Let $G$ be a (not necessarily finite-dimensional) Lie group with Lie algebra $\mathfrak{g}.$ Define the group adjoint $\operatorname{Ad}_\eta:\mathfrak{g}\to\mathfrak{g}$ by $\operatorname{Ad}_\eta v=dL_{\eta}dR_{\eta^{-1}}v$ and the Lie algebra adjoint $\operatorname{ad}_u:\mathfrak{g}\to\mathfrak{g}$ by $$\operatorname{ad}_uv:=\frac{d}{dt}\Big|_{t=0}Ad_{\eta(t)}v,$$ where $\eta:[0,1]\to G$ is a curve in $G$ starting at the identity with $\dot{\eta}(0)=u\in\mathfrak{g}.$ I am having trouble seeing how from these definitions, it follows that $$\frac{d}{dt}(Ad_{\eta(t)})=\operatorname{ad}_{u(t)}Ad_{\eta(t)},$$ where $\eta$ is the Lagrangian flow of the time-dependent vector field $u(t)$, i.e. $\dot{\eta}(t)=dR_{\eta(t)}(u(t))$ and so $u(t)\in\mathfrak{g}$ for every $t\in[0,1]$. The only way I am getting close to making sense of this formula is to somehow use the operator equality $\operatorname{Ad}_{\exp(X)}=\exp(\operatorname{ad}_{X})$ (with $X\mapsto tX$ or $X=u(t)$?) and then differentiate with respect to $t$, but I can not make it work. For example, we can write $\eta(t)=\exp(tX)$ where $\dot{\eta}(0)=X\in\mathfrak{g}$, so that for any $Y\in\mathfrak{g}$, we have $$\frac{d}{dt}(\operatorname{Ad}_{\eta(t)}Y)=\frac{d}{dt}(\operatorname{Ad}_{\exp(tX)}Y)=\frac{d}{dt}\exp(t\operatorname{ad}_{X})Y,$$ where linearity of $\operatorname{ad}_X$ in $X$ was used. Now, I believe we can say that $$\frac{d}{dt}\exp(t\operatorname{ad}_{X})Y=\operatorname{ad}_{X}\exp(t\operatorname{ad}_{X})Y=\operatorname{ad}_{X}\operatorname{Ad}_{\eta(t)}Y,$$ though my choice to place $\operatorname{ad}_{X}$ on the left was arbitrary, and I do know how to get from $\operatorname{ad}_{X}$ to $\operatorname{ad}_{u(t)}.$ Can anyone please assist? Thankful for any insight on what I am doing wrong and/or what I am misunderstanding.
1 Answers
The trick is to write $\eta(t)=\eta(t)\eta(t_0)^{-1}\eta(t_0)$ and use the fact that $\operatorname{Ad}_{gh}=\operatorname{Ad}_g\operatorname{Ad}_h$ as follows. For any $\xi\in\mathfrak{g},$ we have $$\frac{d}{dt}\Big|_{t=t_0}\operatorname{Ad}_{\eta(t)}\xi=\frac{d}{dt}\Big|_{t=t_0}\operatorname{Ad}_{\eta(t)\eta(t_0)^{-1}}\operatorname{Ad}_{\eta(t_0)}\xi=\operatorname{ad}_{u(t_0)}\operatorname{Ad}_{\eta(t_0)}\xi,$$ where the last equality holds by definition of $\operatorname{ad}$ since $\eta(t)\eta(t_0)^{-1}$ is a curve in $G$ starting at the identity when $t=t_0$, with tangent vector exactly $u(t_0)=dR_{\eta(t_0)^{-1}}(\dot{\eta}(t_0))$.
I encountered this trick in the textbook "Geometric Mechanics and Symmetry: From Finite to Infinite Dimensions" by Holm-Schmah-Stoica.
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