I Need help with tensors and group theory

[Haorong Wu]

TL;DR Summary

What is the relation between tensors and group SO(3)?

I am reading Group Theory in a Nutshell for Physicists by A. Zee.

I have big problems when learning chapter IV.1 Tensors and Representations of the Rotation Groups SO(N).

It reads

Mentally arrange the nine objects $T^{ij}$ in a column $\begin{pmatrix} T^{11} \\ T^{12} \\ \vdots \\ T^{33} \end{pmatrix}$. The linear transformation on the nine objects can then be represetned by a 9-by-9 matrix $D\left ( R \right )$ acting on this column.

For every rotation, specified by a 3-by-3 matrix R, we can thus associate a 9-by-9 matrix $D\left ( R \right )$ transforming the nine objects $T^{ij}$ linearly among themselves.

...

The tensor T furnishes a 9-dimensional representation of the rotation group SO(3).

I can understand why $D\left ( R \right )$ is a representation of SO(3), but I hardly can see why the tensor T can be a representation. I thought a representation should be scalar or square matrix, but why a column consisting of tensors can be a representation, as well.

Also, I do not understand why the tensor came in. If I am correct, elements of groups should be some transformations that leave some objects invariant, but I can hardly imagin how tensors become transformations.

I became more frustrated when the following sections are full of tensors, and I got totally lost.

 

[fresh_42]

Maybe these two articles can help:

https://www.physicsforums.com/insights/representations-precision-important/
https://www.physicsforums.com/insights/what-is-a-tensor/

I am not sure why the author calls the nine elements, the coordinates of $SO(3)$ a tensor, but of course we need to consider them as vectors if we construct $GL(so(3))$, which is needed for a (linear) representation. And every vector is automatically a tensor.

So to answer your question, we need to know what you didn't have quoted.

 

[Haorong Wu]

Maybe these two articles can help:

https://www.physicsforums.com/insights/representations-precision-important/
https://www.physicsforums.com/insights/what-is-a-tensor/

I am not sure why the author calls the nine elements, the coordinates of $SO(3)$ a tensor, but of course we need to consider them as vectors if we construct $GL(so(3))$, which is needed for a (linear) representation. And every vector is automatically a tensor.

So to answer your question, we need to know what you didn't have quoted.

Thanks, fresh_42.

The rest content is split into two parts.

First, it explains why $D\left( R \right )$ is a representation of $SO(3)$, i.e., $D\left ( R_1 R_2 \right )=D\left( R_1 \right ) D\left( R_2 \right )$, which is obvious.

And the second part explains why the statement, that a tensor $T^{ij}$ transforms as if it were equal to the product of two vectors $V^i W^j$, is wrong. And this part is not important.

Thus, I skip those two parts.

However, I have found a explanation for myself why the tensor $T$ could furnishes a representation. Here it is.

For every element, that is rotation, in $SO(3)$, it could be associated with a representation $D\left( R \right )$. Meanwhile, we can find a tensor $T$, such that an object in this tensor $T$ must be a linear combination of the nine objects in $T$ under the transformation of $D\left( R \right )$, so $T = D(R) \cdot T$. Then I could associate a given $D\left( R \right )$ to this tensor $T$. Thus, I indirectly associate a rotation with a tensor. Then the tensor $T$ could furnishes a representation of $SO(3)$.

I am not sure whether this explanation is correct or not.

 

[fresh_42]

There is a basic problem I have here. If we consider a representation of $SO(3)$, then we usually speak of a group homomorphism $SO(3) \longrightarrow GL(V)$ with a vector space as representation space $V$. For a rotation this is normally $V=\mathbb{R}^3$, the ordinary matrix representation.

Now what are the tensor elements here? $SO(3)$ is no vector space, and neither is $\operatorname{GL}(V)$. A representation has $\dim SO(3) \cdot \dim \operatorname{GL}(V)=\dim SO(3)\cdot \dim^2V$ many coordinates, which are $27$ in case of $V=\mathbb{R}^3$, not nine.

To get nine, we consider only one specific element of $SO(3)$, say the rotation $R$. Then $R$ maps $3$ coordinates of $\mathbb{R}^3$ on $3$ new coordinates of $\mathbb{R}^3$, a matrix which we can arrange as a column. But how is it a vector, and a tensor is a vector? What is $0$ in this vector space?

 

[Haorong Wu]

There is a basic problem I have here. If we consider a representation of $SO(3)$, then we usually speak of a group homomorphism $SO(3) \longrightarrow GL(V)$ with a vector space as representation space $V$. For a rotation this is normally $V=\mathbb{R}^3$, the ordinary matrix representation.

Now what are the tensor elements here? $SO(3)$ is no vector space, and neither is $\operatorname{GL}(V)$. A representation has $\dim SO(3) \cdot \dim \operatorname{GL}(V)=\dim SO(3)\cdot \dim^2V$ many coordinates, which are $27$ in case of $V=\mathbb{R}^3$, not nine.

To get nine, we consider only one specific element of $SO(3)$, say the rotation $R$. Then $R$ maps $3$ coordinates of $\mathbb{R}^3$ on $3$ new coordinates of $\mathbb{R}^3$, a matrix which we can arrange as a column. But how is it a vector, and a tensor is a vector? What is $0$ in this vector space?

Hi, fresh_42. The book I read go through another way.

For a 2-indixed tensor, whose indices can choose from 1 to 3 for SO(3), there would be 9 objects. But, these 9 objects are not all independent. That means the tensor furnishes a reducible representation. We can decompose the 9 objects into a 5-dimensional irreducible representation, a 3-dimensional irreducible representation, and a 1-dimensional irreducible representation.

It seems that the author chose not to represent SO(3) in a linear space but in a abstract tensor space.

I am still confusing, but I am getting to understand it.

Thanks!

 

[fresh_42]

This sounds as if the author is heading to Young tableaus.

 

[Haorong Wu]

This sounds as if the author is heading to Young tableaus.

Oh, yes! The Young tableaux is introduced shortly. But the author does not talk a lot about it, just mentions that physicists do not concern it.