Algorithmic way to find 2 dimensional sub algebras

Question

I am looking for an algorithmic way to find 2-dimensional sub-algebras of an algebra with commutator relations \begin{align} [X_1, X_2]=-X_3 \end{align} \begin{align} [X_1, X_3] = -X_2 \end{align} \begin{align} [X_2, X_3]= X_1 \end{align}

Answer 1

Here is one way to think about it. Let $Y_1, Y_2$ be a basis of such a subalgebra. We can write $Y_1 = aX_1 + bX_2 + cX_3$. Our goal is to find $a, b, c$. One thing we know is that we can write down the matrix $ad(Y_1)$ w.r.t the standard basis $X_1, X_2, X_3$: it some 3-times-3-matrix with some appearances of $a, b$ and $c$ and a lot of $0$s.

Now $ad(Y_1)$ maps the vector $Y_2$ to some linear combination $pY_1 + qY_2$. We may assume that $q = 1$: it would be 1 when we had been lucky and had picked $(1/q)Y_1$ instead of $Y_1$, but since $Y_1$ is still an 'abstract' basis vector we may in fact assume that we have picked that vector as our $Y_1$.

So in short: $Y_2$ is an element of the kernel of the matrix $ad(Y_1) - I$ where $I$ is the $3\times3$-identity matrix. So an algorihm would be:

1. Write down $ad(Y_1) - I$
2. Check if there is any restriction on $a, b, c$ for this matrix to even have a kernel
3. For each $a, b, c$ satisfying the restriction in 2 we know that $aX_1 + bX_2 + cX_3$ is in some two-dimensional subalgebra and for any $Y_2$ in the kernel of $ad(Y_1) - I$ (which we can compute using standard linear algebra algorithms) we find that $\{Y_1, Y_2\}$ is a basis of such an algebra
4. Find all infinitely many subalgebras generated by the method in 3
5. Find a way to determine which ones on your list are actually the same

Step 5 is a bit vague, admittedly. Here is a better way:

3' Pick one choice of $a, b, c$ satisfying the conditions found in 2 and compute all subalgebras containing the resulting vector $Y_1$ as in 3 above

4' express the condition 'Y_1' does not lie in any of the subalgebras generated in the previous step as a further condition on a, b, c

5' Repeat steps 3 and 4 with the ever increasing list of conditions on $a, b, c$ until no new subalgebras can be found. No guarantee that this loop will ever stop, though!

Answer 2

Over the real numbers this Lie algebra is $\mathfrak{so}_3(\Bbb R)$, whose $2$-dimensional subalgebras have been classified on this site. In particular, there is no $2$-dimensional subalgebra.

Reference:

2-dimensional Lie subgroups of $SO(3)$