# Irreducible representations

I was studying group theory yesterday and i had this question. If i have an irreducible representation D that belongs in G and another irreducible representation F that belongs in G, is it right to say that the direct product $$D {\otimes} F$$ is also an irreducible representation and if yes why? I thought of angular momentum, i.e $$|j_1,m_1>{\otimes}|j_2,m_2>=|j_1,m_1>|j_1,m_1>$$ which is an irreducible representation.

• No. It is typically reducible, cf. Clebsch-Gordan coefficients. – AccidentalFourierTransform Nov 6 '19 at 16:19
• @AccidentalFourierTransform if we assume that F is a 1-dimensional representation of G then is the direct product an irreducible representation of G? – user243882 Nov 6 '19 at 16:34

In general the tensor product of two irreducible representations is reducible. The best example is the coupling of two spin-1/2 states, which give $$\frac{1}{2}\otimes \frac{1}{2}=0\oplus 1\, .$$
If one of the representation is 1-dimensional, then the result will usually remain irreducible. For instance, the alternating representation $${\cal A}$$ of $$S_n$$, when tensored with an irrep $$\{\lambda\}$$ of $$S_n$$, will give the irrep conjugate to $$\{\lambda\}$$. Clearly the trivial irrep tensored with any irrep will give this same irrep back and nothing else.