If $X$ is a symplectic variety with a free Hamiltonian $G$ action and moment map $\mu$, how can I see that the critical locus of the function $W^s: X \times \mathfrak{g} \to \mathbb C$ given by $W^s(x,X) = Tr(\mu(x)X)$ is isomorphic to the zero set of $\mu$ under the projection $X \times \mathfrak g \to X$? Here I suppose $\mu$ must be thought of as a map $X \to \mathfrak g$ using an isomorphism $\mathfrak g \cong \mathfrak g^*$. However in this setting, isn't $W^s$ "isomorphic" to $\mu$ under the trace pairing? Why should we consider $dW^s$?
This is claimed in 2.2.1 of the paper "New quiver like varieties and Lie superalgebras" which can be found on arxiv, unless I am interpreting what they wrote incorrectly. I think this is just a case of me having forgotten how calculus works.