[note dump] Iterations and Actions JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 05/07/2022, 03:01 AM (This post was last modified: 05/07/2022, 03:12 AM by JmsNxn.) Hey, Mphlee! I don't want to comment too much as of yet; I haven't absorbed everything here; but I think this is unbelievably fascinating. Honestly: $$\text{A-Act} = \text{Set}^{\text{BA}}\\$$ Then you say: $$\text{1-Act} = \text{Set}\\$$ Man, I don't know what kind of weed you're smoking, but that is insane. That's Godelian thinking at its finest. The trivial monoid produces the standard theory of sets. If I ever make it to Italy you got to give me your plug. One problem I have, and it's dumb, but: In one of your diagrams $$\mathbb{N} \to \text{A-Act} \to \mathbb{Q} \to \mathbb{R}\to\mathbb{C}$$. Shouldn't $$\text{A-act}$$ come even before $$\mathbb{N}$$. Also, if you have any notes on the topology of these things--the idea of creating a continuous topology to $$A-Act$$, please share them. Because that is insane. I understand now, when you say this relates deeply to tetration. Because we have an explicit example of $$A-Act$$ in a non trivial environment ($$\exp$$); which deriving a continuous extension is almost unforeseeable with the way you've described it in the general case. I'd like to really say, what hits it home for me, is the concept of $$\text{tet}(q)$$ for $$q$$ a quaternion. Despite how absurd this sounds, and when you first wrote it, I thought it was a little whacky. But this expansion perfectly exists. Quaternions work similarly through Taylor expansions (you have to finesse it a bit), but I wouldn't be surprised if there's a Kneser Tetration that works for Quaternions. This certainly works for non-singular square matrices, so I wouldn't be surprised. Locality would be whacked out a bit, and you'd get quaternion whacky branch cuts, but it's probably doable. This is absolutely fascinating. I think the continuous topology problem will probably be the hardest part. Have you read much on Algebraic Topology, and the topology of Algebraic Extensions//Grothendiek shit? This reminds me a lot of the "it's so abstract it's nonsense", "but all the numbers work". Regards, James MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/07/2022, 06:15 PM (!!) 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. Quote: 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: Quote: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. 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: 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 $${\rm End}(Y)$$? 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. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/10/2022, 12:26 AM (This post was last modified: 05/10/2022, 09:41 AM by MphLee.) Comment on the quaternionic extension example. Just thinking about it... I realize all that is missing from my treatment, atm, is the sheaf business: I work globally/algebraically but I don't have a clue yet on how to organize local solutions. But in many cases this means I'm excluding non-trivial solutions. Using sheafs means to consider both local and global solutions and how to patch them. This is what subconsciously one does in analysis and what algebraic geometers and algebraic topologists do with schemes/sheaves and manifolds/sheaves of analytic/holomorphic functions. This is the reason, every time I read old posts in the forum declaring to compute fractional iterates using solutions to abel and superfunction equations or Schroeder and inverse Schroeder functions, I'm puzzled. If you really have a solution of that kind $$\chi(x+1)=f(\chi(x))$$ and you can invert $$\chi$$ globally, this means that the dynamic of the successor is isomorphic to the dynamics of the function $$f$$ you are iterating, e.g. of $$f=\exp$$. Clearly this is not always the case. In fact, you all then start talking about neighborhood of points and equations holding locally on some domain. In the last part I tried to build a narrative filling my notes with more explanations and comments: this time I'll comment less. The following notes continue the meditation on how to travel from functions, $$\mathbb N$$-iterations and $$A$$-iterations and backwards. 2022, april 23 - actions, iterations and generalized elements PART 3 (reboot of old notes) In the previous discussion, given a monoid $$A$$, we were discussing the collection $$A{\rm -Act}\simeq {\rm Set}^A$$ of all the possible pairs $$(Y,\varphi)$$ where $$Y$$ is a set and $$\varphi_a(y)$$ is an $$A$$-action/$$A$$-iteration over the set $$Y$$, for all the possible sets. Now we fix a set $$Y$$, and let the monoid vary. We want to focus only on how to iterate the functions $$f:Y\to Y$$ over $$Y$$. As a convention $${\rm Hom}(A,B)$$ is the set of monoid morphisms $$\varphi:A\to B$$. Definition 1 denote with $$A{\rm -act}_Y$$ the set of $$A$$-actions over $$Y$$. It is the set of object of the fiber of the forgetful functor at $$Y$$ so it results by pulling back the forgetful functor, sending an action to it's underlying set, by the inclusion of $$Y$$ into sets. Question is $$A{\rm -Act}$$ related to the family of $$A{\rm -act}_Y$$ via a something related to the Grothendieck construction? Note that, abusing the notation and only in what regards objects $$A{\rm -Act}=\bigcup_{Y\in{\rm Set}}A{\rm -act}_Y$$ Here we define what we mean with the term $$A$$-iteration over a given set. Definition 2 denote with $${\rm ite}(A;Y)$$ the set of what we want to call $$A$$-iterations over $$Y$$, i.e. it is defined as $${\rm Hom}(A,{\rm End}(Y))$$. $${\rm ite}(A;Y):={\rm Hom}(A,{\rm End}(Y))$$ It is the set of object of the fiber of the forgetful functor at $$Y$$: Remark $${\rm Hom}(A,{\rm End}(A;Y))$$ is also the set of objects of the functor category $${B{\rm End}(Y)}^{BA}$$. Here $$BA$$ is the category naturally associated to the monoid $$A$$ (see delooping). Note that, abusing the notation and only in what regards objects $${B{\rm Set}}^{BA}=\bigcup_{Y\in{\rm Set}}{B{\rm End}(Y)}^{BA}$$ Lemma As expected $$A$$-actions over $$Y$$ are in bijection with $$A$$-iterations over $$Y$$: for every monoid $$A$$ $$A{\rm -act}_Y\simeq {\rm ite}(A;Y)$$ this bijection is natural in a technical sense. We are going to use this fact as a key element of the proof that there is a natural isomorphism of contravariant functors: ^Globe diagram Proposition The presheaf assigning to each monoid the $$A$$-actions over $$Y$$ is a representable presheaf, represented by $${\rm End}(Y)$$: for every monoid morphism $$j:B\to A$$ the following diagram commute Proof: We need to show that given an arbitrary $$j:B\to A$$ and for every $$A$$-action $$\varphi(a,y)$$, for every $$b\in B,\,y\in Y$$ $${\rm curry}_Y(\bar{j}(\varphi))_b(y)=j^*({\rm curry}_Y(\varphi))_b(y)$$ Just unwrap the definitions \begin{align} {\rm curry}_Y(\bar{j}(\varphi))_b(y)&=j^*({\rm curry}_Y(\varphi))_b(y)\\ \bar{j}(\varphi)(b,y)&={\rm curry}_Y(\varphi)_{j(b)}(y) &&&\\ \varphi(j(b),y)&=\varphi(j(b),y)&&& \end{align} Being representable means that iterations and actions are really the same and in a natural way: independently of monoid where the time takes values in. Conjecture (when time is commutative) since $${\rm Act}_Y:{\rm Mon}^{op}\to {\rm Set}$$, $$A\mapsto A{\rm Act}_Y$$ ,is a representable functor $${\rm act}_Y\simeq {\rm ite}(-;Y)$$ I expect it to send direct products to cartesian products, or better to send colimit into limits, when the monoids we are working with are commutative. In other words, let $$L,M$$ be commutative monoids $$(L\times M){\rm -act}_Y={\rm Hom}(L\times M,{\rm End}(Y))\simeq {\rm Hom}(L,{\rm End}(Y))\times {\rm Hom}(M,{\rm End}(Y))=L{\rm -act}_Y\times M{\rm -act}_Y$$ I guess the way to prove amounts to proving that every morphism $$f:L\times M\to A$$ determines, and is determined, uniquely a pair of morphisms $$f_0:L\to A$$ and $$f_1:M\to A$$. Reality check set $$A=\mathbb C$$ the additive abelian group of complex numbers. Since there is an isomorphism of abelian groups $$\mathbb C\simeq \mathbb R\oplus \mathbb R$$, i.e. $$z\mapsto ( \Re(z), \Im(z))$$ and $$(a,b)\mapsto a+ib$$, this means that $${\rm Ite}(\mathbb C;Y)\simeq {\rm Ite}(\mathbb R;Y)\times {\rm Ite}(\mathbb R;Y)$$ This says that every complex-iteration $$\varphi_z$$ is completely determined by $$1$$ and $$i$$, or by any pair $$w_0,w_1$$ of complex numbers that generates -is a base for- all the complex plane. It seems trivial but I'd like to check the details. All the stuff above also provide us the power to:transform every morphism $$A\to\mathbb C$$ into a way to convert $$\mathbb C$$-iterations over $$Y$$ into $$A$$-iterations over $$Y$$ and transform every morphism $$\mathbb C\to A$$ into a way to convert $$A$$-iterations over $$Y$$ into $$\mathbb C$$-iterations over $$Y$$. To be concrete here two trivial examples:The canonical real part surjective morphism $$\Re:\mathbb C\to \mathbb R$$ can convert real iterations over $$Y$$ to complex iterations over $$Y$$ by computing $$(\Re^* \varphi)_z(y):=\varphi_{\Re z}(y)$$; Fix a complex number $$z\in\mathbb C$$, consider the map $$m_z:\mathbb Q\to\mathbb Z$$ that sends the rational $$q\in\mathbb Q$$ to $$zq\in\mathbb C$$. It is a morphism $$m_z:\mathbb Q\to \mathbb C$$ and we can convert complex iterations over $$Y$$ to rational iterations over $$Y$$ by computing $$((m_z)^*\varphi)^{q}(y):=\varphi^{zq}(y)$$; Extracting and repackaging the information of the iterates Philosophy Fix a set $$Y$$, the functor $${\rm ite}(-;Y)$$ assigns to every monoid all the possible iterations with time in that monoid. We can say that $${\rm ite}(-;Y)$$ is a gadget that organizes and hold in a single place all the information of what it means to iterate functions $$f:Y\to Y$$, and with that all the possible way to extend the various iterations and the obstruction to the existence of extensions. It's a very rich gadget. I expect the study of this functor is crucial. we can also visualize this as follows: Crucial observation This makes evident that all the information in this functor is "in some sense contained" by the category of bundles over the monoid $${\rm End}(Y)$$, i.e. the slice category $${\rm Mon}/{\rm End}(Y)$$. Maybe it is possible to prove the isomorphism of categories of the slice to some Grothendiek-related construction. But the really mysterious thing is that the slice construction records a different kind of information: that of divisibility; while the normal category of iterations we record the information about equivariance and preservation of dynamical properties. So in the slice category the relations are relations of divisibility while in the category of actions the relations are superfunctions/equivariances. I'm tempted to describe this as saying that the original view records the dynamical information, regarding addition of time, composition of processes; while the slice construction makes the mutiplicative/fractional iterative information visible. Intrinsic iterates This means that every monoid $$A$$ has not a total freedom on the way it can iterate function over $$Y$$: to be precise it has a limited set of ways. In fact every $$A$$-iteration squishes $$A$$ into $${\rm End}(Y)$$ unless $$a\mapsto \varphi^a$$ is injective as exponentiation, i.e. $$A$$ is a submonoid. So the information on the obstructions on how can we iterated is written in the lattice of submonoids of $${\rm End}(Y)$$ and that is completely determined by the cardinality of $$Y$$ by Cantorian arguments. Observation it is really important to underline the fact that $${\rm ite}(-;Y)$$ is functorial as the monoid time varies but it IS NOT FUNCTORIAL as the set $$Y$$ varies. The meaning of this?it is easy to transform $$A$$-iterations over $$Y$$ to $$B$$-iterations over $$Y$$ in a coherent way; it is not possible to transform and relate $$A$$-iterations over $$Y$$ to $$A$$-iterations over $$X$$ in a coherent way; The slogan can be: we can relate coherently iterations with different times, but not over different sets - change of base is hard. The next part is the 4th and last and is the missing link between the bird's-eye view and the attempt to recover fractional iterates. After that I only have notes about divisibility of iterations, some notes about my first approach to cohomology and the ones about how to generalize Bennet hyperoperations accordingly. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/13/2022, 02:54 AM (This post was last modified: 05/13/2022, 08:22 AM by MphLee.) 2022, april 24 - actions, iterations and generalized elements PART 4 (reboot of old notes) General recap $$A$$-iterations are well defined as monoid homomorphism from the time monoid to the monoid $${\rm End}(Y)$$ of self-transformations of a state space $$Y$$. Every $$A$$ iteration defines what can be thought as a $$A$$-time evolution of the states $$Y$$. Using the term iteration for those is suggestive and is meant to be an heuristic device forcing us to think of them as if they really are iteration of something and we wish that "something" to be a function $$f:Y\to Y$$ even if we don't know if it is always possible. In the case of $$\mathbb N$$-iterations. we have a bijection to elements of $${\rm End}(Y)$$; In the case of $$\mathbb Z$$-iterations we obtain only the bijective elements of $${\rm End}(Y)$$; In the case of $${\mathbb Z}_2$$-iterations we get the involutions over $$Y$$, i.e. particular elements of $${\rm End}(Y)$$ that are inverse to themselves $$f^2={\rm id}_Y$$; In the case of $$({\mathbb Z}_2,\cdot)$$-iterations we get the idempotents over $$Y$$, i.e. particular elements of $${\rm End}(Y)$$ s.t. $$f^2=f$$; In the case of $$A=1$$ then there exist only one $$1$$-iteration over $$Y$$ and it is $${\rm id}_Y\in {\rm End}(Y)$$; For general time monoids $$A$$ we think of $$A$$-iterations as generalized elements in $${\rm End}(Y)$$ or figures of shape $$A$$ in $${\rm End}(Y)$$; It seems that, depending on the chosen monoid of time $$A$$, the $$A$$-iterations corresponds to special self-transformations of the state space. \begin{align} {\rm Hom}(\mathbb N, {\rm End}(Y)) &\simeq {\rm End}(Y) \\ {\rm Hom}(\mathbb Z, {\rm End}(Y)) &\simeq {\rm Aut}(Y) \\ {\rm Hom}(\mathbb Z_2, {\rm End}(Y)) &\simeq {\rm Invl}(Y) \\ {\rm Hom}(1, {\rm End}(Y)) &\simeq 1 \\ {\rm Hom}(A, {\rm End}(Y)) &\simeq ??? \end{align} Example in topology take a topological space $$X$$, let $$\bullet={0}$$ the trivial topological space on the point, $$I=[0,1]$$ the real interval and $$S^1=\{(x,y):x^2+y^2=1\}$$ the circle. Continuous functions $$P:{0}\to X$$ are in bijection with points of $$X$$. Call them figures of shape $$\bullet$$ or just elements of $$X$$; Continuous functions $$\gamma:I\to X$$ are in bijection with paths of $$X$$. Call them figures of shape $$I$$ or just $$I$$-elements of $$X$$; Continuous functions $$\gamma :{}S^1\to X$$ are in bijection with loops in $$X$$. Call them figures of shape $$S^1$$ or just $$I$$-elements of $$X$$; To pursue (difficult) it is possible it makes sense of the previous scheme (points, paths, homotopies,..) in the context of iterations? Is it possible to find a a sequence of abelian groups $$1\to\mathbb N\to A_2\to A_3\to...\to A_n\to ...$$ so that we can define something analogous of points, paths, homotopies, and so on but algebraically and define some kind of iterational homotopy theory of $${\rm End}(Y)$$. Something like $${\bf \pi}_n(Y):={\rm ite}(A_n;Y)/{\simeq}$$. Definition (authentic time) every $$A$$-iteration defines a family of submonoid $$\{\varphi_A:\,a\in A\}\subseteq {\rm End}(Y)$$. Call it $$A_\varphi$$, a kind of quotient of $$A$$ by $$\varphi$$, we can see it as the authentic time of $$\varphi$$. It is the universal monoid making $$\varphi$$ injective (faithful) in the iteration-time argument by factoring it. If $$A$$ is a commutative group $$A_\varphi=A/{\rm ker}\varphi$$ is the quotient of $$A$$ by the anihilator group of $$\varphi$$. Definition (extension) given an "extension" of time $$u:\mathbb N\to A$$, or equivalently a choice of an element $$u(1)\in A$$ (the unit of time), we obtain a "restriction" going from $$A$$-iterations to $$\mathbb N$$-iterations. Since the latter is equivalent to $${\rm End}(Y)$$ we obtain the evaluation of the iteration at $$u(1)\in A$$. This what I'd like to call the underlying function map. This map answer to the question "what is $$\varphi$$ iterating?" $$\varphi$$ is an $$A$$-iteration of $$u^*\varphi=\varphi_u$$ Observation 1 Introduce the language of preimages! The set of extensions of $$f$$ is the preimage of $${\rm ev}_u: {\rm ite}(A;Y)\to {\rm End}(Y)$$ at the value $$f$$. It is the set of divisors $$\varphi$$ , solution of the equation $$\varphi \circ u=f$$ Question (to do) Let's rework question 3 of this post. When the pre-image is empty? When is the restriction not surjective? Is it injective? If $$u^*$$ is injective and $$f\in {\rm im}u^*$$ , then $$f$$ can be extended uniquely to an $$A$$-iteration; The injectivity $$u^*$$ amount to this property: $$u\in A$$ is such that given two $$A$$-iterations $$\varphi,\psi$$ then $$\forall y\in Y.\, \varphi_u(y)=\psi_u(y)=f(y)\quad{\rm implies}\quad \forall a\in A.\,\varphi_a(y)=\psi_a(y)$$ If $$u^*$$ is sujective then $${\rm im}u^*={\rm End}(Y)$$, i.e. every $$f:Y\to Y$$ can be extended to an $$A$$-iteration; The second point can be expressed with the "Wolfram-ian" slogan "All maps over $$Y$$ have $$A$$-flows". To pursue category-theoretically the solutions sets to those division problems (extension-restriction) are known to be functorial (see coslice/slice).The hom-set $$({\rm Mon}/_{{\rm End}(Y)})(f,\varphi)=\{u:\,f=\varphi\circ u\}$$ is the set of $$u\in A$$ such that $$f=\varphi_u$$; The hom-set $$(_{\mathbb N}\setminus{\rm Mon})(u,f)=\{\varphi:\,\varphi\circ u=f\}$$ is the set of $$A$$-iterations $$\varphi$$ extending $$f$$, i.e. st $$\varphi_u=f$$; Denote $${\bf Ex}_u(f):=(_{\mathbb N}\setminus{\rm Mon})(u,f)=\{\varphi :\,\varphi_u(y)=f(y)\}$$, the set of extensions of the integer iteration of $$f$$ to $$A$$ time-iterations that exhibit $$f$$ as $$\varphi_u$$. How this set relates to the solutions sets of the functional equation $$\chi(u+a)=f(\chi (a))$$. Proposition 1 For every $$y_0\in Y$$ and every $$\varphi \in {\bf Ex}_u(f)$$ exists a $${\hat \varphi}_u\in\{\chi:A\to Y\,| \, \chi\lambda_u=f\chi \}$$ $${\hat -}:{\bf Ex}_u(f)\times Y\to{\rm Hom}_{\mathbb N{\rm -Act}}(A^\lambda, Y^f)$$ Proof: we demonstrate that every solution to the iteration problem defines, for every choice of a base point $$y_0$$, a solution to a superfunction problem. Assume we have an action $$\varphi_a(y)$$s.t. $$\varphi_{a+b}(y)=\varphi_{a}(\varphi_{b}(y));\quad \varphi_0(y)=y\quad{\rm and}\quad \varphi_u(y)=f(y)$$ Then, for every $$y_0\in Y$$ we define a map $${\hat \varphi}_{y_0}:A\to Y$$ $${\hat \varphi}_{y_0}(a):=\varphi_a(y_0)$$ we prove we obtain an $$A$$-equivariant map: $$\forall a,b\in A,\,{\hat \varphi}_{y_0}(a+b)=\varphi_a({\hat \varphi}_{y_0}(b))$$ By setting $$a=u$$: $$\,{\hat \varphi}_{y_0}(u+b)=f({\hat \varphi}_{y_0}(b))$$.  $$\quad \square$$ FRACTIONAL ITERATION, DIVISIBILITY AND FUNCTIONAL LOGARITHM Observation 2 assume we have two $$\mathbb N$$-iterations, i.e. two maps $$f(y)$$ and $$g(y)$$, what it means for $$f$$ to be a fractional iterate of $$g$$? In symbols: $$f(f(y))=g(y).$$Now, remember the bijection $${\rm End}(Y)\simeq {\rm ite}(\mathbb N;Y)$$. If we see the phenomenon at the level of $${\rm End}(Y)$$ it says: $$f$$ is a frac. iterate of $$g$$ can be translated in the proposition $$\exists k: f^k=g$$; If instead we see it at the level of $${\rm ite}(\mathbb N;Y)$$ we obtain a factorization problem: $$f$$ is a frac. iterate of $$g$$ is expressed by the proposition "$$\exists k$$ such that for every $$n$$ $${\rm ite}_f(k\cdot n)={\rm ite}_g(n)$$".. The map $$m_k:\mathbb N\to \mathbb N$$ sending $$n\mapsto k\cdot n$$ is an monoid automorphism (distributivity): this produces a factorization Observation 3 (The Power Action)This factorization is deeply linked with the lattice of integers under divisibility and contains a lot of information about the multiplicative structure of the integers (see the crucial observation in this post), hence of extensions to rational numbers. All of this information can be packed naturally into a single algebraic gadget that I call "the power action of $$A$$": the action of $${\rm End}_{\rm Mon}(\mathbb N, +)$$ on $${\rm ite}(A;Y)$$. Proposition 2 every endomorphism $$k:A\to A$$ of the time monoid $$A$$ induces a a map "raising" every $$A$$-iteration to the $$k$$-th power and the result is still an $$A$$-iteration. Proof: functoriality of $${\rm ite}(-;Y)$$ induces a morphism of monoids, thus an action $${\rm pow}(j\circ k,\varphi)={\rm pow}(j,{\rm pow}( k,\varphi))$$ call this the $$k$$-power map, where $$k:A\to A$$ is a monoid morphism. $$\square$$ Corollary 2 if $$A=\mathbb N$$ then  $${\rm End}_{\rm Mon}(\mathbb N, +)\simeq (\mathbb N,\cdot)$$ is the mutiplicative monoid of natural numbers. In fact $$\forall k,j\in\mathbb N.\, f^{jk}=(f^k)^j$$ CLAIM We can use this action to: a) formally define a kind of "functional logarithm" computing naturally for some functions $$\alpha:Y\to Y$$ commuting with $$f:Y\to Y$$, i.e. $$f\alpha =\alpha f$$, their rational height $${\mathfrak {log}}_f(\alpha)\in\mathbb Q$$; b) prove this map always exists and; c) find its domain of definition (what I call the intrinsic iterates of $$f$$). $${\mathfrak {log}}_f(\alpha)=\frac{n}{d}\in\mathbb Q\quad iff \quad \alpha^d=f^n$$ MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 05/13/2022, 03:17 AM Before I over dose on Mphlee mathematics. I'd say you missed an opportunity in your red text. Consider a sequence of accumulation points. Then that admits your theory to holomorphy. I mean, assume you have a sequence identified by its accumulation to $$0$$, that defines the sheaf at zero. I'm just saying this to add that this belongs in your red coloured list. MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/13/2022, 04:17 PM (This post was last modified: 05/13/2022, 04:24 PM by MphLee.) (05/13/2022, 03:17 AM)JmsNxn Wrote: Consider a sequence of accumulation points. Then that admits your theory to holomorphy. I mean, assume you have a sequence identified by its accumulation to $$0$$, that defines the sheaf at zero.I'm sure I don't understand. Pls, if you see something that seems easily translatable in my language from complex analysis give me pointers. I need to study much more before I can make an attempt at extending my rosetta stone to continuity (topological data) and then smoothness (differential data). And what I'm doing is not much more than compiling this rosetta stone. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/14/2022, 11:56 AM While I was typesetting the crucial observation in the part 4 of my series of notes on iterations and actions I realized something very cool about it. The F-grade poetry in the title comes from the fact that this mysterious object puzzled me for many years and only yesterday I understood the real meaning of its morphisms. 2022, May 13 - The tales of the legendary category $${\rm ACT}$$ of Actions (idea) Rememeber and action/iteration with time monoid $$A$$ over a state space $$X$$ is a binary function $$f:A\times A\to Y$$ satisfying $$f(0_A,y)=y,\quad f(a+_Ab)=f(a,f(b,y))$$ Think of this as a structure $$(A,X,f)$$. Limitation There are two main ways to relate two iterations depending on their time monoid and on their state space. The most important relations are the $$A$$-equivariances (state space relations/conjugations). These are possible only if the two iterations have the same time monoid; $$(A,X,f)\overset{\varphi}{\to}(A,X,g) \quad\quad \forall a\in A,x\in X,\,\varphi(f(a,x))=g(a,\varphi(x))$$ The less known restriction/extension, I'll use the umbrella term divisibility relations, (time monoid relations/reparametrizations). These are possible only if the two iterations share the same state space ($$k$$ satisfies $$k(a+_Ab)=k(a)+_Bk(b)$$ and $$k(0_A)=0_B$$); $$(A,X,f)\overset{k}{\to}(B,X,g) \quad\quad \forall a\in A,x\in X,\,f(a,x)=g(k(a), x)$$ Equivariance relates the $$A$$-dynamics over two state spaces: it is possible to learn if they have the same periodic points, if one embeds in the others, or if we can represent part of a unknown dynamics by representing it as a known dynamics, like in the case of Abel and Schroeder equations (linearization). Equivariances tells us that while the sate $$x\in X$$ evolves $$a\in A$$ times the state $$\chi(x)\in Y$$ evolves in the same way ("equivaries"). We can say equivariances are ways to deform the first dynamics in the second dynamics. The biggest limitation is the apparent impossibility to have natural equivariance relations between two iterations using two different time monoids: we can only fix an time $$A$$ and consider all of the iterations with that fixed time $$A{\rm -Act}$$. Question 1 A question that has always eluded me is: how the dynamics of a function $$f$$ compares with the dynamics of its iterates $$f^n$$? What are the solutions of $$\phi (f^n(x))=f^{n+1}(\phi(x)) ?$$ ExamplesIn general $$x\mapsto f^n(x)$$ gets mapped to $$\chi(x)\mapsto g^n(\chi(x))$$; When $$A=\mathbb N$$ we just have superfunctions. $$0\mapsto n$$ gets mapped to $$x_0\mapsto g^n(x_0)$$; In the case of Schroeder functions we have a linearization. The evolution of an unknown map $$x\mapsto f^n(x)$$ gets mapped to scalar multiplication $$\chi(x)\mapsto \lambda^n\cdot \chi(x)$$. Divisibility expresses the relation of two dynamics with different time over the same state space. This is about reparametrizing the orbits of a dynamics $$x_a=f(a,x)\in X$$ get reparametrized by $$b\in A$$ as $$x'_{b}=x_{k(a)}\in X$$, i.e. $$g(b,x)=f(k(a),x)$$. We can say we are globally extending or restricting the orbits of an iteration. In this case the main limitation is: by knowing the divisibility relations alone there is not a clear way to known if it is possible for an iteration over a state space $$X$$ to be the time extension, e.g. from integers to reals, of an iteration over a state space $$Y$$. Everything we can do is: fix a set $$X$$ and consider all of the iterations over $$X$$ as bundles $${\rm Mon}/_{{\rm End}(X)}=\{f:A\to {\rm End}(X)\,|\, A\, is\, a \, monoid\}$$. ExamplesAll the fractional iterations relations are relation of that kind. If $$f^2(x)=g(x)$$ then for every $$n$$ we have $$f^{2n}(x)=g^n(x)$$ so we have a map $$(\mathbb N,X, g)\overset{2\cdot}{\to}(\mathbb N, X,f)$$; All the extension relations are of that kind. If $$i:\mathbb N\to \mathbb C$$ is the inclusion of the naturals inthe complex then $$f(z,x)$$ extends $$g^n(x)$$ to the complex numbers iff for every $$n$$ we have $$f({i(n)},x)=g^n(x)$$ so we have a map $$(\mathbb N,X, g)\overset{i}{\to}(\mathbb C, X,f)$$; Quote:QUESTION is it possible to have an unique place where we have all the various actions and study their interactions (time interactions and equivariant interactions) without caring if they have same time or same state space? In other words, is there an universal category $${\rm ACT}$$ s.t: contains all of the $$A{\rm -Act}$$ for all the monoids $$A$$ and glues to the previous information also all the information of the functors $${\rm Ite}(-;Y)$$? Is there a way to package all the information about equivariant maps (superfunctions ...) and the information of extension/restriction of the time? Definition (ACT) Define the category of monoid actions $${\rm ACT}$$:(Objects) are triples $$(A,X,f)$$ where $$A$$ is a monoid, $$X$$ is a set and $$f(a,x)$$ is an $$A$$-iteration over $$X$$; (Morphisms) $$(A,X,f) \overset{\boldsymbol \chi}{\to} (B,Y,g)$$ are pairs $${\boldsymbol \chi}=(k;\chi)$$ where $$k:A\to B$$ is a monoid morphism, $$\chi:X\to Y$$ is a set function, s.t. $$\chi(f(a,x))=g(k(a),\chi(x))$$ (Composition) of maps $$(A,X,f) \overset{\boldsymbol \chi}{\to} (B,Y,g)\overset{\boldsymbol \phi}{\to} (C,Z,h)$$ is given by  $$(A,X,f) \overset{{\boldsymbol \phi}\circ {\boldsymbol \chi}}{\to} (C,Z,h)$$ where $${\boldsymbol \phi}\circ {\boldsymbol \chi}=(j;\phi)\circ (k;\chi)=(jk;\phi\chi)$$ Proposition The composition defined above is well defined. Proof: assume that $$\chi(f(a,x))=g(k(a),\chi(x))$$ and $$\phi(g(b,y))=h(j(b),\phi(y))$$. We deduce from that $$\phi(\chi( f(a,y) ))=h( j(k(a)) , \phi(\chi(y)) )$$. \begin{align} \phi(\chi( f(a,y) ))&= \phi( g(k(a),\chi(y) ))\\ &= g(j(k(a)),\phi(\chi(y)) ))\\ \end{align} Terminology Every morphism $${\boldsymbol \chi}=(k;\chi)$$ is made by the equivariant part $$\chi$$ and the division part $$k$$. We say the $$A$$-equivariances, e.g. superfunctions, are of purely equivariant nature and of form $$({\rm id}_A;\chi)$$. The divisibility relations (re-parametrizations), e.g. extensions, are action maps  purely of "divisibility nature" $$(k;{\rm id}_X)$$. Examples here some examples of action morpisms that are not of pure nature. Consider the equation $$\chi(f(x))=f^{k}(\chi(x))$$: this is a map $$(m_k,\chi):{}(\mathbb N,X,f)\to(\mathbb N,X,f)$$. Set $$X=\mathbb R$$ and $$f=S$$ the successor. Then $$(m_k;{\rm mul}_k):{}(\mathbb N,\mathbb R,S)\to(\mathbb N,\mathbb R,S)$$ because $${\rm mul}_k(x+1)=S^k({\rm mul}_k(x))=k+{\rm mul}_k(x)$$; consider the action $$(\mathbb N,\mathbb R\,S)$$ and the action $$(\mathbb R,\mathbb C\,f)$$ the $$\mathbb C$$-action on the complex numbers $$f(z,w):=e^z\cdot w$$. We have a map  $$(\mathbb N,\mathbb R\,S) \overset{\boldsymbol \chi}{\rightarrow}(\mathbb R,\mathbb C\,f)$$ where $${\boldsymbol \chi}=(\ln b\cdot;\exp_b)$$ because $$\forall n\in\mathbb N,\,x\in\mathbb R\, \exp_b(n+x)=e^{(\ln b) n}\exp_b(x)$$ Proposition every action map $$\boldsymbol \chi$$ can be factored as a composition of its pure equivariant part and its pure divisibility part. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 05/15/2022, 02:37 AM (This post was last modified: 05/15/2022, 10:07 AM by JmsNxn.) (05/13/2022, 02:54 AM)MphLee Wrote: Example in topology take a topological space $$X$$, let $$\bullet={0}$$ the trivial topological space on the point, $$I=[0,1]$$ the real interval and $$S^1=\{(x,y):x^2+y^2=1\}$$ the circle. Continuous functions $$P:{0}\to X$$ are in bijection with points of $$X$$. Call them figures of shape $$\bullet$$ or just elements of $$X$$; Continuous functions $$\gamma:I\to X$$ are in bijection with paths of $$X$$. Call them figures of shape $$I$$ or just $$I$$-elements of $$X$$; Continuous functions $$\gamma :{}S^1\to X$$ are in bijection with loops in $$X$$. Call them figures of shape $$S^1$$ or just $$I$$-elements of $$X$$; In this sense, we can refer to a sequence $$(x_n,y_n) \to 0$$ of complex numbers $$x_n,y_n \in \mathbb{C}$$ are in bijection with holomorphic functions $$f(z)$$ for $$z$$ in a neighborhood of zero. Though we have to have a certain decay condition $$x_n = \mathcal{O}(1/n^{1+\delta})$$ for $$\delta > 0$$ (this could probably relaxed though).  This essentially creates a bijection between accumulation points and sheafs at zero. This would create an equivalence class of sorts, but it fits perfectly in this list, and I think it would work great for this. I realized I missed a few steps here... Weierstrass products for: $$h(1/x_n) = 1/y_n\\$$ Always exist, and they exist based (with scrutiny), on the behaviour at infinity. But this creates a complex identity theorem (with a bonus Weierstrass construction); which we can say: If $$(x_n, y_n) \to 0$$ and $$f(x_n) = y_n$$, then these objects are bijective to each other (The object of sequences, to holomorphic functions at zero). And all we're asking is an infinite occurrences of $$\bullet$$ that converge.... Damn, Mphlee, I hope this makes sense.... Accumulation points $$\mathcal{A}$$, given as $$(x_n,y_n)$$, maps exactly to sheafs at zero... Remember when I say Weierstrass, we need to add that: $$\sum_{n=0}^\infty |x_n|<\infty\\$$ So just keep that result, and you can make this identification with $$\bullet$$ (infinite point wise operations, as sequences) across $$\mathbb{C}$$ to make holomorphy at $$0$$. Also, when you talk about Weierstrass about zero, you don't talk about Weierstrass' factorization theorem. Where as the latter factorization theorem is designed for $$\mathbb{C}$$, the former is used for simply connected domains. But it is still ultimately, Weierstrass. MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 05/25/2022, 03:32 PM (This post was last modified: 05/25/2022, 04:29 PM by MphLee.) (05/15/2022, 02:37 AM)JmsNxn Wrote: Damn, Mphlee, I hope this makes sense....Moved this discussion HERE (link). Here I begin to save some old pages, illustrations and notes that later evolved into the present thread. This is part 1 of the genesis of the idea of intrinsic iterates. 2015, September - The lattice of divisibility and fractional iterates The integers are ordered into a lattice, the lattice of the divisor relation. The lattice operations are $$\rm gcd$$ and the $$\rm lcm$$. Here a fragment of how it looks ^IMG lcm works in this lattice as vector addition would work, instead integer multiplication is always greater than that. Question how to detect when a function behaves as a fractional iterate? What iscount as a fractional iterate? Let's simplify the problem. A root of $$f:X\to X$$ is a function $$g:X\to X$$ that solves $$g^m=f$$. Consider for each function $$f:X\to X$$ the set $$\langle f\rangle:=\{f^0,f^1,...,f^n,...\}$$. The map associates to every function the submonoid generated by that element. Proposition. $$g$$ is a root of $$f$$ is a relation that is homomorphic to the dual inclusion of submonoids: $$g$$ is a root of $$f$$ iff $$\langle f\rangle \subseteq \langle g\rangle$$. Proof. let $$g^m=f$$. Take $$\alpha \in \langle f\rangle$$ then $$\alpha=f^k=(g^m)^k= g^{km}$$ thus $$\alpha \in \langle g\rangle$$. If $$\langle f\rangle \subseteq \langle g\rangle$$ then $$f\in \langle g\rangle$$. By definition we conclude that exists an $$m$$ st $$f=g^m$$, i.e.  $$g$$ is a root of $$f$$. $$\square$$ Let $${\bf P}\subseteq {\rm Sub}({\rm End}(X))$$ be lattice of submonoids of the monoid $${\rm End}(X)$$ generated a single element , i.e. of the form $$\langle f \rangle$$. The lattice order is given by inclusion, lattice and meet are given by intersection of submonoids and the direct product. Fix a function $$f$$. Consider the sublattice of all the submonoids generated by all its integer iterates: call it $${\bf L}^+_f\subseteq {\bf P}$$. It looks like this:  Observe how the direction of the lattice order is reserved respect to the divisibility relations of the integers. \begin{align} m &| n&& {\textrm m\, divides \, n} & \exists k,\,m\cdot k&=n\\ g &\preccurlyeq f&& {\textrm g\, root\, of\, f} & \exists k,\,g^k&=f\\ \langle g \rangle &\supseteq \langle f \rangle&& & \exists k,\,g^k&=f \end{align} The information of the rational iterations (the roots) is encoded in the lattice of submonoids. We can in fact extend the lattice $${\bf L}_f^+$$ by adjoining all the submonoids that contains some iterate of $$f$$: this process is showed in the next picture as the process of prolonging the lattice under $$f$$. Definition. Define the lattice of roots of $$f$$ as $${\bf L}_f^-\subseteq {\bf P}$$ as $${\bf L}_f^-:=\{\langle g\rangle \in {\bf P}\,|\, \langle f \rangle\subseteq \langle g \rangle\}$$ Philosophy. ideally we can think of an infinitesimal generator of $$f$$ as the smallest iterate $$\lim_{\delta\to 0}f^\delta$$ of $$f$$ that generates all the rational/real iterates, i.e. $$\langle f^{m/n} \rangle\subseteq \langle f^\delta \rangle$$. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 05/26/2022, 10:56 PM (05/25/2022, 03:32 PM)MphLee Wrote: (05/15/2022, 02:37 AM)JmsNxn Wrote: Damn, Mphlee, I hope this makes sense....Moved this discussion HERE (link). Here I begin to save some old pages, illustrations and notes that later evolved into the present thread. This is part 1 of the genesis of the idea of intrinsic iterates. 2015, September - The lattice of divisibility and fractional iterates The integers are ordered into a lattice, the lattice of the divisor relation. The lattice operations are $$\rm gcd$$ and the $$\rm lcm$$. Here a fragment of how it looks ^IMG lcm works in this lattice as vector addition would work, instead integer multiplication is always greater than that. Question how to detect when a function behaves as a fractional iterate? What iscount as a fractional iterate? Let's simplify the problem. A root of $$f:X\to X$$ is a function $$g:X\to X$$ that solves $$g^m=f$$. Consider for each function $$f:X\to X$$ the set $$\langle f\rangle:=\{f^0,f^1,...,f^n,...\}$$. The map associates to every function the submonoid generated by that element. Proposition. $$g$$ is a root of $$f$$ is a relation that is homomorphic to the dual inclusion of submonoids: $$g$$ is a root of $$f$$ iff $$\langle f\rangle \subseteq \langle g\rangle$$. Proof. let $$g^m=f$$. Take $$\alpha \in \langle f\rangle$$ then $$\alpha=f^k=(g^m)^k= g^{km}$$ thus $$\alpha \in \langle g\rangle$$. If $$\langle f\rangle \subseteq \langle g\rangle$$ then $$f\in \langle g\rangle$$. By definition we conclude that exists an $$m$$ st $$f=g^m$$, i.e.  $$g$$ is a root of $$f$$. $$\square$$ Let $${\bf P}\subseteq {\rm Sub}({\rm End}(X))$$ be lattice of submonoids of the monoid $${\rm End}(X)$$ generated a single element , i.e. of the form $$\langle f \rangle$$. The lattice order is given by inclusion, lattice and meet are given by intersection of submonoids and the direct product. Fix a function $$f$$. Consider the sublattice of all the submonoids generated by all its integer iterates: call it $${\bf L}^+_f\subseteq {\bf P}$$. It looks like this:  Observe how the direction of the lattice order is reserved respect to the divisibility relations of the integers. \begin{align} m &| n&& {\textrm m\, divides \, n} & \exists k,\,m\cdot k&=n\\ g &\preccurlyeq f&& {\textrm g\, root\, of\, f} & \exists k,\,g^k&=f\\ \langle g \rangle &\supseteq \langle f \rangle&& & \exists k,\,g^k&=f \end{align} The information of the rational iterations (the roots) is encoded in the lattice of submonoids. We can in fact extend the lattice $${\bf L}_f^+$$ by adjoining all the submonoids that contains some iterate of $$f$$: this process is showed in the next picture as the process of prolonging the lattice under $$f$$. Definition. Define the lattice of roots of $$f$$ as $${\bf L}_f^-\subseteq {\bf P}$$ as $${\bf L}_f^-:=\{\langle g\rangle \in {\bf P}\,|\, \langle f \rangle\subseteq \langle g \rangle\}$$ Philosophy. ideally we can think of an infinitesimal generator of $$f$$ as the smallest iterate $$\lim_{\delta\to 0}f^\delta$$ of $$f$$ that generates all the rational/real iterates, i.e. $$\langle f^{m/n} \rangle\subseteq \langle f^\delta \rangle$$. This looks like number theory, but why does that matter ? Regards tommy1729 « Next Oldest | Next Newest »

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