Let $\frak g$ be a Lie algebra and $\rho$ be a finite dimensional representation of $\frak g$ in a linear space $V$. They define the momentum associated to $\rho$ as the map $\mu: V \otimes V^\ast \to {\frak g}^\ast$ given by $\mu(v \otimes \phi)(x) = \phi(\rho(x)(v))$. Then we're asked to check that $\mu$ interchanges $\rho \otimes \rho^\ast$ with ${\rm ad}^\ast$. in other words, that for all $x \in {\frak g}$ we have this commutative diagram: $\require{AMScd}$ \begin{CD} V \otimes V^\ast @>\mu>> {\frak g}^\ast\\ @A {(\rho \otimes \rho^\ast)(x)} A A @AA {\rm ad}^\ast(x) A\\ V \otimes V^\ast @>>\mu> {\frak g}^\ast \end{CD}
I managed to check that the diagram commutes (just keep a level head and do it), but I don't see why do we need $\dim V < \infty$ here. My first thought would be that $\mu$ would be an isomorphism in that case, but we don't know anything about the dimension of ${\frak g}$, so I guess we can discard that.
(Here $\rho \otimes \rho^\ast$ is a shorthand for $\rho \otimes {\rm id}_{V^\ast}+{\rm id}_{V}\otimes \rho^\ast$)