Timeline for Does the map $\mathrm{SL}(n,\mathbb{Z}/p^2) \rightarrow \mathrm{SL}(n,\mathbb{Z}/p)$ split?
Current License: CC BY-SA 4.0
16 events
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| Mar 29, 2023 at 13:37 | comment | added | David E Speyer | That's right: If $a^2+b^2=-1$, then put $J = \begin{bmatrix} a&b \\ b&-a \end{bmatrix}$ and keep the rest as before. I probably should have written that in the first place. | |
| Mar 29, 2023 at 12:20 | comment | added | Kasper Andersen | @DavidESpeyer I think you are right: Group #12 on the Shephard-Todd list is isomorphic to $\text{GL}_2(\mathbf{F}_3)$ which does lifts all the way to the $3$-adics. | |
| Mar 29, 2023 at 12:20 | comment | added | Dave Benson | I suppose it's defined over any field where $-1$ is a sum of two squares. | |
| Mar 29, 2023 at 12:14 | history | edited | Dave Benson | CC BY-SA 4.0 |
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| Mar 29, 2023 at 12:13 | comment | added | Dave Benson | Thanks, David. Yes, that seems to work. I'll edit it again. | |
| Mar 29, 2023 at 11:44 | comment | added | David E Speyer | I am using the quaternionic row of the character table, not the row that contains a cube root of unity. groupprops.subwiki.org/wiki/… | |
| Mar 29, 2023 at 11:41 | comment | added | David E Speyer | No, $3$-adics really do work. Note that $\sqrt{-2}$ exists $3$-adically. Put $I = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$, $J = \begin{bmatrix} 1&\sqrt{-2} \\ \sqrt{-2}&-1 \end{bmatrix}$ and $K=IJ$. Then $\pm \text{Id}$, $\pm I$, $\pm J$, $\pm K$ and $\tfrac{1}{2}(\pm \text{Id} \pm I \pm J \pm K)$ form a $24$ element group isomorphic to $\text{SL}_2(\mathbb{F}_3)$, and the reduction modulo $3$ is an isomorphism. | |
| Mar 29, 2023 at 9:57 | history | edited | Dave Benson | CC BY-SA 4.0 |
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| Mar 29, 2023 at 8:07 | comment | added | A Stasinski | It should be emphasised that this answer deals only with the case $n=2$. | |
| Mar 28, 2023 at 15:04 | comment | added | BasicQuestionBot | This is very informative, thanks! | |
| Mar 28, 2023 at 14:45 | history | edited | Dave Benson | CC BY-SA 4.0 |
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| Mar 28, 2023 at 14:39 | comment | added | Dave Benson | Sorry, yes, edited. | |
| Mar 28, 2023 at 14:38 | history | edited | Dave Benson | CC BY-SA 4.0 |
Corrected $\frac{1}{2}$ to $\frac{1+i\sqrt{3}}{2}$.
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| Mar 28, 2023 at 13:31 | comment | added | David E Speyer | I think something is wrong in your last statement. $\text{SL}_2(\mathbb{Z}/3 \mathbb{Z})$ contains a copy of the quaternion $8$-group, where as $\text{SL}_2(\mathbb{Z}[1/2])$ embeds in $\text{SL}_2(\mathbb{R})$. I compute that $\text{SL}_2(\mathbb{Z}/3 \mathbb{Z})$ is isomorphic to the unit group of the Hurwitz quaternions, and therefore lifts to $SL_2(R)$ if $R$ splits the quaternions, something which is true for $\mathbb{Z}_3$ but not $\mathbb{Z}$. | |
| Mar 28, 2023 at 8:30 | comment | added | Dave Benson | It appears from other comments that the check that I claimed was easy, is not so easy for some, so here are the details. If $A=\left(\begin{smallmatrix}1+2a&1+2b\\2c&1+2d\end{smallmatrix}\right)$ then modulo four, $A^2$ is $\left(\begin{smallmatrix}1+2c&2+2a+2d\\0&1+2c\end{smallmatrix}\right)$. So $c$ has to be even, and then the determinant of $A$ (ever modulo four) is $1+2a+2d$. So $a+d$ is even, and the result is that $A^2$ is not the identity modulo four. | |
| Mar 28, 2023 at 8:20 | history | answered | Dave Benson | CC BY-SA 4.0 |