Hello,
$g$ is a complex semisimple Lie algebra.
There is the result that $U(g)$ is free over $Z(g)$.
There is another result: If $E$ is a finite dimensional representation of $g$, then $Hom(E,U(g)^{ad})$ is a free $Z(g)$-module of rank equals the multiplicity of the zero weight in $E$ (here $U(g)^{ad}$ denotes $U(g)$ as a $g$-module for the adjoint action $v\cdot u = vu-uv$).
My question is: how can one deduce the second result from the first one?
Thanks, Sasha
The results here go back to a classic paper by Kostant: Lie group representations on polynomial rings, Amer. J. Math. 85 (1963), 327-404. A detailed exposition was given by Dixmier in his 1974 book, translated (with added misprints) into English as Enveloping Algebras and later reprinted by AMS (with some of the misprints listed at the end). Here the relevant sections are 8.2-8.3 for the basic theorem and then its application to the Hom space rank over the center $Z(\mathfrak{g})$ of the enveloping algebra. Note however that your Hom should refer to the space of $\mathfrak{g}$-module maps.
