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
\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:

*

*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}

*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?

*$\mathfrak{su(2)}_{\mathbb{C}}$ 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 $SL(2,\mathbb{C})$

*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 $SL(2,\mathbb{C})$

*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.
 A: 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.
A: 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. 
