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. $\mu(g\cdot p)=\mu(p)\circ\mathrm{Ad}_g$)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$are always linear. So yes, it's enough to check condition 1 on a basis. 2 added 144 characters in body Just so you're aware, not every author insists that a momentum map be infinitesimally equivariant (Prof. Figueroa-O'Farrill's condition 2), although it is part of da Silva's definition (edit: actually, on checking, da Silva requires the slightly stronger condition of equivariance, i.e.$\mu(g\cdot p)=\mu(p)\circ\mathrm{Ad}_g$). The literature isn't uniform - for example, Marsden just requires the first condition. I'm biased towards this definition since Jerry Marsden, who unfortunately passed away a year ago today, was my advisor. To answer your main question, yes the map$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$are always linear. So yes, it's enough to check condition 1 on a basis. 1 Just so you're aware, not every author insists that a momentum map be infinitesimally equivariant (Prof. Figueroa-O'Farrill's condition 2), although it is part of da Silva's definition. The literature isn't uniform - for example, Marsden just requires the first condition. I'm biased towards this definition since Jerry Marsden, who unfortunately passed away a year ago today, was my advisor. To answer your main question, yes the map$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.