On the one hand, the following conjecture seems reasonable, but on the other hand it doesn't seem natural because some objects are being dualised while others are not. I would appreciate if anyone could either confirm the conjecture or provide a counterexample.

Let $G=\mathbb{Z}^d$ and suppose $A_i$ and $B_i$ are sequences of $d\times d$ integral matrices.

Conjecture. If $$\lim_{\rightarrow}(G,A_i)\cong\lim_{\rightarrow}(G,B_i)$$ then $$\lim_{\rightarrow}(G,A_i^T)\cong\lim_{\rightarrow}(G,B_i^T).$$

Here $\displaystyle{\lim_{\rightarrow}}$ is the direct limit of abelian groups over a diagram of shape $\mathbb{N}$ and $M^T$ is the transpose of the matrix $M$.

My initial idea was to rely on the universal property of the colimit, but it's not entirely clear how to interpret the transpose in this setting, especially as it is normally viewed as a contravariant operation on maps, but we are still composing them in the previous (non-reversed) order. I'm not a group theorist so I don't know if this kind of question is one which is generally asked or whether there is any machinery to tackle questions like this.

As far as how flexible I am with regard to the assumptions, I'm happy to consider just the case $d=2$ and for $\{A_i\mid i\in\mathbb{N}\}$ and $\{B_i\mid i\in\mathbb{N}\}$ to be finite sets of non-singular matrices, if these additional conditions force the conjecture to be true.

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    $\begingroup$ Exactly what do you intend to mean by $(G,A)$ where $G=\Bbb Z^d$ and $A\in\Bbb Z^{d\times d}$? $\endgroup$
    – Berci
    Jul 8, 2014 at 20:53
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    $\begingroup$ I'm considering $A$ as a map $T_A\colon \mathbb{Z}^d\to\mathbb{Z}^d$ in the usual way as an $\mathbb{R}$-linear map restricted to the integer sublattice. $\endgroup$
    – Dan Rust
    Jul 8, 2014 at 21:01
  • $\begingroup$ "the following conjecture seems reasonable" - cannot agree with that. $\endgroup$ Jul 9, 2014 at 7:45
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    $\begingroup$ @MartinBrandenburg By reasonable I just mean that the simplest examples don't fail - for instance sequences which are eventually equal. The same can probably also be said if there is a chain of isomorphisms $A_i\to B_i$ making the 'ladder' commute. $\endgroup$
    – Dan Rust
    Jul 9, 2014 at 9:56

1 Answer 1


Let $A=\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$, $B=\left(\begin{array}{cc}0&1\\2&0\end{array}\right)$, $C=\left(\begin{array}{cc}2&0\\0&1\end{array}\right)$, $D=\left(\begin{array}{cc}0&2\\1&0\end{array}\right)$.

Then $ABCD=\left(\begin{array}{cc}1&0\\0&16\end{array}\right)$ and $A^TB^TC^TD^T=\left(\begin{array}{cc}4&0\\0&4\end{array}\right)$.

So if you take the sequence of matrices $A,B,C,D,A,B,C,D,\dots$, then the direct limit will be $\mathbb{Z}\oplus\mathbb{Z}[\frac{1}{2}]$, but the direct limit of the transposes will be $\mathbb{Z}[\frac{1}{2}]\oplus\mathbb{Z}[\frac{1}{2}]$.

But if you takes the sequence $A,A,A,\dots$, then both the direct limit and the direct limit of the transposes are $\mathbb{Z}\oplus\mathbb{Z}[\frac{1}{2}]$.

For a simpler example, but with singular matrices, which perhaps makes it easier to see what's going on, just change all the $2$s to $0$s. That gives an example where the direct limit is $\mathbb{Z}$ but the direct limit of transposes is $0$.


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