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 \(f\). If

- Exists 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.