# Why is $\mathfrak{so}(3,1)_{\mathbb{C}}^\uparrow \cong \mathfrak{su}_\mathbb{C}(2) \oplus \mathfrak{su}_\mathbb{C}(2)$ [duplicate]

I am studying the orthochronous Lorentz algebra $\mathfrak{so}(3,1)^\uparrow$ and it reads

$$[X_i,X_j]=i \varepsilon_{ijk} X_k$$ $$[X_i,Y_j]=i \varepsilon_{ijk} Y_k$$ $$[Y_i,Y_j]=-i\varepsilon_{ijk}X_k$$

If I complexify this by taking complex combinations of the form

$$X_i^\pm = \frac{1}{2}(X_i \pm iY_i)$$

I find that I have two $\mathfrak{su}(2)$ Lie algebras:

$$[X^+_i,X^+_j]=i \varepsilon_{ijk} X^+_k$$ $$[X^-_i,X^-_j]=i \varepsilon_{ijk} X^-_k$$ $$[X^+_i,X^-_j]=0$$

which is two $\mathfrak{su}(2)$ subalgebras, so I should be able to say that $\mathfrak{so}(3,1)_{\mathbb{C}}^\uparrow \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$. However I have read in a lot of places, including here, that $\mathfrak{so}(3,1)_{\mathbb{C}}^\uparrow \cong \mathfrak{su}_\mathbb{C}(2) \oplus \mathfrak{su}_\mathbb{C}(2)$. Why do the $\mathfrak{su}(2)$s have to be complexified? The commutators of $\{ X_i^\pm, Y_j^\pm \}$ are $\mathfrak{su}(2)$, why can I complexify them? and why should they be? Is my decomposition not correct?

• For starters, check the dimensions: The complex Lorentz algebra $\mathfrak{so}(3,1)_{\mathbb{C}}$ has 6 complex dimensions = 12 real dimensions, while $\mathfrak{su}(2)$ has only 3 real dimensions, so a conjectured isomorphism $\mathfrak{so}(3,1)_{\mathbb{C}} \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ cannot possibly be true. – Qmechanic Jun 8 '18 at 18:01
• Essentially a duplicate of physics.stackexchange.com/q/28505/2451. – Qmechanic Jun 8 '18 at 18:11
• @Qmechanic Ah that makes sense, thank you. If they weren't the complexified version too, I would not be able to label my representations by the highest weight because that only applies for complexified Lie algebras doesn't it? – Matt0410 Jun 8 '18 at 18:18
• $\uparrow$ Right. – Qmechanic Jun 8 '18 at 18:19