# Do automorphisms preserve convergence of power series?

Consider a power series

$$f(q)=\sum_{n=0}^\infty a_nq^n.$$

Suppose that $$f$$ has a positive radius of convergence. For an automorphism $$\sigma\in \operatorname{Aut}(\mathbb C)$$ define

$$f^\sigma(q)=\sum_{n=0}^\infty a_n^\sigma q^n.$$

Does $$f^\sigma$$ have a positive radius of convergence?

What if all coefficients $$a_n$$ lie in a number field $$K$$ (that is, a finite extension of $$\mathbb Q$$)?

• It may depend: are you interested in continuous automorphisms, or arbitrary automorphisms? – Cameron Buie Jul 16 at 12:02
• Arbitrary field automorphisms might take a sequence of independent transcendent numbers into an arbitrary sequence of independent transcendent numbers.. – Berci Jul 16 at 12:12
• The answer is no for non-continuous automorphisms : $2-\sqrt{3} \in (0,1)$ and $2+\sqrt{3} > 1$ so that $\sum_{n=0}^\infty (2-\sqrt{3})^{n^2} q^n$ is entire while $\sum_{n=0}^\infty (2+\sqrt{3})^{n^2} q^n$ never converges. Now for L-functions it is different, there is a whole theory (Hecke algebras, Shimura varieties) to find how $L(s,\pi)$ is related to $L(s,\pi^\sigma)$ for certain classes of $\pi$. – reuns Jul 16 at 12:16
• Also you can define a norm on $\overline{\Bbb{Q}}$ as $\|x\|=\sup_\sigma |\sigma(x)|$ and look at the power series with non-zero $\|.\|$ radius of convergence, then $f(q)$ isn't a complex number anymore but an element of the completion of $\overline{\Bbb{Q}},\|.\|$ where the action of $Gal(\overline{\Bbb{Q}}/\Bbb{Q})$ is continuous (note there is a version of $s \mapsto n^{-s}$ and $\zeta(s)$ defined on that thing) – reuns Jul 16 at 12:27
• For the modular forms (with algebraic coefficients) the easiest proof is that $f/E_k$ is a weight $0$ modular function thus it is the root of its minimal polynomial $\in \Bbb{Q}(j)[X]$ and $f^\sigma/E_k^\sigma$ is again a root thus it is a modular function, whence $f^\sigma$ is weight $k$ modular and it suffices to check the "bounded at the cusps" conditions. – reuns Jul 16 at 12:37