07/07/2022, 09:25 PM
Meanwhile I have two minutes. Conj is a field automorphism of the complexes but that doesn't mean that we have to look for it's fractional iterates inside the group \({\rm Aut}(\mathbb C)\) unless we ask for an \(\mathbb R\)-action over \(\mathbb C\) that acts by automorphisms. Sure since conjugation is involution, hence isomorphism, we'd like its fractional iterates to be iso, since composition preserves bijectivity... and also because we'd like fractional iterations to respect the field structure...This however seems too restrictive. We'd get something very rigid and similar to a weakened scalar multiplication...
Let \(\alpha(r,z)\) be an action over the complexes: \(\alpha(0,z)=z\) and \(\alpha(r+s,z)=\alpha(r,\alpha(s,z))\) s.t. \(\alpha(1,z)={\rm conj}(z)\).
First notice that if we ask it to act by automorphisms then we induce a group homomorphisms \(\alpha:\mathbb R \to \rm{ Aut}(\mathbb C) \). The "intrinsic time" of the iteration is given by the group quotient \(\mathbb R / {\rm ker}\alpha\) where \(2\mathbb Z\subseteq {\rm ker}\alpha =\{k\in\mathbb R\, : \,\alpha(r,z)=z \} \), since conjugation being an involution \({\rm conj}^{2n}={\rm id}_{\mathbb C}\). If the kernel is the even numbers then the intrinsic iterates looks like the circle \(S^1 \simeq \mathbb R/2\mathbb Z\).
In other words we are looking for \(S^1\)-actions \(\alpha:{}S^1\times \mathbb C\to \mathbb C\) over \(\mathbb C\). I wonder what this entails at the level of James' remarks. If we consider also wild automorphisms... maybe this wold entails some contradiction... maybe idk. I was thinking about failure of injectivity in \(z\) of iterates \(\alpha(r,z)={\rm conj}^r(z)\) when the time belongs to the circle \(r\in S^1\)... while all of them should be injective since are field morphisms....
this seems fascinating but got no time to study it more.
The second critical point is that we would also have \(\alpha(t,0)=0\) and \(\alpha(r,z+w)=\alpha(r,z)+\alpha(r,w)\), and \(\alpha(t,1)=1\) and \(\alpha(r,zw)=\alpha(r,z)\alpha(r,w)\). This seems too rigid to have any room to play around... probably it is too rigid that ruels out every non-trivial solution....
Just a feeling...
Let \(\alpha(r,z)\) be an action over the complexes: \(\alpha(0,z)=z\) and \(\alpha(r+s,z)=\alpha(r,\alpha(s,z))\) s.t. \(\alpha(1,z)={\rm conj}(z)\).
First notice that if we ask it to act by automorphisms then we induce a group homomorphisms \(\alpha:\mathbb R \to \rm{ Aut}(\mathbb C) \). The "intrinsic time" of the iteration is given by the group quotient \(\mathbb R / {\rm ker}\alpha\) where \(2\mathbb Z\subseteq {\rm ker}\alpha =\{k\in\mathbb R\, : \,\alpha(r,z)=z \} \), since conjugation being an involution \({\rm conj}^{2n}={\rm id}_{\mathbb C}\). If the kernel is the even numbers then the intrinsic iterates looks like the circle \(S^1 \simeq \mathbb R/2\mathbb Z\).
In other words we are looking for \(S^1\)-actions \(\alpha:{}S^1\times \mathbb C\to \mathbb C\) over \(\mathbb C\). I wonder what this entails at the level of James' remarks. If we consider also wild automorphisms... maybe this wold entails some contradiction... maybe idk. I was thinking about failure of injectivity in \(z\) of iterates \(\alpha(r,z)={\rm conj}^r(z)\) when the time belongs to the circle \(r\in S^1\)... while all of them should be injective since are field morphisms....
this seems fascinating but got no time to study it more.
The second critical point is that we would also have \(\alpha(t,0)=0\) and \(\alpha(r,z+w)=\alpha(r,z)+\alpha(r,w)\), and \(\alpha(t,1)=1\) and \(\alpha(r,zw)=\alpha(r,z)\alpha(r,w)\). This seems too rigid to have any room to play around... probably it is too rigid that ruels out every non-trivial solution....
Just a feeling...
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)