Definition of Hamiltonian Action from Ana Cannas da Silva's book

Question

I've been learning about Lie Groups, Lie Algebras and moment maps and my aim is to get to Sympletic Reduction. Currently I'm struggling with a few details of the definition of hamiltonian action from Ana da Cannas Silva's book on Sympletic Geometry. Here's the definition:

Let $(M,\omega)$ be a sympletic manifold, $G$ be a Lie Group, $\mathfrak{g}$ be the Lie Algebra of $G$, $\mathfrak{g}^*$ the dual vector space of $\mathfrak{g}$ and $$\psi:G\to\text{Symp}(M,\omega)$$ a sympletic action. Then $\psi$ is a $\textbf{hamiltonian action}$ if there exists a map $$\mu:M\to\mathfrak{g}^*$$ satisfying:

1. For each $X\in\mathfrak{g}$, let
   • $\mu^X:M\to\mathbb{R},\mu^X(p):=\left<\mu(p),X\right>$ be the component of $\mu$ along $X$ (she a few pages before that $\left<\cdot,\cdot\right>$ is just the natural pairing between $\mathfrak{g}^*$ and $\mathfrak{g}$)
   • $X^\#$ be the vector field on $M$ generated by the one parameter subgroup $\{\exp tX\mid t\in\mathbb{R}\}\subset G$.

Then $$d\mu^X=\imath_{X^\#}\omega$$ i.e., $\mu^X$ is a hamiltonian function for the vector field $X^\#$.

2. $\mu$ is equivariant with respect to the given action $\psi$ of $G$ on $M$ and the coadjoint action $\text{Ad}_g^*$ of $G$ on $\mathfrak{g}^*$: $$\mu\circ\psi_g=\text{Ad}^*\circ \mu$$

For connected Lie Groups, hamiltonian actions can be equivalently defined in terms of the $\textbf{comoment map}$ $$\mu^*:\mathfrak{g}\to C^\infty(M),$$ with the two conditions rephrased as:

1. $\mu^*(X):=\mu^X$ is a hamiltonian function for the vector field $X^\#$,
2. $\mu^*$ is a Lie Algebra homomorphism: $$\mu^*\left[X,Y\right]=\left\{\mu^*(X),\mu^*(Y)\right\}$$ where $\left\{\cdot,\cdot\right\}$ is the Poisson bracket on $C^\infty(M)$

So my questions are:

1. How is $\{\exp tX\mid t\in\mathbb{R}\}$ a subset of $G$? As far as I understood from the places I've read $\exp tX$ is just the flow of $X$. I've seen this more than once, is it just abuse of notation for $\exp tX(e)$ where $e$ is the identity in $G$?
2. Why are the conditions 2. equivalent? I'd also appreciate any intuition on why we want this condition to hold.
3. Why do we need to assume that the Lie Group is connected for the second definition to be valid?

Maybe some/all of these are really obvious but it's a lot of new stuff at once. Any help on any of the questions is greatly appreciated. Thanks in advance.

Answer 1

1. the exponential map $\exp\colon\mathfrak{g}\to G$ works for all Lie group $G$.
2. just write down the Poisson bracket and see.
3. If $G$ is not connected, $\psi$ at the nonidentity components of $G$ can pick out symplectomorphisms that we don't know by just looking at the identity component $G_0$. The equivariance of $\mu$ with respect to $\psi_g$ and $\operatorname{Ad}_g^*$ could break for $g\notin G_0$.