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

Question

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.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952

Comments

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 $\mathbb C$?

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user NiftyKitty95

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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 $su(2)$.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user V. Moretti

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@NiftyKitty95 : Yes, now I understand what you intend to mean, and I guess Moretti has made it clear !!

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952

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@V.Moretti : Also, is the invariant subspace of $V$, the space over which the Casimir Operators of the Lie algebra is defined ?

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user user35952

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If $V$ is irreducible in addition to be invariant, it is an eigenspace of the Casimir operators, indeed.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user V. Moretti

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@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.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user NiftyKitty95

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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 $L(H)$.

This post imported from StackExchange Physics at 2014-04-13 14:49 (UCT), posted by SE-user V. Moretti

Answer 1

$L^\pm$ live in the complexification of the Lie algebra $so(3)$, which is $sl(2)$. The complexification comes with a natural involution $*$ which takes the conjugate of all coefficients of a representation in a basis of the (real) $so(3)$. Thus the $so(3)$ is recovered as the Lie subalgebra of all Hermitian opersators in the complexification. 

The same works in general. indeed, the classification of semisimple Lie algebras is generally done first for the complexified version (where one can introduce raising and lowering operators), and after having the complex Lie algebras one finds the real ones by looking for the possible involutions.