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
\begin{equation} [J_i,J_j]=i \epsilon_{ijk} J_k \end{equation} \begin{equation} [J_i,K_j]=i \epsilon_{ijk} K_k \end{equation} \begin{equation} [K_i,K_j]=- i \epsilon_{ijk} J_k \end{equation}

We can now define new generators with the old ones $N^{\pm}_i= \frac{1}{2}(J_i \pm i K_i)$ which satisfy \begin{equation} [N^{+}_i,N^{+}_j] = i \epsilon_{ijk} N^{+}_k ,\end{equation} \begin{equation} [N^{-}_i,N^{-}_j] = i \epsilon_{ijk} N^{-}_k ,\end{equation} \begin{equation} [N^{+}_i,N^{-}_j] = 0. \end{equation} 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 $\mathfrak{so}(1,3)_+^{\uparrow}$ \begin{equation}(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} = \mathfrak{so}(1,3)_+^{\uparrow}\otimes \mathbb{C} \end{equation}
  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 $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$:
    \begin{equation}\mathfrak{su(2)}_{\mathbb{C}} \simeq (\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C} \end{equation}
    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.

  • $\begingroup$ That's a lot of questions. Perhaps you could break this up into multiple posts? $\endgroup$
    – Jim Belk
    Jan 16, 2014 at 0:59
  • $\begingroup$ Also, what do you mean by $\mathfrak{so}(1,3)_+^\uparrow$? Is this the same as $\mathfrak{so}(1,3)$? $\endgroup$
    – Jim Belk
    Jan 16, 2014 at 1:00
  • $\begingroup$ 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 $\mathfrak{so}(1,3)$. $\endgroup$
    – jak
    Jan 16, 2014 at 6:41
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 28, 2017 at 20:51
  • 1
    $\begingroup$ $\mathfrak{su}(2)$, 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. $\endgroup$ Nov 28, 2017 at 20:54

2 Answers 2


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.

  • 1
    $\begingroup$ Note that your matrices $\sigma_i$ are actually $1/2$ times the Pauli matrices. The Pauli matrices have a factor of $2$ in their commutation relation. $\endgroup$
    – B K
    May 20, 2021 at 12:09

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .