What is the Dimension of a 3-D Rotation Matrix?

[oldmathguy]

I have a similar question about rotation matrices. I'm trying to understand the dimension of the matrix given below which is a 3-D rotation. I think that its dimension is 3 but unsure. Any help appreciated. Thanks, John

[(cosx sin x 0), (-sinx cosx 0), (0 0 1)] with ( ) = row,

 

[Fredrik]

What do you mean by the "dimension" of a matrix? See the first few posts in this thread, where this question was discussed:

https://www.physicsforums.com/showthread.php?p=3180840#post3180840

.

 

[oldmathguy]

Fredrik,

Thanks for your comment. Sorry for not being clearer. I mean the dimension of the vector space of 3 X 3 matrices R in SO(3, R). In other words, the number of elements in the basis. SO(3,R) is nxn real matrices such that RR^T = I & detR = 1.

I think that the answer is 3 but I'm having trouble to list them.

My ultimate goal is to find the dimension of the vector space of the following:
1) SE(3, R) = {g in R^(4x4) | g in [(R r), (000,1)], detg = 1, R in SO(3, R), r in R^3} which, I think, is 6.
2) GL+(3,R)/GL+(2,R) where GL+(n, R) = {M in R^(nxn) | detM >0}

Thanks, John

 

[Fredrik]

SO(3) doesn't have a natural vector space structure, since the sum of two of its members isn't in SO(3). It's a 3-dimensional Lie group (a 3-dimensional manifold that's also a group and satisfies an additional technical requirement).

 

[oldmathguy]

Fredrik,

Thanks for clarifying this. Since a 3-dim. manifold, what do its 3 dimensions represent? By this, I mean are they the rotation matrices (or angles) for rotations about three orthogonal axes ?

SE(3,R) & GL+(3,R) also are Lie groups so how does one get dimensions of these manifolds ?

Sorry to be so clumsy about this !

Thanks, John

 

[Fredrik]

Thanks for clarifying this. Since a 3-dim. manifold, what do its 3 dimensions represent? By this, I mean are they the rotation matrices (or angles) for rotations about three orthogonal axes ?

SE(3,R) & GL+(3,R) also are Lie groups so how does one get dimensions of these manifolds ?

A manifold is always equipped with a bunch of coordinate systems. These are functions from open subsets of the manifold into ℝⁿ for some n. That n is the dimension of the manifold. There are many ways to define a coordinate system on SO(3). One way to do it is to use Euler angles. Three Euler angles specify a rotation uniquely.

 

[oldmathguy]

Fredrik,

Thanks very much for clarifying SO(3, R) using Euler angles. I somewhat understand Euler angles so can see why 3 work.

I found some another good explanation of the dimension of SO(3) by Prof. VVedensky from Imperial College www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter7.pdf (p. 116). He also clarifies SE(3,R) being 6-dimensional.

This leaves (thanks to you both) only GL+(3,R)/GL+(2,R). I realize (think !) that dimensions are 9 for GL(3,R) & 4 for GL(2,R) and that this means 5 for GL(3,R)/GL(2,R). However, I'm wondering if the requirement of a positive determinant reduces the dimensions of GL+(3,R), GL+(2,R), & GL+(3,R)/GL+(2,R).

My work concerns the polar decomposition of the deformation gradient F as in F = [v]R where:
F in GL+(3, R)
[v] = {vu | ux = x, u in GL+(3,R), x in R^3, v in Symm+(3, R) diffeomorphic to GL+(3, R)/SO(3,R)}
[v] in GL+(3, R) / N where N = {u | ux = x, u in GL+(3, R), x in R^3} where N is isometric to GL+(2, R)
[v] are equivalence classes of stretches which include both pure stretch & shear
R in SO(3, R)

So, my goal is to understand the dimension of the manifold GL+(3, R) / N.

Thanks again, very much.

John