[note dump] Iterations and Actions

[MphLee]

the previous: part 1

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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

1. \(f:X\to X\) is compatible with \(E\), i.e, \(xEf(x)\);
   
2. \(f:X\to X\) 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 \[\boxed{f(x)=(\coprod_{b\in B}f_b)(x)=f_{p(x)}(x)}\]