# 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)?

I would very much appreciate your help, thanks!

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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 $$(0,i) = \begin{bmatrix}0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0\end{bmatrix},\quad (0,j) = \begin{bmatrix}0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\end{bmatrix},\quad (0,k) = \begin{bmatrix}0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$$
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: $$F_{i,4} = L_i \qquad F_{i,j} = \epsilon_{ijk} R_k$$