0

I have a fixed plane that takes the projection of a 3D image and we need to prove that all the rotations, fixing the plane, is a subgroup of SO(3).

From basic understanding I know that the rotations, fixing the plane means it is just SO(2) and thus I need to prove SO(2) is a sub group of SO(3), but I am not sure how to write it mathematically.

user1166419
  • 31
  • 1
    No, strictly speaking, you're not proving that $SO(2)$ is a subgroup of $SO(3)$, since these things are very different as sets. Rather, you're proving that that there is a subgroup of $SO(3)$ which is isomorphic to $SO(2)$, by considering all the rotations which fix a given plane. – shalop Apr 04 '15 at 03:18

1 Answers1

2

To check that a subset of a group is a subgroup, you need to check that it is closed under composition and inverses. In other words, you must check that if you are given two rotations $g$ and $g'$ fixing your plane $P$, then $gg'$ and $g^{-1}$ also fix $P$.

[Note that this subgroup is indeed an isomorphic copy of SO(2) sitting inside of SO(3), but you have to be careful in your phrasing, since unless that plane is the $xy$ plane it won't be the 'standard' copy of SO(2) sitting in SO(3).]

Rolf Hoyer
  • 12,399
  • I get your point. But I am still not sure how do I go about this proof. Is there a special name for the group of rotations that fix the plane so that I can read and understand more? – user1166419 Apr 04 '15 at 03:35
  • 1
    The closest terminology I can think of is the stabilizer subgroup, which is the subgroup fixing a particular point for some group action. In this case, you either need to view your subgroup as the intersection of the stabilizers of each point in the plane, or instead view the group as acting on planes in $\mathbb{R}^3$ instead of points. – Rolf Hoyer Apr 04 '15 at 03:46
  • One relatively clean way to handle the proof is to show instead that all the rotations in question fix the $z$ axis and that any rotation that fixes the $z$ axis has its orbits in planes orthogonal to that axis. – Steven Stadnicki Apr 04 '15 at 03:54
  • @user1166419 Rolf Hoyer's solution simply says that if you have two rotations $g, g'$ fixing the plane, then $gg'$ and $g^{-1}$ are also rotations of fixing the plane. If you find this unhelpful, then: What is your definition of a rotation? – Travis Willse Apr 04 '15 at 10:20