Let $\sigma$ be a field automorphism of $\mathbb{C}$ that commutes with the Riemann Zeta function. Can we use Voronin's universality theorem to prove that $\sigma$ is necessarily continuous?

Thanks in advance.

## 1 Answer

I apologize for answering my own question, but it has turned out that the statement I consider can actually be proved without using Voronin's theorem.

Here comes an excerpt from an article of mine entitled "Automorphisms of L-functions" which does the job:

Let $\mathcal{M}$ denote the maximal class of automorphic L-functions belonging to the Selberg class.

**Automorphisms of $\mathcal{M}$**

Let's define the tensor product $F\otimes G$ of any pair $(F,G)$ of elements of $\mathcal{M}$ as in [3]. It follows from Haseo Ki's proof of the strong multiplicity one theorem for $\mathcal{S}$ in [4] that for all prime $p$, $a_{p}(F\otimes G)=a_{p}(F)a_{p}(G)$. We'll say that a bijective map $\Phi\colon\mathcal{M}\to\mathcal{M}$ is an automorphism of $\mathcal{M}$ if and only if it fulfills simultaneously the following three conditions:

1) $\Phi$ maps a primitive element of $\mathcal{M}$ to a primitive element of $\mathcal{M}$

2) $\displaystyle{\Phi(\prod_{i}F_{i}^{e_i})=\prod_{i}\Phi(F_{i})^{e_i}}$

3) $\Phi(F\otimes G)=\Phi(F)\otimes\Phi(G)$

It is worth noticing that condition 2) makes sense as SOC for automorphic $L$-functions (and thus for elements of $\mathcal{M}$) entails unique factorization thereof. As a consequence, the constant function $s\mapsto 1$ is preserved by any automorphism of $\mathcal{M}$. Moreover, as for any $F\in\mathcal{M}, \ \ F\otimes\zeta=F$, where $\zeta$ denotes the Riemann Zeta function, condition 3) implies that $\zeta$ is preserved by any automorphism of $\mathcal{M}$ too. Finally, the three conditions above imply that for any element $F$ of $\mathcal{M}$, the degree of $\Phi(F)$ equals the degree of $F$.

**Automorphisms of an L-function**

Let $\sigma$ be a field automorphism of $\mathbb{C}$. For every $F\in\mathcal{M}$ that commutes with $\sigma$, the map $\Phi_{\sigma}\colon G\mapsto\sigma\circ G\circ\sigma^{-1}$ attached to $\sigma$ is said to be an automorphism of $F$ if and only if $\Phi_{\sigma}$ is an automorphism of $\mathcal{M}$.

**Continuity lemma**

Let $\sigma$ be a field automorphism of $\mathbb{C}$, and $F$ an L-function such that $\Phi_{\sigma}$ is an automorphism of $F$. Then $\sigma$ is continuous.

**Proof**

Let's define the prehilbertian space $\mathcal{J_{M}}$ attached to $\mathcal{M}$ as follows:

To any $F\in\mathcal{M}$ we associate an element $v_{F}$ of the $\mathbb{C}$ linear space $\mathcal{J_{M}}$ such that $F\ne G\Rightarrow v_{F}\ne v_{G}$ and to any automorphism $\Phi$ of $\mathcal{M}$, we associate a linear operator $\tilde{\Phi}$ such that the following diagram commutes:

$\require{AMScd}$

\begin{CD} \prod_{i}F_{i}^{e_{i}} @>{v}>> \sum_{i}e_{i}v_{F_{i}} \\ @V m VV @VV\nu V \\ \prod_{i}\Phi(F_{i}^{e_{i}}) @>\mu>> \sum_{i}e_{i}\tilde{\Phi}(v_{F_{i}}) \end{CD}

Hence $\nu\circ v=\mu\circ m$.

**Inner product**

We endow $J_{\mathcal{M}}$ with the inner product $\langle , \rangle$ such that for any two elements $F$ and $G$ of $\mathcal{M}$, $\langle v_F,v_G\rangle$ is defined as the limit as $x$ tends to infinity of $\displaystyle{\frac{1}{\log\log x}\sum_{p\leq x}\frac{a_{p}(F)\overline{a_{p}(G)}}{p}}$.

As an automorphism $\Phi_{\sigma}$ of $\mathcal{M}$ maps a primitive element of $\mathcal{M}$ to a primitive element of $\mathcal{M}$ and as SOC is true for $\mathcal{M}$, one gets that $\langle v_{F},v_{G}\rangle=\langle\tilde{\Phi}_{\sigma}(v_F),\tilde{\Phi}_{\sigma}(v_G)\rangle$. Hence $\tilde{\Phi_{\sigma}}$ is an automorphism of $\mathcal{J}_{\mathcal{M}}$ and thus $\Phi_{\sigma}$ is continuous.

Now, writing $\langle F,G\rangle_{\mathcal{M}}$ whenever $F$ and $G$ are L-functions as another way to write $\langle v_{F},v_{G}\rangle$, we get:

$\langle F+\epsilon F,F+\epsilon F\rangle_{\mathcal{M}}=\langle\Phi_{\sigma}(F+\epsilon F),\Phi_{\sigma}(F+\epsilon F)\rangle_{\mathcal{M}}$

hence:

$\langle\sigma\circ(F+\epsilon F)\circ\sigma^{-1},\sigma\circ(F+\epsilon F)\circ\sigma^{-1}\rangle_{\mathcal{M}}=\langle F+\epsilon F,F+\epsilon F\rangle_{\mathcal{M}}$

hence:

$\langle \sigma\circ F\circ\sigma^{-1}+\sigma\circ \epsilon F\circ \sigma^{-1},\sigma\circ F\circ\sigma^{-1}+\sigma\circ\epsilon F\circ\sigma^{-1}\rangle_{\mathcal{M}}=\langle F+\epsilon F,F+\epsilon F\rangle_{\mathcal{M}}$

and thus:

$\langle\sigma(\epsilon)F,\sigma(\epsilon)F\rangle_{\mathcal{M}}=\langle \epsilon F,\epsilon F\rangle_{\mathcal{M}}$.

Taking the limit of both sides of the equality as $\epsilon$ tends to $0$, one gets:

$\displaystyle{\lim_{\epsilon\to 0}\sigma(\epsilon)=0}$

hence making $\sigma$ continuous.

[3] Murty, M. Ram and Zaharescu, Alexandru (2002), "Explicit formulas for the pair correlation of zeros of functions in the Selberg class", Forum Math. 14 (2002), no. 1, pp. 65-83

[4] Ki, Haseo, "ON THE STRONG MULTIPLICITY ONE FOR THE SELBERG CLASS", http://web.yonsei.ac.kr/haseo/smo.pdf, submitted for publication