reducing direct products of irreducible representations

From Mathematics wiki

Theorem:

Tensors belonging to a given symmetry class form an invariant, irreducible space. The representation induced on this space is irreducible.

Theorem:

Irreducible representations generated by all symmetry classes are exhaustive.

Therefore, decomposing a direct product of representations into irreducible representations is facilitated by first constructing all the symmetry classes of tensors transforming under the product representation.

1 Example: SO(N)
- 1.1 2-tensor
- 1.2 Fully symmetric 3-tensor
2 References

Example: SO(N)

2-tensor

SO(N) has among its irreducible representations the singlet called the $\mathbf{1}\,$ (dimension 1) as well as the vector representation called the $\mathbf{N}\,$ (dimension $N\,$ ). Consider two vectors $V^i\,$ and $W^i\,$ transforming under the $\mathbf{N}\,$ . Form the 2-tensor

$T^{ij} = V^i W^i\,$ ,

which transforms under the $\mathbf{N}\otimes\mathbf{N}\,$ . I.e., as $V^i \to O^i_j V^j\,$ ,

$T^{ij} \to O^i_l O^j_k T^{lk}\,$ .

This representation is reducible. We can write $T\,$ as the sum of a symmetric and an antisymmetric part, which transform separately. The symmetric part can be written as a traceless part as well as a part containing the trace. Thus,

$T^{ij} = \frac{1}{2}\left( T^{ij}+T^{ji} - \frac{2\operatorname{tr} T}{N} \delta^{ij} \right) + \frac{1}{2}\left( T^{ij}-T^{ji}\right) + \frac{\operatorname{tr} T}{N} \delta^{ij}\,$ .

The trace is invariant under $SO(N)\,$ , since $O^TO = OO^T = 1\,$ . The symmetric and antisymmetric parts of $T\,$ stay symmetric or antisymmetric under $SO(N)\,$ and do not transform into each other. The antisymmetric part has $\frac{N^2 - N}{2} = \frac{1}{2}N(N-1)\,$ independent components. The trace part has $1\,$ component, and therefore the traceless symmetric part has $\frac{1}{2}N(N+1)-1\,$ independent components.

Fully symmetric 3-tensor

Suppose we are given a completely symmetric 3-tensor $T^{ijk}\,$ which transforms under the $\mathbf{N}\otimes\mathbf{N}\otimes\mathbf{N}\,$ . More machinery is needed to handle a general 3-tensor efficiently. Tracing over two indices leaves an object that transforms as a vector, under the $\mathbf{N}\,$ . Because $T^{ijk}\,$ is completely symmetric, any two indices will do, so to make a traceless combination, we must again symmetrize. Then, (summing over repeated indices)

$T^{ijk}\,$

$= \left( T^{ijk} - \frac{\delta^{ij} T^{kll}}{2+N} - \frac{\delta^{ik} T^{jll}}{2+N} - \frac{\delta^{jk} T^{ill}}{2+N} \right) + \left( \frac{\delta^{ij} T^{kll}}{2+N} + \frac{\delta^{ik} T^{jll}}{2+N} + \frac{\delta^{jk} T^{ill}}{2+N} \right)\,$ ,

so that, for instance,

$T^{mjm}\,$	$= \left( T^{mjm} - \frac{ T^{jll}}{2+N} - \frac{N T^{jll}}{2+N} - \frac{T^{jll}}{2+N} \right) + \left( \frac{ T^{jll}}{2+N} + \frac{N T^{jll}}{2+N} + \frac{ T^{jll}}{2+N} \right)\,$ ,
	$= T^{jll} \,$ .

The part we subtracted from $T^{ijk}\,$ to make it traceless is

$\frac{1}{2+N} \left( \delta^{ij} T^{kll} + \delta^{ik} T^{jll} + \delta^{jk} T^{ill}\right)\,$ ,

$= \frac{1}{2+N} \left( \delta^{ij} \delta^{km} + \delta^{ik} \delta^{jm} + \delta^{jk} \delta^{im}\right)T^{mll}\,$ ,

and since the Kronecker deltas are invariant under $SO(N)\,$ , the only remaining index $m\,$ means that the trace transforms as a vector, under the $\mathbf{N}\,$ .