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 $iF \notin \mathfrak{g}$?


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