Timeline for Periodic matrices in SL(3,Z)
Current License: CC BY-SA 3.0
10 events
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| Dec 13, 2011 at 9:56 | comment | added | Geoff Robinson |
An easier way to say it would be that (mod 3), the permutation matrix of a 3-cycle has minimum polynomial $(x-1)^3$, so its Jordan form is a single Jordan block of size $3,$ which is indecomposable.
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| Dec 12, 2011 at 14:29 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
added 70 characters in body
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| Dec 12, 2011 at 9:10 | comment | added | Geoff Robinson | @Tom: A permutation matrix of a $3$-cycle is not conjugate within ${\rm GL}(3,\mathbb{Z})$ to a decomposable matrix, as explained in my comment on the accepted solution. | |
| Dec 12, 2011 at 1:43 | comment | added | Tom Goodwillie | @ Bin Yu: It is true that every periodic automorphism of $\mathbb Z^3$ appears as $(1,0;0,A)$ or $(-1,0;0,A)$ for some two by two matrix $A$. It is not true that if the automorphism has determinant one then $A$ can be chosen to have determinant one. | |
| Dec 11, 2011 at 21:39 | comment | added | Ralph | @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). | |
| Dec 11, 2011 at 21:07 | comment | added | Bin Yu | @Ralph: Yes, both are OK. | |
| Dec 11, 2011 at 21:07 | comment | added | Bin Yu | @Tom: To me, this example isn't a counterexample. | |
| Dec 11, 2011 at 20:57 | comment | added | Will Sawin | @Ralph: You can conjugate the matrices. Shouldn't that help? | |
| Dec 11, 2011 at 18:07 | comment | added | Ralph | 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. | |
| Dec 11, 2011 at 17:43 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |