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This question is a follow-up to Are there infinitely many L-rigs? and to Is an automorphic form of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ determined by its L-function?.

I copy paste a deepl translation of an old answer I got to a question of mine on a French math forum:

The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to \operatorname{Set}$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to \operatorname{Isom}(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=\operatorname{Aut}(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take

• "space $X$" = a field $K$ (we denote $X=\operatorname{Spec}(K)$)
• "covering $Y$ of $X$" = finite separable extension $L/K$
• "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field.

Then the associated "fiber in $x$" functor is the functor $\omega_{x}:L/K\mapsto \operatorname{Hom}_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $\operatorname{Gal}(K^\text{sep}/K)$ of $K$ (Galois group of a separable closure).

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the map that sends an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ to the associated L-function and such a "fiber functor" that would allow to say that the automorphism group/"fundamental group" of $\mathcal{M}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals? Can such an analogy be built on the automorphic representation side to make the map $\pi\mapsto L_{\pi}$ an isomorphism of coverings, so as to view the symmetries of a given L-function as arising from intertwinning operators?

This question is a follow-up to Are there infinitely many L-rigs? and to Is an automorphic form of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ determined by its L-function?.

I copy paste a deepl translation of an old answer I got to a question of mine on a French math forum:

The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to \operatorname{Set}$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to \operatorname{Isom}(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=\operatorname{Aut}(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take

• "space $X$" = a field $K$ (we denote $X=\operatorname{Spec}(K)$)
• "covering $Y$ of $X$" = finite separable extension $L/K$
• "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field.

Then the associated "fiber in $x$" functor is the functor $\omega_{x}:L/K\mapsto \operatorname{Hom}_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $\operatorname{Gal}(K^\text{sep}/K)$ of $K$ (Galois group of a separable closure).

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the sequence of intermediate L-rigs between $\mathcal{L}_{0}$ (being the analogue of $K$) and $\mathcal{M}$ (being the analogue of $\Xi$) that would allow to say that the automorphism group/"fundamental group" of $\mathcal{L_{0}}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals? Can such an analogy be built on the automorphic representation side to make the map $\pi\mapsto L_{\pi}$ an isomorphism of coverings, so as to view the symmetries of a given L-function as arising from intertwining operators?

"The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to Set$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to Isom(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=Aut(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take - space $X$" = a field $K$ (we denote $X=Spec(K)$- "covering $Y$ of $X$" = finite separable extension $L/K$. - "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field. Then the associated "fiber in $x$" functor is the functor $ω_{x}:L/K\mapsto Hom_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $Gal(K^{sep}/K)$ of $K$ (Galois group of a separable closure)."

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the sequence of intermediate L-rigs between $\mathcal{L}_{0}$ (being the analogue of $K$) and $\mathcal{M}$ (being the analogue of $\Xi$) that would allow to say that the automorphism group/"fundamental group" of $\mathcal{L_{0}}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals? Can such an analogy be built on the automorphic representation side to make the map $\pi\mapsto L_{\pi}$ an isomorphism of coverings, so as to view the symmetries of a given L-function as arising from intertwinning operators?

The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to \operatorname{Set}$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to \operatorname{Isom}(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=\operatorname{Aut}(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take

• "space $X$" = a field $K$ (we denote $X=\operatorname{Spec}(K)$)
• "covering $Y$ of $X$" = finite separable extension $L/K$
• "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field.

Then the associated "fiber in $x$" functor is the functor $\omega_{x}:L/K\mapsto \operatorname{Hom}_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $\operatorname{Gal}(K^\text{sep}/K)$ of $K$ (Galois group of a separable closure).

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the map that sends an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ to the associated L-function and such a "fiber functor" that would allow to say that the automorphism group/"fundamental group" of $\mathcal{M}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals? Can such an analogy be built on the automorphic representation side to make the map $\pi\mapsto L_{\pi}$ an isomorphism of coverings, so as to view the symmetries of a given L-function as arising from intertwinning operators?

The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to \operatorname{Set}$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to \operatorname{Isom}(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=\operatorname{Aut}(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take

• "space $X$" = a field $K$ (we denote $X=\operatorname{Spec}(K)$)
• "covering $Y$ of $X$" = finite separable extension $L/K$
• "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field.

Then the associated "fiber in $x$" functor is the functor $\omega_{x}:L/K\mapsto \operatorname{Hom}_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $\operatorname{Gal}(K^\text{sep}/K)$ of $K$ (Galois group of a separable closure).

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the map that sends an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ to the associated L-function and such a "fiber functor" that would allow to say that the automorphism group/"fundamental group" of $\mathcal{M}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals?

"The idea is that to any "space" $X$ one can associate the category of its coverings $R(X)$. If we choose a "geometric point", we have a "fiber in $x$" functor: $\omega_{x}:R(X)\to Set$ which to a covering $f:Y\to X$ associates $f^{-1}(x)$. If we are given a path $\gamma:x_{0}\to x_{1}$, then, for any $y_{0}\in f^{-1}(x_{0})$, it rises to a single path $\tilde{\gamma}$ of origin $y_{0}$. Considering the end of this path, we get a point $y_{1}\in f^{-1}(x_{1})$. And this depends only on the homotopy class of $\gamma$. So we have an application that to a homotopy class of paths associates an isomorphism of fiber functors: $\pi_{1}(X;x_{0},x_{1})\to Isom(\omega_{x_{0}},\omega_{x_{1}})$. Grothendieck's brilliant remark is that this application is an isomorphism! In particular, $\pi_{1}(X,x)=Aut(\omega_{x})$ and the right-hand side term thus gives a purely algebraic definition (we do not use the topology of $\mathbb{R}$) of the fundamental group.

Where we join Galois theory is that, if we take - space $X$" = a field $K$ (we denote $X=Spec(K)$- "covering $Y$ of $X$" = finite separable extension $L/K$. - "geometric point" = embedding $x:K\to\Xi$ into an algebraically closed field. Then the associated "fiber in $x$" functor is the functor $ω_{x}:L/K\mapsto Hom_{K}(L,\Xi)$ which to $L/K$ associates its $K$-embeddings in $\Xi$. The theory applies and we find that the "fundamental group" of $K$ (i.e., the group of automorphisms of the fiber functor $\omega_{\Xi}$) is the absolute Galois group $Gal(K^{sep}/K)$ of $K$ (Galois group of a separable closure)."

So I would like to know if, provided that an L-rig can be extended into a field, one can establish a parallel between the sequence of intermediate L-rigs between $\mathcal{L}_{0}$ (being the analogue of $K$) and $\mathcal{M}$ (being the analogue of $\Xi$) that would allow to say that the automorphism group/"fundamental group" of $\mathcal{L_{0}}$ is an absolute Galois group. If yes, is it provably the absolute Galois group of the rationals? Can such an analogy be built on the automorphic representation side to make the map $\pi\mapsto L_{\pi}$ an isomorphism of coverings, so as to view the symmetries of a given L-function as arising from intertwinning operators?