# Does viewing multiplicative functions as morphisms from $\mathbb{Q}^*\to\mathbb{C}^*$ have any consequnces?

As a lover of number theory, the condition $$f(xy)=f(x)f(y)$$ in group morphisms always reminds me of completely multiplicative functions. Since the natural numbers $$\mathbb{N}$$ do not form a group under multiplication, we will work instead in the non-zero rational numbers $$\mathbb{Q}^*$$ where the value of the multiplicative function at a rational number will be defined as

$$f\left(\frac{a}{b}\right)=\frac{f(a)}{f(b)}$$

which can be easily verified to be consistent under the multiplicativitiy hypothesis. This above representation means that we need to restrict ourselves to multiplicative functions that do not have $$0$$ under their image, but this isn't a huge loss. Collecting what we have above, every completely multiplicative function that does not have $$0$$ in its image induces a natural morphism $$\mathbb{Q^*}\to\mathbb{C}^*$$, and working backwards even morphism $$\mathbb{Q^*}\to\mathbb{C}^*$$ induces a natural multiplicative function $$f$$ that does not have $$0$$ in its image.

It feels very natural to me that the properties of morphisms $$\mathbb{Q}^*\to\mathbb{C}^*$$ have been studied extensively, and so I am wondering if there have been any papers published that utilize our knowledge about these morphisms to state new theorems about multiplicative functions.

• 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 '20 at 4:54
• @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 '20 at 5:43
• @Thomas Browning: As well as a value for $-1$! Nov 11 '20 at 9:21
• 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.
– YCor
Nov 11 '20 at 9:46

So what normally happens is that one tries to come up with a $$p$$-adic analogue of the multiplicative function as well as a version at the infinite place, and then take a product over all places to obtain an adelic version $$f_{\mathbf{A}}: \mathbf{A}_{\mathbb{Q}}^\times \to \mathbb{C}^\times$$ of the multiplicative function $$f$$, where $$\mathbf{A}_{\mathbb{Q}}$$ denotes the ring of adeles of $$\mathbb{Q}$$ (restricted product over $$\mathbb{R}$$ and all $$\mathbb{Q}_p$$).
Why is this the "opposite" of what you suggest? Well, my understanding is that there is no general method to construct $$f_\mathbf{A}$$ from $$f$$. But one case where this is greatly studied is for Dirichlet characters. Here for a Dirichlet character $$\chi$$, the associated adelic version $$\chi_{\mathbf{A}}$$ is trivial on $$\mathbb{Q}^\times \subseteq \mathbf{A}_{\mathbb{Q}}^\times$$!
• For $f$ completely multiplicative $\Bbb{Z}\to \Bbb{C}$ then $f$ extends continuously to $\Bbb{A_Q}^\times$ in the obvious way: $F (a_\infty\prod_p (a_pp^{e_p})_p)=f(sign(a_\infty))\prod_{p,f(p)\ne 0} f(p)^{e_p}$. The point is that for $f=\chi$ a Dirichlet character (modulo a prime power $q^k$) we have another way to extend continuously to the ideles : $G(a_\infty\prod_p (a_pp^{e_p})_p)=\chi(a_q \bmod q^k)$. Then $F(x)/G(x)=1$ for each $x\in \Bbb{Q}^\times$ embedded diagonally in the ideles. $H=F/G$ is an automorphic form on $GL_1(\Bbb{Q})\setminus GL_1(\Bbb{A_Q})$ Nov 11 '20 at 11:27