# Do those manifolds atrached to L-functions give rise naturally to motives? [closed]

Edited after Will Sawin's comment:

Consider the set $\mathcal{M}$ of all automorphic L-functions belonging to the Selberg class. Such a set is closed for the product $.$ and the tensor product $\otimes$ such that $\forall p\in\mathbb{P}, \ \ a_{p}(F\otimes G)=a_{p}(F).a_{p}(G)$ where $a_{n}(H)$ is the $n$-th Dirichlet coefficient of $H$, that is $H(s)=\displaystyle{\sum_{n\gt 0}\dfrac{a_{n}(H)}{n^s}}$ whenever $\Re(s)\gt 1$. This tensor product corresponds, on the automorphic side, to Rankin-Selberg convolution. For the sake of simplicity, the term of 'L-function' will be used to mean any element of $\mathcal{M}$.

Let's define the automorphism group of $\mathcal{M}$ as the group, under composition, of the bijective maps $\Phi$ from $\mathcal{M}$ to itself such that the following properties are simultaneously fulfilled:

A) $\Phi$ maps a primitive L-function to a primitive L-function

B) $\forall (F,G)\in\mathcal{M}^{2}, \ \ \Phi(F\odot G)=\Phi(F)\odot\Phi(G)$ where $\odot\in\{\times, \otimes\}$

Such an automorphism of $\mathcal{M}$ preserves the degree of any L-functions, that is $d_{\Phi(F)}=d_{F}$.

Assuming that for any two L-functions $F$ and $G$, $d_{F\otimes G}=d_{F}.d_{G}$, let's now associate to an L-function $H$ a complex manifold $X_{H}$ of dimension $d_{H}$ such that for any two L-functions $F$ and $G$, $X_{F.G}=X_{F}\oplus X_{G}$ and $X_{F\otimes G}=X_{F}\otimes X_{G}$ so that $F=G\Leftrightarrow X_{F}=X_{G}$. We shall denote the set of all $X_{F}$ where $F$ runs over $\mathcal{M}$ by $\mathcal{M}'$ and any element of $\mathcal{M}'$ will be called an L-manifold.

The automorphism group of $\mathcal{M}'$ is defined in a similar fashion as for the one of $\mathcal{M}$ so that these two groups are isomorphic.

Let's now define the notion of abstract Galois group $\operatorname{Gal}(A/B)$ as the set of all automorphisms of $A$ that preserve $B$ pointwise, and let's associate to any L-function $F$ its ' canonical' representation $(\rho_{F}, V_{F})$ such that the following properties are simultaneously fulfilled:

C) there exists an algebraic number field $K_{F}$ the absolute Galois group of which, denoted by $G_{K_{F}}$, is isomorphic to both $\operatorname{Gal}(\mathcal{M}/<F>)$ and $\operatorname{Gal}(\mathcal{M}'/<X_F>)$, and is such that $\rho_{F}$ is a group homomorphism from $G_{K_{F}}$ to $\operatorname{GL}_{d_{F}}(\mathbb{C})=Aut(V_{F})$ where $<F>=\{\bigodot_{k=0}^{m}F, \odot\in\{\times,\otimes\}, m\in\mathbb{N}_{0}\}$, $<X_{F}>$ is defined in a similar way, and so that $X_{F}$ is locally isomorphic to $V_{F}$.

Assuming $F$ is the L-function attached to an automorphic representation of $\operatorname{GL}_{d_{F}}(\mathbb{A}_{K_{F}})$ where $\mathbb{A}_{K_{F}}$ is the adele ring of $K_{F}$ we require that the considered representation $(\rho_{F}, V_{F})$ is faithful, and that it is irreducible if and only if $F$ is primitive. We also require that $V_{F.G}=V_{F}\oplus V_{G}$ and that $V_{F\otimes G}=V_{F}\otimes V_{G}$.

D) $F(s)=L(s,\rho_{F})$

My questions are:

1) does every L-manifold give rise naturally to a motive?

2) If so, is every L-function motivic? Can one say that any motivic L-function arises from a Galois representation and conversely?

• I think this is a very cool question. Could you explain, or give a reference, to the notion of the L-functions "canonical" representation? Part of this includes the definition of the field $K_F$ as well. – WSL Apr 26 '15 at 12:22
• Okay thanks, but I still would like to know what the canonical representation is. Also, is there a method for attaching to an L-function $H$ an abstract variety, $X_H$, you are refering to? – WSL Apr 26 '15 at 13:33
• There's no such thing as the tensor product of abstract varieties, which you use in the definition of $X_F$. Your second question doesn't really make sense - each variety has a motive attached to it, but you can't interpret a variety as a motive - and your third question seems like a restatement of one of the Langlands conjectures. – Will Sawin Apr 26 '15 at 22:04