Let $ F $ be an element of the Selberg class of degree $ d>1 $ and let $ p $ a prime dividing $ d $ . Does the map $ \phi_{p} : a_{n}(F)\mapsto a_{n}(F)^{1/p} $ where $ F(s)=\sum_{n>0}\frac{a_{n}(F)}{n^s} $ for $ \Re(s)>1 $ defines another element $ \Phi_{p}(F) :s\mapsto\sum_{n}\frac{\phi_{p}(a_{n}(F))}{n^s} $ for $ \Re(s)>1 $ of the Selberg class ? If yes what is its degree ?
Does this map send a member of the Selberg class to another one ?
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number-theory
l-functions
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2The operation isn't well-defined: complex numbers have multiple $p$th roots. Even aside from that, do you have any reason to think this should result in an element in the Selberg class? What about $\sum_{n\ge1} \sqrt{\tau(n)}n^{-s}$, where $\tau$ is the divisor function, so that $\sum_{n\ge1} \tau(n)n^{-s} = \zeta(s)^2$? – 2017-01-29
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0Ok, so let's consider a primitive function $ F $ and $ \phi_{p}(a_n) : =\vert a_{n}\vert^{1/p}e^{2i\pi/p} $. Does the question make more sense now ? – 2017-01-29
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1@Sylvain : No. You need to see how to prove the functional equation for the Dirichlet L-functions, the modular forms, the Dedekind and Artin zeta functions, the rankin selberg convolution, before thinking about creating your own functions of the (extended) Selberg class – 2017-01-29