special relativity - $(1/2, 0)$ representation of the Lorentz Group $SO(1,3)$

Let us consider the $(j, j') = \left(\frac{1}{2}, 0\right)$ representation of $SO(1, 3)\cong SU(2) \otimes SU(2)$.

$j = \frac{1}{2}$ corresponds to $SU(2)$ generated by

$$\tag{1} N_i^+ = \frac{1}{2} \left(J_i + i K_i\right); \quad i =1, 2, 3.$$

$j' = 0$ corresponds to $SU(2)$ generated by

$$N_i^- = \frac{1}{2} \left(J_i - i K_i\right); \quad i =1, 2, 3.$$

For $j' = 0$ representation of $SU(2)$, the generators

$$\tag {2}N_i^- = [0] = 0 \Rightarrow J_i = iK_i$$

Eq.(1) then implies that

$$\tag{3} N_i^+ = \frac{1}{2}(iK_i + iK_i) = iK_i = \frac{1}{2} \sigma_i;$$ where the Pauli matrices $\sigma_i$ are the generators of $SU(2)$ for $j = \frac{1}{2}$.

Therefore $K_i = \frac{-i}{2} \sigma_i$, $J_i = i K_i = \frac{1}{2} \sigma_i$ and an element of $\left(\frac{1}{2} , 0\right) = \exp\left(i \vec{\theta} \cdot \vec{J} + \vec{\phi} \cdot \vec{K} \right)$.

My Question: In Eq. (2), $\quad N_i^-$, $J_i$ and $K_i$ all are $1 \times 1$ matrices. Then how can we substitute $J_i = iK_i$ in Eq. (3), where $N_i^+$ is a $2 \times 2$ matrix? Addition of a number with a $2 \times 2$ matrix is not possible.

Inspiration: This question is inspired by the derivation provided in the book named "Symmetry and the Standard Model" by Matthew Robinson (page: 122).

Answer

This is what happens when physicists try to do group theory but don't bother introducing the proper mathematical notions.

1. 
   

   There is no isomorphism between $\mathrm{SO}(1,3)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$, the former is non-compact, the latter is compact. More around this apparently confusing topic can be found in this answer. Furthermore, using the tensor symbol $\otimes$ is wrong, the product is a direct product, not a tensor product, of groups, see also this question.

   
   


2. 
   

   What is true is that there is an equivalence of finite-dimensional representations of the algebras $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, the latter is the compact real form of the complexification of the former. Indeed, the map between them is given by using $N_i^\pm = \frac{1}{2}(J_i \pm\mathrm{i}K_i)$ as the basis of the latter in terms of the basis $J_i,K_i$ of the former. This is also not an isomorphism of Lie algebras, it is just the case that the finite-dimensional representations of these algebras are equivalent.

   
   


3. 
   

   The argument in the bit that confuses you is supposed to go as follows: You are given the $(1/2,0)$ representation $\rho : \mathfrak{su}(2)\oplus\mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^2)$. Since $\rho$, as a representation, is a Lie algebra homomorphism, you know that $\rho(N_i^-) = 0$ implies $\rho(J_i) = \mathrm{i}\rho(K_i)$. Here, all matrices $N_i^-,J_i,K_i,0$ matrices are two-dimensional matrices on $\mathbb{C}^2$. You know that $\rho(N_i^-) = 0$ as two-dimensional matrices because of how the $(s_1,s_2)$ representation is defined: Take the individual representations $\rho^+ : \mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1})$ and $\rho^- : \mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_2+1})$ and define the total representation map by

   $$\rho : \mathfrak{su}(2)\oplus\mathfrak{su}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1}\otimes\mathbb{C}^{2s_2+1}), h\mapsto \rho^+(h)\otimes 1 + 1 \otimes \rho^-(h)$$ where I really mean the tensor product of vector spaces with $\otimes$. For $s_1 = 1/2,s_2 = 0$, this is a two-dimensional representation where $\rho^-$ is identically zero - and the zero is the two-dimensional zero matrix in the two-by-two matrices $\mathfrak{gl}(\mathbb{C}^2)$.