What is the meaning of the "$i$" featuring in the commutation relations of the Galilean/Poincare group?

Question

I apologize in advance for such a silly question. I am trying to understand the Lie algebra of the Galilean group. Looking at the commutators e.g. on wikipedia, some of the commutation relations give imaginary answers e.g. $[L_{ij},P_k]=i[\delta_{ik}P_j-\delta_{jk}P_i]$. But this should be a real Lie algebra? Where can I find the vector $iP_j$? How should I interpret this? An exceedingly basic answer for this exceedingly basic question would be much appreciated.

Answer 1

As far as I can tell, this is a physics convention. A simpler example is the use of the Pauli matrices to describe $\mathfrak{su}(2)$, which satisfy the commutation relations

$$[\sigma_1, \sigma_2] = 2i \sigma_3$$ $$[\sigma_2, \sigma_3] = 2i \sigma_1$$ $$[\sigma_3, \sigma_1] = 2i \sigma_2.$$

The Pauli matrices have the property that the matrices $i \sigma_1, i \sigma_2, i \sigma_3$ form a basis of the usual copy of $\mathfrak{su}(2)$ inside $\mathfrak{gl}_2(\mathbb{C})$. Physicists work with the Pauli matrices rather than $i$ times the Pauli matrices because they are Hermitian and so define observables in quantum mechanics.

More abstractly, if $\mathfrak{g}$ is a real Lie subalgebra of $\mathfrak{u}(n)$ for some $n$, then you can consider $i \mathfrak{g}$, the subspace of $\mathfrak{gl}_n(\mathbb{C})$ consisting of $i$ times the elements of $\mathfrak{g}$; since $\mathfrak{g}$ consists of skew-Hermitian matrices, $i \mathfrak{g}$ consists of Hermitian matrices, and you can rewrite the commutation relations of $\mathfrak{g}$ using $i \mathfrak{g}$ instead if you want, which introduces a factor of $i$ everywhere.