Friday, November 20, 2015

Explicit computations of some direct limits of ordered groups

The basic definition of direct limits of groups is not so difficult. But concrete computations of these turn out not to be so easy for beginners (like me). I present in this post some computations of this kind.

    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 $\mathbb{Q}$.
We now define $\phi : \mathbb{Z}[\frac{1}{2}] \rightarrow L$ by
$$
  \phi : \frac{a}{2^i} \mapsto [i,a]
$$
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.
We then define the function $\phi : \mathbb{Z}[\tau] \rightarrow L$ by
$$
    \frac{\tau \cdot a + b}{\tau^k} \mapsto [k, \tau\cdot a + b]
$$
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

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