# Lie bracket of Infinitesimal action determined by the adjoint action

I'm trying to prove the following statement:

Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, if $\psi: \mathfrak{g} \rightarrow \mathfrak{X}(\mathfrak{g})$ is the infinitesimal action determined by the adjoint action then $$\psi \left( \left[ X , Y \right] \right) = - \left[ \psi(X) , \psi(Y) \right]\,.$$

Initially I assumed that the adjoint action was the usual $\mathrm{Ad}: G \rightarrow \mathrm{Aut}(\mathfrak{g})$ which would imply that $\psi = \mathrm{ad}$, however in this case, since we have that $\mathrm{ad}(X) (Y) = \left[X,Y \right]$ and then through the Jacobi identity one obtains

$$\mathrm{ad} \left( \left[ X,Y \right] \right) (Z) = \left[ \mathrm{ad}(X) ,\mathrm{ad}(Y) \right] (Z)$$

which does not agree with what is asked. I guess I'm probably looking at the wrong thing but I can't understand what could also be meant by the adjoint action in that case.

Any pointers on what is $\psi$ (if indeed it is something different than what I interpreted it to be) or otherwise on what I might be doing wrong, would be greatly appreciated.

• Strange. Maybe $\psi = -ad$? Then the relation holds. – Torsten Schoeneberg Jan 8 '18 at 3:39
• Here $\mathfrak{g}$ is viewed as a smooth manifold, $\mathfrak{X}(\mathfrak{g})$ is the set of vector fields on $\mathfrak{g}$, and $\psi$ maps each $X\in\mathfrak{g}$ to the complete vector field on $\mathfrak{g}$ whose flow is $(t,Z)\mapsto\operatorname{Ad}_{\exp(tX)}Z$. – Spenser Jan 8 '18 at 9:27
• Under the identification of $\mathfrak{X}(\mathfrak{g})$ with the set of smooth functions $\mathfrak{g}\to\mathfrak{g}$, the infinitesimal action $\psi$ takes $X\in\mathfrak{g}$ to the smooth function $\psi(X):\mathfrak{g}\to\mathfrak{g}$ given by $Z\mapsto [X,Z]$, or in other words, $\psi=\mathrm{ad}$, as you have guessed. However, the Lie bracket of vector fields on $\mathfrak{X}(\mathfrak{g})$ is not the one you expected but can be computed easily. – Spenser Jan 8 '18 at 10:07

## 1 Answer

This is a an attempt at a solution based on the previous suggestions:

For any given $X,Z \in \mathfrak{g}$, the flow of $\psi(X)$ is given by $\widehat{\phi}^t_{\psi(X) }(Z) = \mathrm{Ad}_{\exp(tX)} (Z)$. Hence, for any $X,Y,Z \in \mathfrak{g}$ we have $\psi(X)(Y) = \frac{d}{dt}\big|_{t=0} \mathrm{Ad}_{\exp(tX)} Y$. Thus, $$\psi\left(\left[X,Y \right] \right) (Z) = \frac{d}{dt} \Big|_{t=0} \mathrm{Ad}_{\exp\left(\left[X,Y \right] \right)} Z = \left[X,Y \right](Z) - Z \left( \left[X,Y \right]\right) = \left[\left[X,Y\right],Z \right]\,.$$

On the other hand, denoting by $\phi^t_X(g)= g \exp(tX)$ as the flow of $X$, we also have

\begin{eqnarray*} \left[\psi(X) ,\psi (Y) \right](Z) &=& \frac{d}{dt}\Big|_{t=0}\mathrm{Ad}_{\exp(-\sqrt{t}Y)} \circ\mathrm{Ad}_{\exp(-\sqrt{t}X)} \circ \mathrm{Ad}_{\exp(-\sqrt{t}Y)} \circ \mathrm{Ad}_{\exp(-\sqrt{t}X)} (Z) \\ &=& \frac{d}{dt}\Big|_{t=0} \exp\left(-\sqrt{t} Y \right)\exp\left(-\sqrt{t} X \right)\exp\left(\sqrt{t} Y \right)\exp\left(\sqrt{t} X \right) Z (\dots) \\ &=& \frac{d}{dt}\Big|_{t=0} \left(\phi^{-\sqrt{t}}_{-X} \circ \phi^{-\sqrt{t}}_{-Y}\circ \phi^{\sqrt{t}}_{-X} \circ \phi^{\sqrt{t}}_{-Y}\right) (Z) \left(\phi^{-\sqrt{t}}_{-X} \circ \phi^{-\sqrt{t}}_{-Y}\circ \phi^{\sqrt{t}}_{-X} \circ \phi^{\sqrt{t}}_{-Y}\right)^{-1} \\ &=& \left[- Y, -X \right] (Z) - Z \left(\left[- Y, -X \right] \right)= \left[ \left[Y,X\right] Z\right] \\ &=& - \psi\left( \left[X,Y\right] \right) (Z)\,. \end{eqnarray*}