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There are two claims in McDuff-Salamon's Introduction to Symplectic Topology, 3rd edition on p. 202 that I've been trying to figure out but haven't been able to.

Let $G$ be a Lie group acting symplectically on symplectic manifold $(M,\omega)$; this means that we have a smooth group morphism $G \to \text{Symp}(M,\omega), g \mapsto \psi_g$. Here, $\psi_g$ is a symplectomorphism. Then, we get a Lie algebra morphism $\mathfrak{g} \to \mathcal{X}(M,\omega)$ on which assigns to each $\xi \in \mathfrak{g}$, a symplectic vector field $X_\xi$. The prominent property of this vector field is that the contraction $\iota_{X_\xi} \omega$ is a closed 1-form.

More explicitly, we may define this $X_\xi$.

$$X_\xi:= \left.\frac{d}{dt}\right|_{t=0} \psi_{\exp(t\xi)}.$$

The authors claim that it is a straightforward calculation to show the following for $\xi,\eta \in \mathfrak{g}, g \in G$:

  • Letting $g^{-1}\xi g:= \text{Ad}(g^{-1})\xi := \frac{d}{dt}|_{t=0} \;g^{-1}\exp(t\xi)g$, we have that $X_{g^{-1}\xi g} = \psi^*_g X_\xi$.
  • $X_{[\xi,\eta]} = [X_\xi,X_\eta]$.

I do not have much fluency with Lie groups so I barely know where to start. Any help is appreciated.

inkievoyd
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1 Answers1

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(I am not careful in keeping track of the $\pm $ sign. Everything might be off by a $\pm$, depending on the definition of the Lie brackets/derivatives)

Note that $X_{\xi}$ is a vector fields, so $\psi_g^*X_\xi$ does not make sense, since you cannot pullback a vector fields.

Let's do the calculation to see what should we expect: By definition,

\begin{align*} X_{g^{-1}\xi g} & = X_{\operatorname{Ad}_{g^{-1}} \xi} \\ &= \left.\frac{d}{dt}\right|_{t=0} \psi_{\exp(t\operatorname{Ad}_{g^{-1}} \xi)}\\ &=\left.\frac{d}{dt}\right|_{t=0} \psi_{\exp(\operatorname{Ad}_{g^{-1}} (t\xi))}\\ \tag{1}&= \left.\frac{d}{dt}\right|_{t=0} \psi_{g^{-1} \exp (t\xi) g}\\ &= \left.\frac{d}{dt}\right|_{t=0} \psi_{g^{-1}}\circ \psi_{\exp (t\xi)}\circ \psi_ g \\ &= (\psi_{g^{-1}})_* X_\xi (\psi_g (\cdot)) \end{align*}

Here (1) follows from a property of Lie groups: exponential map and adjoint action.

The second property follows from the first one: Let $\eta, \xi \in \mathfrak{g}$. Then

$$ [\eta, \xi] = \operatorname {ad}_{\eta} \xi = \frac {d}{dt}\bigg|_{t=0} \operatorname {Ad}_{\exp (t\eta)} (\xi).$$

Thus \begin{align*} X_{[\eta, \xi]} & = \frac{d}{dt}\bigg|_{t=0} X_{\operatorname{Ad}_{\exp(t\eta)} \xi} \\ &= \frac{d}{dt}\bigg|_{t=0}(\psi_{\exp(t\eta)})_* X_\xi (\psi_{\exp(-t\eta)} (\cdot)) \\ &= \mathscr L_{X_{\eta}} X_\xi \\ &= [X_\eta, X_\xi]. \end{align*}

Arctic Char
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