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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?

Did I get the ideas right? I am an undergraduate, so please don't use advanced terms.

In case, $N_{i}^+$ and $N_{i}^-$ are $J_{i}$ + i$K_{i}$ and $J_{i}$ - i$K_{i}$ apart from some constant factor.

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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 $v^+ \otimes v^-$.

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    $\begingroup$ isn't (0) a one dimensional vector space ? It has dimension (2j+1). $\endgroup$ – user183683 Jan 9 at 8:53
  • $\begingroup$ Sorry for mistakes. I have corrected it. $\endgroup$ – Eric Yang Jan 9 at 8:55
  • $\begingroup$ 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. $\endgroup$ – user183683 Jan 9 at 8:59
  • $\begingroup$ 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. $\endgroup$ – Eric Yang Jan 9 at 9:16
  • $\begingroup$ But aren't direct sum and direct product same for finite dimensional vector spaces? $\endgroup$ – user183683 Jan 9 at 9:25

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