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In two dimensions, any rotation can be decomposed into two reflections. But no reflection can be done by any number of rotations. (I mean these operations are carried with centre at origin.)

I can also almost see that in three dimensions too, any rotation can be decomposed into two reflections. Is it correct?

If correct, does the same go for higher dimensions as well?

This is how I view it (please tell if some fixing is needed): Rotations and inversions are two different operations, with rotations being continuous and inversions being discontinuous (closely related to the concept of discrete and continuous symmetries is in physics), any of which can't be derived from any number of compositions of the other. Correct? (Note: I'm not considering reflections as inversions. I define inversions as $(x_1,\ldots, x_n)\mapsto(-x_1,\ldots, -x_n)$. And I’m considering reflections with respect to any nonzero subspace of $\Bbb{R}^n$.) In the middle sits reflection. It is a combination of rotation and inversion, a roto-inversion. (Following is the part I'm dubious about.) Any rotation in any dimensional Euclidean space can be realized by certain number (which if true, I suspect to be two) of reflections. I also think the same holds for inversions. So reflections are fundamental in such issues.

Correct?

(These questions popped up when I was studying homogenous coordinates for two dimensional plane in Linear Algebra.)

Well, I am taking just a first course in Linear Algebra and have never had an exposure to Group Theory before. So please bear with me.

Atom
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  • I tried to explain a few things here. The answers to your question depend somewhat on what exactly you mean by a rotation, a reflection, and an inversion (the last one in particular feels a bit strange in this context - may be I am unfamiliar with your use of the term). – Jyrki Lahtonen May 12 '19 at 06:08
  • Anyway, if by a rotation you mean a linear transformation that preserves lengths as well as handedness, and by a reflection you mean an orthogonal reflection of $\Bbb{R}^n$ with respec to an $(n-1)$-dimensional hyperplane through the origin, then some answers are 1) Yes, in 3D a rotation is always a combination of two reflections. Both planes of reflection can be chosen in such a way that their intersection is the axis of rotation. 2) No, in higher dimensions you need more reflections to achieve the same. More precisely, two reflections per each pair of eigenvalues $\neq1$. – Jyrki Lahtonen May 12 '19 at 06:13
  • The gist of my linked answer is that a generic rotation from $SO(n)$ stabilizes a sequence of 2-dimensional subspaces such that its restriction to those subspaces is a usual rotation of a plane, and in addition, the rotation leaves the orthogonal complement of the direct sum of thow 2d-subspaces pointwise invariant. – Jyrki Lahtonen May 12 '19 at 06:16
  • But, what is an "inversion" to you? A linear transformation like $(x,y,z)\mapsto (x,y,-z)$? For me an inversion is a planar transformation aka as a reflection with respect to a circle, but that doesn't seem to fit here (unless I missed something). – Jyrki Lahtonen May 12 '19 at 06:19
  • A caveat is that me needing more than two reflections in higher dimensions is a consequence of the definition of a reflection that I used. If any reflection fixes a subspace of codimension one, then the composition of two such fixes a subspace of codimension two. But there are rotations that violate this. So the precise answer to your question also depends on what you mean by a reflection. – Jyrki Lahtonen May 12 '19 at 06:25
  • @JyrkiLahtonen I don’t know anything about the Group Theory and $SO(n)$ things. Can you please explain in more layman terms? – Atom May 12 '19 at 12:06
  • Also, I’ve made what I mean by reflections and inversions more clear. Please have a look at the edits. – Atom May 12 '19 at 12:22
  • Also does it imply that if some physical symmetry is reflective (that is discontinuous) in nature, then it should necessarily be continuous also? – Atom May 12 '19 at 12:22
  • Does time reversal symmetry then imply time translational invariance? I guess no, since here it is an inversion not reflection and hence can’t be composed in any way to obtain the continuous time translation. – Atom May 12 '19 at 12:27
  • Hmm. Ok. So for you the mapping $(x,y)\mapsto(y,x)$ would not be a reflection even though according to my definition it would be a reflection w.r.t. the line $y=x$? And for you an inversion simply negates all the coordinates? – Jyrki Lahtonen May 12 '19 at 12:48
  • Also, if you only allow coordinate negations as reflections, how do you get something like a $27$ degree rotation as a composition of reflections? The argument I had in mind (in 2D) is that the compositiion of reflections w.r.t. two lines through the origin, forming an angle $\theta$, is rotation by $2\theta$. – Jyrki Lahtonen May 12 '19 at 12:50
  • Oh, my bad! I messed up in explaining what I mean. Let me do it again. By inversion, I mean inversion about origin, i.e. $(x_1,\ldots, x_n)\mapsto(-x_1,\ldots, -x_n)$. By reflection, I mean physical reflection with respect to any nonzero subspace of $\Bbb{R}^n$ and it includes $(x,y)\mapsto(y,x)$. And this form of reflection can be extended to $\Bbb{R}^n$ I think. – Atom May 12 '19 at 13:32
  • Ok, sounds reasonable. Got it. – Jyrki Lahtonen May 12 '19 at 13:33
  • Anything to say? – Atom May 13 '19 at 03:31

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