Oscillation on lower traces and Moore's modular colouring

Let $F$ be a finite set of ordinals in increasing order and let $s,t:F\to \omega$. For a nonminimum $\xi \in F$, write ${\xi }^{-}$ for its immediate predecessor in $F$. The oscillation set of $s$ and $t$ on $F$ is

$\mathrm{Osc}(s,t;F)=\{\xi \in F\setminus \{\min F\}:s({\xi }^{-})\le t({\xi }^{-})\text{ and }s(\xi )>t(\xi )\}\mathrm{.}$

If $F$ is empty, this set is empty without evaluating $\min F$; if $F$ is a singleton, it is empty because there is no predecessor. For $\alpha <\beta <{\omega }_1$, define

$\mathrm{Osc}(\alpha ,\beta )=\mathrm{Osc}(e_{\alpha },e_{\beta };L(\alpha ,\beta )),\mathrm{osc}(\alpha ,\beta )=\mathrm{\mid }\mathrm{Osc}(\alpha ,\beta )\mathrm{\mid }\mathrm{.}$

Both restrictions are defined because $L(\alpha ,\beta )\subseteq \alpha$, and they are finite by construction. Coherence from The minimal-walk functions are coherent and finite-to-one is a later structural control on these comparisons, not a prerequisite for the finite count itself.

The labelled lower trace of Minimal-walk weights, labelled lower traces, and the functions e-beta gives the stronger integer-valued colouring used here. We count labels only at oscillation points:

$o(\alpha ,\beta )={\sum }_{q\in \omega \setminus \{0\}}(\mid \{\xi \in \mathrm{Osc}(\alpha ,\beta ):\mu (\alpha ,\beta ;\alpha )(\xi )=q\}\mid <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q)\mathrm{.}$

This oscillation-supported formula is the variant for which the block lemma's labelled new oscillations give exact changes of the summands. Moore's printed Section 5 formula takes the inverse image on the entire labelled lower trace; clauses (2)--(4) of his Lemma 4.1 do not control labels at the other newly adjoined trace points, so that stronger formula is not used here.

Only finitely many summands are nonzero because the evaluated trace has finite domain. The value $q=0$ is excluded, so reduction modulo zero never occurs; for $q=1$ its contribution is zero.

Enumerate the primes increasingly as $p_0=2,p_1=3,\dots$. Define $\ast :\omega \to \omega$ by $\ast (0)=0$ and, for $m>0$,

$\ast (m)=\min \{n<\omega :p_n\nmid m\}\mathrm{.}$

The minimum exists because a positive integer has only finitely many prime divisors. Put

$o^{\ast }(\alpha ,\beta )=\ast (o(\alpha ,\beta ))\mathrm{.}$

This transform can take values larger than $1$. The binary colouring used by the topology is defined later as $c(\alpha ,\beta )=o(\alpha ,\beta )<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$; the finite-pattern theorem controls both maps but does not conflate them.