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In Matthew Robinson's book Symmetry and the Standard Model he explains that we have generators for rotations $J$ and for boosts $K$. To analyse the group structure, we will look at $N^\pm = J \pm i K$ though and find, that both $N^+$ and $N^-$ form a Lie algebra, which means they generate a Lie group.

Why can I look at a linear combination of generators? Is any linear combination of generators a new generator?

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    $\begingroup$ "Is any linear combination of generators a new generator?" - Yes! A Lie algebra is a vector space so you can add and subtract any two vectors in any way you want! $\endgroup$ Commented Jul 7, 2021 at 13:54

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The generators of a given Lie group (describing the symmetry in this case) form a vector space (the tangent space, which is at $0$ the lie algebra of the Lie Group), hence you can take linear combinations of them. Maybe you are confused since you think, that boost and rotations are forming two separate groups and therefore is it unnatural to take linear combinations of generators of rotations with generators of boosts. However, although the rotations form indeed a group as their own separately, the Lorentz boosts do not form a group as their own, they are coupled to the rotations, which can be seen by considering the commutator of two Boost-Generators $K^i,K^j$. It is given by $[K^i,K^j]=-i\epsilon^{ijk}J^k$. Hence, if one wishes to describe the structure describing the transformations generated by the boosts generators, one needs naturally to consider also rotations. So there is a bigger structure combine both together. The generators of this bigger structure, containing both types of generators $J$ and $K$, are forming a vector space, hence you can take linear combinations.

Usually, the vector space is taken over the field $\mathbb{R}$(real scalar multiplication is allowed), however, you can also generalize this to the field $\mathbb{C}$ (also complex scalar multiplication is allowed).

It turns out that the step going to the complexified Lie-Algebra is even necessary since you want to look at representations of the Poincare-group on the projective Hilbert space (QM state are defined up to a global phase). One can show that the unitary projective representations of a Lie group $G$ can be expressed as the unitary representations of the double cover of $G$ (which will be reconstructed via the complexified Lie-Algebra) on the Hilbert space itself. You can find more about this via looking at "Bargmann's Theorem"

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    $\begingroup$ "Maybe you are confused since you think, that boost and rotations are forming two separate groups. They indeed form groups as their own separately" - Rotations from a group, but Lorentz boosts do not form a group on their own as $[K, K] \sim J$. The OP's linear combinations $N^\pm$ are needed to form subgroups using the boost generators. $\endgroup$ Commented Jul 7, 2021 at 19:12
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    $\begingroup$ Thank you! I changed the answer accordingly. $\endgroup$ Commented Jul 8, 2021 at 10:02
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Given a Lie group $\mathcal{G}$ the generators of the transformations in the group $\mathcal{G}$, lives in the Lie algebra $\mathfrak{g}$ associated. A Lie algebra as you can tell, is a particular kind of an algebra. An algebra $A$ over a field $\mathbb{K}$ is a $\mathbb{K}$-vector space with an additional structure, that is a $\mathbb{K}$-bilinear map (in the Lie algebra case is the Lie bracket [,], with its additional properties). So the algebra has the structure of a vector space, which allows to sum its elements.

The non trivial part is that usually the Lie algebra is defined over a field $\mathbb{R}$, and here in defining $N^+$ $N^-$ you are multiplying the generators by $i$. Basically you are complexifying the Lie algebra, and finding the following chain of isomorphisms. $\mathfrak{so}(3,1)_{\mathbb{C}}\simeq\mathfrak{su}(2)_{\mathbb{C}}\oplus\mathfrak{su}(2)_{\mathbb{C}}\simeq\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$, where the underlying $\mathbb{C}$ in $\mathfrak{g}_{\mathbb{C}}$ means you have complexified the Lie algebra.

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