An infinitesimal moment map is automatically equivariant

Define $\mu :M\to {\mathfrak{g}}^{\ast }={\mathbb{R}}^2$ by $\mu (x,y)=(y,-x)$, so that ${\mu }^{(a,b)}(x,y)=ay-bx$. Then $d{\mu }^{(a,b)}=ady-bdx$, while by [F2] $-{\iota }_{{\xi }_M}\omega =-(-{\iota }_{a{\mathrm{\partial }}_x+b{\mathrm{\partial }}_y}\omega )={\iota }_{a{\mathrm{\partial }}_x+b{\mathrm{\partial }}_y}\omega =ady-bdx$; hence the component equations hold for every $\xi$ and $\mu$ is an infinitesimal moment map.

The Lie algebra $\mathfrak{g}={\mathbb{R}}^2$ is abelian, so the coadjoint action is trivial and $\mu$ would be equivariant only if it were constant; it is not. Hence $\mu$ is not equivariant, and by [F4] its defect cannot vanish identically.

The defect is computed directly: for $\xi =(1,0)$ and $\eta =(0,1)$, ${\mu }^{\xi }=y$ and ${\mu }^{\eta }=-x$, with $\{y,-x\}=-\{y,x\}=-\omega (X_y,X_x)$. Since ${\iota }_{{\mathrm{\partial }}_x}\omega =dy$ and ${\iota }_{{\mathrm{\partial }}_y}\omega =-dx$, one has $X_y={\mathrm{\partial }}_x$ and $X_x=-{\mathrm{\partial }}_y$, so $\{y,-x\}=-\omega ({\mathrm{\partial }}_x,-{\mathrm{\partial }}_y)=\omega ({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)=1$; meanwhile $[\xi ,\eta ]=0$ and ${\mu }^0=0$. Thus $c(\xi ,\eta )=1\mathrm{\ne }0$, and $\mu$ is an infinitesimal moment map that is not equivariant.