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This will probably turn out to be a really simple issue, like bad notation in one of my sources confusing me, but here goes.

In Groups, Representations, and Physics (Jones) it's stated that the generators of rotations $X_i$, and boosts $Y_i$ (Jones' notation, I know they're usually written as $J_i$ and $K_i$) can be combined to give $\vec{X}^{(\pm)} = \frac{1}{2}(\vec{X}\pm i\vec{Y})$, and that these obey the commutation relations of spin, $[X^{(\pm)}_i,X^{(\pm)}_j]=i\varepsilon_{ijk}X^{(\pm)}_k$ and $[X^{(+)}_i,X^{(-)}_j]=0$. That's easy to confirm using the commutation relations of the $X$ and $Y$. So far, so good.

I've also seen it claimed (e.g. Physics of the Lorentz group by Baskal, Kim, and Noz) that the rotation and boost generators can be written in $2\times 2$ form as $X_i = \frac{1}{2}\sigma_i$ and $Y_i = \frac{i}{2}\sigma_i.$ However this seems to create a problem, to me, because we then get $$X^{(+)}_i = \frac{1}{2}(X_i + iY_i)=\frac{1}{4}(\sigma_i + (i^2)\sigma_i)=0.$$ So yes, it obeys the commutation relations of spin, but only trivially, as $[0,0]=0$. Conversely $$X^{(-)}_i = \frac{1}{2}(X_i - iY_i)=\frac{1}{4}(\sigma_i - (i^2)\sigma_i)=\frac{1}{2}\sigma_i = X_i.$$

I would have thought the $X^{(\pm)}$ would be non-trivial, and at least distinct from the rotation generators. What am I missing?

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  • $\begingroup$ Review this. The (+) and (-) generators act on completely disjoint spaces, and X and Y act on the tensor product. thereof.... $\endgroup$ Commented Apr 19, 2023 at 15:09

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Seems like all your conclusions are correct. $X_i = \frac{1}{2}\sigma_i$ and $Y_i = \frac{i}{2}\sigma_i$ indeed provide one of the representations of the algebra of the Lorentz group, because they obey the commutation relations. The same expressions for generators are given in "Symmetry and the Standard Model" by Matthew Robinson, page 122, which I found in a question.

The definition of a Lie algebra representation (see here) does not prohibit the matrices $X_i^{(+)}$ or $Y_i^{(+)}$ to be zero or to be equal to rotation generators. Even more, for every Lie algebra (and for Lorentz algebra in particular), there exists a trivial representation in which all the matrices are zeros.

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  • $\begingroup$ Well firstly, I must say that book (Robinson) which you mentioned turned out to provide the clearest, best description I've come across of the matter. It makes everyone else look like they're intentionally trying to be obtuse, or have just given up on explaining the subject clearly. So thank you for pointing me to it. Also I realised that I was forgetting the distinction between group and algebra, so of course having $X^{(\pm)}=0$ is fine because zero exponentiates to give the trivial group. So I feel stupid! But your answer helped my recognise my stupidity, so that's good! $\endgroup$
    – StarWombat
    Commented Apr 20, 2023 at 22:39

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