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

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In summary, the advantage of asserting that a rotation matrix is a member of the SO(3) group, when all you really need is to know the inverse and closure properties, is that it is a shorter way of stating the same thing.
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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?
 
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Trying2Learn said:
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.##
 
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Trying2Learn said:
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.
 
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Trying2Learn said:
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?
 
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fresh_42 said:
##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.
 
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vanhees71 said:
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)
 
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Trying2Learn said:
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.
Trying2Learn said:
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.

Trying2Learn said:
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/

Trying2Learn said:
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.

Trying2Learn said:
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.##
 
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Thank you, everyone... again. I learn so much here!
 

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

What is SO(3)?

SO(3) is a mathematical group known as the special orthogonal group in three dimensions. It consists of all the possible rotations in three-dimensional space that preserve the orientation of an object.

Why is it important to know that something is in a group?

Knowing that something is in a group allows us to understand and analyze its properties and behavior. It also helps us to make predictions and solve problems more efficiently.

What are the advantages of studying group theory?

Group theory provides a powerful framework for understanding the structure and symmetry of objects and systems. It also has many practical applications in fields such as physics, chemistry, and computer science.

How does knowing something is in a group help in problem-solving?

Group theory provides a systematic approach to problem-solving by breaking down complex problems into simpler components. It also allows us to identify patterns and relationships that can lead to more efficient solutions.

Can knowing something is in a group help in real-life situations?

Yes, group theory has many real-life applications, such as in cryptography, coding theory, and the design of experiments. It also helps us to understand the underlying principles behind natural phenomena and make predictions about their behavior.

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