Let $\mathcal{S}=\{ s_i,\dots,s_n \}$ be a set of elements of the matrix Lie algebra $su(d)$, and let $\mathcal{L}(\mathcal{S})$ be the Lie subalgebra of $su(d)$ that is obtained by closing $\mathcal{S}$ under the commutator $[A,B]=AB-BA$. One can obtain a basis for $\mathcal{L}(\mathcal{S})$ by first obtaining all depth-1 commutators of the form $[s_i,s_j]\equiv d_{ij}$, then obtaining all depth-2 commutators $[d_{ij},s_k]$, and so on until no new linearly independent operators are obtained.
My question is: are there any upper bounds on the depth at which this process terminates? In other words, what is the longest sequence of commutators of elements of $\mathcal{S}$ we need in order to generate the whole Lie subalgebra?
I am particularly interested in the case where $s_i$ are $k$-fold tensor products of Pauli matrices (Pauli strings), such that they are elements of $su(2^k)$. In this case, $[s_i,s_j]$ is always proportional to another Pauli string (or equal to 0). Furthermore, I am interested in the case where $\mathcal{L}(\mathcal{S})=su(2^k)$, which means that, by repeatedly taking commutators of elements of $\mathcal{S}$, I can obtain all Pauli strings of length $k$, which constitute a basis for $su(2^k)$. Then the question is: given a set of Pauli strings which generates all Pauli strings using commutators, what is the maximum number of commutators needed to generate an arbitrary Pauli string? In my observations, this number scales linearly with $k$. Is this true in general?
