# Chevalley restriction theorem for exterior algebras

Suppose $G$ is semisimple Lie group, $\mathfrak{g}$ is its Lie algebra, $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, and $W$ is the correspondent Weyl group.

Chevalley restriction theorem says that there is isomorphism of algebras of invariants $\mathbb{C}[\mathfrak{g}]^{G}\simeq \mathbb{C}[\mathfrak{h}]^{W}$ given by restriction. Here $\mathbb{C}[\mathfrak{g}]$ and $\mathbb{C}[\mathfrak{h}]$ denote the algebras of polynomial functions on correspondent algebras viewed as vector spaces. In other words, $\mathbb{C}[\mathfrak{g}]=Sym(\mathfrak{g}^*)$ is a symmetric algebra.

My question is: is there a similar result for algebra of $G$-invariants of the exterior algebra $\bigwedge(\mathfrak{g}^*)$?

Thank you very much!

• Edit:

See lemma 3.3 in the first paper: $C^\infty$ and polynomial (by Solomon, cited in the paper) $W$-invariant differential forms on $\mathfrak h$ correspond to horizontal $G$-invariant forms on $\mathfrak g$.

I just noted: You are asking for $G$-invariants of $\bigwedge \mathfrak g^\star$, i.e., constant differential forms.
These describe the de Rham cohomology of a compact form of $G$, by the theorem of Chevalley and Eilenberg, and can be described as the set of primitive elements. See books, for example,

• MR0400275 (53 #4110) Reviewed Greub, Werner; Halperin, Stephen; Vanstone, Ray Connections, curvature, and cohomology. Volume III: Cohomology of principal bundles and homogeneous spaces. Pure and Applied Mathematics, Vol. 47-III. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1976. xxi+593 pp.

• MR1379333 (97j:57057) Reviewed Onishchik, A. L.(RS-YAR) \cyr Topologiya tranzitivnykh grupp preobrazovaniĭ. (Russian. English, Russian summary) [Topology of transitive transformation groups] Fizmatlit Nauka'', Moscow, 1995. 384 pp. ISBN: 5-02-014724-9

There is a translation into English which cannot be found in MathRev:

• Onishchik, Arkadi L. Topology of transitive transformation groups. (English) Zbl 0796.57001 Leipzig: Johann Ambrosius Barth. xv, 300 p. (1994).
• Can you, please, explain to me, how to get the formula for $\bigwedge(\mathfrak{g}^*)$? I thought you should take $M=\mathfrak{g}$, $G$ acts on $M$ by conjugation. Then the tangent space to the orbit through $x\in M$ at $x$ will be $[\mathfrak{g},x]\subset \mathfrak{g}$, right? But then horizontal form $\omega$ is a form that annihilates all such tangent spaces. This means that $\omega$ annihilates $[\mathfrak{g},\mathfrak{g}]$. But if $\mathfrak{g}$ is semisiple, then $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so $\omega=0$. Can you, please, explain what I am doing wrong? – Sasha Patotski Oct 20 '13 at 21:24