# Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$

I'm desperatly confused by notations and formulations so if someone could clarify the following things a little Í would be deeply grateful. The Lie algebra $\mathfrak{so}(1,3)_+^{\uparrow}$ of the proper orthochronous Lorentz group $SO(1,3)_+^{\uparrow}$ is given by
$$[J_i,J_j]=i \epsilon_{ijk} J_k$$ $$[J_i,K_j]=i \epsilon_{ijk} K_k$$ $$[K_i,K_j]=- i \epsilon_{ijk} J_k$$

We can now define new generators with the old ones $N^{\pm}_i= \frac{1}{2}(J_i \pm i K_i)$ which satisfy $$[N^{+}_i,N^{+}_j]=i \epsilon_{ijk} N^{+}_k ,$$ $$[N^{-}_i,N^{-}_j]=i \epsilon_{ijk} N^{-}_k ,$$ $$[N^{+}_i,N^{+}_j]= 0.$$ where we can see that $N^{+}_i$ and $N^{-}_i$ make up a copy of the Lie algebra $\mathfrak{su}(2)$ each. My problem is to get what is going one here mathematically precise. Are the following statements correct and if not why:
1. When we build the new operators from the old generators we complexified $\mathfrak{so}(1,3)_+^{\uparrow}$ $$(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} = \mathfrak{so}(1,3)_+^{\uparrow}\otimes \mathbb{C}$$ 2. We saw that $\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C}$ is isomorph to two copies of the complexified Lie algebra of $\mathfrak{su(2)}$: $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq \mathfrak{su(2)}_{\mathbb{C}} \oplus \mathfrak{su(2)}_{\mathbb{C}}$. Where exactly did we need that $\mathfrak{su(2)}$ is complexified here? The Lie algebras defined by $N^{\pm}_i$ are exactly those of $\mathfrak{su(2)}$ and we never use complex linear combination of $N^{\pm}_i$ or am I wrong here?
3. $\mathfrak{su(2)}_{\mathbb{C}}$ is isomorph to $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$:
$$\mathfrak{su(2)}_{\mathbb{C}} \simeq (\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$$
Here $(\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C}$ denotes the complexified Lie algebra of $SL(2,\mathbb{C})$
4. Is $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{R}$ correct? Here $(\mathfrak{sl}(2, \mathbb{C}))_\mathbb{R}$ denotes the real Lie algebra of $SL(2,\mathbb{C})$
5. Is $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C} \oplus (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C}$ correct?

I looked this topic up in different books and each seemed to state something different. One book even used three differrent versions of $\mathfrak{sl}(2,\mathbb{C})$ namely: $\mathfrak{sl}(2,\mathbb{C})$, $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$ and $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{R}$. Wikipedia states simply that $\mathfrak{sl}(2,\mathbb{C})$ is the complexification of $\mathfrak{su(2)}$ without making any reference to $SL(2,\mathbb{C})$ which does not help me either. Any help would be great.

• That's a lot of questions. Perhaps you could break this up into multiple posts? – Jim Belk Jan 16 '14 at 0:59
• Also, what do you mean by $\mathfrak{so}(1,3)_+^\uparrow$? Is this the same as $\mathfrak{so}(1,3)$? – Jim Belk Jan 16 '14 at 1:00
• Although these are indeed a lot of questions i hoped they are short to answer for someone with more expertise than i have, but maybe i have a lot more wrong than i thougt. I named the Lie group $\mathfrak{so}(1,3)_+^\uparrow$ because i derived it from the known representation of proper orthocronous Lorentz group $SO(1,3)_+^{\uparrow}$, but the Lie algebra is, as far as i know, the same as $\mathfrak{so}(1,3)$. – jak Jan 16 '14 at 6:41
• The upshot of all the answers is that the $\mathbb{R}$-Lie algebra $\mathfrak{so}(1,3)$ (which has dimension six over $\mathbb{R}$) is isomorphic to the scalar restriction of $\mathfrak{sl}_2(\mathbb{C})$ to $\mathbb{R}$. That is, the three-dimensional $\mathbb{C}$-Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ viewed as a six-dimensional Lie algebra over $\mathbb{R}$. For algebraists and reductive group people, the Satake-Tits diagram of these isomorphic algebras consists of two vertices, without an edge, but with an arrow between them. – Torsten Schoeneberg Nov 28 '17 at 20:51
• $\mathfrak{su}(2)$, on the other hand, is the compact real Lie algebra (three-dimensional), whose complexification is $\mathfrak{sl}_2(\mathbb{C})$. Its Satake-Tits diagram consists of one black vertex and nothing else. – Torsten Schoeneberg Nov 28 '17 at 20:54

I have been thinking about this the past few days in preparation for an exam at EPFL as a result of some really shitty course notes. My familiarity with the subject is thus rather poor but at least I sympathize with your plight for clarity.

1 . I think that key to working with this problem is to first make concrete what the complexification of $$\mathfrak{su}(2)$$, $$\mathfrak{su}(2)_\mathbb{C}$$, really is and what its algebra is. We know that the natural basis of the $$\mathfrak{su}(2)$$ are the Pauli matrices $$\{\sigma_1, \sigma_2, \sigma_3\}$$ with the familiar Lie Bracket $$[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k$$. This is a REAL vector space and the complexification is a particular complex vector space where the Lie bracket is essentially what we expect it to be when treating the bracket as if it is linear over $$i$$ as well

$$\mathfrak{su}(2)_\mathbb{C}$$ is the Lie algebra of formal sums $$u + iv$$ where $$u,v \in \mathfrak{su}(2)$$ and where the complexified Lie-bracket expressed in terms of the real Lie bracket is $$[x + iy, u + iv]_{\mathbb{C}} = ([x,u] - [y,v]) + i([x,v] + [y,u])$$ I wont write the complex sign again as its easy to take as implicit. Now that we hopefully agree on the definition I am probably going to annoy you by viewing complexified algebras as real algebras of twice the dimension because I find this situation to be more transparent. I am free t view my complexified algbra as a real algebra and in this picture the most natural basis we can come up with is $$\sigma_1, \sigma_2, \sigma_3, i \sigma_1, i\sigma_2, i\sigma_3$$

I check the resulting Lie brackets and we end up with $$[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k \\ [\sigma_i, i\sigma_j] = i \varepsilon_{ijk}(i\sigma_k) \\ [i\sigma_i, i \sigma_j ] = -i \varepsilon_{ijk}\sigma_k$$

We easily see a correspondence $$J_j \leftrightarrow \sigma_j \qquad K_j\leftrightarrow i\sigma_j$$ and conclude $$\mathfrak{so}(1,3) \simeq \mathfrak{su}(2)_\mathbb{C}$$ thus to be it looks like it is the REAL $$\mathfrak{so}(1,3)$$ which is isomorphic to the complexification of $$\mathfrak{su}(2)$$ (but also viewed as a REAL Lie algbera, of real dimension $$6$$). I find this to be a much more transparent way of arriving at the isomorphism rather than going via the complexification.

2. To me this looks like it will imply $$\mathfrak{so}(1,3)_\mathbb{C} \simeq (\mathfrak{su}(2)_\mathbb{C})_\mathbb{C} \simeq \mathfrak{su}(2)_\mathbb{C} \oplus_\mathbb{C}\mathfrak{su}(2)_\mathbb{C}$$

I have to admit I don't know how to make sense of going via the complexification of $$\mathfrak{so}(1,3)$$ neither. I had an argument planned out but it collapsed and I reverted to the one above. Maby I'll try to fix this if you come back and discuss it with me.

3. I started thinking about this but I think you actually mean $$\mathfrak{sl}(2,\mathbb{R})_\mathbb{C} \simeq \mathfrak{sl}(2,\mathbb{C}) \simeq \mathfrak{su}(2)_\mathbb{C}$$? $$\mathfrak{sl}(2,\mathbb{C})$$ is a real vector space made up of traceless complex matrices so the 6 most obvious basis matrices are $$\alpha_1 = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \alpha_2 = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \alpha_3 = \begin{pmatrix}0& 0 \\ 1 & 0\end{pmatrix}, \; \text{and} \; i\alpha_1, i\alpha_2, i\alpha_3$$ From this we can find an explicit change of basis to the complexified Pauli matrices $$\sigma_1 = \alpha_2 + \alpha_3, \quad \sigma_2 = i\alpha_1 - i\alpha_3, \quad \sigma_3 = \alpha_1\\ i \sigma_1 = i\alpha_2 + i\alpha_3, \quad i\sigma_2 = \alpha_1 - \alpha_3, \quad i\sigma_3 = i\alpha_1$$ and since the the bracket is the commutator we see that the Lie-structures of these two Lie algebras are the same meaning they are the same.

4. To me it looks like we will have $$\mathfrak{so}(1,3) \simeq \mathfrak{sl}(2,\mathbb{C})$$ (where the latter is viewed as a $$6$$-dimensional real Lie algbera) which kind of surprises me.

5. Well if 4. holds then it should hold.

I came up with the same question and this link helped me a lot https://en.wikiversity.org/wiki/Representation_theory_of_the_Lorentz_group#The_Lie_algebra. Look at the isomorphism chain in (A1) and read those passages. Also, Brian C. Hall's book "Lie groups, lie algebras, and representations." helps.

I think the confusion comes from not distinguishing between complex-linear (C-linear) and real-linear (R-linear) representations of the algebras. Every representation I will be talking about below is of finite dimension, V is a complex vector space.

• R-linear representation of [the real lie algebra] su(2) over V has 1-1 correspondence with C-linear representation of complexification of [the real lie algebra] su(2).

i.e. R-linear rep. of [real lie algebra] su(2) over V has 1-1 correspondence with C-linear rep. [complex lie algebra] sl(2,C) over V.

• R-linear rep. of [real lie algebra] so(1,3) over V has 1-1 correspondence with C-linear rep. of complexification of [real lie algebra] so(1,3) over V.

Complexification of [real lie algebra] so(1,3) is isomorphic to the direct sum of two copies of [complex lie algebra] sl(2,C), which in turn is isomorphic to the complexification of [complex lie algebra] sl(2,C).

• C-linear rep. of complexification of [complex lie algebra] sl(2,C) over V has 1-1 correspondence with R-linear rep. of the decomplexification of [complex lie algebra] sl(2,C) over V.

Hence, R-linear rep. of [real Lie algebra] so(1,3) over V has 1-1 correspondence with R-linear rep. of the decomplexification of [complex lie algebra] sl(2,C) over V. For me this makes sense, but if anyone could verify it as well it would be great.