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

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,

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

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.

• "The commutation relations (1) form the real Lie algebra $so(3,\mathbb{R})$ in the physics conventions." , But why real? What is real about it when there is an '$i$' sitting in the commutation relation (I am talking about the physics convention itself)? @Qmechanic
– SRS
Jun 8 '20 at 14:33
• The i there in the comm.relations comes from considering representations by essentially self-adjoint generators. You can drop the i, as mathematicians do, but those Ls are no longer interpreted as angular momenta in QM. Jun 8 '20 at 14:42
• In general, given a Lie algebra structure [such as $(1)$ or $(2)$], how will I figure out whether it is a real Lie algebra or complexified?
– SRS
Jun 8 '20 at 14:57
• Use the mathematics convention. Jun 8 '20 at 15:03

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 $$\mathfrak{su}(2)\oplus \mathfrak{su}(2)$$).

• Thanks! The last paragraph was particularly helpful. But before that, I must say that I am not yet at ease with " what a real Lie algebra supposed to mean?" & "How do we identify whether a given algebra such as $(1)$ or $(2)$ is real or complexified (with or without any reference to representation)?". It bugs me (the content of Q$1$). First I need to understand why $(1)$ is real & $(2)$ is complex, remaining confined within physicists' convention. When we say $(1)$ is real algebra, we surely do not mean $J_{1,2,3}$ are all real. That's not possible if $(1)$ has to hold. @ZeroTheHero
– SRS
Jun 8 '20 at 16:02
• About the last paragraph: So far as I understand, one of the reasons why representations of complexified Lie algebra must be distinguished from the representations of real Lie algebra is that, in certain cases, it may so happen that representations which are actually reducible irreducible under reals become reducible under complex. I think you have self-dual and self-antidual representations in mind. Do I get it right? @ZeroTheHero
– SRS
Jun 8 '20 at 16:15
• @SRS the kind of nomenclature like "self-dual" and "self-antidual" I find confusing so I cannot answer your query. Jun 8 '20 at 16:28
• For example, $F_{\mu\nu}$ is an irreducible representation of $\mathfrak{so(3,1)}$ but reducible into $F_{\mu\nu}+i\tilde{F}_{\mu\nu}$ and $F_{\mu\nu}-i\tilde{F}_{\mu\nu}$ under $\mathfrak{so(3,1)}_\mathbb{c}$. What about the first comment? Am I asking a dumb question?
– SRS
Jun 8 '20 at 16:33
• see also Campoamor-Stursberg, R., de Guise, H. and de Montigny, M., 2013. su (2)-expansion of the Lorentz algebra so (3, 1). Canadian Journal of Physics, 91(8), pp.589-598 where there's a discussion of such issues but it's behind paywall and no arXiv that I can see. Jun 8 '20 at 17:01