to impose the condition ${{\boldsymbol{G}}}(t) = {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} + {\text{h.c.}}$ is to remove the redundant degree of freedom: It is easy to observe that ${\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}$ is redundant and does not play any role in $\dot{{{\boldsymbol{U}}}}(t)$. Thus, we might as well let ${\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \equiv 0$. Further if we assume ${\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t){\mathcal{P}}_{{{\boldsymbol{U}}}(t)} \equiv 0$, then it directly implies that ${{\boldsymbol{U}}}^{\dagger}(t) \dot{{{\boldsymbol{U}}}}(t) = 0$, which is used as orthogonal constraint in [@Lubich07]. Conversely, if ${{\boldsymbol{U}}}^{\dagger}(t) \dot{{{\boldsymbol{U}}}}(t) = 0$, then ${\mathcal{P}}_{{{\boldsymbol{U}}}(t)} {{\boldsymbol{G}}}(t) {\mathcal{P}}_{{{\boldsymbol{U}}}(t)} = 0$, which shows that such constraint is similar to the constraint ${{\boldsymbol{U}}}^{\dagger}(t) \dot{{{\boldsymbol{U}}}}(t) = 0$ for matrix ODEs.

    Another reason comes from the above proof. With such constraint, the first order stationary condition with respect to $\delta {{\boldsymbol{G}}}$ yields a unique expression for ${{\boldsymbol{G}}}(t)$. This indicates that redundancy has been removed via the above constraint.

-   The main reason for imposing constraint in tangent space of ${{\mathcal{M}}_{{\mathbb{C}}^r}}$ is that Fokker-Planck type generator automatically helps to preserve the positivity of measure. In fact, the whole tangent space is rather big and too opaque to handle, since it might involve generators for other stochastic processes, [*e.g.*]{}, jump processes. The mixture of jump process and diffusion makes it 