[note dump] Iterations and Actions JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 05/26/2022, 11:01 PM (This post was last modified: 05/26/2022, 11:02 PM by JmsNxn.) (05/26/2022, 10:56 PM)tommy1729 Wrote: This looks like number theory, but why does that matter ? Regards tommy1729 Mphlee is mostly building up his approach to fractional ranks, and to do that he's note dumping his categorical interpretation of iteration theory! There's a method to the madness, Tommy. We can't all be drowning in analysis, lol. MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 05/26/2022, 11:20 PM (This post was last modified: 05/27/2022, 12:19 AM by MphLee.) Hy Tommy, Quote:This looks like number theory you are right. To be precise, it's not totally number theory but just the part about divisibility. Why this kind of math appears when we study iteration? I'd explain it as follows. Divisibility is basically the math of ideals in rings. Ideals are mutiplicative objects, i.e. they relates to the multiplicative structure of a ring. Primality, irriducibility and stuff like that are ideal theoretic in nature. The passage is a bit tricky: integer itaration is about the action of the monoid of positive integers, fractional iteration is a bout actions of the abelian group of rational numbers (under addition), continuous iteration (aka dynamics) is about continuous actions of the additive ab. grp. of real numbers... and so on. In general we could say that dynamics/iteration is about the action of a monoid over something: the monoid is the place where the time lives. But to be less exotic we can restrict ourselves to commutative and reversible time (abelian and group). Now the point is, when the monoid is an abelian group, its endomorphisms form a non commutative ring where addition is defined pointwise from the addition of the group, and the multiplication is the composition (endomorphism=distributivity). Eg. the ring of endomorphisms of the abeliang group integers under addittion is essentially the ring of integer numbers (here is where number theory kicks in). ${\rm End}(\mathbb Z,+)\simeq (\mathbb Z, +,\cdot)$ Another tricky passage is needed to have the full picture. The multiplication of the ring of endomorphism of the abelian group of time $A$... acts on "$A$-timed" iterations by something that in, the special case of integer iteration, is the power map $f\mapsto f^{\circ k}$. In other words, the complete information of the $A$-iteration of something is contained in the ring of endomorphisms of $A$. The divisibility relation in the ring is reflected somehow in the root relation between the iterates. Notice Tommy, that the purpose of this serie of posts is just writing down some of my old notes I want to share, some of these are just incomplete ideas. The value of this is subjective I guess. I like to save some of these ideas so that maybe I can extract something valuable later, when I have more time. Also, I find usefull to unify known algebraic facts about iteration using an unique language. Regards. ps. I Just hope we are not at this point already 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: 374 Threads: 30 Joined: May 2013 07/14/2022, 07:12 PM (This post was last modified: 07/15/2022, 02:42 PM by MphLee.) This exposition was written to make clear some of the mechanism used for decomposing actions used in the thread by Leo Qs on extension of continuous iterations from analytic functs to non-analytic 2022, June 26 - Decomposing actions/iterations part 1 - Extensions preserves coproducts Here I share some notes on the concept of decomposing $\mathbb N$-iterations in order to obtain extensions to arbitrary time monoids $A$. First: everything is purely algebraic... so pretty simple and mechanical... no black magic involved, only set theory and composition. Second, we are talking about dynamical systems $(X,f)$ think of them as equivalent to a group action ${\bar f}:\mathbb N\times X\to X$. The two objects are dual descriptions of the same thing, as I've shown in my note dump thread here. The question we are discussing is this one: let $a:\mathbb N\times X \to X$ be an arbitrary action, so that $a(1,-):X\to X$ has no nice properties, no analyticity nor continuity... how do we extend it to an action ${\hat a}:\mathbb R\times X\to X$? Considering the duality actions/iterations, this is equivalently the question: let $f:X\times X$ be a shitty random function, can we $\mathbb R$-iterate it? Leo proceeds considering the case $X=\mathbb C$ and observes that some functions $f$, not analytic, seem to suggest a way to be naturally iterated... as if they are built nicely enough that it is possible to to extend them because they can be decomposed into smaller actions that possess a canonical/natural extensions to the reals. Ok, lets start from the core idea: summing/decomposing dynamical system. As often happens math procedures can be studied in two directions.Start with two functions $f:X\to X$ and $g:Y\to Y$ defined on two disjoint sets $X,Y$. The universal properties of coproduct (read disjoint union), produces from $f$ and $g$ an unique function $X+Y\to X+Y$. It is called the coproduct of the two functions and here is the standard notation for it $f\sqcup g:X\sqcup Y\to X\sqcup Y$ Start with a function $f:X\to X$. Is $f$, a sum/coproduct of dynamical systems? Can we decompose it in smaller pieces? The answer is yes of we can find at least two nonempty and disjoint sets $A$, $B$, st $X=A\sqcup B$, i.e. a partition of $X$, that is compatible with $f$. We say (def) the partition $A\sqcup B$ is compatible  with $f$ iff $f|_A$ lands in $A$, and $f|_B$ lands in $B$. The idea. Assume we have a system $(X,f)$, it is an $\mathbb N$-iteration. We don't know how to extend it to an $\mathbb R$-iteration but we know it can be decomposed into two subsystems $(X_0,f_0)$, $(X_1,f_1)$ that forms a partition of it: we have $f_0:X_0\to X_0$ and $f_1:X_1\to X_1$ s.t. $X=X_0\sqcup X_1$ and $f=f_0\sqcup f_1$, i.e $f(x)=\begin{cases} f_0(x), & \text{if}\, x\in X_0\\ f_1(x), & \text{if}\, x\in X_1 \end{cases}$ Also, we are lucky enough to know very well how to extend those two subsystems from $\mathbb N$ to $\mathbb R$-actions, i.e. we know a natural way to $\mathbb R$-iterate $f_0$ and $f_1$ giving as output $\hat f_0$ and $\hat f_1$. Given an $\mathbb R$-action over $X_0$ and one over $X_1$ glue them back to the total space: we can define an $\mathbb R$-action over $X_0\sqcup X_1$ as follows ${\hat f}(t,x)=\begin{cases} {\hat f}_0(t,x), & \text{if}\, x\in X_0\\ {\hat f}_1(t,x), & \text{if}\, x\in X_1 \end{cases}$ Example. consider the sets $X_0=\mathbb R_{\neq 0}:=\mathbb R\setminus \{0\}$ and $X_1=i\mathbb R_{\neq 0}$. Consider two functions $f_i:X_i\to X_i$ where  $f_0(x)=k_0x$ and $f_1(x)=k_1 x$ for $k_0,k_1\in(0,+\infty)$. The total space $X=X_0\sqcup X_1 \subseteq \mathbb C$ is a subset of the complex numbers and we can define on it an endofunction. $f(z)=\begin{cases} k_0z, & \text{if}\, z\in \mathbb R_{\neq 0}\\ k_1z, & \text{if}\, z\in i\mathbb R_{\neq 0} \end{cases}$ Can we iterate $f:X\to X$? We do it by decomposing it as $f=f_0\sqcup f_1$, then we iterate both of them in the natural way using exponentiation $f(t,z)=\begin{cases} k_0^tz, & \text{if}\, z\in \mathbb R_{\neq 0}\\ k_1^tz, & \text{if}\, z\in i\mathbb R_{\neq 0} \end{cases}$ The general result can be stated informally as follows: assume a $\mathbb N$-action $(X,f)$ can be decomposed as a sum of disjoint subsystems that are 'naturally' extendable to $A$-actions then, $(X,f)$ can be extended to an $A$-action. Proposition (piecewise extension proposition) let $A$ be a monoid, $u:\mathbb N \to A$ a distinguished element of $A$ called unit (of time) and $(X,f)$ be an $\mathbb N$-iteration over $X$, denoted by an abuse of notation as $f$. IfExists a decomposition of $f=\coprod_{i \in I} f_i$ s.t. for every $f_i$ exists an $A$-iteration $g_i$ extending it, i.e. s.t. $u^* g_i= g_i \circ u=f_i$ Then, exists an $A$-iteration $(X,g)$, denote it simply by $g$, that extends $(X,f)$ from $\mathbb N$ to $A$, in symbols $u^* g=f$  Here $u^*$ is the restriction of scalar/time functor defined here. Proof: we prove that given a family of $A$-actions $g_i$, then there exists an $A$-action $g$ such that $u^*(g)= \coprod_{i\in I}u^*(g_i)$. I claim that $g=\coprod_{i\in I}g_i$ is the $g$ we are looking for: let's compute it. Assume wlog $x \in X_j$, for some $j$ and $n\in \mathbb N$. We have $u^*(\coprod_{i\in I}g_i)(n,x)=\coprod_{i\in I}g_i (n\cdot u,x)=g_j(n\cdot u, x)=u^*(g_j)/(n,x)$ but this is the same as computing $\coprod_{i\in I}u^*(g_i) (n,x)$ thus $u^*(\coprod_{i\in I}g_i)= \coprod_{i\in I}u^*(g_i)$ $\square$ Corollary: the restriction scalars/time functor $u^*:A{\rm -Act}\to \mathbb N{\rm -Act}$ preserves coproducts. This is the same as saying that if ${\bf Ex}_u(f_i)$, defined here (as extension), is not the empty for every $i\in I$ then ${\bf Ex}_u(\coprod_{i\in I}f_i)$ is non-empty. 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: 374 Threads: 30 Joined: May 2013 07/15/2022, 04:08 PM the previous: part 1 2022, June 26 - Decomposing actions/iterations part 2 - On bundles and equivalence relations The previous discussion uses the possibility of decomposing the state space $X=\coprod_{i\in I}X_i$ of an endofunction $f:X\to X$ in such a way that the resulting components $f|_{X_i}$ are still endofunctions. Decomposing into a disjoint partition is equivalent to the choice of an equivalence relation. But not all the choices of equivalence relations will do the job. We need only the equivalence relations $E$ that are compatible with the action of $f$ in such a way that the equivalence classes are $f$-stable. Definition 1. An equivalence relation $E\subseteq X\times X$ is compatible with $f:X\to X$ iff each eq. class $[x]_E$ is $f$-stable, i.e. $f([x]_E)\subseteq [x]_E$ Interesting fact. for an eq. rel $E$ on $X$ the following are equivalent statements$E$ is compatible with $f:X\to X$; for every $x\in X$, $x\,E\,f(x)$ Proof: assume every $E$-class is $f$-stable, then for every $x\in[x]_E$ we have $f(x)\in [x]_E$, and this implies $xEf(x)$. In the other direction, assume that for every $x$ we have $xEf(x)$. Take an arbitrary $x$.If $y\in [x]_E$ then $yEx$ by definition.We deduce from this that $[x]_E=[y]_E$ and by our assumption $f(y)\in [y]_E=[x]_E$: the class $[x]_E$ is $f$-stable. $\square$ I'll introduce the concept of bundle to put some order into this and find a third equivalent way of treating compatible decompositions/compatible partitions that gives a tidy presentation of $f=\coprod_I f_i$ in term of its decomposition. Note (on bundles). here with bundle I mean something very abstract and simple, it is just a map $p:X\to B$. The exotic term stands for how we see this map. We see it as specifying a partition of $X$ indicized by $B$. It is a family of preimages bundled together to form $X$, the total space. I'll call the preimages $p^{-1}(b)=\{x\in X\,:\, p(x)=b\}$ the fibers of the bundle at $b$. It is possible to prove that every equivalence relation $E$ defines a surjective bundle $X\to X/E$ and every bundle $p:X\to B$ defines an equivalence rel. on $X$, namely ${\rm ker}p$. When the map is surjective we get precisely equivalence relations, i.e. set partitions. Example. (Covering, Riemann surfaces and others) The first example is rather trivial. The $\mod 2$ projection $[\cdot]_2:\mathbb Z\to \mathbb Z/2\mathbb Z$ over $\mathbb Z/2\mathbb Z$, its subdivides the integers in two fibers: of odd and even numbers. The second is the fractional part bundle $\{\cdot\}:\mathbb R\to S^1$, over over the circle, it can be visualized as winding the real line on itself The third is the floor function bundle ${\rm floor}:\mathbb R\to\mathbb Z$ over $\mathbb Z$, of subdivides the reals in the sum of its length one intervals. Proposition 1. any bundle $p:X\to B$ presents the total space $X$ as a sum of smaller pieces. $X=\bigcup_{b\in B} p^{-1} (b)$ I omit the proof because it is trivial. I can add it on request. Here's the table of how the decomposition applies to the three example given. Crucial. now that our mind can switch from equivalence relations to bundles easily, we ask: when an eq. relation over $X$ is compatible with an endofunction $f:X\to X$? How this condition translates to bundles? If every bundle $p:X\to B$ is the same as describing a partition/eq.rel over $X$, when a bundle $p$ is compatible with an endofunction  $f:X\to X$? The answer is simple, when $f$ respects the fibers, i.e. the decomposition of $X$. Remember that the bundle map sends elements belonging to the same fiber $X_b=f^{-1}(b)$ to the same index $b\in B$, so compatibility of $f$ with the bundle means that elements in the same fiber $X_b$ stay in the same fiber after the action of $f$. If $p(x)=b$ then $p(f(x))=b$ I.e. $p(f(x))=p(x)$ In fact we can prove that if an equivalence relation $E$ and a bundle $p$ induce the same partition over $X$, then the eq. relation $E$ is compatible with $f$ if and only if $p$ is compatible with $f$. Proposition 2. Let $X$ be a set, $E\subseteq X\times X$ be an eq. relation over $X$ and $p:X\to B$ a bundle (surjective) s.t. The partition $X/E$ in eq. classes induced by $E$ coincides with the decomposition induced by $p$ on $X$. In technical terms, if $E={\rm ker} p$, i.e. $xEy \iff p(x)=p(y)$ then the following propositions are equivalent $f:X\to X$ is compatible with $E$, i.e, $xEf(x)$; $f:X\to X$ is compatible with the bundle $p:X\to B$, i.e. $p\circ f=p$ ($f$ is a bundle endomorphism of $p$). Proof. our assumption is $E={\rm ker}p$. If $E$ and $f$ are compatible, i.e. $xEf(x)$, then $x{\rm ker}p\, f(x)$ by our assumption, thus $p(x)=p(f(x))$ by definition of kernel. Since $x$ was arbitrary we deduce $p\circ f=p$. Every step was a biconditional so we can go in the other direction. $\square$. We conclude by using these facts to obtain a presentation of the endofunction $f$ in terms of a compatible bundle function $p$ and the components $f_b$ it induces. Proposition 1 extends: any bundle $p$ compatible with and endofunction $f$ presents $f$ as a coproduct of endofunctions over the components. Proposition 3. given an endofunction $f:X\to X$ and a compatible bundle $p:X\to B$, let $f_b:p^{-1}(b)\to p^{-1}(b)$ be the induced components of the partition. We have $\boxed{f(x)=(\coprod_{b\in B}f_b)(x)=f_{p(x)}(x)}$ MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ « Next Oldest | Next Newest »

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