I read years ago in the great French book "l'aventure des nombres" (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to "the" algebraic closure of $Q$.

In other words, denoting by $\mathcal{M}$ the maximal "L-rig", i.e. rig whose elements are L-functions, I expect $\operatorname{Aut}(\mathcal{M})$ to be isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. The key idea is to define the symmetry group $\operatorname {Sym}(F)$ of a given element $F$ of $\mathcal{M}$ as made of elements of $\operatorname {Aut}(\mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $\operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip. That way, one has to restrict the potential candidates in $\operatorname{Aut}(\mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements of finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?

I read years ago in the great French book "l'aventure des nombres" (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to "the" algebraic closure of $Q$.

In other words, denoting by $\mathcal{M}$ the maximal "L-rig", i.e. rig whose elements are L-functions, I expect $\operatorname{Aut}(\mathcal{M})$ to be isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. The key idea is to define the symmetry group $\operatorname {Sym}(F)$ of a given element $F$ of $\mathcal{M}$ as made of elements of $\operatorname {Aut}(\mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $\operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip, preserving this multiset. This group consists of isometries of the complex plane, where we identify two such isometries $f$ and $g$ if any non trivial zero has the same image under the action of $f$ or of $g$. Such a group is finite. RH is equivalent to this group being trivial iff $F$ is non self-dual, and of order $2$ otherwise. 

That way, one has to restrict the potential candidates in $\operatorname{Aut}(\mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements of finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH eliminating discontinuous automorphisms of $C$? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?

I read years ago in the great French book "l'aventure des nombres" (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to "the" algebraic closure of $Q$.

In other words, denoting by $\mathcal{M}$ the maximal "L-rig", i.e. rig whose elements are L-functions, I expect $\operatorname{Aut}(\mathcal{M})$ to be isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. The key idea is to define the symmetry group $\operatorname {Sym}(F)$ of a given element $F$ of $\mathcal{M}$ as made of elements of $\operatorname {Aut}(\mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $\operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip. That way, one has to restrict the potential candidates in $\operatorname{Aut}(\mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements if finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?

I read years ago in the great French book "l'aventure des nombres" (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to "the" algebraic closure of $Q$.

In other words, denoting by $\mathcal{M}$ the maximal "L-rig", i.e. rig whose elements are L-functions, I expect $\operatorname{Aut}(\mathcal{M})$ to be isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. The key idea is to define the symmetry group $\operatorname {Sym}(F)$ of a given element $F$ of $\mathcal{M}$ as made of elements of $\operatorname {Aut}(\mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $\operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip. That way, one has to restrict the potential candidates in $\operatorname{Aut}(\mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements of finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?