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Then ##I_{+}^{R_1 \otimes R_2} |u \bar d \rangle = \frac{1}{2} \left( \sigma_1^{R_1 \otimes R_2} \pm i \sigma_2^{R_1 \otimes R_2}\right)|u \bar d \rangle = \frac{1}{2}\left( \sigma_1^{R_1} |\bar d\rangle \otimes \text{Id} |u \rangle + \text{Id} |\bar d \rangle \otimes \sigma_1^{R_2} |u \rangle \pm i(1 \leftrightarrow 2)\right)##

1)My first question is if I take ##|u \rangle \rightarrow (1,0), |\bar d \rangle = (0,1)## then I have $$I_+^{R_1 \otimes R_2} |\pi^+ \rangle = I_+^{R_1 \otimes R_2} |u \bar d \rangle $$ This is equal to, $$ \frac{1}{2} \left( \sigma_1^{R_1} | \bar d \rangle \otimes \text{Id}_{2\times 2} |u \rangle + \text{Id}_{2 \times 2} |\bar d \rangle \otimes \sigma_1^{R_2}|u \rangle + i(1 \rightarrow 2) \right)$$ Inputting relevant matrices for sigma_i I get the zero vector as I should since pi^+ is the state of greatest weight and I am applying the raising operator. I just wondered would I get this result independent of the choice of what I take for ##| u \rangle ## and ##|d \bar \rangle## basis vectors? As long as the choice for u and d transforming in fundamental and ubar and dbar transforming in conjugate fundamental are linearly independent?

2) Alternatively, I could just construct the representations for ##I_{\pm}^{R_1 \otimes R_2}## and I would end up with ##4 \times 4## matrices. But what would the ##4 \times 1## objects that these operators act on represent? Would a generic vector be something like ##(u, d, \bar u, \bar d)## so for example I would write ##u = (1,0,0,0)## and ##\bar d = (0,0,0,1)## for example?