# SO(3) -- What is the advantage of knowing something is in a group?

Trying2Learn
TL;DR Summary
What is the advantage of knowing something is in a group.
Good Morning!

I know that Rotation matrices are members of the SO(3) group.
I can prove some useful properties about it:
The inverse is the transpose;
Closure properties;

However, what is the advantage of asserting that a rotation matrix is a member of the SO(3) group, when all I really need is to know the inverse and closure properties?

## Answers and Replies

Mentor
2021 Award
TL;DR Summary: What is the advantage of knowing something is in a group.

However, what is the advantage of asserting that a rotation matrix is a member of the SO(3) group, when all I really need is to know the inverse and closure properties?
##A\in \operatorname{SO}(3)## is definitely shorter than listing all properties. E.g., the group is a symmetry group of several processes in nature. This fact isn't obvious from ##A^{-1}=A^\tau.##

topsquark, malawi_glenn and vanhees71
Gold Member
However, what is the advantage of asserting that a rotation matrix is a member of the SO(3) group, when all I really need is to know the inverse and closure properties?
In addition to those, an important property of 3d rotations is that their order affects the result, i.e. it's a non-Abelian group.

topsquark, malawi_glenn and vanhees71
Gold Member
2021 Award
TL;DR Summary: What is the advantage of knowing something is in a group.

Good Morning!

I know that Rotation matrices are members of the SO(3) group.
I can prove some useful properties about it:
The inverse is the transpose;
Closure properties;

However, what is the advantage of asserting that a rotation matrix is a member of the SO(3) group, when all I really need is to know the inverse and closure properties?
But then you have a group? Why shouldn't one use the standard definitions of mathematics?

topsquark
Trying2Learn
##A\in \operatorname{SO}(3)## is definitely shorter than listing all properties. E.g., the group is a symmetry group of several processes in nature. This fact isn't obvious from ##A^{-1}=A^\tau.##

I like your response... May I follow up with one more?

I know the properties of SO(3)
I know there is an associated algebra so(3)

Do all Groups have an associated algebra?
What is the meaning of this? (I am sorry, I do not know how to be more specific)

For example, if R is a rotation matrix, Rt is the transpose and R-dot is the time derivative, then we obtain

omega-skew = Rt * R-dot

Is it expected that all groups have an associated algebra?
Is there a way I could have anticipated the form of the associated algebra?

Please (if I may) do not go too far into abstract algebra.

Trying2Learn
But then you have a group? Why shouldn't one use the standard definitions of mathematics?
Also for you: followup below (if you have the time)

Homework Helper
Gold Member
topsquark
Mentor
2021 Award
I like your response... May I follow up with one more?

I know the properties of SO(3)
I know there is an associated algebra so(3)

Do all Groups have an associated algebra?
All groups for which multiplication and inversion are differentiable.
What is the meaning of this? (I am sorry, I do not know how to be more specific)
Meaning of what?

The idea from ##\operatorname{SO}(3)## to ##\mathfrak{so}(3)## is the same as from a surface (group) to its tangent space (algebra). Plus a bit of mathematics, but basically that.

The symmetry group of natural processes? Imagine looking in a mirror. It swaps left and right but not up and down. But it is the same whether you rotate yourself or not.

For example, if R is a rotation matrix, Rt is the transpose and R-dot is the time derivative, then we obtain

omega-skew = Rt * R-dot
I'm afraid I don't know what you mean.
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

Is it expected that all groups have an associated algebra?
See above. All groups which can be considered a Riemannian manifold, or shortly: with differentiable multiplication and inversion.

Is there a way I could have anticipated the form of the associated algebra?
##\mathfrak{so}(3)## is a Lie algebra. Its multiplication is defined by ##(X,Y)\longmapsto [X,Y]=X\cdot Y- Y\cdot X## which produces a Lie algebra. The associativity fails here since the defining property is ##X^\tau=-X## and ##(XY)^\tau=Y^\tau X^\tau=YX\neq -(XY)##. But we have the Leibniz rule from the differentiation as property ##[X,[Y,Z]]=[Y,[X,Z]]+[[X,Y],Z].##

It is all about differentiation with all its consequences. The most important one is, that differentiation is a local property. It happens at a point and makes a statement about points nearby. Here (at the beginning) is an example of a local Lie group:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

It is roughly differentiation that transforms group elements to algebra elements: chose a path in the group ##t \longmapsto X(t),## differentiate it at ##t=0## and get the tangent vector in the algebra. Consequently, we have to solve a differential equation if we want to go from tangent space to group. And as usual, when it comes to solving differential equations, we let the exponential function do the work. The details are a bit complicated. Maybe you could read a little bit more from the insight article I just linked to, or the others I have written. Many about Lie theory.

If I am asked to explain it without differentiation along paths in the group then I would say:

• ##1\in \operatorname{SO}(3)## turns into the origin of the vector space ##0\in \mathfrak{so}(3)##
• inversion in the group becomes a minus sign in the algebra
• transposition remains the same
• multiplication in the group becomes addition in the algebra
• determinant becomes trace
So ##X^{-1}=X^\tau\, , \,X\cdot X^\tau=1## gets ##-X=X^\tau\, , \,X+X^\tau=0## and ##\det X = 1## gets ##\operatorname{trace}X=0.##

Last edited:
vanhees71, dextercioby and topsquark
Trying2Learn
Thank you, everyone... again. I learn so much here!