meer intro

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 first two actions are \emph{left-multiplication} and \emph{right-multiplication} respectively.
-The latter is called conjugation.
+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, \NC if $\sigma(i) = 0$ \NR
-  \NC a_{i+1}, \NC if $\sigma(i) = 1$ \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