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



Kostant's theorem about U(g) being free over Z(g) and a corollary of it
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.


