Timeline for Do these Zariski-dense subgroups of $\operatorname{SO}_{6}(\mathbb C)$ have non-empty intersection with this subset?
Current License: CC BY-SA 4.0
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| Oct 6, 2019 at 4:20 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 6, 2019 at 0:54 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 23:57 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 22:50 | history | edited | LSpice | CC BY-SA 4.0 |
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| Oct 5, 2019 at 22:41 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 21:02 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 19:47 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 19:47 | comment | added | Ami | @YCor yes thanks. | |
| Oct 5, 2019 at 19:37 | comment | added | YCor | Since (passing to a smaller finite index subgroup) you can suppose that $G$ is normal in $\mathrm{SO}_6(\mathbf{Z})$ (and hence normalized by $e_{1,1}$), you can write $G$ instead of $A^{-1}GA$ in your new question. | |
| Oct 5, 2019 at 19:34 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 19:11 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 13:41 | history | edited | YCor | CC BY-SA 4.0 |
fixed question (according to answer in comment)
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| Oct 5, 2019 at 13:35 | comment | added | Ami | @YCor Thank you yes I did miss this simple fact as Misha pointed, in regards to the second question I believe that $Be_{1,1}B$ is a Bruhat cell (which is not the big cell) | |
| Oct 5, 2019 at 13:33 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 5, 2019 at 6:01 | comment | added | YCor | Clearly $H\cap\mathrm{GL}_6(\mathbf{Z})$ is finite (of cardinal $4$) and does not contain the identity matrix. Hence there is a finite index subgroup with empty intersection with $H$ (e.g., kernel of reduction modulo 2). | |
| Oct 5, 2019 at 5:58 | history | edited | YCor | CC BY-SA 4.0 |
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| Oct 5, 2019 at 2:25 | comment | added | Ami | I use the definition $\operatorname{SO}_{2n}(k)=\{A\in\operatorname{GL}_{2n}(k)|A^TJ_{2n}A=J_{2n}\}$ where $J_{2n}$ is the identity matrix flipped 90 degrees | |
| Oct 5, 2019 at 1:34 | comment | added | LSpice | Since your specific choice of $H$ (and the requirement that the upper-triangular matrices form a Borel subgroup) means that the form of $\operatorname{SO}_6$ matters, you should specify what form you use. | |
| Oct 5, 2019 at 1:33 | history | edited | LSpice | CC BY-SA 4.0 |
Fixed some language (I think—preview looks strange)
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| Oct 4, 2019 at 23:25 | comment | added | Misha | You should look more closely at what's written and revise again: Since $G$ is supposed to have integer matrix entries, if the intersection is nonempty then $x=\pm 1, y=\pm 1$. Hence, $G\cap H=\emptyset$ for most congruence subgroups. I am not sure what happens for double cosets with Borel. | |
| Oct 4, 2019 at 23:14 | comment | added | Ami | @Misha thank you, I edited my question a bit. | |
| Oct 4, 2019 at 23:13 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 4, 2019 at 22:59 | history | edited | Ami | CC BY-SA 4.0 |
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| Oct 4, 2019 at 22:54 | comment | added | Misha | You can even take $G$ to be a "generic" Zariski dense 2-generated subgroup in $SO_6({\mathbb C})$. Then the intersection with $H$ will be empty. Here generic means that the two generators are chosen generically. | |
| Oct 4, 2019 at 20:40 | review | First posts | |||
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| Oct 4, 2019 at 20:36 | history | asked | Ami | CC BY-SA 4.0 |