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?