7
$\begingroup$

Suppose we have the following sequence of $\mathbb{Z}$-modules $G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, G\,\,\overset{M}{\longrightarrow}\,\, \cdots,$

where each $G:=(\frac{\mathbb{Z}}{2\mathbb{Z}})^5\oplus\mathbb{Z}^5$, and the bonding map $M$ is given by the square matrix \begin{eqnarray*} M:=\left( \begin{array}{cccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 3 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 & 2 \\ \end{array} \right). \end{eqnarray*} For example, if $t=(t_1,\ldots,t_{10})\in \mathbb{Z}^{10}$, then the element $([t_1]_2,\ldots,[t_5]_2,t_6,\ldots,t_{10})\in G$ is mapped to $([(Mt)_1]_2,\ldots,[(Mt)_5]_2,(Mt)_6,\ldots,(Mt)_{10})$. What is the direct limit of such sequence?

  • 1
    If the lower right corner of your matrix is diagonalizable as Lukas suggests, then it is not difficult to see that your direct limit is isomorphic to $\mathbb Z[1/6]\oplus \mathbb Z^4$. Use the fact that the direct limit of a system of the form $\mathbb Z\overset{n}{\to}\mathbb Z\overset{n}{\to}\dots \overset{n}{\to}\mathbb Z\overset{n}{\to}\dots$ is isomorphic to $\mathbb Z[1/n]$, where $\mathbb Z[1/n]$ is the subring of $\mathbb Q$ generated by $1/n$.2012-12-03

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