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Background

In this post, I have shown that a plausible visual representation of a group $K$ in $\operatorname{Sym}(K)$ can be established, where $\operatorname{Aut}(K) \setminus \lbrace \iota_{\operatorname{Sym}(K)} \rbrace$ is an annular area whose innermost region - the only overlapping $\Theta\Gamma$, $\Theta$ and $\Gamma$ being left and right homomorphic images of $K$ in $\operatorname{Sym}(K)$ - is precisely $\operatorname{Inn}(K) \setminus \lbrace \iota_{\operatorname{Sym}(K)} \rbrace$. (Incidentally, therein I noted that center of $K$ is another "visually appropriate" name, indeed.)

With this background, I've been brought to look for a place in this arena for another structure with a strongly visual traditional name, namely the orbits of an action. The following is an attempt to do that, but I can't "close the loop" (literally), as my final open points make evident.


Given a group $G$ and a normal subgroup $H \unlhd G$, the map $(g,h) \mapsto g \cdot h:=g^{-1}hg$ is a $G$-action on $H$.

Definition. We say that $h' \stackrel{\cdot}{\sim} h \stackrel{(def.)}{\Longleftrightarrow} \exists g' \in G \mid h'=g' \cdot h = g'^{-1}hg'$.

$\stackrel{\cdot}{\sim}$ is an equivalence relation in $H$.

Definition. $O_h := [h]_{\stackrel{\cdot}{\sim}} = \lbrace h' \in H \mid h'\stackrel{\cdot}{\sim} h \rbrace = \lbrace g'\cdot h, g' \in G \rbrace$.

$O:=\lbrace O_h, h \in H \rbrace$ is a partition of $H$.

$O_h$ is then obtained by allowing all the instances of $G$ to act left-side on $h$. Distinct instances of $G$ may "move" $h$ to one same $h'$. Indeed:

Definition. We say that $g' \stackrel{h}{\sim} g \stackrel{(def.)}{\Longleftrightarrow} g' \cdot h = g \cdot h$.

$\stackrel{h}{\sim}$ is an equivalence relation in $G$.

Definition. $S_g(h) := [g]_{\stackrel{h}{\sim}} = \lbrace g' \in G \mid g'\stackrel{h}{\sim} g \rbrace$.

$S(h):=\lbrace S_g(h), g \in G \rbrace$ is a partition of $G$.

Theorem (Orbit-Stabilizer). The map:

\begin{alignat*}{1} \chi:S(h)&\longrightarrow O_h\\ [g]_{\stackrel{h}{\sim}}&\longmapsto \chi([g]_{\stackrel{h}{\sim}}):= g \cdot h \end{alignat*}

is well-defined and bijective.

Proof.

  • Good definition: $g' \in [g]_{\stackrel{h}{\sim}} \Rightarrow \chi([g']_{\stackrel{h}{\sim}}) = g' \cdot h = g \cdot h = \chi([g]_{\stackrel{h}{\sim}})$.
  • 1-1: $\chi([g']_{\stackrel{h}{\sim}})=\chi([g]_{\stackrel{h}{\sim}}) \Rightarrow g' \cdot h = g \cdot h \Rightarrow g' \stackrel{h}{\sim} g \Rightarrow [g']_{\stackrel{h}{\sim}}=[g]_{\stackrel{h}{\sim}}$.
  • Onto: by definition, $O_h=\lbrace \chi([g']_{\stackrel{h}{\sim}}), g' \in G \rbrace$.

$\blacksquare$

Proposition. $S_g(h)=C_G(h)g$, the right coset by $g$ of the centralizer of $h$ in $G$.

Proof. $S_g(h)=\lbrace g' \in G \mid g'\stackrel{h}{\sim} g \rbrace = \lbrace g' \in G \mid g'^{-1}hg'=g^{-1}hg \rbrace = \lbrace g' \in G \mid hg'g^{-1}=g'g^{-1}h \rbrace = \lbrace g' \in G \mid g'g^{-1} \in C_G(h) \rbrace = \lbrace g' \in G \mid g' \in C_G(h)g \rbrace = C_G(h)g$.

$\blacksquare$


Now, take $G$ and $H$ to be, respectively, $\operatorname{Aut}(K)$ and $\operatorname{Inn}(K)$ of the Background. There are basically two "topological options" to partition an annular region, say by means of radial fibres or by means of orbits. My idea was to use somehow the OST to show that the only plausible "topological matchings" are $S_g(h) \leftrightarrow$ radial fiber and $O_h \leftrightarrow$ orbit, whence the traditional name for these latter would be justified. Accordingly, I've tried thinking of the possible "crossings" between the "curves" represented by the elements of these two partitions, but I haven't come to any conclusive point. So, as far as I know, an $O_h$ could be as much a "radial fiber" as an "orbit".

Can this approach be kept somehow and led to a conclusion?

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  • $\begingroup$ This is, to me, a high-quality question, since it contains the necessary definitions & theorems and even reads like a blog post (so +1). One criticism I have, though, is that Can this approach be kept somehow and led to a conclusion? is a bit open-ended. But it's fine if you ask me. $\endgroup$ – Shaun Mar 3 at 9:27

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