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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.

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user35952
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  • I might be misunderstanding something here, so let me raise a point: Without judging if the operators do or do not lie in the algebra, why does your question arise anyway? In my ear, it sounds similar to "I want to study the properties of consecutive derivatives and people use abstract algebra to do it. How is that justified?" Why not? If you study how $a\mapsto\mathrm{e}^{i\phi}a$ affects elements of $\mathbb C$, is there a reason you would you restrict your study by demanding not to use complex conjugation on $\mathbb C$? – Nikolaj-K Apr 04 '14 at 08:38
  • Sorry, I do not understand why $L_\pm |l :m \rangle = C_\pm |l: (m\pm 1)\rangle$ should require that $L_\pm$ belongs to a representation of the (real) Lie algebra of $so(3)$ or $su(2)$. – Valter Moretti Apr 04 '14 at 08:39
  • @V.Moretti : Neither do I ! But I am not able to convince myself that if they don't belong this, then how can I use them in construction of the representation ?? – user35952 Apr 04 '14 at 08:43
  • @NiftyKitty95 : Thanks for that point, although your analogy has not gotten on me yet. Will ponder again with this. – user35952 Apr 04 '14 at 08:45
  • Ok, so does it mean that when I construct a representation of this algebra using its operation on a Linear Vector space(LVS), only few legitimate operators on this LVS belong to algebra and not all the operators defined over the LVS ? – user35952 Apr 04 '14 at 08:50
  • @user35952: My points is, e.g., if you study the multiplication of the number $7$ by the number $5$ in $\mathbb N$, there is no reason to write this as $7\mapsto(1-2i),7,\overline{(1-2i)}$, if you think that's useful. – Nikolaj-K Apr 04 '14 at 08:55
  • Of course! Usually the representation is constructed over a complex vector space $H$, so the algebra of operators over that space $L(H)$ has a natural complex structure. Nevertheless the representation of a (real) Lie algebra is defined only in a real subspace of $L(H)$. – Valter Moretti Apr 04 '14 at 08:56
  • @NiftyKitty95 : Yes, now I understand what you intend to mean, and I guess Moretti has made it clear !! – user35952 Apr 04 '14 at 16:44
  • @V.Moretti : Also, is the invariant subspace of $V$, the space over which the Casimir Operators of the Lie algebra is defined ? – user35952 Apr 04 '14 at 17:00
  • If $V$ is irreducible in addition to be invariant, it is an eigenspace of the Casimir operators, indeed. – Valter Moretti Apr 04 '14 at 17:13

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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.)

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