Need help with tensors and group theory

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Summary:

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

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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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.
 
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  • #3
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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.
 
  • #4
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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?
 
  • #5
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28
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!
 
  • #7
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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.
 

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