Symplectic group action

Question

Let $(M,\omega)$ be a symplectic manifold. We say that a group action $\phi: G \times M \rightarrow M$ is symplectic if each $\phi(g,.)$ is a symplectomorphism.

Now, I am going through some lecture notes that are not available online, where I have difficulties to understand some details.

So we define the vector field $\psi$ on the Lie Algebra ($\eta \in \mathfrak{g},p \in M$) such that $$\psi(\eta)(p) = \frac{d}{dt}|_{t=0} (\phi(e^{t \eta},p)) \in T_pM$$

So far so good. Now the author says: Assume that for each $\psi(\eta)$ there is a global Hamilton function $H_{\eta}: M \rightarrow \mathbb{R}$ such that $dH_{\eta} = \omega(\psi_{\eta},.).$ (So $\psi_\eta$ is the Hamiltonian vector field to $dH_{\eta}.)$ So far so good:

Now the author wants to show that

$H_{\xi} ( \phi(g,p)) = H_{Ad_{g^{-1}}(\xi)}(p)$ for all $g \in G$ and $\xi \in \mathfrak{g}.$

In order to do this he claims that it is sufficient to prove this for all $1-$parameter subgroups $g(t) = e^{t \eta}, \eta \in \mathfrak{g}.$

I have no idea why this is sufficient cause I don't see that the exponential map is surjective in general (despite, he may refer here to a special case. at least, this is what I suspect, maybe you can help me identifying this one.)

Then he writes $H_{\xi} ( \phi(g,p)) = H_{Ad_{g^{-1}}(\xi)}(p)$ if and only if

$H_{\xi} ( \phi(e^{t \eta},p)) = H_{Ad_{e^{-t\eta}}(\xi)}(p).$

and after that he again claims that it is sufficient (I have not the slightest idea why) to show this for $\frac{d}{dt}|_{t=0}$ so he differentiates both sides and gets for the right hand-side

$$\frac{d}{dt}|_{t=0}H_{Ad_{e^{-t\eta}}(\xi)}(p) = H_{ad_{-\eta}(\xi)}(p).$$ Actually, I have even difficulties to understand why you can differentiate this function and leave $H$ as it is (no chain rule).

Later on, he claims that any $T_p(orb_G(p))$ can be written as $\psi_{\eta}(p)$ for some $\eta \in \mathfrak{g}$ where $orb_G(p)$ is the orbit of $p$ under the group action $g$. I think this is similar to the previous issues.

Either he likes to proof only special cases or I am missing a point here, but I don't see where all his sufficiency arguments come from? Is there anybody who looks through this or could comment on this?

I can talk to the author soon, but I don't want to contact him without being sure that there is really something missing here.

Answer 1

You should complain to them, since this result is simply wrong. Consider how $\mathbb{R}^2$ acts on itself by translating with the usual symplectic form. The functions $x$ and $y$ are Hamiltonian functions for generators of the Lie algebra (you can work this out; I'll just get bogged down into notation), and since the adjoint action is trivial, you should get that these are invariant, which is obviously false. The correct theorem is that $H_\xi(\phi(g,p))$ is a Hamiltonian function for $\mathrm{Ad}_{g^{-1}}\xi$, but you can't consistently choose these in a way that's equivariant. But I don't see this showing up in any of the issues you point out.

I have no idea why this is sufficient cause I don't see that the exponential map is surjective in general (despite, he may refer here to a special case. at least, this is what I suspect, maybe you can help me identifying this one.)

This is sufficient because the image of the exponential map generates the group; so an arbitrary group element is a product of exponentials, and you can use this calculation multiple times.

after that he again claims that it is sufficient (I have not the slightest idea why)

Because differentiating at $t=a$ is the same as replacing $p$ by $e^{a\eta}p$, and doing the differentiation.

Actually, I have even difficulties to understand why you can differentiate this function and leave H as it is (no chain rule).

There's no chain rule because there's no differentiation in the space direction.