# 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

Let $$X,Y,Z \in \mathfrak{g}$$. Define $$\phi^t_X = \mathrm{Ad}_{exp(tX)}$$ and similarly for $$Y$$. Furthermore $$\psi(X) = \frac{d}{dt}\big\vert_{t=0} \phi^t_X$$. Note that $$\psi(X), \psi(Y) \in \mathfrak{X}(\mathfrak{g})$$. Hence, $$\left[\psi(X), \psi(Y) \right] = \mathcal{L}_{\psi(X)} \psi(Y)$$, with $$\mathcal{L}$$ the Lie derivative. Hence, we find
\begin{align*} \left[\psi(X), \psi(Y) \right]\vert_{Z} &= \frac{d}{dt}\Big \vert_{t=0} \left(d \phi^X_{-t}\right) \psi(Y)_{\phi^X_t (Z)} \\ &= - \mathrm{ad}_X \left(\mathrm{ad}_Y|_Z \right) + \mathrm{ad}_Y \left(\mathrm{ad}_X|_Z \right) \\ &= - \mathrm{ad}\left([X,Y]\right)|_Z = -\psi\left([X,Y]\right) \,, \end{align*} where we used the Jacobi identity in the third line.