Does the Lie closure of a set of Hamiltonians describe all unitaries you can generate with them?

Suppose we have $$k$$ Hermitian matrices $$H_1, ..., H_k \in M_d(\mathbb{C})$$. Define $$\mathfrak{g} = \langle iH_1, ... iH_k \rangle_{\text{Lie}}$$ to be the Lie algebra generated by these Hermitian matrices (i.e. the vector space spanned by these matrices and all possible nested commutators). Assume also that $$\operatorname{dim} \mathfrak{g} < d^2$$, so that these matrices don't generate the entire $$U(d)$$.

Denote $$G \subset U(d)$$ as the group generated by matrices of the type $$e^{itH_j}$$. Denote also as $$e^{\mathfrak{g}}$$ the set of all unitary matrices that are exponents of some elements in $$\mathfrak{g}$$.

My question is, how do $$G$$ and $$e^{\mathfrak{g}}$$ relate to each other?

1. $$e^{\mathfrak{g}} \subseteq G$$. This direction is apparently shown by Lloyd (1995) by showing that $$e^{[iA, iB]t}$$ can be performed to arbitrary precision.
2. $$G \subseteq e^{\mathfrak{g}}$$. This is where it gets tricky. For small enough $$\epsilon$$, we know that $$e^{i\epsilon H_l} e^{i\epsilon H_m}$$ will belong to $$e^\mathfrak{g}$$ from the Baker-Campbell-Hausdorff formula, but for general case it's unclear. In other words, is it possible, by repeated multiplication of matrices of the type $$e^{iH_j t}$$, to obtain a unitary $$U = e^{iF}$$ such that $$iF \notin \mathfrak{g}$$?
• – glS
Jun 23 at 13:26
• I found a related question on math.SE: math.stackexchange.com/questions/3847555. Will come back to this question when I make sense of it Jun 27 at 14:03