Let $G$ be a connected Lie group and let $\gamma : [0,1]\to G$ be a smooth path starting at the identity $\gamma(0)=e$.
Let $R_g :G\to G$ be the right-translation by $g$, i.e., $$R_g(h)=gh.$$
Then $\gamma$ naturally satisfies a differential equation $$\gamma'(s)=[R_{\gamma(s)}]_{\ast e} X(s),\quad X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s)\in \mathfrak{g}\simeq T_eG.\tag{1}$$
Suppose now $U : G\to {\rm U}({\cal H})$ is a unitary representation of $G$ on a Hilbert space $\cal H$. We also know that $U$ induces a representation of the Lie algebra $\mathfrak{g}$ by means of the derived representation construction. We simply set, for $\Psi\in {\cal H}^\infty_U$ the smooth vectors,
$$D(X)\Psi=\dfrac{d}{ds}\bigg|_{s=0}U(\exp sX) \Psi.\tag{2}$$
My question is: can we translate Eq. (1) in the group $G$ into a differential equation for $U(\gamma(s))$ in the group ${\rm U}({\cal H})$?
My take is that the equation will necessarily involve the Lie algebra representation $D$ and will provide a way to "integrate $D$ to $U$ along paths in $G$".
My Attempt
My initial idea has been to use $[R_{\gamma(s)}]_{\ast e}:\mathfrak{g}\to T_{\gamma(s)}G$ to induce on $T_{\gamma(s)}G$ a Lie algebra structure and induce a representation $D_{\gamma(s)}:T_{\gamma(s)}G\to \operatorname{End}({\cal H})$ by means of $$D_{\gamma(s)}(X_{\gamma(s)})=D([R_{\gamma(s)}]_{\ast e}^{-1}X_{\gamma(s)})$$
and then apply $D_{\gamma(s)}$ to Eq. (1). The issue seems to be how to extract $$\dfrac{d}{ds} U(\gamma(s))$$
out of all of this, which we certainly need to appear if we want a differential equation for it.
Best Answer
Since $U:G\to \operatorname{U}(\mathcal{H})$, we have $U_*:TG\to T\operatorname{U}(\mathcal{H})$, and in particular $U_{*g}:T_gG\to T_{U(g)}\operatorname{U}(\mathcal{H})$. The fact that $U$ is a representation implies that $$ U\circ R_g = R_{U(g)}\circ U $$ ($R$ denotes right multiplication in $G$ on the lhs, and in $\operatorname{U}(\mathcal{H})$ on the rhs). Taking the derivative at $e\in G$ on both sides, and applying to $\xi\in \mathfrak{g}\simeq T_eG$ gives $$ U_{*g}([R_g]_{*e}(\xi)) = [R_{U(g)}]_{*I}(U_{*e}(\xi)) $$ or in more suggestive notation $$ U_{*g}(\xi\cdot g) = U_{*e}(\xi)\cdot U(g). $$ You can use this equation to translate the ODE $\gamma'(s) = X(s)\cdot\gamma(s)$ on $G$ into an ODE on $\operatorname{U}(\mathcal{H})$, namely $$ \frac{d}{ds}U(\gamma(s)) = U_{*\gamma(s)}(\gamma'(s)) = U_{*\gamma(s)}(X(s)\cdot \gamma(s))= U_{*e}(X(s))\cdot U(\gamma(s)). $$ $U_{*e}$ is what you call $D$ in your question, while $U_{*g}(\cdot)\cdot U(g)^{-1}$ is what you call $D_g$.