The SO(3) group in Group Theory

[sophiatev]

TL;DR

Confusion about the elements of the SO(3) group

In Griffith's Introduction to Elementary Particles, he provides a very cursory introduction to group theory at the start of chapter four, which discusses symmetries. He introduces SO(n) as "the group of real, orthogonal, n x n matrices of determinant 1 is SO(n); SO(n) may be thought of as the group of all rotations in a space of n dimensions", pg. 118. Based on this description, it seems that the elements of the SO(3) group would be the set of 3 x 3 orthogonal rotation matrices. Then later, on page 119 he asserts that "Actually, you have already encountered several examples of group representations, probably without realizing it: an ordinary scalar belongs to the one-dimensional representation of the rotation group, SO(3), and a vector belongs to the three-dimensional representation". What exactly does he mean by this? Clearly neither is a matrix, so I don't quite see how this is true. And how can there be a one-dimensional representation of a group that seems to inherently be in 3D space? (Sorry if this is a question with a very simple answer, this is my first time reading about group theory.)

 

[martinbn]

A representation here means a vector space on which the group acts (linearly). The word might make you think that it means a way of representing (as in expressing it in some way) the group itself, but it only means way the group acts on the space. For example SO(3) rotates vectors in three space, so the 3D space is a representation of the group. The group acts by multiplication on the space of all 3x3 matrices. And many others.

 

[fresh_42]

What exactly does he mean by this? Clearly neither is a matrix, so I don't quite see how this is true. And how can there be a one-dimensional representation of a group that seems to inherently be in 3D space?

A (linear) representation of a group $G$ on a vector space $V$ is a group homomorphism $\varphi\, : \,G\longrightarrow GL(V)$.

There is no a priori restriction for the dimension of $V$. In the one dimensional case we could e.g. consider the left multiplication with the determinant: $\varphi(g)\, : \,(\lambda \longmapsto \det(g)\cdot \lambda)$. Not very interesting for $g\in G=SO(3)$, admitted, but it fulfills the required conditions.

In case $\dim V=3$ we have the matrix application on vectors as natural representation (= group homomorphism = operation) on $V=\mathbb{R}^3$, namely $\varphi(g)(\vec{v})=g \cdot \vec{v}$.

There is always the trivial representation $\varphi(g)=\operatorname{id}_V$ which maps any group element (here orthogonal matrix) on the identity in $GL(V)$. This shows that for any dimension of $V$ there is at least one representation. The one dimensional example above with the determinant is indeed the trivial representation, since all determinants are equal $1$, so $\lambda$ maps onto $\lambda$, regardless which $g\in SO(3)$ we choose.

 

[sophiatev]

Thank you, this makes more sense to me now!

 

[PeroK]

Thank you, this makes more sense to me now!

I'm working my way through that book. In my opinion there is too little in the book to give you a working knowledge of the group theory underpinning the subject. Either you take time out to learn that separately or, what i did, was focus on the specific symmetries inherent in the material. And not worry about the abstract maths behind the scenes.

 

[sophiatev]

I'm working my way through that book. In my opinion there is too little in the book to give you a working knowledge of the group theory underpinning the subject. Either you take time out to learn that separately or, what i did, was focus on the specific symmetries inherent in the material. And not worry about the abstract maths behind the scenes.

Yeah, I've had to take the latter approach as well. Maybe I'll have time in the future to really learn group theory and then revisit the concepts in more depth.