Understanding real and complexified Lie algebras of ${\rm SO(3)}$

Question

In the form, $$[J_i,J_j]=i\epsilon_{ijk}J_k\tag{1}$$ the Lie algebra of ${\rm SO(3)}$, denoted by $\mathfrak{so}(3)$, is called real Lie algebra.

By taking complex linear combinations $J_{\pm}=J_1\pm iJ_2$, $(1)$ can be written in the form $$[J_3,J_{\pm}]=\pm 2J_{\pm},~~~ [J_+,J_-]=2J_3.\tag{2}$$ Now, it is called the complexified Lie algebra of ${\rm SO(3)}$, denoted by $\mathfrak{so}(3)_{\mathbb{C}}$.

Question $1$ In what sense the algebra $(1)$ is real but $(2)$ is complex(ified)? Essentially, I am asking, what was so real about $(1)$ that has become complex in $(2)$?

Addendum The issue is, given a Lie algebra structure [such as $(1)$ or $(2)$], how does one figure out whether it is a real Lie algebra of the group or a complexified one?

Question $2$ From the point of view of representation theory (as applied to physics), why is it necessary to differentiate real and complexified Lie algebras?

I did look at a couple of similar posts, in particular,

"How does complexifying a Lie algebra $\mathfrak{g}$ to $\mathfrak{g}_\mathbb{C}$ help me discover representations of $\mathfrak{g}$?" and,

"Motivating Complexification of Lie Algebras?".

But I think, here I am asking a more elementary question than these posts seem to deal with.

Answer 1

1. The commutation relations (1) form the real Lie algebra $so(3,\mathbb{R})$ in the physics conventions, where the Lie algebra elements are chosen Hermitian.

   In contrast in the mathematics convention, where the Lie algebra elements are chosen anti-Hermitian, there is no explicit imaginary unit $i$ in the commutation relation (1) for $so(3,\mathbb{R})$. In other words, the structure constants are real. This explains why it's a real Lie algebra. See also my related Phys.SE answer here.

   The complexification is isomorphic to $so(3,\mathbb{C})$.

2. The commutation relations (2) form the real Lie algebra $sl(2,\mathbb{R})\cong so(1,2;\mathbb{R})$ in the mathematics convention. See also this related Phys.SE post.

   Their complexification is isomorphic to $so(3,\mathbb{C})$.

The above is a good example why it is important to distinguish between real and complex Lie algebras.

Answer 2

A reasonably simple way to disentangle this is to start from the group. Surely a rotation by an angle $\theta$ about $\hat z$ would be represented by the real matrix \begin{align} R_z(\theta)&= \left(\begin{array}{ccc} \cos\theta & \sin\theta & 0 \\ -\sin\theta &\cos\theta &0 \\ 0&0&1\end{array}\right)\, \tag{1} \end{align} etc. Note that of course (1) is NOT a diagonal matrix with complex entries, but a real matrix which cannot be made diagonal without introducing complex numbers.

The generator of infinitesimal rotation (defined without the "i" as is traditional in physics) \begin{align} \hat {\mathbb{L}}_z=\frac{d}{d\theta}R_z\bigl\vert_{\theta=0} \end{align} would be the real antisymmetric matrix \begin{align} \hat {\mathbb{L}}_z = \left(\begin{array}{ccc} 0 & 1 & 0 \\ -1 &0 &0 \\ 0&0&0\end{array}\right)\, \tag{2} \end{align} and NOT hermitian.

You see how the physics convention would differ as the generators are defined with an $i$ in it: \begin{align} \hat {{L}}_z=-i\frac{d}{d\theta}R_z\bigl\vert_{\theta=0}\, . \end{align}

The introduction complex numbers is required at some point because of the insistence on using diagonal operators. The eigenvectors of (2) are complex combination of the basis vectors $\hat{\boldsymbol{e}}_{x,y,z}$.

The factor of "$i$" is of course not an issue if you are dealing with matrices with complex entries, such as $SU(2)$.

In dealing with real form and complex extensions, the mathematics way of doing things is less confusing although not familiar to physics. The only math/phys. book I know who consistently follows the math convention is

Cornwell, J.F., 1984. Group theory in physics. 2 (1984). Acad. Press.

If you deal with compact groups, then one can complexify and decomplexify without second thoughts. If you are dealing with non-compact groups (v.g. Lorentz), then one has to be careful as representations that are irreducible under the reals may become reducible over the complex (v.g. Lorentz again: if you're not allowed to take the combo $K\pm iL$ then the adjoint is irreducible and does not break into $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$).