# Solving the Direct (or Inductive) Limit

Suppose we wish to compute the direct limit of the following.

$$G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, \cdots,$$

where each $G=\mathbb{Z}^d$, $d\in\mathbb{N}$, and the bonding map $M$ is given by a $d\times d$ square matrix. For example, how to find the direct limit, if $G=\mathbb{Z}^2$ and the bonding map is given by $M=\left[\begin{array}{cr} 0 & -2 \\ 3 & 6 \end{array}\right]$?

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Really just the start of an answer.

In the case $d=1$ and $M=m\neq 0\in \mathbb Z$, the result is the group $\mathbb Z[\frac{1}{m}]$, the set of rationals of the form $n/m^k$ for some $n\in \mathbb Z$ and $k\in \mathbb N$.

Analogous with the case $d=1$, if $M$ is $1-1$, this is the set of vectors in $\mathbb Q^d$ which can be written in the form $M^{-n}g$ for some $n>0$ and $g\in G$. It is a subset of $(\mathbb Z[\frac{1}{\det M}])^d$. It seems to depend on the eigenvalues of $M$.

For example, a simple case of $M$ being diagonal with $M_{ii}=d_i$ would give the limit $\prod_i \mathbb Z[\frac{1}{d_i}]$. If $M$ is diagonalizable with integer eigenvalues $d_i$, is this still true? I think so. What if it is diagonalizable but with arbitrary complex eigenvalues? I have no idea.

What if $M$ is non-diagonalizable? Non-invertible?

For the specific $M$, we can see that $M^2-6M+6=0$, so $M^{-2}(M-1) = \frac{1}{6}I$. That means that if $(x,y)$ is in the limit, then $\frac{1}{6}(x,y)$ is in the limit, so the limit is all of $(\mathbb Z[\frac{1}{6}])^2$.

This will always be the case for $2\times 2$ matrices whose trace is a multiple of its determinant - then the limit is $(\mathbb Z[\frac{1}{\det M}])^2$.

The same is true for $d\times d$ matrices, $M$, if all the coefficients of the characteristic polynomial, $p(x)=\det (xI-M)$, other than the degree $d$ term, are divisible by $\det M$. You lighten this restriction: if all the coefficients are divisible by $r(\det M)$, where $r(n)$ is the product of the distinct prime factors of $n$, then the limit is $(\mathbb Z[\frac{1}{\det M}])^d$.

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Thank you very much. Now I will study this. Cheers! – Eric Nov 25 '11 at 13:28
Hi there @Thomas Andrews! I am reading this again and I am really thankful for your answer. Now, I have a better "grasp" of things. Just a question though, do you have a reference, which you can recommend? Most of the materials I see on direct limit rely on the formal definition and basic properties of the direct limit, but not really on actual examples. Thanks! – Eric Nov 22 '13 at 19:12

I don't have time to check the details right now, but I think the answer is as follows for the case when $M$ is nonsingular.

Let $V_n = M^{-n}\mathbb{Z}^d\subseteq \mathbb{Q}^d$ for $n\geq 0$. Then $V_n\subseteq V_{n+1}$: if $A\in V_n$, then $A = M^{-n}B$ for $B\in\mathbb{Z}^d$, so $A = M^{-n}B = M^{-(n+1)}MB\in V_{n+1}$ because $M:\mathbb{Z}^d\to\mathbb{Z}^d$. The direct limit is then $\displaystyle \bigcup_n V_n$, the set of all rational vectors which multiplication by $M$ eventually maps to integer vectors.

The idea is that each $V_n$ is isomorphic to $G$, but by replacing the $n^{\text{th}}$ copy of $G$ with $V_n$, we replace the map $M$ from the $n^{\text{th}}$ to $(n+1)^{\text{st}}$ copy of $G$ with the inclusion map. A direct limit of sets where each map is an inclusion of subsets is just a union.

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Thanks, I will look into this. Cheers! – Eric Nov 25 '11 at 13:28