Can Lie algebra $sl(2,\mathbb{C})$ be decomposed to direct sum of two $sl(2,\mathbb{R})$?

Question

The number of generators of Lie algebra $sl(2,\mathbb{C})$ is 6, and $sl(2,\mathbb{R})$ has 3 generators, Can Lie algebra $sl(2,\mathbb{C})$ be decomposed to direct sum of two $sl(2,\mathbb{R})$? Say \begin{equation} sl(2,\mathbb{C})=sl(2,\mathbb{R}) \oplus sl(2,\mathbb{R}) ~? \tag{1} \end{equation} If this holds, can you give one explicit representation of those generators?

By the way there is a similar relation which I know is hold \begin{equation} so(4)=su(2) \oplus su(2). \tag{2} \end{equation}

Answer 1

1. On one hand, as 6-dimensional real Lie algebras, we have the following isomorphisms $$\begin{align}so(4;\mathbb{R})~&\cong~su(2)\oplus su(2),\qquad\qquad\text{(compact form)}\\ so(2,2;\mathbb{R})~&\cong~sl(2,\mathbb{R})\oplus sl(2,\mathbb{R}),\qquad\qquad\text{(split form)}\\ so(3,1;\mathbb{R})~&\cong~sl(2,\mathbb{C}).\qquad\qquad\text{(simple Lie algebra)}\end{align}$$ In particular, OP's suggested decomposition (1) is not possible$^1$ as real Lie algebras.

2. On the other hand, as 6-dimensional complex Lie algebras, we have the following isomorphism $$so(p,q;\mathbb{C})~\cong~sl(2,\mathbb{C})\oplus sl(2,\mathbb{C}), \qquad p+q~=~4.$$ See also this and this related Phys.SE posts.

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$^1$ In fact, the Lie algebra

$$ sl(2,\mathbb{C})~:=~\left\{\sigma\in{\rm Mat}_{2\times 2 }(\mathbb{C})\mid {\rm tr}(\sigma)=0\right\}~=~\left\{\left.\begin{pmatrix} a&b\cr c&-a\end{pmatrix}\in{\rm Mat}_{2\times 2 }(\mathbb{C})\right| a,b,c\in\mathbb{C}\right\} $$

is a simple Lie algebra, because every non-zero Lie algebra element

$$\sigma_0~\in~ sl(2,\mathbb{C})\backslash\{0\}$$

is a cyclic vector.

Sketched proof: By simularity transformations, we may assume that $\sigma_0$ is on Jordan normal form. There are two cases.

1. Case $\sigma_0= \begin{pmatrix} \lambda &0\cr 0&-\lambda\end{pmatrix}$ is diagonal where $\lambda\in\mathbb{C}\backslash\{0\}$. Then $$[\sigma_0,\sigma]~=~\left[\begin{pmatrix} \lambda &0\cr 0&-\lambda\end{pmatrix}, \begin{pmatrix} a&b\cr c&-a\end{pmatrix} \right] ~=~\begin{pmatrix} 0&2\lambda b\cr -2\lambda c&0\end{pmatrix}.$$ In other words, we can generate all off-diagonal matrices. In particular, we can generate the seed matrix for the other case 2.

2. Case $\sigma_0= \begin{pmatrix} 0 &1\cr 0&0\end{pmatrix}$ is nilpotent. Then $$[\sigma_0,\sigma]~=~\left[\begin{pmatrix}0 &1\cr 0&0\end{pmatrix},\begin{pmatrix} 0&0\cr \lambda&0\end{pmatrix} \right] ~=~\begin{pmatrix} \lambda &0\cr 0&-\lambda\end{pmatrix}.$$
   In other words, we can generate all traceless diagonal matrices. In particular, we can generate the seed matrix for the other case 1.

Altogether we can generate all traceless matrices. $\Box$

Answer 2

$\mathfrak{sl}(2,\mathbb{C})$ is indeed equal to $\mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{sl}(2,\mathbb{R})$. The famous representation is given by $L_{-1}$, $L_0$, $L_1$ of the Virasoro algebra for the first copy of $\mathfrak{sl}(2, \mathbb{R})$, and by their conjugations $\overline{L}_{-1}$, $\overline{L}_0$, $\overline{L}_1$ for the second one.