Isomorphism between $\mathfrak o(4,\mathbb R)$ and $\mathfrak o (3,\mathbb R) \oplus\mathfrak o (3,\mathbb R) $ I've been trying to find a Lie algebra isomorphism $$\mathfrak o(4,\mathbb R)\cong\mathfrak o (3,\mathbb R) \oplus\mathfrak o (3,\mathbb R) $$
but haven't managed so far. I have written down the values of the Lie brackets on the canonical bases and played around with that a bit, but I couldn't find an appropriate basis of $\mathfrak o(4,\mathbb R)$ that would naturally correspond to the canonical basis of $\mathfrak o (3,\mathbb R) \oplus\mathfrak o (3,\mathbb R) $.
So I would like to ask, whether someone knows a reference where I could find such an isomorphism written down (or someone might be able to come up with one)?
 A: See this answer for some insight into the nature of this isomorphism.  Roughly speaking, the Lie algebra $\mathfrak{o}(3)$ can be viewed as the collection of all quaternions of the form
$$
ai+bj+ck,\quad a,b,c\in\mathbb{R}
$$
with the Lie bracket being half the commutator
$$
[q,r] = \frac{qr-rq}{2}.
$$
(The factor of $1/2$ is for normalization. Just the plain commutator works as well.)  Given this description, the standard matrix representation for $\mathfrak{o}(3)$ can be obtained from the adjoint action of $\mathfrak{o}(3)$ on itself:
$$
i = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{bmatrix},
\qquad
j = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{bmatrix},
\qquad
k = \begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
$$
Geometrically, $i$, $j$, and $k$ represent infinitesimal counterclockwise rotations about the $x$, $y$, and $z$ axes.
The Lie algebra $\mathfrak{o}(4)$ is isomorphic to $\mathfrak{o}(3)\times\mathfrak{o}(3)$.  In particular, each element of $\mathfrak{o}(4)$ is an ordered pair $(q,r)$ of quaternions in the form given above.  From this point of view, the action of $\mathfrak{o}(4)$ on $\mathbb{R}^4$ is defined by
$$
(q,r)\cdot v \,=\, qv + vr
$$
where the element $v\in\mathbb{R}^4$ is thought of as a quaternion.  That is, $(q,0)$ acts as left-multiplication by $q$, while $(0,r)$ acts as right-multiplication by $r$.  Using the basis $\{1,i,j,k\}$ for $\mathbb{R}^4$, one can obtain $4\times 4$ matrices for this action:
$$
(i,0) = \begin{bmatrix}0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0\end{bmatrix},\quad
(j,0) = \begin{bmatrix}0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{bmatrix},\quad
(k,0) = \begin{bmatrix}0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}
$$
and
$$
(0,i) = \begin{bmatrix}0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0\end{bmatrix},\quad
(0,j) = \begin{bmatrix}0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\end{bmatrix},\quad
(0,k) = \begin{bmatrix}0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}
$$
A: Suppose $L_i$, $R_i$ for $i=1,2,3$, are generators of two copies of $\mathfrak{o}(3,\mathbb{R})$ with
$$
   \left[ L_i, L_j \right] = \epsilon_{ijk} L_k \qquad\qquad
   \left[ R_i, R_j \right] = \epsilon_{ijk} R_k
$$
6 generators  of $\mathfrak{o}(4,\mathbb{R})$ are arranged in an anti-symmetric $4 \times 4$ matrix:
$$
  F_{i,4} = L_i \qquad F_{i,j} = \epsilon_{ijk} R_k
$$
A: Here is an explicit construction of the Lie algebra isomorphism:

*

*The Lie algebra
$$\begin{align}o(4,\mathbb{R})
~:=~&\{A\in {\rm Mat}_{4\times 4}(\mathbb{R}) \mid A^{t}=-A\}\cr
~=~& o(3,\mathbb{R})_+\oplus o(3,\mathbb{R})_-\end{align} \tag{1}$$
consists of real antisymmetric $4\times 4$ matrices.


*The two Lie algebra copies
$$o(3,\mathbb{R})_{\pm}~:=~ \{ A\in o(4;\mathbb{R}) \mid \star A = \pm A \} \tag{2}$$
consist of selfdual (anti-selfdual) real antisymmetric $4\times 4$ matrices, respectively. Here $\star$ denotes the Hodge star.


*To check that the two copies indeed commute, since the dimension is relatively small, perhaps the simplest is to just explicitly calculate all the relevant Lie-brackets:
$$ \begin{align} [A_{12}\pm A_{34}, A_{14} \pm A_{23}]~&=~2(A_{42}\pm A_{13}) \cr  [A_{12}\pm A_{34}, A_{14} \mp A_{23}]~&=~0 \cr
&~\vdots \end{align} \tag{3}$$
and so forth.
