# Why are these general representations of the Lorentz Lie algebra irreducible?

I am reading about the irreducible representations of the Lorentz Lie algebra. The author states that the general irreducible representation is given by

$d^{(j_1,j_2)}(J_i) = d^{(j_1)}(T_i) \otimes I + I \otimes d^{(j_2)}(T_i)$,

$d^{(j_1,j_2)}(J_i) = id^{(j_1)}(T_i) \otimes I -i I \otimes d^{(j_2)}(T_i)$,

where $d^{(j)}$ is the spin $j$ representation of $\mathfrak{su}(2)$, $T_i=-\frac{1}{2}\sigma_i$ generate $\mathfrak{su}(2)$ and the $J_i$, $K_i$ are generators for the Lorentz Lie algebra.

I don't understand why this is irreducible. I believe a representation is irreducible if there is no invariant subspace. Isn't the subspace $V_1\otimes{0}$ invariant under $d^{(j_1,j_2)}$ and hence under the group element $\exp\left(t d^{(j_1,j_2)}\right)$?

• The subspace $\text{anything}\otimes0$ is the zero subspace. Jul 31, 2014 at 0:31

In the $(j_{1},j_{2})$ representation, one has the 6 matrices :
$$M_{i} = d^{(j_{1})}(T_{i})$$ $$N_{i} = d^{(j_{2})}(T_{i}) ; \ i = 1,2,3$$ Out of these, one can choose, WLG, $N_{3}$ and $M_{3}$ to be simultaneously diagonal in a suitable basis. The $M_{i}$ and $N_{i}$ individually form an $su(2)$ lie algebra while at the same time commuting with each other.
The dimension of this representation is $(2j_{1}+1)(2j_{2}+1)$. Now given any non-trivial invariant subspace $S$ of the above representation, it is generally true that the matrices $M_{3}$ and $N_{3}$ can be simultaneously diagonalised on $S$. So there exists a simultaneous eigenvector $|m_{1},m_{2}\rangle$ in $S \subset V$ where $V$ is the vector space of the $(j_{1},j_{2})$ representation. Following the properties of the $su(2)$ lie algebra, and using the fact that the $M_{i}$ and $N_{i}$ commute with each other, it follows that all other simultaneous, linearly independent eigenvectors can be generated By suitable actions of matrices of the form $M_{i}N_{j}$ on $|m_{1},m_{2}\rangle$. This is a standard result about the representations of $su(2)$ (colloquially called raising/lowering operators).
The total number of such linearly independent eigenvectors is precisely the dimension of V $= (2j_{1}+1)(2j_{2}+1)$ and all of them must lie in $S$ since $S$ is an invariant subspace. This means that $S=V$ (equality of dimensions).
Thus the only non-trivial invariant subspace of $V$ is $V$ itself and the representation is irreducible.