Little note: before I forget it.
I believe that some key ingredient is missing in the "iteration machine", how Gottfried calls it, before we can go fully equivariant with it.
First notice that Goodstein Hyperoperations are just a faithful extension of the Peano Arithmetic Axioms.... In other words Peano Theory, extended via Goodstein, builds up arithmetical operations just by superfunctions... but since we are over \(\mathbb N\), we automatically obtain \(\mathbb N\) iterations....
In the case of \(T\) a more general monoid we have that superfunctions identities do no coincide anymore with having \(T\)-iterations, iterations monoids with time in, parametrized by, \(T\).
Since a Goodstein sequence follows the defining mechanism of Peano, extending it, maybe a truly equivariant extension of Goodstein can be unlocked only after we upgrade the Peano arithmetic core itself.
Attempt 1: first fix a monoid \(M\), we take it to be the set of time, every iteration will have iteration index in there. We think of it as if it was \(\mathbb N\) in the Peano Arithmetic.
The classical recursion principle implies that given an action \(f:(\mathbb N)\to \mathbb N^\mathbb N\) over \(\mathbb N\), i.e. \(f^n(m)\) we can define an unique sequence of functions \({\sf R}f:\mathbb N\to \mathbb N^\mathbb N\) \[({\sf R} f)(n)=(f^n(0),f^n(1),...,f^n(m),...)\]
i.e. the recursion operator is a map of sets \({\sf R}_m:{\rm Mon}(\mathbb N,{\rm End}_{\rm Set}(\mathbb N))\to {{\rm Hom}_{\rm Set}(\mathbb N)}\) that takes monoid morphisms, discrete iterations, and gives set endomaps of the monoid. We can write
\[({\sf R} S)(n)=(n,1+n,2+n,...,m+n,...)\]
Let \(H_{1}(n)=b+n\) then \(({\sf R} H_{1})(n)=(H_{2}(n),1+bn,2+bn,...,m+bn,...)\);
where \(H_{2}(n)=bn\) and then \(({\sf R} H_{2})(n)=(0,H_{3}(n),2b^n,...,mb^n,...)\), and so on.
Let's change the game for a monoid \(T\). Define the recursive operators \({\sf R}_s\) for \(s\in T\). It takes a monoid morphism \(\varphi:T\to T^T\) as an input and gives the monoid map \({\sf R}_s\varphi : T\to T\) as output.
\[{\sf R}_s\varphi(t)=\varphi_t(s)\]
\[{\sf R}_m:{\rm Mon}(T,{\rm End}_{\rm Set}(T))\to {{\rm Hom}_{\rm Set}(T)}\]
where \(({\sf R}_s\varphi)(r+t)=\varphi_{r}(\varphi_t(s))=\varphi_{r}({\sf R}_s\varphi)(t))\). i.e. it is a \(T\)-equivariant function from the regular \(T\)-action to the \(T\)-action \(\varphi\)
\[{\sf R}_a: {\rm Set}^T\to T/{\rm Set}^T \]
As we can see, the recursion operator takes \(T\)-actions over \(T\) and gives not another \(T\)-action over \(T\), as it instead happens for \(T=\mathbb N\) (because an action for the integers is just an endufunction, i.e. both notions coincide) but gives a mere endofunction of the support set of the monoid \(T\).
Question: given a \(T\)-iteration over \(T\), call it \(\varphi_t:T\to T\) we can associate its sequence of recursions over \(T\)
\[{\sf R}\varphi (t)|_s=\varphi_t(s)\]
can we extract naturally from this data a new \(T\)-iteration \((\uparrow \varphi)_t:T\to T\)? What properties should we ask for it?
It seems that the solution doesn't exists and if it does it has to do with finding a mysterious operation
\[\uparrow: T/{\rm Set}^T \to {\rm Set}^T\]
Explicitly we should define a process that takes as input a function \(\Psi:T\to T\) that for some \(\varphi_t\) satisfies \(\Psi(r+t)=\varphi_r(\Psi(t))\), and give as output a something that we can see as a \(T\)-iteration of \(\Psi\), i.e. \[\Psi_{s+r}(t)=\Psi_{s}(\Psi_{r}((t));\,\,\,\Psi_u(t)=\Psi(t)\]
Second attempt: let's follow the previous philosophy.
We assume we start with a commutative monoid \(T\). We sees it as the base number system where \(0_T\) plays the role of the Peano's zero.
Rank=0 iteratively
We define the monoid operation \(+_T\) as the rank zero Goodstein hyperoperation and we build up from this axiom. It is associated with the regular \(T\)-action of the monoid over itself: the identity monoid morphism \(i_0:T\to T\) where the action is given by \(s[0]_T t=i_0(s)+_T t=s+_T t\)
Rank=1 equivariantly
For each \(s\in T\) we define rank one functions as the function \(h_s(1,-):T\to T\) that satisfy \[h_s(1,0_T)=s;\,\,\,h_s(1,r+t)=r+h_s(1,t)\]
This means that we are taking as \(T\)-equivariant rank zero the regular \(T\) action and as rank one maps turning the regular \(T\) action in itself,
\[h_s(1,t)=t+h_s(1,0_T)=t+s\]
Rank=1 iteratively
In order to repeat the process having an equivariant function is not enough, we must define a \(T\)-iteration that behaves as rank one. An iteration must be a monoid morphism of the form \(i_1:T\to T^T\). In this case we want to \(T\)-iterate the monoid operation itself. But how?
How to jump from rank 1 to rank 2? We look for maps \(h_s(2,-):T\to T\) satisfying \[h_s(2,0_T)=0_T;\,\,\,h_s(2,r+t)=h^r_s(1,h_s(2,t)t)\]
\[h_s(2,t)=h^t_s(1,0_T)\]
I believe that some key ingredient is missing in the "iteration machine", how Gottfried calls it, before we can go fully equivariant with it.
First notice that Goodstein Hyperoperations are just a faithful extension of the Peano Arithmetic Axioms.... In other words Peano Theory, extended via Goodstein, builds up arithmetical operations just by superfunctions... but since we are over \(\mathbb N\), we automatically obtain \(\mathbb N\) iterations....
In the case of \(T\) a more general monoid we have that superfunctions identities do no coincide anymore with having \(T\)-iterations, iterations monoids with time in, parametrized by, \(T\).
Since a Goodstein sequence follows the defining mechanism of Peano, extending it, maybe a truly equivariant extension of Goodstein can be unlocked only after we upgrade the Peano arithmetic core itself.
Attempt 1: first fix a monoid \(M\), we take it to be the set of time, every iteration will have iteration index in there. We think of it as if it was \(\mathbb N\) in the Peano Arithmetic.
The classical recursion principle implies that given an action \(f:(\mathbb N)\to \mathbb N^\mathbb N\) over \(\mathbb N\), i.e. \(f^n(m)\) we can define an unique sequence of functions \({\sf R}f:\mathbb N\to \mathbb N^\mathbb N\) \[({\sf R} f)(n)=(f^n(0),f^n(1),...,f^n(m),...)\]
i.e. the recursion operator is a map of sets \({\sf R}_m:{\rm Mon}(\mathbb N,{\rm End}_{\rm Set}(\mathbb N))\to {{\rm Hom}_{\rm Set}(\mathbb N)}\) that takes monoid morphisms, discrete iterations, and gives set endomaps of the monoid. We can write
\[({\sf R} S)(n)=(n,1+n,2+n,...,m+n,...)\]
Let \(H_{1}(n)=b+n\) then \(({\sf R} H_{1})(n)=(H_{2}(n),1+bn,2+bn,...,m+bn,...)\);
where \(H_{2}(n)=bn\) and then \(({\sf R} H_{2})(n)=(0,H_{3}(n),2b^n,...,mb^n,...)\), and so on.
Let's change the game for a monoid \(T\). Define the recursive operators \({\sf R}_s\) for \(s\in T\). It takes a monoid morphism \(\varphi:T\to T^T\) as an input and gives the monoid map \({\sf R}_s\varphi : T\to T\) as output.
\[{\sf R}_s\varphi(t)=\varphi_t(s)\]
\[{\sf R}_m:{\rm Mon}(T,{\rm End}_{\rm Set}(T))\to {{\rm Hom}_{\rm Set}(T)}\]
where \(({\sf R}_s\varphi)(r+t)=\varphi_{r}(\varphi_t(s))=\varphi_{r}({\sf R}_s\varphi)(t))\). i.e. it is a \(T\)-equivariant function from the regular \(T\)-action to the \(T\)-action \(\varphi\)
\[{\sf R}_a: {\rm Set}^T\to T/{\rm Set}^T \]
As we can see, the recursion operator takes \(T\)-actions over \(T\) and gives not another \(T\)-action over \(T\), as it instead happens for \(T=\mathbb N\) (because an action for the integers is just an endufunction, i.e. both notions coincide) but gives a mere endofunction of the support set of the monoid \(T\).
Question: given a \(T\)-iteration over \(T\), call it \(\varphi_t:T\to T\) we can associate its sequence of recursions over \(T\)
\[{\sf R}\varphi (t)|_s=\varphi_t(s)\]
can we extract naturally from this data a new \(T\)-iteration \((\uparrow \varphi)_t:T\to T\)? What properties should we ask for it?
It seems that the solution doesn't exists and if it does it has to do with finding a mysterious operation
\[\uparrow: T/{\rm Set}^T \to {\rm Set}^T\]
Explicitly we should define a process that takes as input a function \(\Psi:T\to T\) that for some \(\varphi_t\) satisfies \(\Psi(r+t)=\varphi_r(\Psi(t))\), and give as output a something that we can see as a \(T\)-iteration of \(\Psi\), i.e. \[\Psi_{s+r}(t)=\Psi_{s}(\Psi_{r}((t));\,\,\,\Psi_u(t)=\Psi(t)\]
Second attempt: let's follow the previous philosophy.
We assume we start with a commutative monoid \(T\). We sees it as the base number system where \(0_T\) plays the role of the Peano's zero.
Rank=0 iteratively
We define the monoid operation \(+_T\) as the rank zero Goodstein hyperoperation and we build up from this axiom. It is associated with the regular \(T\)-action of the monoid over itself: the identity monoid morphism \(i_0:T\to T\) where the action is given by \(s[0]_T t=i_0(s)+_T t=s+_T t\)
Rank=1 equivariantly
For each \(s\in T\) we define rank one functions as the function \(h_s(1,-):T\to T\) that satisfy \[h_s(1,0_T)=s;\,\,\,h_s(1,r+t)=r+h_s(1,t)\]
This means that we are taking as \(T\)-equivariant rank zero the regular \(T\) action and as rank one maps turning the regular \(T\) action in itself,
\[h_s(1,t)=t+h_s(1,0_T)=t+s\]
Rank=1 iteratively
In order to repeat the process having an equivariant function is not enough, we must define a \(T\)-iteration that behaves as rank one. An iteration must be a monoid morphism of the form \(i_1:T\to T^T\). In this case we want to \(T\)-iterate the monoid operation itself. But how?
How to jump from rank 1 to rank 2? We look for maps \(h_s(2,-):T\to T\) satisfying \[h_s(2,0_T)=0_T;\,\,\,h_s(2,r+t)=h^r_s(1,h_s(2,t)t)\]
\[h_s(2,t)=h^t_s(1,0_T)\]
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)