Timeline for Conjugacy for p-adic matrices of finite order II
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2011 at 6:24 | comment | added | Junkie | Here is a general idea of the example. Take $p=3$ and $A=C(\Phi_9)\oplus C(\Phi_1)$ and $B=C(\Phi_9\Phi_1)$ where $C(f)$ is the companion matrix. Then $A$ and $B$ are conjugate in $GL_7(Q_3)$, but $\bar A$ and $\bar B$ are not conjugate in $GL_7(F_3)$. These can be expanded to include any product of distinct $\Phi_{3^j}$, and the preponderant question of Tim Dokchitser appears to spin on whether repeats are allowed. | |
| Aug 10, 2011 at 7:25 | comment | added | Tim Dokchitser | Sorry, Konstantin, the question was meant to ask whether conjugacy over ${\mathbb F}_p$ implies that over ${\mathbb Q}_p$. I fixed it now. | |
| Aug 10, 2011 at 7:12 | comment | added | Geoff Robinson | Yes: for general $p$, you take a matrix for a $p$-cycle as one of the matrices, and the direct sum of the companion matrix for $1 + x + \ldots + x^{p-1}$ with the $1 \times 1$ matrix $(1)$. | |
| Aug 10, 2011 at 6:55 | comment | added | Junkie | So you have as an example $T=[-1,1,0; 0,-1,1; 1,1,1]$ for $TAT^{-1}=B$, and $T$ has determinant 3, ergo not reducing to $GL_3(F_3)$. I agree the question posing was obscure in intent. | |
| Aug 10, 2011 at 6:48 | comment | added | Geoff Robinson | @Junkie: One matrix is $\left( \begin{array}{clcr} 0 &1 & 0\\0 & 0 * 1\\1&0&0 \right) \end{array}$ and the other is $\left(\begin{array}{clcr} 0 & 1 & 0\\-1&-1&0\\0&0&1 \end{array}$. Another explanation is that the first corresponds to a reduction (mod $p$) which is projective, and the reduction (mod $p$) of the second is not projective (speaking of them as representations of the cyclic group of order $p$). I think the question was probably not posed as intended. The implication in the other direction is less obvious (does same reduction (mod $p$) imply same character?). | |
| Aug 10, 2011 at 4:07 | comment | added | Junkie | Can you write the matrices $A$ and $B$ down explicitly for $p=3$? | |
| Aug 9, 2011 at 21:43 | comment | added | user91132 | Well then the answer is "yes" because the algebra $\mathbb{Z}_p[C_m]$ splits up into a direct sum of blocks corresponding to the $Gal(\overline{\mathbb{F}_p} / \mathbb{F}_p)$-orbits of linear characters of $C_m$. So the $\mathbb{F}_p$-representation theory and the $\mathbb{Q}_p$-representation theory of $C_m$ is the same if $p \nmid m$. In fact this is true for any finite group $G$ of order coprime to $p$. | |
| Aug 9, 2011 at 20:41 | history | edited | user91132 | CC BY-SA 3.0 |
added 129 characters in body
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| Aug 9, 2011 at 20:36 | comment | added | Dror Speiser | (only if there's a short answer, ) what happens if $m$ is prime to $p$? | |
| Aug 9, 2011 at 20:25 | history | answered | user91132 | CC BY-SA 3.0 |