About the decomposition of a rank 2 tensor into its irreducible components

A rank 2 tensor $$T_{ij}$$ of 3D rotation group $$SO(3)$$ is a reducible representation. It has the decomposition $$9=5+3+1$$ where 5 is the symmetric traceless tensor, 3 is the vector and 1 is the scalar. When the rotation group acts on a vector, the representative matrix is a $$3\times 3$$ orthogonal matrix with unit determinant.

After we make this decomposition into irreducible tensors by block diagonalization of a $$9\times 9$$ matrix into a $$5\times 5$$, $$3\times 3$$ and $$1\times 1$$ blocks, I have some questions.

• What kind of matrices are the $$9\times 9$$ square matrices which act on the $$9\times 1$$ general tensor $$T_{ij}$$?

• What kind of matrices are the $$5\times 5$$ square matrices which act on the $$5\times 1$$ symmetric traceless tensor?

When I ask what kind I am asking if they are also orthogonal and having unit determinant. I guess not. In that case, what kind they are. I am quite sure the $$3\times 3$$ & $$1\times 1$$ matrices that act on $$3$$ and $$1$$ are respectively orthogonal with determinant one and the number 1 respectively?

Take the $$3\times 3$$ rotation matrix $$R(\Omega)$$ and tensor it with itself: $$R(\Omega)\otimes R(\Omega)$$ will be a $$9\times 9$$ matrix acting on the components of the $$9\times 1$$ $$T_{ij}$$.
As you have guessed, $$R(\Omega)\otimes R(\Omega)$$ is reducible. The $$5\times 5$$ block contains the Wigner D-matrices for $$L=2$$, i.e the $$D^2_{MM'}(\Omega)$$ functions. The $$3\times 3$$ block will contain $$D^{1}_{mm'}(\Omega)$$ matrices and the $$1\times 1$$ block is the scalar rep with $$D^0=1$$. All these are representations of $$SO(3)$$ so are unitary with determinant $$+1$$.
• I did not understand. The block matrices are orthogonal or unitary? You said unitary with determinant +1. That confuses me. I understand why an orthogonal matrix with unit determinant should act on a vector representation. Because its magnitude has to remain unchanged. But why the $5\times 5$ block matrix also need to be orthogonal and unit determinant? Also, can you tell the nature of the $9\times9$ reducible matrix? Is it unitary/orthogonal and halso having determinant 1? What is $\Omega$ by the way? – mithusengupta123 Mar 12 '20 at 14:30
• They are complex because one uses complex combinations of components, v.g $x+iy$, or because the $Y^L_M$ are complex: the eigenstates of $L_z$ are complex combinations of cartesian components. All reps of SO(3) are equivalent to unitary ones, and the complex matrices with elements $D^L_{mm'}$ are entries in a unitary matrix. The properties of the $5\times 5$ follow because the decomposition of a tensor product of two irreps with det=+1 contains irreps with det=+1. Basically the tensor product of SO(3) irreps decomposes into a sum of SO(3) irreps. $\Omega$ labels the elements in SO(3). – ZeroTheHero Mar 12 '20 at 14:44