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Periodic matrices in SL(3,Z) will be conjugated to product of periodic matrices in SL(2,Z) by +- indentity on a third integer direction. Is this true?


Sorry, following your comments, maybe something I said is misleading. I state the original question: Consider a periodic automorphism $\phi$ on $Z^3$, can we find a coordinate on $Z^3$, such that $\phi$ is either (1,0;0,A) or (-1,0;0,A).

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Note that such a matrix has order $1,2,3,4$ or $6$, and has $1$ as an eigenvalue. –  Geoff Robinson Dec 11 '11 at 11:52
    
Conjugate in what? –  HJRW Dec 11 '11 at 12:44
    
@HW: conjugate in SL(3,Z) –  Bin Yu Dec 11 '11 at 13:53
    
@Geoff: I just know in SL(2,Z), periodic matrix always is conjugate to one of the following seven cases: (1,0;1,0); (-1,0;0,-1); (0,1;-1,-1); (0,1;-1,0); (0,-1;1,1); (1,0;0,-1);(1,1;0,-1). the order of such matrices just is $1,2,3,4,6$. –  Bin Yu Dec 11 '11 at 14:06

2 Answers 2

up vote 2 down vote accepted

ok, let me expand Geoff's suggestion. Let $A\in SL(3,\mathbb{Z})$ be such that $A^n=Id$ for some positive integer $n$. Since the characteristic polynomial of $A$ is a cubic polynomial of the form $-t^3+\cdots +1$, it has a positive real root; and since all roots of $A$ are roots of the unit, 1 is an eigenvalue of $A$. Moreover, since $A^n=Id$, $A$ is semisimple over $\overline{\mathbb{Q}}$ and so its minimal polynomial over $\mathbb{Q}$ is of the form $(t-1)p(t)$ with $p(t)\in \mathbb{Z}[t]$ a cyclotomic polynomial of degree at most 2 with $p(1)\neq 0$. This leaves only the following possibilities: $1$, $t+1$, $t^2+t+1$, $t^2+1$ or $t^2-t+1$, corresponding to $A$ having period $1,2,3,4$ or $6$ respectively. The period 1 case is trivial. For the period 6 case one can work as follows: we have a splitting of $\mathbb{Q}^3$ as $V\oplus W$, where $V$ and $W$ are $A$-stable $\mathbb{Q}$-vector spaces, with $\dim_{\mathbb{Q}}(V)\geq 1$ and with $A$ acting as the identity on $V$. Moreover, since $p(1)=1$, we can find polynomials $a(t)$ and $b(t)$ in $\mathbb{Z}[t]$ such that $a(t)p(t)+b(t)(t-1)=1$. This implies that $\mathbb{Z}^3=(\mathbb{Z}^3\cap V)\oplus (\mathbb{Z}^3\cap W)$. The abelian subgroup $\mathbb{Z}^3\cap V$ of $\mathbb{Z}^3$ is free since subgroups of free abelian groups are free, and has rank $\dim_{\mathbb{Q}}V$, since $V$ is a $\mathbb{Q}$-subspace of $\mathbb{Q}^3$. So we can find a $\mathbb{Z}$-basis $B_V$ for $\mathbb{Z}^3\cap V$ consisting of $\dim_{\mathbb{Q}}V$ elements. The same for $\mathbb{Z}^3\cap W$. The two basis $B_V$ and $B_W$ together are a $\mathbb{Z}$-basis of $\mathbb{Z}^3$ and up to multiplying by $-1$ one of the vectors in this basis we may assume that the change of basis matrix $P$ from the standard basis of $\mathbb{Z}^3$ to the basis $B_V\cup B_W$ is an element of $SL(3;\mathbb{Z})$. By construction, $PAP^{-1}$ is a block-diagonal matrix in $SL(3;\mathbb{Z})$, with an upper $1\times 1$ block $(1)$ and a lower $2\times 2$ block in $SL(2;\mathbb{Z})$.

This method, however, clearly does not apply to the remaining three cases.

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1) I guess $A$ in the 2nd sentence should be 3x3. 2) I don't see that such $a,b$ actually exist. Take $A$ the permutation matrix of the cycle (123). Its minimal polynomial is $x^3-1 = (x-1)(x^2+x+1)$, i.e. $p(t) = x^2+x+1$ and $p(1) = 3$. –  Ralph Dec 11 '11 at 17:47
    
Hi Ralph. Sure, what a stupid mistake! I was confused by $p(0)$ while writing. I'll now edit my answer and correct that. Thanks. –  domenico fiorenza Dec 11 '11 at 17:53
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The finite subgroups of $SL_3(\mathbb{Z})$ are well-known: See Tahara's paper projecteuclid.org/…. The only open case is the permutation matrix in my comment to Tom's answer. –  Ralph Dec 11 '11 at 18:23
    
Is not also the following order 4 matrix from Tahara's paper $$ \pmatrix{ 1&0&1\\ 0&0&-1\\ 0&1&0 } $$ an open case with respect to this MO question? –  domenico fiorenza Dec 11 '11 at 18:57
    
Yes, I would say it is. –  Ralph Dec 11 '11 at 19:26

The matrix

$0\ 1\ 0$

$1\ 0\ 0$

$0\ 0 \-1$

in $SL_3(\mathbb Z)$ is not conjugate to any block sum of an $SL_2(\mathbb Z)$ matrix and $+1$. And of course it is not conjugate to any block sum of an $SL_2(\mathbb Z)$ matrix and $-1$, either.

For $GL_3$ and $GL_2$ the answer is yes.

EDIT This last statement is wrong; see Geoff Robinson's comments.

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I'm not sure how the question is to be interpreted: Are only products of matrices of one of the following two form as are allowed or of either form ? $$\begin{pmatrix} A & \\ & 1 \end{pmatrix} \quad\quad \begin{pmatrix} 1 & \\ & A \end{pmatrix}$$ In the first case a simple counterexample is $\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. In the latter case, however, the problem seems to be harder. –  Ralph Dec 11 '11 at 18:07
    
@Ralph: You can conjugate the matrices. Shouldn't that help? –  Will Sawin Dec 11 '11 at 20:57
    
@Tom: To me, this example isn't a counterexample. –  Bin Yu Dec 11 '11 at 21:07
    
@Ralph: Yes, both are OK. –  Bin Yu Dec 11 '11 at 21:07
    
@Will: I just wanted to point out that it was possible to interped the question in its original version in two ways, namely: Is every periodic matrix from $SL_3(\mathbb{Z})$ $SL_3(\mathbb{Z})$-conjugate to a matrix of the group $U_i$ where $$U_1 = \langle \begin{pmatrix} A & \\ & 1 \end{pmatrix} \mid A \in SL_2(\mathbb{Z}) \rangle$$ $$U_2 = \langle \begin{pmatrix} A & \\ & 1 \end{pmatrix}, \begin{pmatrix} 1 & \\ & B \end{pmatrix}\mid A, B \in SL_2(\mathbb{Z}) \rangle$$ In the meanwhile the OP has clarified that he is interested in $U_1$ (I think the $U_2$-case would be harder). –  Ralph Dec 11 '11 at 21:39

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