Timeline for Can we construct an isomorphism between $\mathrm{BS}(1,n)$ and $\mathbb{Z}[1/n]\rtimes\mathbb{Z}$ such that it preserve the order?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| 3 hours ago | history | edited | Corentin B | CC BY-SA 4.0 |
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| 3 hours ago | history | edited | Corentin B | CC BY-SA 4.0 |
added 815 characters in body
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| 4 hours ago | comment | added | navashree chanania | Let us continue this discussion in chat. | |
| 4 hours ago | comment | added | Corentin B | The easiest way to think about this order is through the isomorphism. You have an isomorphism with a left-ordered group, you can just "pullback" the order. | |
| 4 hours ago | comment | added | navashree chanania | Will it hold if we take only positive element? | |
| 4 hours ago | comment | added | navashree chanania | Can you tell me more about ordering in BS(1,n)? In $\mathbb{Z}[1/n]\rtimes\mathbb{Z}$ it is easy to see when one element is lesser than another, (like (r,m)<(r',m') if either m>m'or r'>r). But I am not sure how to see this in BS(1,n). Is it a little bit complicated or simple? | |
| 4 hours ago | comment | added | Corentin B | But yes, $\phi$ is an isomorphism regardless of the sign of $n$. | |
| 4 hours ago | comment | added | navashree chanania | Is this isomorphism true if we take $n<-1$, I did not get any article where it is proved for negative $n.$ Almost everywhere this isomorphism is proved for $n$ positive and when $n\geq2$ this group $BS(1,n)$ is an ordered group. Am I right? | |
| 4 hours ago | history | answered | Corentin B | CC BY-SA 4.0 |