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It is well known that the Lorentz algebra can be written as two $SU(2)$ algebras. By defining

$$N_i=\frac{1}{2}(J_i+iK_i), \qquad N^{\dagger}_i=\frac{1}{2}(J_i-iK_i)$$

we have

$[N_i,N_j]=i\epsilon_{ijk}N_k,\qquad [N^{\dagger}_i,N^{\dagger}_j]=i\epsilon_{ijk}N^{\dagger}_k, \qquad[N_i,N^{\dagger}_j]=0$

We can find representations in the same way as the spin angular momentum operators of non-relativistic QM, with representations labeled by half-integers. The $(\frac{1}{2},0)$ representation corresponds to left-handed spinors. Similarly we could have a representation $(1,\frac{1}{2})$.

I'm a bit confused by this. In this scenario $N_3$ is a 3x3 matrix and $N^\dagger_3$ is a 2x2 matrix, then $J_3=N_3+N^\dagger_3$ which is a 3x3 matrix + a 2x2 matrix which doesnt make any sense.

Even the $(\frac{1}{2},0)$ representation could be seen as $J_3$ being a 2x2 matrix plus a scalar.

Where am I going wrong?

Qmechanic
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Okazaki
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    It may be well-known, but it is technically false, see Qmechanic's answer here. Representations of $\mathrm{SO}(1,3)$ may correspond to representations of two copies of $\mathrm{SU}(2)$, but the Lorentz algebra does not contain two $\mathfrak{su}(2)$ algebras. – ACuriousMind May 26 '15 at 13:14

1 Answers1

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Apart from the issues of $SU(2)$ vs. $SL(2,\mathbb{C})$, complexification, etc, which is covered e.g. in this Phys.SE post and links therein, it seems OP's core question (v2) is essentially the following question.

How does the Lie algebra generator $J_3=J^L_3+J^R_3$ act on a vector space $V_L\otimes V_R$ for a tensor representation?

The short answer is: By Leibniz rule: $$J_3(v_L\otimes v_R)~=~(J^L_3v_L)\otimes v_R+v_L\otimes (J^R_3v_R),\qquad v_L\in V_L,\qquad v_R\in V_R.$$ Extend the definition to a general element of the tensor product $V_L\otimes V_R$ via linearity.

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