From 364e464d20c8958b599c8dc957a1485570442cdc Mon Sep 17 00:00:00 2001 From: Joshua Moerman Date: Tue, 16 Oct 2018 08:39:56 +0200 Subject: [PATCH] meer intro --- content/introduction.tex | 78 +++++++++++++++++++++++++++++++++++----- 1 file changed, 69 insertions(+), 9 deletions(-) diff --git a/content/introduction.tex b/content/introduction.tex index 21dd72b..d6e6b68 100644 --- a/content/introduction.tex +++ b/content/introduction.tex @@ -235,8 +235,8 @@ There are three different interesting actions one can define: \NC \pi \cdot_{c} \sigma \NC = \pi \sigma \pi^{-1} \NR \stopalign\stopformula Here the group multiplication is written by juxtaposition. -The first two actions are left-multiplication and right-multiplication respectively. -The latter is called conjugation. +The first two actions are \emph{left-multiplication} and \emph{right-multiplication} respectively. +The latter is called \emph{conjugation}. For each of them, one can verify the requirements. \stopitemize \stopexample @@ -256,12 +256,20 @@ The relation of \quotation{being in the same orbit} is an equivalence relation ( This relation partitions the set $X$ in a collection of orbits: \startformula X = \bigcup_{x \in X} \orb(x). \stopformula -For a trivial $G$-set $X$, each element defines its own orbit, since $\orb(x) = \{ g x \mid g \in G \}$ is a singleton set. +We can picture orbits in the following way. +\todo{PLAATJE} +As we wish to represent such sets (in order to run algorithms on them), we are especially interested in orbit-finite sets. +For such sets, we can represent the whole set by a collection of its orbits. +What remains to be represented are the orbits themselves. +An easy way to do is, is to choose a representative of the orbit $x \in \orb(x)$. (Any element will do as the other elements can be constructed via the group action.) +\todo{PLAATJE} \startexample -We will describe the orbits for some non-trivial $\Pm$-sets. +We will describe the orbits for some $\Pm$-sets. \startitemize \item +For a trivial $G$-set $X$, each element defines its own orbit, since $\orb(x) = \{ g x \mid g \in G \}$ is a singleton set. +\item The $\Pm$-set $\atoms$ only has \emph{one orbit}. To see this, take two (distinct) elements $a, b \in \atoms$ and consider the bijection $\pi = \swap{a}{b}$. Then we see that $\pi \cdot a = b$, meaning that $a$ and $b$ are in the same orbit. @@ -286,24 +294,76 @@ This quantity grows exponential in $k$.) This shows that the set $\atoms^{*} = \bigcup_k \atoms^{k}$ has countably many orbits. \item The set $\atoms^{\omega}$ has \emph{uncountably many orbits}. -To see this we will order the elements of $\atoms = \{ a_0, a_1, a_2, \ldots \}$. +To see this, fix two distinct elements $a, b \in \atoms$. Now, let $\sigma \in 2^{\omega}$ be an element of the Cantor space. We define the following sequence $x^{\sigma} \in \atoms^{\omega}$: \startformula\startalign -\NC x^{\sigma}_0 \NC = a_0 \NR +\NC x^{\sigma}_0 \NC = a \NR \NC x^{\sigma}_{i+1} \NC = \startmathcases - \NC a_i, \NC if $\sigma(i) = 0$ \NR - \NC a_{i+1}, \NC if $\sigma(i) = 1$ \NR + \NC a, \NC if $\sigma(i) = 0$ \NR + \NC b, \NC if $\sigma(i) = 1$ \NR \stopmathcases \NR \stopalign\stopformula Now for two distinct elements $\sigma, \tau \in 2^{\omega}$, the elements $x^{\sigma}$ and $x^{\tau}$ are different. -More importantly, their orbits $\orb(x^{\sigma})$ and $\orb(x^{\tau})$ are different. +More importantly, their orbits $\orb(x^{\sigma})$ and $\orb(x^{\tau})$ are different too. This shows that there is an injective map from $2^{\omega}$ to the orbits of $\atoms^{\omega}$. This concludes that $\atoms^{\omega}$ has uncountably many orbits. \stopitemize \stopexample +Having finitely many orbits is not enough for a finite representation which we can use algorithmically. +We need an additional finiteness on the elements of a $G$-set, +namely the existence of a \emph{finite support}. +In order to define this, we need the notion of a data symmetry. + +\startdefinition +A \emph{data symmetry} is a pair $(\mathcal{D}, G)$, where $\mathcal{D}$ is a structure and $G \leq \Sym(\mathcal{D})$ is a subgroup of the automorphism group of $\mathcal{D}$. +\stopdefinition + +\startdefinition +Let $X$ be a $G$-set and $x \in X$. +A set $C \subset \mathcal{D}$ \emph{supports} $x$ if for all $g \in G$ with $g|_C = \id|_C$ we have $g \cdot x = x$. + +A $G$-set $X$ is called \emph{nominal} if every element has a finite support. +\stopdefinition + +In a way, if an element is supported by a finite set $C$, it means that the element is somehow constructed from only the elements in $C$. +We can see this from the definition, as changing any element outside of $C$ will leave the element $x$ fixed. + +\startexample +\startitemize +\item +The sets $\atoms$, $\atoms^{k}$, $\atoms^{*}$ are all nominal. +For an element $a_1 a_2 \ldots a_k \in \atoms^{*}$, its support is simply given by $\{a_1, a_2, \ldots, a_k\}$. +\item +The set $\atoms^{\omega}$ is \emph{not} nominal. +To see this, let us order the elements of $\atoms$ as $\atoms = \{ a_1, a_2, a_3, \ldots \}$. +Now the element $a_1 a_2 a_3 \in \atoms^{\omega}$ is not finitely supported. +\todo{fs subset van $\atoms^{\omega}$?} +\item +The set $\{ X \subset \atoms \mid X \text{ is not finite nor co-finite} \}$ (with the group action given by direct image) is a single orbit set, but it is not a nominal set. +\stopitemize +\stopexample + +These examples show that being orbit-finite and nominal are orthogonal properties. +There are $G$-sets which are orbit-finite, but non-nominal. +Conversely, there are nominal sets which are not orbit-finite. + +\startremark +The last example above needs a bit more clarification. +In the book of \citet[Pitts13], the group of permutations is defined to be +\startformula +G_{< \omega} = \{ \pi \in \Perm \mid \pi(x) \neq x \text{ for finitely many } x \}. +\stopformula +This is the subgroup of $\Pm$ of \emph{finite} permutation. +The set $\{ X \subset \atoms \mid X \text{ is not finite nor co-finite} \}$ has infinitely many orbits when considered as a $G_{< \omega}$-set, but only one orbit as a $\Pm$-set. +This poses the question which group we should consider (for example, \citet[DBLP:journals/corr/BojanczykKL14] use the whole group $\Pm$). +For nominal sets, there is no difference: nominal $G_{< \omega}$-sets and nominal $\Pm$-sets are equivalent, as shown by \citet[Pitts13]. +It is only for non-nominal sets that we can distinguish them. +We will mostly work with the set of all permutations $\Pm$. +\stopremark + \stopsubsection \stopsection