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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:

http://www.springerlink.com/content/wur75t1t65371812/

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

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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 springerlink.com/content/p03325375750vu6m 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
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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.

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Thanks! VERY interesting and helpful answer. –  Dick Palais Dec 11 '10 at 16:25
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