Representation Theory – Concept of Irreducible Tensors Explained

Question

This might be a little naive question, but I am having difficulty grasping the concept of irreducible tensors. Particularly, why do we decompose tensors into symmetric and anti-symmetric parts? I have not found a justification for this in my readings and would be happy to gain some intuition here.

Answer 1

You can decompose a rank two tensor $X_{ab}$ into three parts:

$$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$

The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets denote symmetrization). n is the dimension of the vector space.

Now under, say, a rotation $X_{ab}$ is mapped to $\hat{X}_{ab}=R_{a}^{c}R_{b}^{d}X_{cd}$ where $R$ is the rotation matrix. The important thing is that, acting on a generic $X_{ab}$, this rotation will, for example, take symmetric trace free tensors to symmetric trace free tensors etc. So the rotations aren't "mixing" up the whole space of rank 2 tensors, they're keeping certain subspaces intact.

It is in this sense that rotations acting on rank 2 tensors are reducible. It's almost like separate group actions are taking place, the antisymmetric tensors are moving around between themselves, the traceless symmetrics are doing the same. But none of these guys are getting rotated into members "of the other team".

If, however, you look at what the rotations are doing to just, say the symmetric trace free tensors, they're churning them around amongst themselves, but they're not leaving any subspace of them intact. So in this sense, the action of the rotations on the symmetric traceless rank 2 tensors is "irreducible". Ditto for the other subspaces.