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

Main Question or Discussion Point

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?

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
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
Mentor
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
Staff Emeritus
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.