Timeline for A "multi-adic" absolute value / topology?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2021 at 11:27 | comment | added | Piotr Achinger | You can also consider the supremum of the $2$-adic and the $3$-adic absolute value. This gives a non-multiplicative nonarchimedean ring norm on $\mathbf{Q}$, and the completion is $\mathbf{Q}_2\times \mathbf{Q}_3$. | |
| Jun 10, 2021 at 9:19 | history | edited | Jérôme Poineau | CC BY-SA 4.0 |
mutli -> multi in the title
|
| Jun 10, 2021 at 9:18 | comment | added | Jérôme Poineau | For any integers $m,n$, $2^m$ and $3^n$ are coprime, so you can write $2^m u+3^n v=1$. If you want to have a basis of neighborhoods of 0 stable by addition, then you will be forced to have 1 in all those neighborhoods. I am not sure what kind of structure you would like to preserve but this looks quite problematic to me. | |
| Jun 10, 2021 at 2:08 | history | edited | MCS | CC BY-SA 4.0 |
added 465 characters in body
|
| Jun 10, 2021 at 2:05 | comment | added | MCS | @BenjaminDickman Well then, is there a non-metrizable topology that I can get out of this? | |
| Jun 10, 2021 at 1:17 | comment | added | Benjamin Dickman | Leaving as a comment in case it is already known to the OP: The only (non-trivial) absolute values on $\mathbb{Q}$ are the standard (Archimedean) one and the $p$-adic ones; this is the content of Ostrowski's Theorem. So, I think finagling a metric completion of the rationals based on your "absolute value" will not be easy . . . | |
| Jun 10, 2021 at 0:42 | comment | added | paul garrett | yes, but that feature is what makes the triangle inequality hold... I think you've got a fundamental conflict in desiderata... | |
| Jun 10, 2021 at 0:22 | comment | added | MCS | I need a metric (in the figurative, English-language sense of the word) that makes things small whenever they are divisible by large powers of one or more (but not necessarily all!) of the elements of $S$. Adding the $p$-adic absolute values gives us a "metric" that is only small when our quantity is divisible by large powers of all the elements of $S$. | |
| Jun 10, 2021 at 0:09 | comment | added | paul garrett | Why not the sum, rather than product? | |
| Jun 10, 2021 at 0:05 | history | asked | MCS | CC BY-SA 4.0 |