Timeline for Does viewing multiplicative functions as morphisms from $\mathbb{Q}^*\to\mathbb{C}^*$ have any consequnces?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 11, 2020 at 15:45 | vote | accept | Milo Moses | ||
| Nov 11, 2020 at 9:46 | comment | added | YCor | This gives a "classification" but it says nothing about the analytic behavior... for instance given a multiplicative function (determined by values $(a_p)_{p\text{ prime}}$ and $a_{-1}=\pm 1$), one can wonder for which topologies on $\mathbf{Q}$ is the resulting map is continuous, etc. Also describing a function in this way has the inconvenient of being based on the prime factor decomposition, which theoretically sounds mechanical but practically is not obvious to compute. | |
| Nov 11, 2020 at 9:21 | comment | added | Daniel Loughran | @Thomas Browning: As well as a value for $-1$! | |
| Nov 11, 2020 at 9:21 | answer | added | Daniel Loughran | timeline score: 3 | |
| Nov 11, 2020 at 5:43 | comment | added | Milo Moses | @ThomasBrowning yes, that is true, in the same way that completely multiplicative functions are the same as picking their values at primes. The interesting ideas come from examining the structure of these morphisms/multiplicative functions. For example, the kernel of a morphism $\mathbb{Q}^*\to\mathbb{C}^*$ directly depends on the subgroup of the rational numbers for which $f(p)=1$. Does this mean that multiplicative functions will have special properties depending on what values of $p$ we have $f(p)=1$? Maybe so, maybe not. | |
| Nov 11, 2020 at 4:54 | comment | added | Thomas Browning | Morphisms $\mathbb{Q}^\ast\to\mathbb{C}^\ast$ are the same as choosing an element $a_p\in\mathbb{C}^\ast$ for each prime $p$. | |
| Nov 11, 2020 at 0:06 | history | asked | Milo Moses | CC BY-SA 4.0 |