[note dump] Iterations and Actions

[MphLee]

(!!) The previous post was updated/completed with the missing information.

2022, May 07 - on the method+UPDATE

Those are really really good questions. Thanks for the engagement. I hope I can enlighten you even more with my answers.
I know it is not good practice but I'll include the answers to your technical points as comments and remarks inside the previous post, together with some grammar fixes (as many as I can). I want all of this to be tidy before I can continue the transcription of my old notes: there is a part of me that would like to expand endlessly on every single detail... but sadly I have not much time and I have to use it wisely.

You are correct! Everything I'm trying to do is, I hope, pure Grothendieck. It's like going after the geometry of iteration. Yeah, I know, I shouldn't even pronounce his name. He was a giant among giants. He was taller than mathematicians that were/are 100 times the average smartest math student of one's local university. And to be precise, I was not even near the smartest student at my local university.
So what I'm mean when I say I'm trying to go full Grothendieck on this? Let's be realistic: on a simple level let's say I'm trying to use abstract algebraic (undergrad algebra/geometry courses) methods to make sense of iteration, with an eye towards categorical methods. I'll use more and more category theoretical methods as soon as I learn and understand them myself. All of this is done with the philosophy of being able, at the end, to understand and apply modern Grothendieck's-style functorial-geometric methods to the problem of iteration. But this is just a mean to an end: the ultimate goal is to gain enough insight to perform effortlessly the final assault on non-integer ranks. That's the rising sea.

I'm not the kind of mathematician that can face directly a problem and emerge victorious, maybe my kung-fu is too weak. I prefer to build around it, redefine the basic terms, gain philosophical understanding, and then let my prefrontal cortex work on it while I study other related things, until the problem solves automatically and becomes trivial.

Many talk about abstract nonsense. Nowadays is said not too seriously or with negative meaning, I know, but it helps another quote from Grothendieck here:

The introduction of the digit 0 or the group concept was general nonsense too, and
mathematics was more or less stagnating for thousands of years because nobody was
around to take such childish steps...
— A. Grothendieck, [R. Brown and T. Porter, Analogy, concepts and methodology, in mathematics]

How can all of this work? It is, for me, one of the mystery of the logico-philosophical "quadrivium" of universal/particular, and abstract/concrete.

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2022, april 23 - actions, iterations and generalized elements PART 2 (reboot of old notes)

Among all the monoids \(\mathbb N\) plays a special role: \(\mathbb N\)-iterations of functions are equivalent to functions: this happens only for the natural numbers! The importance of this is fundamental for the concept of generalized element. Let's make some details and proof explicit.

Definition Recursion defines a function \({\rm ite}_{-}:{\rm End}(Y)\to {\rm End}(Y)^\mathbb N\). Define recursively for every \(f:Y\to Y\) the function \({\rm ite}_f:\mathbb N\to {\rm End}(Y)\) as
\[{\rm ite}_f(0):={\rm id}_Y,\quad\quad {\rm ite}_f(n+1):=f \circ {\rm ite}_f(n)\]
By recursion theorem that function is uniquely defined.

We prove this is bijective and and its outputs always land in \({\rm Hom}_{\rm Mon}(\mathbb N,{\rm End}(Y))\subseteq{\rm End}(Y)^\mathbb N\). To prove the result we need first this lemma:

Lemma: the integer iterates of a function commutes with the function they are iterates of. In symbols \(\forall n,\, f\circ {\rm ite}_f(n)={\rm ite}_f(n)\circ f\), i.e. the iterates are in the centralizer of \(f\) \[f^n\in C_{{\rm End}(Y)}(f) \]
Proof: By induction, the case \(n=0\) is evident. Assume \(f\circ {\rm ite}_f(n)={\rm ite}_f(n)\circ f\) for a given \(n\), we prove the identity for \(n+1\).
\[\begin{align}
f\circ {\rm ite}_f(n+1)&=f\circ(f\circ {\rm ite}_f(n))\\
&=f\circ({\rm ite}_f(n)\circ f)\\
&=(f\circ {\rm ite}_f(n))\circ f\\
&=({\rm ite}_f(n+1)\circ f\\
\end{align}\]
by the induction this holds for every \(n\).  \(\square\)

Crucial observation: The lemma states that the commutative(!) monoid \(\langle f\rangle =\{{\rm id}_Y,f,f^2,...,f^n,...\}\) generated by \(f\) is a submonoid of the centralizer \(C_{{\rm End}(Y)}(f)=\{\alpha:Y\to Y\,|\, \alpha f= f\alpha\}\) and we have the chain of monoid extensions:
\[\langle f\rangle \subseteq C_{{\rm End}(Y)}(f) \subseteq {\rm End}(Y)\]
In other words, natural iterates are in the centralizer, and we expect that the centralizers is the place where also all the non-integer iterates live because we WANT TO regard them as iterates only when they commute with the function they are iterates of.

Theorem: Let \(Y\) be a set and \({\rm End}(Y)\) be the monoid of functions \(\{f:Y\to Y\}\) under function composition. By recursion theorem a choice of an element \(f\in {\rm End}(X)\) is the same as a choice of a group homomorphism \(\phi:\mathbb N\to {\rm End}(X)\) and vice versa. In symbols: endofunctions over \(Y\) are in bijection with \(\mathbb N\)-iterations over \(Y\).
\[{\rm End}(Y)\simeq {\rm Hom}_{\rm Mon}(\mathbb N,{\rm End}(Y))\]
Proof: we prove: a) that \({\rm ite}_{-}:{\rm End}(Y)\to {\rm Hom}_{\rm Mon}(\mathbb N,{\rm End}(Y)) \) and b) it is has an inverse, i.e. is a bijection.
a) We prove induction that each \(f\) is sent to a monoid homomorphism. For every \(n\) and for \(m=0\) the base of the induction holds
\[\begin{align}
{\rm ite}_f(n+0)&={\rm ite}_f(n)\circ {\rm id}_Y\\
&={\rm ite}_f(n)\circ{\rm ite}_f(0)
\end{align}\]
For the step of the induction assume \({\rm ite}_f(n+m)={\rm ite}_f(n)\circ{\rm ite}_f(m)\) for a fixed \(m\). We deduce the result for \(m+1\).
\[\begin{align}
{\rm ite}_f(n+(m+1))&={\rm ite}_f((n+m)+1)&&&\\
&=f\circ {\rm ite}_f(n+m)&&&\\
&=f\circ  {\rm ite}_f(n)\circ {\rm ite}_f(m)&&&by \, inductive\, hypothesis\\
&={\rm ite}_f(n)\circ f\circ  {\rm ite}_f(m)&&&by \, commuting\, lemma\\
&={\rm ite}_f(n)\circ  {\rm ite}_f(m+1)&&&
\end{align}\]
Conclude by induction principle: \({\rm ite}_f\) is a monoid homomorphism for every \(f\).

b) we need an inverse of iteration: it is the evaluation at \(1\in \mathbb N\)
\[{\rm End}(Y)\leftarrow {\rm Hom}_{\rm Mon}(\mathbb N,{\rm End}(Y)):{\rm ev}_{1}\]
Clearly \({\rm ev}_{1}({\rm ite}_{f})={\rm ite}_{f}(1)=f\) obtaining a left inverse. Is it a right inverse too? Yes, given a monoid morphism \(\phi:\mathbb N\to {\rm End}(Y)\) we derive
\[\phi(n+1)=\phi(1+n)=\phi(1)\circ \phi(n)\]
The equation above has a single solution, by the recursion theorem, and it is \({\rm ite}_{\phi(1)}\) and this forces \(\phi={\rm ite}_{{\rm ev}_1(\phi)}\)  \(\square\)

Notation: we usually define the scripture \(f^{n}:={\rm ite}_f (n)\).
Corollary: the bijectivity implies if two functions have the same \(\mathbb N\)-iteration \({\rm ite}_f={\rm ite}_g\) then they are the same function \(f=g\) (injectivity) and there aren't \(\mathbb N\)-iterations that don't come from some function (surjectivity).

CLOSING REMARKS

Ok, this was tedious but this made precise and explicit why we want to talk about monoid morphisms \(\phi:A\to {\rm End}(Y)\) as "generalized iterations". In the case of \(\mathbb N\) they literally are iterations of a function \(f:Y\to Y\).

The questions:

1. If \(A\neq \mathbb N\), what in the hell, exactly, is \(\phi:A\to {\rm End}(Y)\) iterating? Is an \(A\)-iteration an "iteration" of something that lives in \({\rm End}(Y)\)?
   
2. In the opposite direction we may ask: given a \(f\in {\rm End}(Y)\) how to identify the \(A\)-iteration \(A\neq \mathbb N\) that, in some sense, is iterating \(f\)?
   

In other words, how we can relate fill the question marks in \[{\rm End}(Y)\simeq {\rm Hom}_{\rm Mon}(\mathbb N,{\rm End}(Y) )\overset{???}{ \longleftrightarrow} {\rm Hom}_{\rm Mon}(A,{\rm End}(Y) )\]

Part 3 will be about how to classify and find relationships between \(A\)-iterations.