# About $(0,1/2)$ representations

While studying representations of Lorentz group, we get the generators to be $$J_{i}$$ - rotations and $$K_{i}$$ - boosts. We define $$N_{i}^+$$ and $$N_{i}^-$$ operators and these operators obey the same Lie algebra as the $$SO(3)$$. Hence we conclude that we use these representations for $$N_{i}^+$$ and $$N_{i}^-$$. If we look at these representations, how can we make $$N_{i}^+$$ and $$N_{i}^-$$ act on different dimension vectors, which we do while studying $$(0,1/2)$$ or $$(1/2,0)$$ representations?

In case, $$N_{i}^+$$ and $$N_{i}^-$$ are $$J_{i}$$ + i$$K_{i}$$ and $$J_{i}$$ - i$$K_{i}$$ apart from some constant factor.
The vectors in representation $$(0,1/2)$$ can be seen as a tensor product of vectors in $$(0)$$ and $$(1/2)$$, where $$(0)$$ is a one dimensional space and $$(1/2)$$ is a two-dimensional space, i.e., a spinor. The operator $$N_i^+$$ is actually $$N_i^+ \otimes I$$, and $$N_i^-$$ is $$I \otimes N_i^{-}$$, and they act on vectors $$v^+ \otimes v^-$$.
• Anyway, that's not what I wanted to ask. So (0,1/2) representations act on 3 component vector space, where $N_{i}^+$ acts on the first two components and the third one doesn't change under Lorentz transformations as it (0). Have I got it right? So we neglect writing the third component and say the right spinor has 2 complex components. – user183683 Jan 9 at 8:59
• The dimension of the direct product of a $m$-dimensional space and a $n$-dimensional space is $m \times n$, not $m+n$. For example, if the base vector of $(0)$ is $e_1$ and the base vector of $(1/2)$ is $j_1$ and $j_2$, the base vector for (0,1/2) will be $e_1 \otimes j_1$ and $e_1 \otimes j_2$. Direct product is not direct sum. – Eric Yang Jan 9 at 9:16