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I stumbled over the following question:

First, let me give the basic definition of a symplectic group action: Let $(M, \omega)$ be a symplectic manifold and $G$ a Lie group. A smooth action $\Phi:G \times M \rightarrow M$ is symplectic if each $\Phi_g$ is a symplectomorphism, i.e. $\Phi_g^* \omega = \omega.$

Now, it is clear that if $\Phi$ is symplectic, then the Lie derivative $L_{\phi_\chi} \omega =0$ for all $ \chi \in \mathfrak{g},$ where $\phi_{\chi}(x) = \frac{d}{dt}|_{t=0} \Phi_{e^{t \chi}}(x)$ are called the infinitesimal generators.

This is easy to see as $L_{\phi_{\chi}}(\omega) = \frac{d}{dt}|_{t=0} (\Phi_{e^{t \chi}})^* \omega = 0 $ by definition.

What is a priori unclear to me is whether this ($L_{\phi_\chi} \omega =0$ for all $ \chi \in \mathfrak{g}$) is already sufficient to conclude that $\Phi$ is a symplectic group action.

Does anybody know when (maybe under which additional conditions) this is the case?

I should say that I stumbled over this when reading this book, where it is called the infinitesimal version of this equation:

read the reference on google books.

But maybe the term "infinitesimal version" has a different meaning in symplectic geometry and it is not in general possible to go back to the "global version."

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  • $\begingroup$ Is $G$ connected? (I assume it is.) $\endgroup$ – Jason Starr Jul 26 '15 at 18:21
  • $\begingroup$ @JasonStarr could you explain why this property does it (maybe in an answer), because even then the exponential map is not necessarily onto? $\endgroup$ – Zlatan12 Jul 26 '15 at 18:29
  • $\begingroup$ I am not saying that connectedness is sufficient. I am saying that connectedness is necessary. $\endgroup$ – Jason Starr Jul 26 '15 at 18:33
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    $\begingroup$ For a connected LIe group, this is really a standard argument in Lie theory since any neighbourhood of the identity generates the connected component of the identity. In the nonconnected case you can not say anything reasonable: think of a discrete Lie group with non-symplectic group action... $\endgroup$ – Stefan Waldmann Jul 27 '15 at 6:16
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I think it is sufficient to assume that $G$ is a connected Lie group. It is not necessary to assume that the exponential map is surjective (intuitively because one can follow broken one-parameter groups).

In fact, the set $H$ consisting of all $g\in G$ such that $\Phi_g^*\omega=\omega$ is a closed subgroup of $G$. A closed subgroup of a Lie group is a Lie group with the induced topology. What is the Lie algebra of $H$? The assumption that implies that the Lie algebra of $H$ equals the Lie algebra of $G$. In case $G$ is connected, this implies that $H=G$.

Edit: Another way to think, perhaps more elementary. $H$ is an (abstract) group. The assumption that $L_{\phi_\chi}(\omega)=0$ for all $\chi\in\mathfrak g$ implies that $H$ contains a neighborhood of the identity, and $G$ (if connected) is generated by any neighborhood of the identity.

To see that $H$ contains a neighborhood of the identity, one uses that the exponential map is onto a neighborhood of the identity and computes that (denote $\varphi_t=\Phi_{e^{t\chi}}$):

$$\frac{d}{dt}\Big|_{t=s}\varphi_t^*\omega=\frac{d}{dt}\Big|_{t=s}(\varphi_{t-s}\varphi_s)^*\omega=\frac{d}{du}\Big|_{u=0}(\varphi_u\varphi_s)^*\omega=\varphi_s^*\frac{d}{dt}\Big|_{u=0}\varphi^*_u\omega=0$$

for all $s\in\mathbb R$, so $\varphi_t^*\omega=\omega$ for all $t\in\mathbb R$.

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