MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $G$ is a semi-simple Lie group that acts smoothly (i.e., $C^\infty$) on a smooth, finite dimensional manifold $M$. Does it follow that the action of $G$ is linearizable near any stationary point of the action? This was conjectured by Steve Smale and myself in 1965, and was proved for the case that $M$ and the action were analytic by Bob Herman and Guillemin and Sternberg in two papers from long ago:

Hermann, R.: The formal linearization of a semi-simple Lie algebra of vector fields about a singular point. Trans. Am. Math. Soc. 130, 105-109 (1968)

Guillemin, V., Sternberg, S.: Remarks on a paper of Hermann. Trans. Am. Math. Soc. 130,110-116 (1968)

I have not heard whether any progress has been made since then and I would be interested to hear from anyone who has heard of a proof or a counter-example. The reason is not just idle curiousity; this is the missing step in a proof that what I call The Principle of Symmetric Criticality is valid for smooth finite dimensional actions of a semi-simple group: see (particularly page 29 of) the paper downloadable here:

for more details on this principle and why it is important, particularly in mathematical physics.

share|cite|improve this question
I just want to mention that the analytical linearization was independently proved by A. G. Kushnirenko (and was published before the paper by Guillemin and Sternberg) in his paper "Linear-equivalent action of a semisimple Lie group in the neighborhood of a stationary point" (Functional Analysis and Its Applications 1967, Volume 1, Number 1, 89-90). (The translation to English is very bad as even I can see.) See Kushnirenko's paper was submitted 15 days later then the G&S's paper. – Petya Dec 3 '10 at 21:39
Thanks, Petya, I wasn't aware of the Krushnirenko's paper. – Dick Palais Dec 3 '10 at 21:58
up vote 9 down vote accepted

There are smooth counter-examples by Cairns and Ghys [Ens. Math. 43, 1997], for instance a smooth non-linearizable action of $SL(2,\mathbb{R})$ on $\mathbb{R}^3$ (fixing the origin) or of $SL(3,\mathbb{R})$ on $\mathbb{R}^8$. By contrast, they show that any $C^k$ action of $SL(n,\mathbb{R})$ on $\mathbb{R}^n$ (same $n$, fixing the origin) is $C^k$-linearizable. Here is a link to their paper.

share|cite|improve this answer
Thanks! VERY interesting and helpful answer. – Dick Palais Dec 11 '10 at 16:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.