So I'm reading the part in Ana Cannas da Silva's book "Lectures on Symplectic Geometry" available (on her website) about hamiltonian group actions on a symplectic manifold. She starts by defining $\mathbb R$-actions and $\mathbb S^1$ actions by saying that the vector field on $M$ that they generate must be hamiltonian. Then she defines hamiltonian $\mathbb T^n$-actions (p.154) by the requirement that the restriction of the action to each circle $\{1\} \times\ldots\times\mathbb S^1\times\ldots\times\{1\}$ be hamiltonian (plus the requirement that each of the $n$ corresponding hamiltonian functions be invariant under the action of the rest of $\mathbb T^n$)


And then, finally, she defines a hamiltonian action of a general Lie group G as one having a "moment map". This is a natural generalisation because the existence of a moment map is equivalent (I believe!) to the fact that for each $X\in\operatorname{Lie}(G)$, the vector field $X^\* $ on $M$ induced by $X$ is hamiltonian (i.e. $ X^\*\_p=\frac{d}{dt}(\exp(tX)\cdot p)(0)$).


For indeed, if that it that case, then on can just define the moment map $\mu : M \rightarrow\operatorname{Lie}(G)^\* $ by setting $\langle\mu(p),X\rangle:=\mu^X(p)$, where $\mu^X$ is a hamiltonian function for $X^\* $ chosen so that $\mu$ is G-equivariant with respect to the coadjoint action on $\operatorname{Lie}(G)^\* $. (Note that in the case where $G$ is commutative such as $G=\mathbb T^n$, this last condition boils down to $\mu$ being $G$-invariant.)


**The question:** I am trying to prove that the ad-hoc definition implies the general definition in the $\mathbb T^n$ case. The problem I am having is that we only know that n vector fields $ X\_1^\*,\ldots,X\_n^\* $ (one for each subcircle $ 1\times\ldots\times\mathbb S^1\times\ldots\times 1 \subset\mathbb T^n $) are hamiltonian. Knowing that the $X\_i$'s form a basis of $\operatorname{Lie}(\mathbb T^n)$, does this imply that every $X\in\operatorname{Lie}(\mathbb T^n)$ induces a hamiltonian $X^\* $. Does something like $\color{red}{(X+Y)^\* =X^\* +Y^\*}$ hold? 


More generally, given a Lie group $G$ acting on a symplectic manifold $M$, is it necessary to check that each $X\in\operatorname{Lie}(G) $ induce a hamiltonian vector field on $M$, or is it sufficient to check this for a basis of $\operatorname{Lie}(G) $ ? 


Thanks.



Question about the definition of hamiltonian group action.
Just so you're aware, not every author insists that a momentum map be infinitesimally equivariant (Prof. Figueroa-O'Farrill's condition 2), although it is part of da Silva's definition (edit: actually, on checking, da Silva requires the slightly stronger condition of equivariance, i.e. $\mu(g\cdot p)=\mu(p)\circ\mathrm{Ad}\_{g^{-1}}$). The literature isn't uniform - for example, Marsden just requires the first condition. I'm biased towards this definition since Jerry Marsden, who unfortunately passed away a year ago today, was my advisor.


To answer your main question, yes the map $X\in\mathfrak{g}\mapsto X^\* \in\mathfrak{X}(M)$ is always linear, regardless of whether the action is Hamiltonian. To see this explicitly, let $\Phi^p:G\rightarrow M$ be the map $g\mapsto g\cdot p$. By definition $X^\* \_p$ is precisely $T\_e\Phi^p(X)$ (you can see this agrees with your definition by writing $\exp(tX)\cdot p$ as $\Phi^p(\exp(tX))$, and using the chain rule to calculate $\frac{d}{dt}$), and derivative maps $T\_xf$ are always linear. So yes, it's enough to check condition 1 on a basis.


