This question came up while trying to learn about the Riemann zeta function - as is the guilty pleasure of many of us.

A Dirichlet $L$-function is notated as $$L\{z,\chi\}(z) = \sum_{n=0}^{\infty}\frac{\chi(n)}{n^z}$$ Where it is usually supposed that $z$ is complex and $\chi$ is a multiplicative function: $\chi(a)\chi(b)=\chi(ab)$

One can further show that $$L\{z,\chi\}(z)=\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)}{p^z}\Big)^{-1}$$ Where $\prod_{p\in\mathbb{P}}$ means take a product over the prime numbers.

Next I tried multiplying two $L$-functions as follows: $$L\{z,\chi\}(z)\cdot L\{\omega,\phi\}(\omega)$$ $$||$$ $$\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)}{p^z}\Big)^{-1}\Big(1-\frac{\phi(p)}{p^\omega}\Big)^{-1}$$ $$||$$ $$\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)p^\omega + \phi(p)p^z - \chi(p)\phi(p)}{p^{z+\omega}}\Big)^{-1}$$ So let's choose a new function for this product that we might clean it up a little $$M(a)=\chi(a)a^\omega + \phi(a)a^z - \chi(a)\phi(a)$$ Now we may say $$ L\{z,\chi\}(z)\cdot L\{\omega,\phi\}(\omega) =\prod_{p\in\mathbb{P}}\Big(1-\frac{M(p)}{p^{z+\omega}}\Big)^{-1} =L\{z+\omega,M\}(z+\omega) $$ And we (almost) have another Dirichlet $L$-function. The catch is that $M$ has to be multiplicative.

I noticed that $$M(1)=\chi(1)1^\omega + \phi(1)1^z - \chi(1)\phi(1)=(1)(1)+(1)(1)-(1)(1)=1$$ So it seems reasonable that there exists some choice of $\chi$ and $\phi$ that make $M$ multiplicative as well. It would be best if this choice $\chi$ and $\phi$ could be done for all $z,\omega \in \mathbb{C}$ but I suspect that $z$ and $\omega$ will have to be fixed to some particular values.

So finally

For what $\chi, \phi, z,$ and $\omega$ is $M(a)=\chi(a)a^\omega + \phi(a)a^z - \chi(a)\phi(a)$ a multiplicative function.

  • $\begingroup$ Have you tried restricting to $z=\omega$? $\endgroup$ – Somos Aug 30 '17 at 21:48
  • $\begingroup$ Now if $\chi_1,\chi_2$ are Dirichlet characters then so is $\psi(n) = \chi_1(n) \chi_2(n)$ $\endgroup$ – reuns Aug 31 '17 at 21:35

First off, you are using highly nonstandard notation. In number theory, a Dirichlet $L$-function is written as $L(s,\chi)$ (or occasionally $L(\chi,s)$), where $s$ is a complex variable and $\chi$ is a Dirichlet character, not $L\{z,\chi\}(z)$. (Why is $\{z,\chi\}$ in brackets? Why does $z$ appear twice?) This has the Euler product \[L(s,\chi) = \prod_p \frac{1}{1 - \chi(p) p^{-s}},\] where the product is over all primes $p$. (There is no need to write $p \in \mathbb{P}$; it is always implicitly understood that the product is over all primes.)

In any case, if $\chi_1, \chi_2$ are Dirichlet characters, then $L(s_1,\chi_1) L(s_2,\chi_2)$ is never equal to some Dirichlet $L$-function $L(s_3,\chi_3)$ for some Dirichlet character $\chi_3$ and complex variable $s_3$. This is basically because \[L(s_1,\chi_1) L(s_2,\chi_2) = \sum_{n = 1}^{\infty} \sum_{ab = n} \frac{\chi_1(a) \chi_2(b)}{a^{s_1} b^{s_2}},\] whereas \[L(s_3,\chi_3) = \sum_{n = 1}^{\infty} \frac{\chi_3(n)}{n^{s_3}},\] and then one can easily check that the arithmetic function $\frac{\chi_3(n)}{n^{s_3}}$ is completely multiplicative (i.e. $\frac{\chi_3(m)}{m^{s_3}} \frac{\chi_3(n)}{n^{s_3}} = \frac{\chi_3(mn)}{(mn)^{s_3}}$), but that $\sum_{ab = n} \frac{\chi_1(a) \chi_2(b)}{a^{s_1} b^{s_2}}$ is not (take $n$ to be a prime power).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.