Q:

Where do $L_+$ and $L_-$ live, if not in $\mathfrak{so(3)}$?

This question is continuation to the previous post. The lie algebra of $ \mathfrak{so(3)} $ is real Lie-algebra and hence, $ L_{\pm} = L_1 \pm i L_2 $ don't belong to $ \mathfrak{so(3)} $. 
However, when constructing a representation for $\mathfrak{so(3)} $, one uses these operators and take them to be endomorphisms (operators) defined on some vector space $V$. Let $\left|lm \right> \in V $,then
$$ L_3\left|lm \right>  = m \left|lm \right> \;\;\;\;\; L_{\pm}\left|lm \right> = C_{\pm}\left|l(m\pm1) \right>  $$
Now, how do we justify these two things ? If $L_{\pm} \notin \mathfrak{so(3)}$, then how is this kind of a construction of the representation possible ?
I belive similar is the case with $\mathfrak{su(n)}$ algebras, where the group is semi simple and algebra is defined over a real LVS.

A:

They do not lie in $\mathfrak{so}(3)$ but they lie in its complexification, which would be $A_1$ in the usual mathematical classification.  Much of Lie representation theory is set up this way: you work at the level of the complexification then go back to the real form. For compact groups it's not a big deal; for non-compact groups extra care is needed. 
So while $L_{\pm}$ do not make sense as elements of $\mathfrak{so}(3)$, they make sense in the complexification.  You can revert back to $\mathfrak{so}(3)$ by using $L_1=\frac{1}{2}(L_++L_-)$ and $L_2=\frac{1}{2i}(L_+-L_-)$.  (Be careful: the basis where $L_0$ is diagonal is a complex combination of the real basis vectors.)

