Question about the irreducible representation of a rank 2 tensor under SO(3)

[TroyElliott]

When discussing how a rank two tensor transforms under SO(3), we say that the tensor can be decomposed into three irreducible parts, the anti-symmetric part, traceless-symmetric part, and a 1-dimensional trace part, which transforms as a scalar. How do we know that the symmetric and anti-symmetric parts are truly irreducible, i.e. that they cannot be further block diagonalized via some change of basis?

 

[Orodruin]

How do we know that the symmetric and anti-symmetric parts are truly irreducible, i.e. that they cannot be further block diagonalized via some change of basis?

The same way you show that any irrep is irreducible. Show that there are no invariant subspaces.

 

[fresh_42]

The same way you show that any irrep is irreducible. Show that there are no invariant subspaces.

You seem to understand the question. As it is in the math section and the wording is mathematical nonsense "the tensor can be decomposed into three irreducible parts", can you translate the question into math?

 

[Orodruin]

You seem to understand the question. As it is in the math section and the wording is mathematical nonsense "the tensor can be decomposed into three irreducible parts", can you translate the question into math?

Given the fundamental representation of SO(3) on $\mathbb R^3$, there is a natural representation on $\mathbb R^3 \otimes \mathbb R^3$. This representation is reducible to one copy each of the 1, 3, and 5 dimensional irreps of SO(3). How do we know that these representations are irreducible?

Edit, Alternatively: A representation $\rho$ of SO(3) on the vector space of real 3x3 matrices is given by $\rho(g) A = g A g^{-1}$, where $g \in SO(3)$ and $A \in \mathbb R^{3\times 3}$. This representation is reducible to the representation on matrices proportional to the unit matrix, the representation on anti-symmetric matrices, and the representation on traceless symmetric matrices. How do we know that these representations are irreducible.

 

[TroyElliott]

The same way you show that any irrep is irreducible. Show that there are no invariant subspaces.

The book that I am following, "Group Theory in a Nutshell", basically says since a rank-two tensor can be decomposed into a direct sum of 1, 3, and 5 dimensional representations, we will call them irreducible. Could you suggest a way to prove that the 3 and 5 dimensional anti-symmetric and symmetric-traceless tensors cannot be further reduced with a change of basis?

In general, given some tensor, I am not really sure how to show that it has no invariant subspaces under some transformation.

 

[samalkhaiat]

In general, given some tensor, I am not really sure how to show that it has no invariant subspaces under some transformation.

Split the tensor $T$ into symmetric $S$ and anti-symmetric $A$ parts, then show that $S$ and $A$ do not mix under the transformations of the group in question, i.e., they belong to different invariant subspaces. For example, under $SO(3)$ transformations $R_{i}{}^{j}$, you can easily show that $$S_{ij} \to \bar{S}_{ij} = R_{i}{}^{l}R_{j}{}^{k} \ S_{lk} ,$$$$A_{ij} \to \bar{A}_{ij} = R_{i}{}^{l}R_{j}{}^{k} \ A_{lk} .$$ And if you try the transformation $$S_{ij} \to \bar{A}_{ij} = M_{i}{}^{l}M_{j}{}^{k} \ S_{lk} ,$$ you can show that there exists no such matrix $M$.