Timeline for Periodic matrices in SL(3,Z)
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2011 at 23:20 | comment | added | Ralph | Nice argument. Thank you very much. In fact $$\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$ | |
| Dec 11, 2011 at 23:11 | comment | added | Geoff Robinson | The copy of $S_4$ is the group of all integral monomial matrices of determinant $1$ (recall that momonmial matrices have one non-zero entry in each row and one non-zero entry in each column). The Klein $4$-group is normal, and the factor group is isomorphic to $S_3$. | |
| Dec 11, 2011 at 22:40 | comment | added | Geoff Robinson | I mean $A 0;0 1$ and $1 0;0 B$ together with that Klein 4-group generate a subgroup isomorphic to $S_4.$ | |
| Dec 11, 2011 at 22:34 | comment | added | Geoff Robinson | Yes, I think that the group $U_2$ contains a subgroup isomorphic to $S_4.$ It contains a Klien 4 group consisting of all diagonal matrices with entries $1$ or $-1$ and determinant $1$. Now take $A = B$ to be a matrix of order $4$ of the form $0,-1;1,0$. The $A,B$ and the given Klein $4$-group generate a group isomorphic to the symmetric group $S_4$, which contains a $3$-cycle permutation matrix. | |
| Dec 11, 2011 at 22:20 | comment | added | Ralph | Geoff, thanks for conferming. Do you have an idea, if that permutation matrix could be contained in the group $U_2$ (defined in my second comment in Tom's answer) ? | |
| Dec 11, 2011 at 22:04 | comment | added | Geoff Robinson |
Well, it is true that the $3$-dimensional representation of the cyclic group of order $3$ obtained by representing a generator by a permutation matrix associated to a $3$-cycle is indecomposable as an integral representation, because its reduction (mod $3$) is projective and necessaily indecomposable. Hence, even in ${\rm GL}(3,\mathbb{Z})$, such a matrix is not equivalent to a decomposable matrix.
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| Dec 11, 2011 at 21:30 | comment | added | Bin Yu | @domenico and @Ralph: I'm sure that such a matrix isn't a counterexample. | |
| Dec 11, 2011 at 21:08 | vote | accept | Bin Yu | ||
| Dec 11, 2011 at 21:08 | vote | accept | Bin Yu | ||
| Dec 11, 2011 at 21:08 | |||||
| Dec 11, 2011 at 21:05 | comment | added | Bin Yu | @domenico and @Ralph: you mean such a example is a conterexample? how you see that? | |
| Dec 11, 2011 at 20:46 | history | edited | domenico fiorenza | CC BY-SA 3.0 |
exposition slightly improved
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| Dec 11, 2011 at 19:26 | comment | added | Ralph | Yes, I would say it is. | |
| Dec 11, 2011 at 18:57 | comment | added | domenico fiorenza | 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? | |
| Dec 11, 2011 at 18:23 | comment | added | Ralph | 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. | |
| Dec 11, 2011 at 18:14 | history | edited | domenico fiorenza | CC BY-SA 3.0 |
a typo fixed; deleted 58 characters in body
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| Dec 11, 2011 at 17:53 | comment | added | domenico fiorenza | 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. | |
| Dec 11, 2011 at 17:47 | comment | added | Ralph | 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 $p(1) = 3$. | |
| Dec 11, 2011 at 17:19 | history | answered | domenico fiorenza | CC BY-SA 3.0 |