1
$\begingroup$

I've read that orbits are the equivalence classes of the equivalence relation induced by the existence of $g$ $\in$ $G$ such that for every $x$, $y$ $\in$ $\Omega$, $x^g = y$ (here, exponentiation is used to denote group action).

Now, if $G$ is transitive on $\Omega$, then the action of $G$ on $\Omega$ will induce just one orbit. In my understanding, this will also imply that for all $x, y \in \Omega$, $x \sim y$?

Furthermore, if we now let $G$ act on $\Omega × \Omega$, using the natural action $(x,y)^g = (x^g, y^g)$, will the definition of orbit mentioned above still apply? $G$ is not not necessarily transitive on $\Omega × \Omega$, so we now expect to see distinct orbits. How do we now determine which elements of $\Omega × \Omega$ are related?

Thank you very much.

$\endgroup$
  • $\begingroup$ Yes to your first question. It is not possible to say without more information (about the group, about $\Omega$, and about the action) what the orbits of the second action will be like. In general we can say say $(x_1, y_1) \sim (x_2, y_2)$ if there is a $g \in G$ such that $x_2 = x_1^g$ and $y_2 = x_2^g$. $\endgroup$ – Abhiram Natarajan Feb 12 '18 at 15:30
  • $\begingroup$ If G is transitive on $\Omega$ then there will be only one orbit. Not sure in the latter case, more information is needed. $\endgroup$ – Osama Ghani Feb 12 '18 at 15:30
  • $\begingroup$ Has the expression "doubly-transitive action" been mentioned where you are reading/hearing this stuff? $\endgroup$ – Clément Guérin Feb 12 '18 at 15:31
  • $\begingroup$ @AbhiramNatarajan Then, in that case, will $(x_1, y_1)$ and $(x_2, y_2)$ be in the same orbit? $\endgroup$ – sugoi_overload Feb 13 '18 at 2:38
  • $\begingroup$ @ClémentGuérin $G$ is doubly-transitive if $\Omega$ has 2 orbits, if I'm not mistaken? $\endgroup$ – sugoi_overload Feb 13 '18 at 2:42
1
$\begingroup$

As some of the comments mention, it's true that if $G$ acts transitively on $\Omega$, then, by definition, the action has a single orbit, namely, $\Omega$ itself.

One can say at least some things about the natural action $(x, y)^g := (x^g, y^g)$ of $G$ on $\Omega \times \Omega$. For example, for any element of the diagonal $\Delta := \{(x, x) : x \in \Omega\} \subset \Omega \times \Omega$, we have $(x, x)^g = (x^g, x^g) \in \Delta$, so $\Delta$ is a union of orbits, and hence so is $(\Omega \times \Omega) - \Delta$. Thus (if $|\Omega| > 1$) the action of $G$ on $\Omega \times \Omega$ is never transitive.

Can you show using the definition of the action on $\Omega \times \Omega$ that, since $G$ acts transitively on $\Omega$, $\Delta$ is a single orbit?

On the other hand, the decomposition of $(\Omega \times \Omega) - \Delta$ into orbits depends on the nature of the original action, and various behaviors are possible. We acan nalyze the $G$-orbit structure on $\Omega \times \Omega$ just like any other group actions. For any $(x, y) \in \Omega \times \Omega$, $(x, y), (x', y')$ are in the same orbit iff there is a $g \in G$ such that $(x', y') = (x, y)^g$, or, unwinding the definition, such that $x' = x^g$ and $y' = y^g$.

Two "extremal" behaviors are exhibited in the following examples; working out the orbit structure of $G$ on $\Omega \times \Omega$ for both would give you some sense of the possible behaviors:

  1. $G = S_{\Omega}$, the usual permutation action. For concreteness, you might like to take $\Omega = \{1, \ldots, n\}$.
  2. $\Omega = G$, the left regular action (i.e., $h^g := gh$).

1. In this case, $(\Omega \times \Omega) \setminus \Delta$ is a single orbit, and so we say that the action of $G$ on $\Omega$ is doubly transitive. 2. In this case, $(g, h) \sim (g', h')$ implies that there is a $k$ such that $g' = kg, h' = kh$, and so $g'g^{-1} = h' h^{-1}$ (and the converse holds too). Thus the orbits are the sets $H_h := \{(g, hg) : g \in G\}$. In this case, only the identity $e_G \in G$ fixes any element of $\Omega$, so we say that the action of $G$ on $\Omega$ is free.

$\endgroup$
  • $\begingroup$ Hi, does the choice of $g$ have anything to do with which orbit an element belongs to? Say, for $S_3$ acting on $T= \{1,2,3\}$ x $\{1,2,3\}$, the identity permutation would produce the diagonal orbit which would contain all elements of $T$? $\endgroup$ – sugoi_overload Feb 13 '18 at 14:18
  • $\begingroup$ I don't think either part of the question makes sense as written. The orbit of $\omega \in \Omega$ is the set $\{\omega^g : g \in G\}$ of the all of the elements of $\Omega$ one can "reach" via the action of $G$. The orbit decomposition depends only on the action itself, there's no choice of an element $g \in G$ in the picture. $\endgroup$ – Travis Feb 13 '18 at 15:08
  • $\begingroup$ The action of $S_3$ on $\Omega := \{1, 2, 3\}$ by permutations is transitive, so we know that this action has a single orbit, $\Omega$. This means that the induced action on $\Omega \times \Omega$ acts transitively on the diagonal $\Delta=\{(1,1),(2,2),(3,3)\}$, because if $g$ maps $x$ to $x'$, then it also maps $(x,x)$ to $(x',x')$. $\endgroup$ – Travis Feb 13 '18 at 15:11
  • $\begingroup$ Oh, okay. So, in this case, the $S_3$ has 2 orbits on $T$, the diagonal $\Delta$ and $T$ \ $\Delta$, since these are the two "unique" sets produced by applying each $g \in G$ to the elements of $T$? $\endgroup$ – sugoi_overload Feb 13 '18 at 23:07
  • $\begingroup$ That's correct. In particular, $\Delta \setminus T$ is an an orbit because for any pairs $(a, b)$, $(c, d)$, where $a$ and $b$ are distinct and so are $c$ and $d$, there is a permutation mapping $(a, b) \mapsto (c, d)$, i.e., mapping $a \mapsto c$ and $b \mapsto d$. $\endgroup$ – Travis Feb 14 '18 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.