9
$\begingroup$

We know that every $2\times 2$ matrix in $PGL(2, \mathbb{Z})$ of order $3$ is conjugate to the matrix $$ \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array} \right) $$.

I am interested in finding out to what extent this holds for $3\times 3$ integer invertible matrices.

In other words how many conjugacy classes of order 3 matrices in $PGL(3, \mathbb{Z})$ are there?

$\endgroup$
  • $\begingroup$ In the $2 \times 2$ case, such a matrix should have trace -1, not 1. There are probably two conjugacy classes of elements of order $3$ in ${\rm GL}(3, \mathbb{Z}),$ (certainly at least two), but some work would be needed. $\endgroup$ – Geoff Robinson Oct 9 '13 at 11:50
  • 2
    $\begingroup$ The matrix you have written down has order 6. $\endgroup$ – Derek Holt Oct 9 '13 at 12:21
  • 1
    $\begingroup$ Perhaps the OP means $PGL(2)$? $\endgroup$ – Marc Palm Oct 9 '13 at 12:38
  • $\begingroup$ @Geoff Robinson, Can you please give me any references in this direction. $\endgroup$ – user41076 Oct 9 '13 at 14:04
  • $\begingroup$ @Derek Holt Sorry for not mentioning $PGL(2, Z)$ $\endgroup$ – user41076 Oct 9 '13 at 14:04
13
$\begingroup$

I will work in ${\rm GL}$ instead of ${\rm PGL}$.

The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of rank 3 over $\mathbb{Z}$ there are up to isomorphism, where $G$ is the cyclic group of order 3. Any such module is a direct sum of indecomposable modules, and those have been classified in I. Reiner, Integral representations of cyclic groups of prime order, Proc. Amer. Math. Soc. 8 (1957), 142–146. There are three indecomposable $\mathbb{Z}[G]$-modules:

  • the trivial module of $\mathbb{Z}$-rank 1, $\Gamma_1$,
  • the augmentation ideal of $\mathbb{Z}[G]$, which has $\mathbb{Z}$-rank 2, $\Gamma_2$,
  • the regular module $\Gamma_3=\mathbb{Z}[G]$ itself.

So there are two isomorphism classes of faithful modules of rank 3:

  • $\Gamma_1\oplus \Gamma_2$,
  • $\Gamma_3$.

One should beware, that, in general, the Krull-Schmidt Theorem fails for $\mathbb{Z}[G]$-modules, but in this case it is easy to see that the two guys are not isomorphic, since e.g. in one of them the trivial isotypical component is a direct summand, and in the other one it isn't. Alternatively, the two are not isomorphic over $\mathbb{Z}_3$, and Krull-Schmidt does hold over local rings.

${}^1$ To remove the word "essentially", one needs to check that any matrix obtained in this way is conjugate to its inverse. This is true because each of the indecomposable modules listed above can be extended to a module under the symmetric group $S_3$, as is easy to check.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ It is also worth pointing out that each of the two modules occurs as a submodule of finite index in the other. They are equivalent over ${\mathbb Q}$. $\endgroup$ – Derek Holt Oct 9 '13 at 15:38
  • $\begingroup$ @DerekHolt: indeed, Derek! Otherwise proving that they are non-isomorphic would be even easier of course. It is also worth mentioning that this sort of analysis will solve this question in any given dimension (the list of indecomposables will be the same), although in even dimension one would need to complement it by a similar analysis of modules under the cyclic group of order 6 if one wanted an answer in ${\rm PGL}$, rather than in ${\rm GL}$. $\endgroup$ – Alex B. Oct 9 '13 at 16:19
  • 3
    $\begingroup$ @Alex: no, the center of $GL(3,\mathbb{Z})$ is not trivial. $\endgroup$ – YCor Oct 9 '13 at 17:28
  • $\begingroup$ @AlexB.: I thought that the centre of $GL(3,\mathbb Z)$ was $\mathbb Z_2$. $\endgroup$ – SashaKolpakov Oct 9 '13 at 22:46
  • 1
    $\begingroup$ @Yves Sorry, I was being silly and was thinking of SL. $\endgroup$ – Alex B. Oct 9 '13 at 23:17
8
$\begingroup$

The finite subgroups of $GL_3(\mathbb{Z})$ are known in the literature:

$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209.

In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are $$U_1=\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad U_2=\langle\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$

Added: In $PGL_3(\mathbb{Z})=GL_3(\mathbb{Z})/\langle -I\rangle$ there are also exactly two conjugacy classes of subgroups of order 3 and representants are $\bar{U}_1,\bar{U}_2$.

For, let $V_i=\langle x_i\rangle \le G := GL_3(\mathbb{Z}),i=1,2$ be subgroups of order 3. If $\bar{V}_i$ are conjugate in $\bar{G} :=PGL_3(\mathbb{Z})$ then there is $g \in G$ s.t. $x_2=(\pm I)gx_1^kg^{-1} (k=1,2)$ and hence $(\pm I)^3=I$ and $x_2=gx_1^kg^{-1}$, i.e. the $V_i$ are conjugated in $G$.

Conversely, let $\bar{V}$ be a subgroup of $\bar{G}$ of order three. It's preimage in $G$ has order 6. Hence there is $V \le G$ of order three that maps to $\bar{V}$ and by the above $V$ is conjugated to some $U_i$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @tj: no, $GL_3(\mathbb{Z})$ is not isomorphic to $PGL_3(\mathbb{Z})$, but to $PGL_3(\mathbb{Z})\times\{\pm 1\}$. $\endgroup$ – YCor Oct 9 '13 at 17:44
  • $\begingroup$ @Yves: Thanks for pointing out that $GL_3(\mathbb{Z}) \not\cong PGL_3(\mathbb{Z})$. I understand $GL_3(\mathbb{Z}) \cong SL_3(\mathbb{Z})\times \lbrace \pm 1\rbrace$. But do we really have $GL_3(\mathbb{Z}) \cong PGL_3(\mathbb{Z}) \times \lbrace \pm 1\rbrace$ ? $\endgroup$ – tj_ Oct 9 '13 at 19:11
  • 4
    $\begingroup$ yes because the obvious homomorphism $SL_3(\mathbb{Z})\to PGL_3(\mathbb{Z})$ is an isomorphism $\endgroup$ – YCor Oct 9 '13 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.