Lie Algebra of the Lorentz Group $SO(1,3)^{\uparrow}$

I'm trying to get my head around the Lie algebra of the Lorentz group once and for all, but have got tied up in knots. Where is my error in the following?

The universal covering group of the Lorentz group is $SL(2,\mathbb{C})$. Therefore the Lie algebra of the Lorentz group is $\frak{sl}(2,\mathbb{C})$, a real Lie algebra of dimension 6. Its complexification is again $\frak{sl}(2,\mathbb{C})$, since that's already a complex Lie algebra!

The problem is this doesn't agree with Wikipedia here which suggests that the complexification of the Lie algebra of the Lorentz group is $\frak{sl}(2,\mathbb{C})\oplus \frak{sl}(2,\mathbb{C})$.

Where do they get this extra term from and/or why am I missing it? Apologies if it's something obvious - it's been a long day and is nearly midnight here!

In your second paragraph, you are viewing $SL(2,\mathbb C)$ as a real Lie group. Its Lie algebra is $\mathfrak{sl}_2(\mathbb C)$ viewed as a real Lie algebra. The complexification is therefore $\mathfrak{sl}_2(\mathbb C)\otimes_{\mathbb R}\mathbb C$. This is not the complex Lie algebra $\mathfrak{sl}_2(\mathbb C)$: for one thing, its dimension is $6$ while the latter has dimension $3$.
Now, if $g$ is a complex Lie algebra and we let $g_{\mathbb R}$ be $g$ viewed as a real Lie algebra, wwe always have that the complexification of $g_{\mathbb R}$ is $g\oplus g$. Proving this is an instructive exercise.
• Is it not that when you complexify a real vector space $V$ say, via $V^{\Bbb C}=V \otimes_{\Bbb R} \Bbb C$. This is now a real vector space of twice the dimension, however to complete the complexification you must change the base field to $\Bbb C$, and define scalar multiplication by a complex number via $\lambda \in \Bbb C$, and $v\otimes z \in V^{\Bbb C}$, then $\lambda \cdot (v\otimes z)=v\otimes (\lambda\cdot z)$ and $\mathrm{dim}_{\Bbb C}V^{\Bbb C}=\mathrm{dim}_{\Bbb R}V$. – snulty Nov 12 '15 at 12:39