$

(Notational note: what Zhuravlev calls $\delta(n)$ is related to my $g^{-1}(n)$ via $-\delta(n)/\varphi = \overline{g^{-1}(n)}$. Zhuravlev's Fibonacci sequence is off by $1$ compared to Knuth.)

Proof. We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we can produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick a generator in $T$ for each prime ideal. Since all elements of $T$ and all chosen representatives are positive, we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$

Lemma. Every nonzero element $x \in \mathbf Z[\varphi]$ can be written as $\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.

The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include the (very easy) argument here.

Proof. Since $\lvert \overline{\varphi}\rvert < 1$, there exists $n \in \mathbf N$ such that $t:=\varphi^n x$ satisfies $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$. Since both $x$ and $\varphi^n$ have nonnegative coefficients, so does $\varphi^n x$. Thus the lemma above gives $t \in T$, so $x = \varphi^{-n} \cdot t$ as desired. $\square$

Lemma. Every positive element $x \in \mathbf Z[\varphi] \cap \mathbf R_{>0}$ can be written as $\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.

The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include an argument here.

Proof. Let $M \colon \mathbf Z[\varphi] \to \mathbf Z[\varphi]$ be multiplication by $\varphi$, so $M$ has eigenvalues $\lambda_1 = \varphi$ and $\lambda_2 = \overline\varphi$. If $x = a + b\varphi$ with $a,b \in \mathbf Z$, then we can write the vector $\big(\begin{smallmatrix}a \\ b\end{smallmatrix}\big) \in \mathbf R^2$ as $\mathbf v_1 + \mathbf v_2$ for real eigenvectors $\mathbf v_1,\mathbf v_2$ of $M$ with eigenvalues $\lambda_1$ and $\lambda_2$ respectively. Since $\lvert \lambda_1 \rvert > 1 > \lvert \lambda_2 \rvert$, we have $M^nx - \varphi^n \mathbf v_1 \to 0$ as $n \to \infty$. But $\mathbf v_1$ is given by $$\mathbf v_1 = c\begin{pmatrix}\varphi-1 \\ 1\end{pmatrix}$$ for some $c \in \mathbf R$, which is nonzero since $\mathbf v_2 \not\in \mathbf Z^2$. Since $x>0$, we conclude that $t:=\varphi^n x$ has positive coordinates for all $n \gg 0$. On the other hand, we have $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$ since $\lvert \overline{\varphi}\rvert < 1$. Taking $n$ large enough therefore gives $t \in T$ by the previous lemma, so $x = \varphi^{-n} \cdot t$ as desired. $\square$