Galilei algebra has a simple subalgebra?

Question

The Galilei algebra has the following generators $$J_i = \epsilon_{ijk}x^j \frac{\partial}{\partial x^i}, \qquad P_i = \frac{\partial}{\partial x^i} \qquad K_i = t\frac{\partial}{\partial x^i} \qquad H = \frac{\partial}{\partial t}$$ The commutators are given by

$\begin{alignat}{2} &[J_i, J_j] = \epsilon_{ijk}J_k \qquad \qquad &&[J_i,P_j] = -\epsilon_{ijk}P_k\\ &[J_i, K_j] = -\epsilon_{ijk}K_k \qquad \qquad &&[J_i,H] = 0\\ &[P_i, H] = 0 \qquad \qquad &&[P_i,P_j] = 0\\ &[K_i, H] = -P_i \qquad \qquad &&[K_i,K_j] = 0\\ &[P_i, K_j] = 0 \end{alignat}$

I'm convinced that there is no simple subalgebra of this algebra, but my course notes say otherwise. I know that there is a solvable subalgebra (namely generated by $\{K_i, P_i, H\}$), but I don't seem to see the simple algebra.

Thanks!

So the subalgebra generated by ${J_i}$ is the simple subalgebra... — Geigercounter, Jan 18 '23 at 21:07
See here. The antisymmetric matrices form the orthogonal Lie algebra, which is simple. — Dietrich Burde, Jan 18 '23 at 21:30

score 1 · Answer 1 · answered Jan 19 '23 at 10:25

Let me give a further reference with more details, so that the question is not only answered in the comments. There is an article On the Classification of Subalgebras of the Galilei Algebras by Leonid Barannyk, published $2013$ in the Journal of Nonlinear Mathematical Physics. It gives much more details on the subalgebras.

Furthermore there is a post at physics stackexchange called What is the Lie algebra of the Galilean group and what is the structure of it? It has an answer and several comments. In particular it is pointed out that the rotations form a simple Lie subalgebra (the orthogonal Lie algebra).

Galilei algebra has a simple subalgebra?

1 Answers1