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)?

Solution 1:

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 $$ (0,i) = \begin{bmatrix}0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0\end{bmatrix},\quad (0,j) = \begin{bmatrix}0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\end{bmatrix},\quad (0,k) = \begin{bmatrix}0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix} $$

Solution 2:

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: $$ F_{i,4} = L_i \qquad F_{i,j} = \epsilon_{ijk} R_k $$

Solution 3:

Here is an explicit construction of the Lie algebra isomorphism:

1. 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.

2. 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.

3. 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.