I was reading this answer and I don't quite understand how the $\rho$ homomorphism works. The generators of the two copies of $\mathfrak{su}(2)$ in $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ are given by $N_i^+ = \frac{1}{2}(J_i+\mathrm{i}K_i)$ , $N_i^- = \frac{1}{2}(J_i-\mathrm{i}K_i)$ respectively. The $J_i$'s are the generators corresponding to rotations in $SO^+(1,3)$ and $K_i$'s are the generators corresponding to boosts in $SO^+(1,3)$. This is the excerpt where it is defined.

"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 $\mathfrak{gl}(\mathbb{C}^2)$."

1) If $h = N_i ^+$, $\rho ( N_i ^+) = \rho ^+ (N_i^+)\otimes1 + 1 \otimes \rho^-(N_i^+)$ , why is $\rho^-$ defined on $N_i^+$?

My guess of how it ends: $\rho (N_i^+) = (\sigma_i /2)\otimes1 + 1 \otimes {0} = \sigma_i /2$

2)If $h = N_i ^-$, $\rho ( N_i ^-) = \rho ^+ (N_i^-)\otimes1 + 1 \otimes \rho^-(N_i^-)$ , same, why is $\rho^+$ defined on $N_i^-$?

I guess $\rho^+(N_i^-) = 0$ so that $\rho (N_i^-) = 0\otimes1 + 1 \otimes {0} = 0$ , but I am not sure why. Thanks in advance.

  • $\begingroup$ Define your notation... What is $N^{\pm}_i$? What is $h$? What is $\rho^{\pm}$? $\endgroup$
    – JamalS
    Nov 20, 2017 at 21:37
  • $\begingroup$ Ok, I'm sorry, I'll edit it and add it. Everything comes from the answer I linked. $\endgroup$
    – Cami77
    Nov 20, 2017 at 21:38
  • $\begingroup$ That's fine but keep in mind all posts should be self-contained, with the exception of 'obvious' notation like $\mathfrak{su}$. $\endgroup$
    – JamalS
    Nov 20, 2017 at 21:40

1 Answer 1


The total representation map would look less confusing if we express $h \in su_1(2)\oplus su_2(2)$ in terms of the basis $\{N_1^+, N_2^+, N_3^+, N_1^-,N_2^-,N_3^-\}$ of $su_1(2)\oplus su_2(2)$. We recall $\{N_1^+, N_2^+, N_3^+\}$ is the basis of $su_1(2)$ and $\{N_1^-,N_2^-,N_3^-\}$ the basis of $su_2(2)$.

$h=a_1N_1^+ + a_2N_2^+ + a_3N_3^+ +b_1 N_1^-+ b_2N_2^-+ b_3N_3^-$, where $a_i, b_i \in C$

Moreover, let $h=X+Y$ such that

$X=a_1N_1^+ + a_2N_2^+ + a_3N_3^+$

$Y=b_1 N_1^-+ b_2N_2^-+ b_3N_3^-$.

(Notice $X\in su_1(2)$ and $Y\in su_2(2)$)

Also $su_1(2)\cong su_C (2)$ and similarly $su_2(2)\cong su_C (2)$, where $su_C (2)$ is the complexification of the real Lie algebra $su(2)$. Therefore, complex-linear representations of $su_C(2)$ have one-to-one correspondence with complex-linear representations of $su_1(2)$ and $su_2(2)$. This means both $\rho^+$ and $\rho^-$ are complex-linear, where $\rho^+ : \mathfrak{su_1}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1})$ and $\rho^- : \mathfrak{su_2}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_2+1})$

So $$ \rho : \mathfrak{su_1}(2)\oplus\mathfrak{su_2}(2)\to\mathfrak{gl}(\mathbb{C}^{2s_1+1}\otimes\mathbb{C}^{2s_2+1}) $$ $$h\mapsto \rho^+(X)\otimes I_{2s_2 +1} + I_{2s_1+1} \otimes \rho^-(Y).$$ OBS: $\rho$ is complex-linear as a consequence of $\rho ^+$ and $\rho^-$ being complex-linear.


1) If $h = N_i ^+$, for $s1=1/2$, $s2=0$

$\rho ( N_i ^+) = \rho ^+ (N_i^+)\otimes I_1 + I_2 \otimes \rho^-(0) = (\sigma_i /2)\otimes I_1 + I_2 \otimes {0} = \sigma_i /2$

2) If $h = N_i ^-$

$\rho ( N_i ^-) = \rho ^+ (0)\otimes I_1 + I_2 \otimes \rho^-(N_i^-) = 0 \otimes I_1 + I_2 \otimes {0} = 0$


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