Posts: 377
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Joined: May 2013
Before I forget it, let me post here a note for the future.
As defined formally by me elsewhere (document in preparation), formal in the sense of deprived of interpretation/representation as endofunction, a formal (pre-)Goodstein sequence inside a pointed non-commutative monoid \((M,s)\) with ranks belonging to an \(\mathbb N\)-iteration \((J,{(-)}^+)\) is a function \({\bf h}:J\to M\) satisfying the system of \(\mathbb N\)-equivariance condition (aka superfucntion equations)
\[{\bf h}_{j^+}s={\bf h}_{j}{\bf h}_{j^+}\]
Since after the latest discussions on the forum I have started to internalize and understand fully the relationship between being a superfucntion, being a family of superfunctions and being an iteration I believe that the previous version of the Goodstein f.equation shows itself as just the \(\mathbb N\)-equivariant version of a more general \(A\)-equivariant Goodstein f.equation.
Definition (\(A\)-equivariant (pre-)Goodstein equation): Fix the monoid of time \(A\) and an unit of time \(u\in A\). Take an \(A\)-pointed non-comm. monoid \((M,s)\), it will be our support and monoid morphism \(s:A\to M\) will be called the seed. Let \((J,{(-)}^+)\) be a an \(\mathbb N\)-action, called the space of ranks. An \(A\)-equivariant (pre-)Goodstein map is a map \({\bf h}:J\to {\rm Hom}_{\rm Mon}(A,M)\), i.e. a sequence of \(A\)-iterations/monoid homomorphisms \({\bf h}_j:A\to M\), over \(M\) indexed by \(J\) that satisfies the \(A\)-equivariant Goodstein f.equation over the seed \(s\) wrt the unit of time \(u\):
\[\forall a\in A.\,{\bf h}_{j^+}(u)s(a)={\bf h}_{j}(a){\bf h}_{j^+}(u)\]
Example: look for the special case \(A=\mathbb R\) adn \(u=1\), then this means the \(\mathbb R\)-equivariant goodstein functional equation doesn't ask the next hyperoperation to be just a superfunction of the previous but also to respect \(\mathbb R\)-iterations of the previous. This means the new definition is more close to the naive expectation of what we would like Goodstein hyperoperations to be. This means we have a sequence of \(\mathbb R\)-iterations \(f_j^t\) and that \[f^{\circ 1}_{j^+}\circ s^{\circ a}=f^{\circ a}_{j}\circ f^{\circ 1}_{j^+}\]
Open problem. some can clearly see that this is not perfection. The rank variable still belongs to the world of \(\mathbb N\)-iterations/actions. The ultimate Goodstein functional equation should be \(B\)-equivariant also i the rank variable... but how? The only way I can think of is by iterating group conjugation. We need to ask \(M\) to be a group. In this way, maybe we can find to make \(A\)-equivariant (pre-)Goodstein map \({\bf h}:J\to {\rm Hom}_{\rm Mon}(A,M)\) into a \(B\)-equivariant map, for some monoid \(B\) acting on the space of ranks.... but how?
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 377
Threads: 30
Joined: May 2013
07/18/2022, 05:40 PM
(This post was last modified: 07/22/2022, 01:11 AM by MphLee.)
Additional note. the problem of extending \(B\)-equivariance to the space of ranks need to be analyzed as follows. We need to keep track of where our maps lives. There are six cases of increasing complexity
\[\begin{array}[|ccc|cc|]
&&&&{\rm ranks\, equiv.}&\\
\hline
{\rm set\, theoretic\,}{\bf h}&over\, monoid\, M&{\rm satisfying\, goodstein} &B=1& A=\mathbb N\\
{\rm set\, theoretic\,}{\bf h}&over\, monoid\, M&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B=1& A\\
\hline
\mathbb N{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying\, goodstein} &B=\mathbb N&A=\mathbb N\\
\mathbb N{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B=\mathbb N&A\\
\hline
B{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying\, goodstein} &B&A=\mathbb N\\
B{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B&A\\
\hline
\end{array}\]
Classical case over monoids. In the case of \(M\) a monoid \(\mathbb N\)-equivariant goodstein map is a map that lives in the category \(1{\rm -Act}={\rm Set}\)
\[{\bf h}\in {\rm Hom}_{\rm Set}(J,{\rm Hom}_{\rm Mon}(\mathbb N,M))\]
that satisfies additional condition \({\bf h}_{j^+}s={\bf h}_{j}{\bf h}_{j^+}\), i.e. each \({\bf h}_j\) is a map in \(\mathbb N{\rm -Act}={\rm Set}^{B\mathbb N}\).
Equivariant case over monoids. In the case the case of \(A\)-equivariant goodstein maps is a map that still lives in \(1{\rm -Act}={\rm Set}\)
\[{\bf h}\in {\rm Hom}_{\rm Set}(J,{\rm Hom}_{\rm Mon}(A,M))\]
but this time they satisfies \(\forall a\in A.\, {\bf h}_{j^+}(u)s(a)={\bf h}_{j}(a){\bf h}_{j^+}(u)\), i.e. each \({\bf h}_j\) is a map in \(A{\rm -Act}={\rm Set}^{BA}\).
In the case of \(M=G\) a group an incredible simplification becomes available for \(\mathbb N\)-equivariant goodstein maps: since \({\rm Hom}_{\rm Mon}(\mathbb N,G)\simeq G\) and on this set we can define the subfunction map \(\Sigma_s:G\to G\), sending \(g\in G\mapsto{ }gsg^{-1} \) we can upgrade the object \({\rm Hom}_{\rm Mon}(\mathbb N,G)\) from the category \(1{\rm -Act}={\rm Set}\) making it into an object \(\Sigma^G_S:=(G,\Sigma_s)\) of \(\mathbb N{\rm -Act}\), the same place where the space of ranks \(J=(J,(-)^-)\) lives. It turns out that the goodstein can be sinthetized by asking for maps that live in the category \(\mathbb N{\rm -Act}\).
Classical case over groups. In the case of \(M=G\) a group, a \(\mathbb N\)-equivariant goodstein map is just map that lives in the category \(\mathbb N{\rm -Act}\) of discrete dynamical systems
\[{\bf h}\in {\rm Hom}_{\mathbb N{\rm -Act}}(J,\Sigma_s^G)\]
Nothing more should be asked! This already implies it satisfies \({\bf h}_{j}s={\bf h}_{j^-}{\bf h}_{j}\).
Equivariant case over groups. Here we have a big obstacle since \({\rm Hom}_{\rm Mon}(A,G)\) is not equivalent to \(G\) in general and on this space of group homorphisms is not closed under the map \(\Sigma_s\) unless \(G\) is abelian... but in this case everything becomes trivial: even if it was it is not clear this is what we need to enforce the equivariant goodstein condition. The question is: what is the endofunction over \({\rm Hom}_{\rm Mon}(A,G)\) that we should consider and why? This is a big problem.
Maybe we can face this from a synthetic point of view. We need an object \(\mathfrak S(A,G)\in \mathbb N{\rm -Act}\) of the category \(\mathbb N{\rm -Act}\) associated with the group \(G\) that behaves as if it were the space of \(A\)-iterations over \(G\) and closed under taking subfunctions... something that is not possible as stated. We then look for maps
\[{\bf h}\in {\rm Hom}_{\mathbb N{\rm -Act}}(J,\mathfrak S(A,G) ) \]
And such that, given enough extra structure, we can somehow reconstruct the condition \(\forall a\in A. \Sigma_{s(a)}({\bf h}_{j}(u))={\bf h}_{j^-}(a)\).
Two more cases are to be studied and that completes the study of goodstein maps: the case with equivariant ranks but classical maps, and the case with everything equivariant.
\(B\)-Equivariant ranks and classical Goodstein over group. The first is again pretty straightforward. Just take any \(B\)-iteration of the map \(\Sigma_s:G\to G\), i.e. a map \({\boldsymbol \Sigma}:B\to G^G\) s.t. \( \forall \alpha,\beta\in B.\, \, {\boldsymbol \Sigma}(\alpha+\beta,g)={\boldsymbol \Sigma}(\alpha,{\boldsymbol \Sigma}(\beta,g))\) s.t. for some \(\upsilon \in B\) we have \({\boldsymbol \Sigma}(\upsilon,g)=gsg^{-1}\). And as space of ranks take an object \(J\in B{\rm -Act} \)
We look for maps \[{\bf h}\in {\rm Hom}_{B{\rm -Act}}(J,{\boldsymbol \Sigma})\]
This is actually iterating conjugation... it is really the way to access authentic non-integer ranks as started by Trappmann in his 2007 2008 thread: non-natural operation ranks.
But is hard... and we really should be interested when \(M\) is not a group... thus making everything harder.
\(B\)-Equivariant ranks and \(A\)-equivariant Goodstein over group. Here the model becomes highly non-trivial... I still don't know how to treat this... Probably the way to go is to upgrade this discussion, that is really at set theoretic level to categories. If we turn every object into a category... and every map into a functor... maybe all the problem presented here will dissolve.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 377
Threads: 30
Joined: May 2013
Let's switch to the symbolic dynamic version, aka the shift goodstein condition \(\sigma({\bf h})\cdot s={\bf h}\cdot \sigma({\bf h})\).
Maybe we the right way to go is by obtaining the \(A\)-equivariant shift goodstein as a normal equation inside the structure \({\rm Hom}_{\rm Mon}(A,M)^J\): \({\rm ev}_u(\sigma({\bf h}))\cdot s={\bf h}\cdot {\rm ev}_u(\sigma({\bf h}))\).
A quick check \(({\rm ev}_u(\sigma({\bf h}))\cdot s)_j=({\rm ev}_u(\sigma({\bf h})))_j\cdot s_j \) the problem is that \(({\rm ev}_u(\sigma({\bf h})))_j\) is not defined in any meaningful way unless we move to another codomain by transporting the equation along the function \[u^*: {\rm Hom}_{\rm Mon}(A,M)^J\to{\rm Hom}_{\rm Mon}(\mathbb N,M)^J\]
but then it just is a rephrasing of the previous definition that doesn't add anything of value... it seems
\[\forall a\in A.\, u^*(\sigma({\bf h}))\cdot a^*(s)=a^*({\bf h})\cdot u^*(\sigma({\bf h}))\]
or maybe isn't... since there is a canonical map \(EV({\bf x},a)=a^*({\bf x})\)
\[EV: {\rm Hom}_{\rm Mon}(A,M)^J\times A\to{\rm Hom}_{\rm Mon}(\mathbb N,M)^J\]
define then for \(M=G\) the map \(\mathcal F ({\bf h}, a)=EV(\sigma{\bf h},u)\cdot EV(s,a)\cdot EV(\sigma{\bf h},u)^{-1} \) and ask for \(\bf h\) solutions of
\[\forall a.\,\mathcal F({\bf h},a)=EV({\bf h},a)\]
Apply curry to the second variable, obtain that \(A\)-equivariant goodstein maps are solutions of \[{\bar {\mathcal F}}({\bf h})=\bar{EV}({\bf h})\]
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 377
Threads: 30
Joined: May 2013
The last attempt seems too complicated to give low hanging fruits
I think it is better to start from this table
\[\begin{array}[|ccc|cc|]
&&&&{\rm ranks\, equiv.}&\\
\hline
{\rm set\, theoretic\,}{\bf h}&over\, monoid\, M&{\rm satisfying\, goodstein} &B=1& A=\mathbb N\\
{\rm set\, theoretic\,}{\bf h}&over\, monoid\, M&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B=1& A\\
\hline
\mathbb N{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying\, goodstein} &B=\mathbb N&A=\mathbb N\\
\mathbb N{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B=\mathbb N&A\\
\hline
B{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying\, goodstein} &B&A=\mathbb N\\
B{\rm -Act}\, {\rm theoretic\,}{\bf h}&over\, group\, G&{\rm satisfying}\, A{\rm -equivariant\, goodstein} &B&A\\
\hline
\end{array}\]
and refine it to a big picture.
Here the general scheme is: we take a map inside a certain category, from an object of ranks to an object of the support, and ask it to satisfy a condition, be it goodstein condition or some equivariant goodstein, or variant of it. All of this must be fixed by appropriate forgetfull functors \(U\) sending objects to the appropriate categories. From now on I'll identify the categories \(A{\rm - Act}\) and the functor (co-presheaves) category \([A,{\rm Set}]\) whenever \(A\) is a monoid and I'll omit the \(B\)-functor in order to keep notation clear. I'll use the common practice of denoting the hom-sets of a category \(\mathcal C\) by \(\mathcal C(X,Y)\) instead of \({\rm Hom}_{\mathcal C}(X,Y)\) except in the case of sets.
Set theoretic goodstein maps
This time I'll break the scheme differently, leaving aside everything that is \(A\)-equivariant since it need a qualitative upgrade that is intermediate to going categorical.
\[\begin{array}[|cccc|]
f{\rm maps\, in}&{\bf category}&{from\,the\,\,{\bf ranks}\,\, object}&{to\,the\,\,{\bf support}\,\, object}&&{\rm satisfying}&\\
\hline
{\bf h}\in{\rm Hom}(UJ,UM)&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{+}{\to}J)\in \mathbb N{\rm - Act}& M\in{\rm Mon}&&\forall j\in UJ.\, {\bf h}_{j^+}s={\bf h}_{j}{\bf h}_{j^+}&{\bf goodstein\, equation\,\, (GE)}\\
{\bf h}\in{\rm Hom}(UJ,UM)&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{-}{\to}J)\in \mathbb N{\rm - Act}& M\in{\rm Mon}&&\forall j\in UJ.\, {\bf h}_{j}s={\bf h}_{j^-}{\bf h}_{j}&{\bf anti}\!-\!{\rm GE\,\, (GE^-)}\\
{\bf h}\in{\rm Hom}(UJ,UG)&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{+}{\to}J)\in \mathbb N{\rm - Act}& G\in{\rm Grp}&&{\bf h }\in {\rm fix}((\Sigma^G_S)_* \circ (+)^* )&{\bf GE\,\, over\, grps}\\
{\bf h}\in{\rm Hom}(UJ,UG)&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{-}{\to}J)\in \mathbb N{\rm - Act}& G\in{\rm Grp}&&\forall j\in UJ.\, \Sigma^G_s({\bf h}_{j}^{-1})={\bf h}_{j^-}&{\bf GE^-\,\, over\, grps}\\
\hline
\end{array}
\]
Special cases. since we just need monoid and groups that are non-abelian we can study set theoretic goodstein map on nice groups/monoids to obtain particular subtheories where is easier to obtain explicit results/computations. I believe I have proven that the theory over finite groups is trivial, but the problem is open over finite monoids. The theory over the monoid \(\mathbb N^\mathbb N\) is basically extends recursion theory and gives hyperoperations. Also we can investigate the theory over multiplicative groups or rings or \(R\) algebras, i.e. \(R\)-modules \(X\) equipped with bilinear unital/associative maps \((-,-)_X:X\times X\to X\). Or, the most interesting case, by multiplication arising as multiplication in general linear groups of \(n\)-order matrices over a field (eg. finite fields). Here four interesting special cases that seems promising.
\[\begin{array}[|cccc|]
f{\rm maps\, in}&{\bf category}&{from\,the\,\,{\bf ranks}\,\, object}&{to\,the\,\,{\bf support}\,\, object}&&{\rm satisfying}&\\
\hline
A\in{\rm Hom}(U\mathbb N,{\rm End}(\mathbb N))&{in\,\,}1{\rm -Act} \,(of\,sets) & (\mathbb N\overset{S}{\to}\mathbb N)\in \mathbb N{\rm - Act}& \mathbb N\in{\rm Set}&&A(n+1,x+1)=A(n,A(n+1,x))&{\bf Ackermann\, equation\,\,}\\
{\bf f}\in{\rm Hom}(UJ,UX)&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{+}{\to}J)\in \mathbb N{\rm - Act}& X\in R{\rm -Alg}&& ({\bf f}_{j},s)_X=({\bf f}_{j},{\bf h}_{j^+})_X&\\
f\in{\rm Hom}(UJ,R^{\times})&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{+}{\to}J)\in \mathbb N{\rm -Act}& R\in{\rm Ring}&&f(j^+)\cdot s=f(j)\cdot f(j^+)&{\bf GE\,\, over\, non-comm.\,rings}\\
M\in{\rm Hom}(UJ,U{\sf GL}_n(k))&{in\,\,}1{\rm -Act} \,(of\,sets) & (J\overset{+}{\to}J)\in \mathbb N{\rm -Act}& k\in{\rm Field}&& M_{j^+}SM_{j^+}^{-1}=M_j&k{\bf -linear\, GE\,\,}\\
\hline
\end{array}
\]
The set theoretic approach seems very rich yet limited. It boils down to these 4 cases, they can be reduced by two if the ranks dynamics is invertible because at that point goodstein and anti-goodstein equations give same solutions.
Dynamical goodstein maps
The first generalization appears spontaneously if we restrict the previous theory to groups.
Before generalizing further. Notice that the support object needs to induce a monoid operation over the Hom-set so we are pretty limited in extending the category from where we pick the support object. We are also limited in the choice of the ranks object. It can't just be a set because the goodstein equation ask us for a procedure that gives "the next rank", so it has to be at least a discrete dynamical system, or something equipped with an action of something that can be restricted to a discrete action: an object of a category equipped with an appropriate forgetfull functor \(F:\mathcal C\to\mathbb N{\rm -Act}\).
Generalizing. as I've noticed before, groups makes a cool phenomenon to appers. The anti-goodstein one does reduce in such a way that we can make it hold by structural means, without asking for it. We then use instead of the forgetfull functor, the functor \((G,s\in G) \mapsto \Sigma^G_s\) from pointed groups to dynamical systems sending a group and an element to the sub-function by s (it is functorial in \((G,s)\)).
This way we can naturally extend the scheme to \({\mathbb N}\)-equivariant goodstein maps with \(B\)-equivariant ranks. We take the ranks to be equipped with a \(B\)-action \(J\), for \(B\) a monoid equipped with an unit of time \(\upsilon\in B\), and as support object we take a \(B\)-action over the group \(G\) that when restricted to the natural numbers gives back \(\Sigma^G_s\): it is an element of the preimage category \({\boldsymbol \Sigma}\in (\upsilon^*)^{-1}(\Sigma^G_s)\subseteq B{\rm -Act}\) also expressible as the pullback, i.e. the fiber of the restriction functor bundle \(\upsilon^*\) that we can denote more comfortably as \(B{\rm -Act}_{\Sigma^G_s}\).
\[\begin{array}[|cccc|]
f{\rm maps\, in}&{\bf category}&{from\,the\,\,{\bf ranks}\,\, object}&{to\,the\,\,{\bf support}\,\, object}&&{\rm satisfying}&\\
\hline
{\bf h}\in\mathbb N{\rm -Act}(J,\Sigma^G_s)&{in\,\,}\mathbb N{\rm -Act} & J\in B{\rm -Act}& G\in{\rm Grp}&¬hing&{\bf GE^-\,\,}by\, default\\
{\bf h}\in B{\rm -Act}(J,{\boldsymbol \Sigma})&{in\,\,}\mathbb N{\rm -Act} & J\in B{\rm -Act}& {\boldsymbol \Sigma}\in B{\rm -Act}_{\Sigma^G_s}&&i.e. \, \forall \beta \in B,\,j\in J.\, {\bf h}_{\beta j}={\boldsymbol \Sigma}^\beta ({\bf h}_j)&B{\bf -equiv.\, ranks\, GE^-\,\,}by\, default\\
\hline
\end{array}
\]
This route seems even more promising but is much more limited than the previous one... I believe it is a dead end or maybe something bringing us to a totally different theory: the theory of group conjugation and its meaning.
I believe we should go another way and turning everything categorical.
Set-theoretic \(A\)-equivariant-goodstein maps
Here is the point where things gets harder at first but I've seen an opening for going functorial. Let \(A\) be a monoid and \(u\in A\) the unit of time.
The concept is simple: goodstein equation impose pointwise the condition of being \(\mathbb N\)-equivariant. We use the abstract identification of points \(x\in M\) of the support monoid to \(\mathbb N\)-iterations over \(M\), i.e. \[M\simeq {\rm Mon}(\mathbb N,M)\] and we use the same philosophy for seeing \(A\)-equivariant goodstein maps as selecting many \(A\)-iterations over \(M\) that by the \(A\)-equivariant goodstein equation (\(A\)-EGE) have their \(A\) equivariance imposed pointwise.
\[\begin{array}[|cccc|]
f{\rm maps\, in}&{\bf category}&{from\,the\,\,{\bf ranks}\,\, object}&{to\,the\,\,{\bf support}\,\, object}&&{\rm satisfying}&\\
\hline
{\bf h}\in {\rm Hom}(UJ,{\rm Mon}(\mathbb N,M))&{in\,\,}1{\rm -Act} & J\in \mathbb N{\rm -Act}& M\in{\rm Mon}&&\forall j\in UJ.\, {\bf h}_{j^+}s={\bf h}_{j}{\bf h}_{j^+}&{\bf GE\,\,}\\
{\bf h}\in {\rm Hom}(UJ,{\rm Mon}(A,M))&{in\,\,}1{\rm -Act} & J\in \mathbb N{\rm -Act}& A,M\in{\rm Mon}&&i.e. \, \forall j\in UJ,\,a\in A.\, {\bf h}_{j^+}(u)s(a)={\bf h}_{j}(a){\bf h}_{j^+}(u)&A{\bf -EquivGE\,\,}\\
\hline
\end{array}
\]
Road to category theoretic goodstein maps. Here it is where we can spot the opening. First, note that until now the most lazy of our parameters was the object of ranks but since we can send dynamical system to sets and sets to discrete categories, embedding the category of sets in the category of categories is all we need. Secondly, the category of monoids can be enriched in cat: this means that the set of monoid homorphims is itself a category, a functor category \([A,M]\): namely \[{\rm Ob}([A,M])={\rm Mon}(A,M)\] The reason I think this happens is because monoids themselves can be seen as one-object categories, monoid morphims as functors and natural transformations between two functors \(f,g:A\to M\) are exactly \(A\)-equivariances.
I find this spectacular. In symbols: let \(M(x,y)=\{\phi\in M:\, \phi x=y\phi\}\) then we have a bijection \[ M(x,y)\simeq {\rm Nat}_{[\mathbb N,M]}(x^\bullet,y^\bullet) \] where \(x^\bullet:\mathbb N\to A\) is the monoid morphism defined as \(x^0=1_M\) and \(x^{n+1}:=xx^n\).
In the same way, let \(f,g:A\to B\) be functor between monoid seen as categories. natural transformations \(\phi:f\implies g\) are elements \(\phi\in M\) that satisfy \(\forall a.\, \phi f(a)=g(a)\phi\)... so if \(M^A(f,g)\) is the set of such \(\phi\)s then we have a bijection \[M^A(f,g)\simeq {\rm Nat}_{[A,M]}(f,g) \]
The question remains... if we use functor categories as support objects... how we define on them the monoid operation on objects? t seems to be possible only when \(M\) is abelian or for some special \(A\) like the integers or the naturals.
TO BE CONTINUED
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 377
Threads: 30
Joined: May 2013
12/19/2022, 07:31 PM
(This post was last modified: 12/19/2022, 10:28 PM by MphLee.)
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 \[h_s(2,0_T)=0_T;\,\,\,h_s(2,r+t)=h^r_s(1,h_s(2,t)t)\]
\[h_s(2,t)=h^t_s(1,0_T)\]
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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12/20/2022, 03:21 AM
(This post was last modified: 12/20/2022, 04:28 AM by JmsNxn.)
Hey, Mphlee.
I haven't devoured this post fully yet. I'll definitely comeback to it; but you have me intrigued by "upper rank iterations respect lower rank iterations". This is absolutely true for the bounded analytic hyperoperators. And I think it's not as strict of an idea as you are suggesting. Any good iteration should have some level of this happening. That absolutely this happens; and probably in a good general sense.
For example, let:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{n+1} z &= \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\,\circ k} (1)\frac{w^k}{k!}\\
\alpha \uparrow^{n,\,\circ y} z &= \frac{d^{z}}{dw^z}\Big{|}_{w=0} \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y} k \frac{w^k}{k!}\\
\alpha \uparrow^{n,\circ y} \left(\alpha \uparrow^{n+1} z\right) &= \alpha \uparrow^{n+1}( z+y)\\
\end{align}
\]
And this holds for all \(\Re(y) > 0\) & \(\Re(z) > 0\)--where the trouble is getting this to work for \(s \Big{|}_{\mathbb{N}} = n\) and \(\Re(s) > 0\). But the moment we have fractional hyper operators/semi-operators/bounded analytic semi-operators... your equation definitely holds up. It's a built in feature from a generation ago
Also note that for \(s \Big{|}_{\mathbb{N}}\)--each of these expressions are convergent integral transforms. And upon which we can even define things like:
\[
H(w,u) = \sum_{y=0}^\infty \sum_{k=0}^\infty \alpha \uparrow^{n,\circ y}(k)\frac{u^yw^k}{y!k!}\\
\]
The action \(\frac{d^y}{du^y}\) pulls out the iteration of the rank at hand (finite iterations of tetration), the action \(\frac{d^z}{dw^z}\) pulls out the iteration of the rank previous (it gives the tetration value, as iterations of exponentiation).
Edit:
This is very incorrect notation, but I believe the notation can be descriptive and informative. When doing the bounded analytic approach:
\[
\alpha\uparrow^{(s_0,k_0),(s_1,k_1),...,(s_n,k_n)} z = \frac{d^{s_0}}{dw_0}...\frac{d^{s_n}}{dw_n^{s_n}}\Big{|}_{w_0,...,w_n=0} \sum_{j_1,...,j_n=0}^\infty \alpha \uparrow^{(j_0,k_0),(j_1,k_1),...,(j_n,k_n)} z \frac{w_0^{j_0}\cdots w_n^{j_n}}{j_0! \cdots j_n!}\\
\]
Where:
\[
\alpha \uparrow^{(s_n,k_n)} z = f(z)\\
\]
Where,
\[
g(z) = \alpha \uparrow^{k_n} z\,\,\text{for}\,\,k_n \in \mathbb{N}\\
\]
And:
\[
f(z) = g^{\circ s_n}(z)\\
\]
Whereby, the chaining is:
\[
\alpha \uparrow^{(s_n,k_n),(s_{n+1},k_{n+1})} z = \alpha \uparrow^{k_n,\circ s_n} \left(\alpha \uparrow^{k_{n+1}, \circ s_{n+1}} z\right)\\
\]
THIS ALWAYS CONVERGES AS I'VE WRIT IT!
To describe the chaining is very difficult (I.e: \((s_n,k_n),(s_{n+1},k_{n+1})\)). We actually have a good amount of cancellation. I do not know the rules, or how to do it. I've just seen the cancellation in effect. So although the notation I just posited is garbage. It'll look something like this. Where the "depth of iteration" is really just, how many Differintegrals we need.
And when nesting these things we are following your "Upper ranks are affected by lower ranks" rule.
Anyway, I drank too much now--I'm a little tipsy and I should probably shut up. Just trying to say the rules you seem to be describing; bounded analytic hyper operators follow them to a tee
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Your notations destroy my brain cells... every line is killing 50/100 neurons. Idk how you do, it seems like your brain works without definitions ... damn.
Anyways... it is interesting that your operators satisfy the, let's call it like that "equivariant Goodstein condition".
Maybe we could fine tune that as a new property to ask for our iteration methods... maybe its a first step of an ascending sequence finer, algebraically motivated, conditions that at the end will give us natural constructions.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 1,214
Threads: 126
Joined: Dec 2010
12/21/2022, 11:39 PM
(This post was last modified: 12/22/2022, 03:59 AM by JmsNxn.)
Ya, I always used to get in trouble for that. I would jump too fast. I do most of the math in my head, I only write down things if I have to. I'll give you some examples of what I mean by this notation:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{0,\circ m} z &= \alpha^m \cdot z\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^{1,\circ n} z &= \exp_{\alpha}^{\circ n} z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^1 z &= \alpha^m \cdot \alpha^z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^{1, \circ n} z &= \alpha^m \cdot \exp_{\alpha}^{\circ n} z\\
\end{align}
\]
Where my point was, when we use regular iteration (the schroder iteration/the fractional calculus iteration), we get your rule:
\[
\alpha \uparrow^{n,\circ y} \alpha \uparrow^{n+1} z = \alpha \uparrow^{n+1} z+y\\
\]
But for god's sake! Do we need all these uparrows? Let's get rid of one:
\[
\alpha \uparrow^{(n,\circ y),n+1} z=\alpha \uparrow^{n+1} z+y\\
\]
So there's a correspondence between this notation and a translation algebra.
So I write:
\[
\alpha \uparrow^{(k_1,\circ m_1),...,(k_n,m_n)} z = \alpha \uparrow^{(k_1,\circ m_1)}\cdots \alpha \uparrow^{(k_n,\circ m_n)} z\\
\]
Which just saves time because it gets rid of \(n-1\) up arrows and I'm not constantly retyping that out, lol. It's not the prettiest, but it gets the job done.
Additionally, if you were confused; \(s\Big{|}_{\mathbb{N}}\) means the variable \(\Re(s) >0\) is now restricted to \(s \in \mathbb{N}\)--which is standard notation, though not very popular. I like it though.
EDIT:
I really think the fractional calculus is the approach for you though. It works off of \(\mathbb{N}\), but constructs it for \(T = \mathbb{R}^+\). Which is to say, your PEANO idea, is totally satisfied by the bounded analytic hyper-operators. Albeit, I don't get everything you're saying. But the points you seem to be iterating, seem to point exactly to the bounded analytic operators. The sole difference, I'd say, is that we start with the seed functions:
\[
\begin{align}
&\alpha+x\\
&\alpha \cdot x
\end{align}
\]
Which are analytic for \(1 < \alpha < \eta\) and \(x > 0\). From here we can construct hyper operators; and they are bounded and analytic. Additionally, I believe they are an example of the monoid goodstein sequence you keep saying. Correct me if I'm wrong. I apologize, as confused you are by my work; I can empathize with the feeling I have for some of your work
But I think the bounded analytic hyper-operators are at least an example of "something that looks like this monoid/\(\mathbb{N}\)/goodstein thing you have". But maybe I'm misinterpreting.... I think this is super important because; everything in this scenario relates to Mellin transforms and Generating functions. And you know what's next door to the mellin transform... the fourier transform. What's next door to that is Hilbert spaces. Next thing you know we're doing categorical fourier transforms, lmao (Which are a thing!).
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12/22/2022, 09:48 AM
(This post was last modified: 12/22/2022, 10:55 AM by MphLee.)
(12/21/2022, 11:39 PM)JmsNxn Wrote: Ya, I always used to get in trouble for that. I would jump too fast. I do most of the math in my head, I only write down things if I have to. I'll give you some examples of what I mean by this notation:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{0,\circ m} z &= \alpha^m \cdot z\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^{1,\circ n} z &= \exp_{\alpha}^{\circ n} z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^1 z &= \alpha^m \cdot \alpha^z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^{1, \circ n} z &= \alpha^m \cdot \exp_{\alpha}^{\circ n} z\\
\end{align}
\] I see man! That was pretty clear to me... in fact this kind of notation is not only a notation but the key to unlocking the full algebraic treatment of the topic: that has to do with the major update on my research on ranks theory. What was confusing was how you were using this insider your auxilliary funtions thing... that is what breaks my brain xd the iterated application of partial differintegral xd hahah sorry... that's more a problem on my side...
I can't spoil to much, but it is too exciting. This is not just a notation but is an extension of what I call a Goodstein map into a full monoid homomorphism. This extension is induced naturally by a well known process in group theory (that holds in monoid theory and universal algebra too) and is called an adjunction.
In fact that move is about extending the Goodstein map into a monoid homomorphism from the set of ranks, closed under the "+1" operation, to the free monoid generated by the ranks. This is also where universal algebraic geometry kicks in. I'm hyped by this but I can't spoil... Now that I have one week free I want to write a good introduction to this idea: in that way I can make the terminology crystal clear.
Quote:Where my point was, when we use regular iteration (the schroder iteration/the fractional calculus iteration), we get your rule:
\[
\alpha \uparrow^{n,\circ y} \alpha \uparrow^{n+1} z = \alpha \uparrow^{n+1} z+y\\
\]
But for god's sake! Do we need all these uparrows? Let's get rid of one:
\[
\alpha \uparrow^{(n,\circ y),n+1} z=\alpha \uparrow^{n+1} z+y\\
\]
So there's a correspondence between this notation and a translation algebra.
So I write:
\[
\alpha \uparrow^{(k_1,\circ m_1),...,(k_n,m_n)} z = \alpha \uparrow^{(k_1,\circ m_1)}\cdots \alpha \uparrow^{(k_n,\circ m_n)} z\\
\]
Which just saves time because it gets rid of \(n-1\) up arrows and I'm not constantly retyping that out, lol. It's not the prettiest, but it gets the job done.
Exactly, this is all implicit in my Goodstein maps yoga. What you write with your second last expression I write as
\[{\bf h}^y(n){\bf h}^1(n^+)={\bf h}^1(n^+){\bf h}^y(0)\]
That is all there is about my equivariant upgrade of Goodstein. But, if we expand the ranks \(n\in J\), where \(J\) is a discrete dynamical system, to the free monoid \({\sf F}(J)\) generated on the set of ranks as symbols, we have that a word \(w\in {\sf F}(J)\) is, by definition of free monoid, is just a finite sequence in the \(J\)-symbols \(w=\prod_{i=0}^d (j_i,n_i)\) where \(j_i\in J\) and where \((j,m)(j,n)=(j,n+m)\). By general abstract nonsensense we have an universal extension of the Goodstein map \({\bf h}:J\to {\rm End}(X)\) to a monoid homomorphism \({\bf h}^\dagger:{\sf F}(J)\to {\rm End}(X)\) that satisfies
\[{\bf h}^\dagger(w)={\bf h}^\dagger(\prod_{i=0}^d (j_i,n_i))=\Omega_{i=0}^d{\bf h}^\dagger((j_i,n_i))=\Omega_{i=0}^d{\bf h}(j_i)^{\circ n_i}\]
Where the omega is to be undesrtood as an abstract iterated composition of functions, that's why I omit the bullett: \(\Omega_{i=0}^d{\bf h}(j_i)^{\circ n_i}={\bf h}(j_0)^{\circ n_0}\circ {\bf h}(j_1)^{\circ n_1}\circ ...\circ {\bf h}(j_d)^{\circ n_d}\). Clearly we have \(\alpha \uparrow^{(n,\circ y),n+1} z=\alpha \uparrow^{n+1} z+y\) but I write at the function composition level as
\[{\bf h}^\dagger ((n,y)(n^+,1))={\bf h}^\dagger ((n^+,1)(0,y))\]
Here if you are comfortable with group presentations, ring theory, ideals and coordinate rings in AG (Algebraic Geometry=Alexander Grothendieck ), you can see where this is going.... DDDD Notice that the elegant and minimal expression is given by choosing the successor to be the seed and the seed to coincide with the zerot-th rank... note that in you construction the seed is still the successor but it is not the zero rank anymore.
The point is: what is not implicit in my yoga of Goodstein maps is how to extend to the monoid freely genereated by the ranks to something that allows \(y\in \mathbb R\) in the expression \(\uparrow^(j,y)\)... and your construction gives a model of the extended theory.
Quote:Additionally, if you were confused; \(s\Big{|}_{\mathbb{N}}\) means the variable \(\Re(s) >0\) is now restricted to \(s \in \mathbb{N}\)--which is standard notation, though not very popular. I like it though.
EDIT:
I really think the fractional calculus is the approach for you though. It works off of \(\mathbb{N}\), but constructs it for \(T = \mathbb{R}^+\). Which is to say, your PEANO idea, is totally satisfied by the bounded analytic hyper-operators.
Yea thanks you xD... that was one of the points that confused me.
Quote:I apologize, as confused you are by my work; I can empathize with the feeling I have for some of your work
But I think the bounded analytic hyper-operators are at least an example of "something that looks like this monoid/\(\mathbb{N}\)/goodstein thing you have". But maybe I'm misinterpreting.... I think this is super important because; everything in this scenario relates to Mellin transforms and Generating functions. And you know what's next door to the mellin transform... the fourier transform. What's next door to that is Hilbert spaces. Next thing you know we're doing categorical fourier transforms, lmao (Which are a thing!).
In some sense you are right. But the material in this thread is not really my work but just some speculations... I'm working the foundations for tackling the problems I'm describing in this thread.
It is not your fault if you don't get the terminology right... I need to do a good work explaining all of this from zero... I'll do this. It is so much easier that your differintegral/transform stuff...
It is like the difference between the Cappella Sisitina (your work) and something like the Giza pyramids (my work).
Notice that all of this algebra of ranks, and the remark about translation algebra, was foreseen by you back in August 2013 and we probably came at this almost independently (see a private message I sent you back in June 2013 ) before our research diverged. It is hard to tell who came there first since I was thinking about the algebra of ranks almost since the beginning of my journey... back in summer 2010, but I was deeply influenced by this November 2011 thread by you... at the point that it made me see how that idea I had back in 2010 could work... I really mean this... that thread made me have a clear grand vision... something that grew so big that in 2013 I felt the need to join the forum and send a PM to you. We can say that the last 12 years were just about getting the right tools to develop that vision.
The map \(n\mapsto [n]\) is just the prototype of Goodstein map I'm talking about, \(h(n)=[n]\)! And the free monoid nonsense is about making satisfy \([(s,n)(t,m)]=[(s,n)][(s,m)]\) where \([(s,n)]=[s]^{\circ n}\). Then we have \(T^m[s]=[s][0]^{\circ m}\) and clearly \([s][s+1]=T[s+1]=[s+1][0]\). All that Goostein theory is about studying in full generality maps \(h:J\to M\) that behaves as \([-]:\mathbb N\to (\mathbb N\times \mathbb N \to\mathbb N)\).
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Posts: 1,214
Threads: 126
Joined: Dec 2010
I am very excited Mphlee! I do believe you and I have a very big overlap in how we view this problem. There's no need to credit me for something you came to just as independently as myself. Hopefully if we find the solution to quantum gravity, or something, this doesn't turn into a Newton vs Leibniz situation! (Though we all know Newton was just a dick who couldn't handle someone else figuring out the same result as him )
I'm a little giddy, because I too have some more Mellin transform stuff up my sleeve. I haven't had time lately to work on it--but I believe the Mellin transform results should hold for all multipliers \(\lambda \neq 0\); thanks to the deep dive into Gottfried's work. (Previously we had to restrict \(|\lambda| \neq 0,1\).) So I am super pumped for what the future holds with tetration/iteration/ and *fingers crossed* semi-operators!
Regards, James
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