**1. Direct limit - definition and construction**

The general definition of direct limit can be found on wikipedia. Here, I will focus on direct limits of partially ordered abelian groups (poa-groups). Recall that a poa-group is simply an abelian group $(G,+,0)$ equipped with a partial order $\le$ which is translation invariant: $a \le b$ implies $a+g \le b+g $ for all $g$. The subset $G_+$ of group elements $a \ge 0$ is the

*positive cone*of $G$. A morphism $f : G \rightarrow H$ of poa-groups is a group homomorphism that agrees with the partial orders: $a \le b$ in $G$ implies $f(a) \le f(b)$ in $H$.

The ingredients for a direct limit are: a directed set $(I,\le)$ of ``indices'', a family $(G_i)_{i \in I}$ of poa-groups indexed by $I$, and a family $f_{i,j} : G_i \rightarrow G_j$ of morphisms for every pair $i \le j$ in $I$. These morphisms have to satisfy a compatibility condition, namely, for any $i \le j \le k$ in $I$, $f_{i,k} = f_{j,k} \circ f_{i,j}$. This data forms what is called a

*direct system of poa-groups*. Then there is a universal poa-group $L$ and morphisms $\phi_i : G_i \rightarrow L$ such that for every pair $i \le j$ in $I$, $\phi_j = f_{i,j} \circ \phi_i$. The term universal refers to the usual property in category theory of ``being the most general entity satisfying the given constraints''. Thanks to this universality, the poa-group $L$ is unique up to isomorphism, and is usually denoted by

$$

L = \lim G_i \xrightarrow{f_{i,j}} G_j

$$

Because I want to perform concrete computations, I won't insist on the universal characterization. Instead, I present a (classical) explicit construction of $L$. We first take the disjoint union $U = \bigsqcup_{i \in I} G_i$. As a set, $U$ is made of elements $(i,a)$ with $i \in I$ and $a \in G_i$. We now define an equivalence relation: $(i,a) \sim (j,b)$ iff there exists $k \ge i,j$ such that $f_{i,k}(a) = f_{j,k}(b)$. Intuitively, two elements are equivalent iff they

*eventually*agree. We denote by $[i,a]$ the equivalence class of $(i,a)$. The poa-structure is defined as follows:

- The zero element is defined by $0 = [i,0]$ for any $i \in I$. The choice of $i$ does not matter.
- The addition is defined by $[i,a] + [j,b] = [k, f_{i,k}(a) + f_{j,k}(b)]$ for some $k \ge i,j$. The choice of the representatives and $k$ does not matter.
- The partial order is defined by: $[i,a] \le [j,b]$ iff for some $k \ge i,j$, $f_{i,k}(a) \le f_{j,k}(b)$ in $G_k$. Again, the choice of the representatives and $k$ does not matter.

In the following examples, we will consider an even more restricted settings. Indeed, any poa-group morphism $f : G \rightarrow G$ yields a direct system $f_{i,j} : G_i \rightarrow G_j$ where the directed set is the set of natural integers $I = \mathbb{N}$ (with the usual order), each $G_i$ is a copy of $G$, and $f_{i,j} = f^{j-i}$ is the $j-i$-th iterate of $f$.By ``computing the direct limit'', I mean finding a poa-group isomorphic to the direct limit, but which is easier to work with.

**2. Dyadic rationals**

We consider the direct limit generated by the multiplication by $2$ on integers, denoted by $\mathbb{Z} \xrightarrow{2} \mathbb{Z}$.

$$

L = \lim \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \dots

$$

The intuition goes as follows. By the definition given above, we have $[i,a] = [i+1,2 \cdot a]$, i.e., each time we move one step forward, we multiply the data by $2$. Therefore, intuitively, moving one step backward amounts to dividing by $2$. Abusing the notations, we could write $[i,a] = [i-1,a/2]$, and thus $[i,a] = [0,a/2^i]$. This suggests considering the poa-group $\mathbb{Z}[\frac{1}{2}]$ of dyadic rationals:

- Its elements are the fractions $\frac{a}{2^i}$ in $\mathbb{Q}$ with $a \in \mathbb{Z}$ and $i \in \mathbb{Z}$.
- The poa-structure is the one induced by $\mathbb{Q}$.

$$

\phi : \frac{a}{2^i} \mapsto [i,a]

$$

We show that $\phi$ is an isomorphism of poa-groups. First, it is well defined: if $a/2^i = b/2^j$ in $\mathbb{Z}[\frac{1}{2}]$, then for any $k \ge i,j$, we have $2^{k-i} \cdot a = 2^{k-j}\cdot b$, whence $[i,a] = [j,b]$. Second, $\phi$ agrees with addition since

$$

\frac{a}{2^i} + \frac{b}{2^j} = \frac{2^{k-i}\cdot a + 2^{k-j}\cdot b}{2^k}

$$

Third, $\phi$ agrees with the partial order since

$$

\frac{a}{2^i} \le \frac{b}{2^j} \Leftrightarrow \frac{2^{k-i}\cdot a}{2^k} \le \frac{2^{k-j}\cdot b}{2^k}

$$

Finally, $\phi(a/2^i) = 0 = [i,0]$ implies that $a = 0$. Since $\phi$ is obviously surjective, $\phi$ is an isomorphism of poa-groups.

**3. ``Fibonacci'' integers**

I do not know if this name is appropriate, but it turns out that the construction below is related to the famous Fibonacci sequence; yet, I will not cover this topic here.

Consider the poa-group $\mathbb{Z}^2$ with $(a,b) \le (c,d)$ iff $a \le b$ and $c \le d$, and the multiplication $\mathbb{Z}^2 \xrightarrow{A} \mathbb{Z}^2$ by the matrix

$$

A = \left(\begin{array}{cc}

1 & 1 \\

1 & 0 \\

\end{array}\right)

$$

We compute the direct limit $L = \lim \mathbb{Z}^2 \xrightarrow{A} \mathbb{Z}^2 \dots$. As in the previous section, the idea is consider the element $[k,u]$ as the informal element $u/A^k$. To give a coherent meaning to this element, we ``notice'' the following. Let $\tau = (1+\sqrt{5})/2$ denote the golden mean. We have the $\tau^2 = \tau + 1$. Therefore, the group $G = \mathbb{Z}[\tau]$ of integral combinations of powers of $\tau$ decomposes as $G = \mathbb{Z}\tau + \mathbb{Z}$. If we identify the vectors $(1,0)$ and $(0,1)$ in $\mathbb{Z}^2$ with $\tau$ and $1$ in $\mathbb{Z}[\tau]$ respectively, then multiplication by $A$ on $\mathbb{Z}^2$ translates into multiplication by $\tau$ in $\mathbb{Z}[\tau]$. Also, the order structure on $G$ is defined by: $\tau\cdot a+ b \le \tau\cdot c + d$ iff $a \le c$ and $b \le d$. By the matrix form, we see that multiplication by $\tau$ agrees with this order: $u \le v$ implies $\tau\cdot u \le \tau\cdot v$. Thanks to this trick, the direct limit can be written (is isomorphic to)

$$

L = \lim G \xrightarrow{\tau} G \dots

$$

and we can compute it as in the case of dyadic integers. We consider the poa-group defined as follows:

- Its elements are $\frac{\tau\cdot a + b}{\tau^k}$ (the quotient being taken in $\mathbb{R}$) with $a,b,k \in \mathbb{Z}$.
- Its poa-structure is the one induced by $\mathbb{R}$.
- Since $1/\tau = \tau-1$, this poa-group is actually $\mathbb{Z}[\tau] = \tau\mathbb{Z} + \mathbb{Z}$ with the order induced by the one of $\mathbb{R}$. Note that it is important to distinguish $G$ and $\mathbb{Z}[\tau]$ although they have the same underlying group structure. The only difference is between their order relations.

$$

\frac{\tau \cdot a + b}{\tau^k} \mapsto [k, \tau\cdot a + b]

$$

As in the case of dyadic integers, we verify that $\phi$ is a poa-isomorphism. First, it is well defined: if $(\tau\cdot a + b)/\tau^k = (\tau\cdot c + d)/\tau^l$, then $\tau^{m-k}\cdot(\tau\cdot a + b) = \tau^{m-l}\cdot(\tau\cdot c + d)$ for some $m \ge k,l$, and $[k,\tau\cdot a+b] = [l,\tau\cdot c + d]$. Second, $\phi$ agrees with addition since

$$

\frac{\tau\cdot a + b}{\tau^k} + \frac{\tau\cdot c + d}{\tau^l} = \frac{\tau^{m-k}\cdot(\tau\cdot a + b) + \tau^{m-l}\cdot(\tau\cdot c + d)}{\tau^m}.

$$

The fact that $\phi$ agrees with the order is less trivial. Since $\phi$ agrees with addition, it suffices to check that $\phi$ sends the positive cone of $\mathbb{Z}[\tau]$ to the positive cone of $L$. This amounts to prove that if $\tau\cdot a + b \ge 0$ in $\mathbb{R}$ with $a,b \in \mathbb{Z}$, then there exists $k\in \mathbb{Z}$ and two non-negative integers $a',b' \in \mathbb{N}$ such that

$$

\tau\cdot a + b = \frac{\tau\cdot a' + b'}{\tau^k} ~~~~(\bigstar)

$$ To prove this, we shall turn back to the matrix form. In the base $(\tau,1)$, multiplication by $\tau$ is modeled by the matrix $A$. Consider the action of $A$ on the plane $\mathbb{R}^2$. Let $\Delta$ denote the line $\tau\cdot x + y = 0$, and $\Delta^+$ the half-plane $\tau\cdot x + y \ge 0$. The proof of $(\bigstar)$ amounts to show that iterating $A$ on any point of $\Delta^+$ eventually leads to a point of the positive quadrant $\{(x,y)~|~ x,y \ge 0\}$.

Basic matrix algebra shows that the eigenvalues of $A$ are $\tau$ and $\overline{\tau} = (1-\sqrt{5})/2$. The eigenspace associated with $\tau$ is $\nabla ~:~ \overline{\tau}\cdot x + y = 0$, while the eigenspace associated with $\overline{\tau}$ is $\Delta ~:~ \tau\cdot x + y = 0$. We have $|\tau| > 1$ and $|\overline{\tau}| < 1$, so $A$ dilates $\nabla$, while $A$ contracts $\Delta$. By Figure 1, we see that iterating $A$ sufficiently enough moves any point of the half-plane into the positive quadrant.

Fig. 1 - Action of $A$ |

Therefore, we just showed that the direct limit $L$ is isomorphic to $\mathbb{Z}\tau + \mathbb{Z}$, with positive cone $\{a\cdot\tau + b \ge 0 ~|~ a,b \in \mathbb{Z}\}$.

pb