What are the orbits of the following group action of $(\mathbb{R}^+,\times)$ on the plane $\mathbb{R^2}$ where $a ^ {(x,y)} = (ax,y)$.

I have found

  • $Orb(1,0) = (r,0)$ for $r \geq 0,$
  • $Orb(0,a) = (0,a)$ for $a \in \mathbb{R},$
  • $Orb(-1,0) = (-r,0)$ for $r \geq 0,$

These three orbits indeed cover the entire plane $\mathbb{R}^2$ but are supposed to be disjoint while first orbit and last one are not $(0,0)$ appears in both.

What am I doing wrong?


$(0,0)$ is a fixed point so it is not in any orbit.

The orbits you describe does not fully cover $\mathbb R^2$, for example $(2,2)$ is not in any of your orbits. Maybe a picture is helpful.

A total description of the orbits is $O_{+,y} = \{ (a,y) : a > 0 \}$, $O_{-,y} = \{ (a,y) : a < 0 \}$ and $ O_{0,y} = \{(0,y)\}$. The orbits $O_{+,y}$ $O_{-,y}$ are (open) horizontal half-lines in opposite directions and the orbits $O_0$ represent the point between these two half-lines.

  • $\begingroup$ Many thanks - I came to the same conclusion a few minutes ago! $\endgroup$ – ZeroCool Apr 12 '17 at 9:44
  • $\begingroup$ While I start grasping the use of Orbits as in the counting theorem - Any other use of this fantastic tool? $\endgroup$ – ZeroCool Apr 12 '17 at 9:48
  • 1
    $\begingroup$ You have many many use of the orbits ! Probably one of the most simple and beautiful applications is the classification of finite subgroups of $SO(3)$, which are fully determined by the "exceptional orbits". Here is a link : staff.city.ac.uk/maud.devisscher.1/GS/classif_rotation.pdf $\endgroup$ – user171326 Apr 12 '17 at 9:53
  • $\begingroup$ Will give it a try - Many thanks for the link! $\endgroup$ – ZeroCool Apr 12 '17 at 11:35

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