The title says it all : assuming all L-functions are objects of a symmetric bimonoidal category $ (\mathcal{C},\oplus,\otimes,s\mapsto 1,\zeta) $ , where $ \oplus $ stands for the usual product, what can we prove about L-functions from general theorems about symmetric bimonoidal categories that turns out to have analytic or number theoretic interest ? I know some people here might get angry to see me ask this question but still, it may be insightful.
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5$\begingroup$ Why would you expect this? What’s the tensor product? What’s Homs from the Riemann zeta to itself? $\endgroup$– Noah SnyderCommented Nov 21, 2017 at 0:23
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3$\begingroup$ Nothing; with only the given hypotheses $C$ could be a discrete category with only identity morphisms, and then $C$ with this extra structure could be any commutative semiring. $\endgroup$– Qiaochu YuanCommented Nov 21, 2017 at 0:23
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1$\begingroup$ Under the automorphy and Langlands conjectures, isn't the operator $\sum_{n=1}^\infty a_n n^{-s} \mapsto\sum_p a_p p^{-s}$ injective on the products and quotients of normalized L-functions and its image is a ring with the pointwise addition and multiplication of functions $\mathcal{P} \to \mathbb{C}$ $\endgroup$– reunsCommented Nov 21, 2017 at 5:32
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$\begingroup$ Of functions $\mathcal{P} \to \mathbb{C}$ quotiented by "equal on all but finitely many primes" (so that it doesn't contain non-primitive Dirichlet L functions) $\endgroup$– reunsCommented Nov 21, 2017 at 5:41
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$\begingroup$ @Qiaochu Yuan : I'd be interested in knowing what happens with as many morphisms as possible given this double structure, for different definitions of the tensor product. $\endgroup$– Sylvain JULIENCommented Nov 21, 2017 at 10:47
1 Answer
Let $\mathcal{D} = \{ \sum_{n=1}^\infty a_n n^{-s} = \prod_{j=1}^l L(s,\chi_j)^{e_j}, e_j \in \mathbb{Z}\}$ the set of products and quotients of Dirichlet L-functions and $$T : \mathcal{D} \to \ldots, \qquad T(\sum_{n=1}^\infty a_n n^{-s}) = \sum_p a_p p^{-s}$$ By multiplicate-ness $\scriptstyle\sum_{n=1}^\infty a_n n^{-s} =\prod_p (1+\sum_{k \ge 1}a_{p^k} p^{-sk})= \prod_p (\prod_{j=1}^l (1-\chi_j(p) p^{-s})^{-e_j})$ $\scriptstyle= \prod_p (\prod_{j=1}^l (1+\chi_j(p) p^{-s}+\mathcal{O}(p^{-2s}))^{e_j}) =\prod_p (\prod_{j=1}^l (1+e_j \chi_j(p) p^{-s}+\mathcal{O}(p^{-2s})))$ $\scriptstyle=\prod_p (1+\sum_{j=1}^l e_j \chi_j(p) p^{-s}+\mathcal{O}(p^{-2s}))$ thus $a_p = \sum_{j=1}^l e_j \chi_j(p)$ and $$T(\prod_{j=1}^l L(s,\chi_j)^{e_j}) = \sum_p p^{-s} \sum_{j=1}^l e_j \chi_j(p)$$
Thus $T\mathcal{D}$ is an abelian group under addition, and exploiting the fact products of Dirichlet characters are Dirichlet characters, there is a natural ring structure on $T\mathcal{D}$ :
$$(\sum_p p^{-s} \chi_1(p))\times (\sum_p p^{-s} \chi_2(p)) =\sum_p p^{-s} \chi_1(p)\chi_2(p)$$
Then we quotient $T \mathcal{D}$ by the equivalence relation "equal on all but finitely many primes". We then obtain that $$T \mathcal{D} = \{ \sum_p p^{-s} \sum_{j=1}^l e_j \chi_j(p), \quad e_j \in \mathbb{Z},\quad \chi_j \text{ primitive Dirichlet charaters}\}$$ Almost by definition, the set of primitive Dirichlet characters is a group $G\cong \lim_{n \to \infty} (\mathbb{Z}/n!\mathbb{Z})^\times$ thus $T \mathcal{D} = \mathbb{Z}[G]$ as a ring (and it is a subring of $\mathbb{Z}[\hat{\mathbb{Z}}^\times]$ whatever it means in term of L-functions)
The question is if all this works when assuming some Langlands conjectures and replacing the Dirichlet L-functions by the automorphic normalized L-functions $L(s,\pi)$, I'd say yes, since $T (L(s,\pi_1)) \times T(L(s,\pi_2)) = T(L(s,\pi_1 \otimes \pi_2))$.
What would be the ring structure of $T \mathcal{L}$ ? How would you define $T^{-1}$ ?
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$\begingroup$ I'll read your answer tomorrow cause I need to sleep but I doubt "multiplicate-ness" exists in English. I guess you meant "multiplicativity". $\endgroup$ Commented Nov 21, 2017 at 22:57
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$\begingroup$ By the way, what is $\mathcal{L}$? $\endgroup$ Commented Nov 22, 2017 at 16:02
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$\begingroup$ @SylvainJULIEN What is your guess.. ? $\endgroup$– reunsCommented Nov 22, 2017 at 20:29
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$\begingroup$ The set of supposedly automorphic L-functions ? $\endgroup$ Commented Nov 22, 2017 at 21:06
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$\begingroup$ @SylvainJULIEN Yes. Here what we need is : Euler product, that $F(s) \mapsto T(F(s))$ is injective (ie. functional equation + Ramanujan conjecture + ??), and that the set is closed under $(F,G) \mapsto F \otimes G(s) \cong \sum_n a_n(F) a_n(G) n^{-s}$ (where $\cong$ means "equal for almost every prime"). About the rest : anything unclear to you ? In particular, do you know the Chinese remainder theorem, the structure of $(\mathbb{Z}/p^e \mathbb{Z})^\times$ and how to construct all the Dirichlet characters modulo $q$ ? (start with $q=p$, then $q = p_1p_2$) $\endgroup$– reunsCommented Nov 22, 2017 at 21:13