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Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism.

Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is hyperbolic if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.

Then Proposition 3.6 in Smale's paper (here) says that:

  • Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

  • Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results?

Thanks!


After Vladimir Dotsenko:

$$(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma) v].$$

Applying above to the pair $\hat{u}=\lambda^i\phi^j(\phi-\lambda)^{a}u$ and $\hat{v}=\gamma^k\phi^l(\phi-\gamma)^{b}v$ we have $$(\phi-\lambda\gamma)[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]= [(\phi-\lambda)\hat{u},\phi \hat{v}]+[\lambda \hat{u},(\phi-\gamma) \hat{v}]$$ $$=[\lambda^i\phi^j(\phi-\lambda)^{a+1}u,\gamma^k\phi^{l+1}(\phi-\gamma)^bv] +[\lambda^{i+1}\phi^j(\phi-\lambda)^au, \gamma^k\phi^l(\phi-\gamma)^{b+1}v].$$

Tracing the indices we get $$(i,j,a;k,l,b)\overset{\phi-\lambda\gamma}{\to}(i,j,a+1;k,l+1,b)\cup (i+1,j,a;k,l,b+1),$$ and in particular $(a,b)\overset{\phi-\lambda\gamma}{\to}(a+1;b)\cup (a;b+1)$. Then $$(\phi-\lambda\gamma)^{m+n}[u,v] =\sum_{a+b=m+n}[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=0$$ since either $a\ge m$ or $b\ge n$.

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Care to post the hints? –  MTS Mar 12 '12 at 6:43

3 Answers 3

up vote 4 down vote accepted

First, if $x$ and $y$ are generalised eigenvectors of $\phi$ with generalised eigenvalues $\alpha$ and $\beta$, that is $(\phi-\alpha\mathrm{Id}_\mathfrak{g})^N(x)=(\phi-\beta\mathrm{Id}_\mathfrak{g})^M(y)=0$, then we observe that $\mathrm{ad}(x)^K(y)=0$ for $K>>0$, just noticing that $\mathrm{ad}(x)^K(y)$ is a generalised eigenvector with generalised eigenvalue $\alpha^K\beta$, so since $\alpha$ is not a root of unity, not all of these have nonzero generalised eigenvectors. Thus, if $x$ is a generalised eigenvector of $\phi$, then $\mathrm{ad}(x)$ is nilpotent.

Here I got stuck and ended up checking Bourbaki; they say that one has to use the following result (which I simplify substantially compared to the general statement of Ex.11 in the same volume): "If $L$ is a subset [not necessarily a subspace!] of the matrix algebra closed under commutators and we know that $L$ consists of nilpotent matrices, then the associative subalgebra of the matrix algebra generated by $L$ is nilpotent." (This seems fairly easy, I am just a bit lazy to type the proof.) This result should, of course, be applied to the subset of $\mathrm{End}(\mathfrak{g})$ consisting of all linear operators $\mathrm{ad}(x)$, where $x$ is a generalised eigenvector of $\phi$.

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Thank you! I got some feeling of the propositions. But I can only prove the trivial case $N=M=1$ since $(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma v)]$ I do not how to prove '$\mathrm{ad}(x)^K(y)$ is a generalised eigenvector with generalised eigenvalue $\alpha^K\beta$' for general cases. –  Pengfei Mar 12 '12 at 13:24
1  
You did all the job already! From $(\phi-\lambda\gamma)[u,v]=[(\phi-\lambda)u,v]+\lambda[u,(\phi-\mu)v]$, you can deduce by induction $(\phi-\lambda\gamma)^L[u,v]=\sum_{i=0}^L\lambda^{L-i}[(\phi-\lambda)^iu,(\phi-\‌​mu)^{L-i}v]$, so taking $L=N+M$ would show that if $u$ and $v$ are generalised eigenvectors, then their bracket is too, with eigenvalue being the product of eigenvalues. –  Vladimir Dotsenko Mar 12 '12 at 13:39
    
$\gamma$ should be $\mu$, or vice versa - your choice of notation clashed with my notational habits, sorry! –  Vladimir Dotsenko Mar 12 '12 at 13:45
    
I see! Just realized that the identity is valid for all $u$ and $v$. So we can iterate. –  Pengfei Mar 13 '12 at 2:01

The statement (as in Bourbaki) is equivalent to: every (finite dimensional) Lie algebra with an invertible derivation is nilpotent.

Since every derivation of a semisimple Lie algebra is inner (this is elementary, see e.g. http://amathew.wordpress.com/2010/01/30/derivations-of-semisimple-lie-algebras-and-the-abstract-jordan-decomposition/), you already know the Lie algebra $\mathfrak{g}$ is solvable. Now from the derivation you can define a semidirect product $\mathfrak{h}=\mathfrak{g}\rtimes\mathfrak{a}$, where $\mathfrak{a}$ is the one-dimensional Lie algebra. Since the derivation is invertible, the derived subalgebra $\mathfrak{h}'$ is equal to $\mathfrak{g}$. Since the derived subalgebra of any solvable Lie algebra is nilpotent (consequence of Ado's Theorem [Edit: no it's not needed, see below]), it follows that $\mathfrak{g}$ is nilpotent.

[Edit: consider the adjoint representation of any solvable Lie algebra \mathfrak{h}; triangulate it (over an algebraic closure), so that $\mathfrak{h}'$ is mapped to algebra of upper nilpotent matrices. Since the adjoint representation has central kernel, this shows that $\mathfrak{h}'$ is nilpotent. Also as pointed out by Jim, I assume characteristic zero in the whole argument.]

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This is interesting but gets complicated, so maybe it's useful to supplement Vladimir's answer with more details.

1) Smale is applying a fairly elementary but not quite standard algebraic fact about Lie algebras to show that a certain real Lie algebra is nilpotent (and therefore is the Lie algebra of a nilpotent Lie group, by the usual correspondence). As he notes in his 1967 paper, Armand Borel suggested this exercise from the 1960 Chapter 1 in Bourbaki's treatise; Borel was himself active in Bourbaki, of course.

2) Chapter 4 of Bourbaki is devoted to finite dimensional nilpotent Lie algebras over an arbitrary field. This includes Engel's Theorem: a Lie algebra is nilpotent iff every operator ad $x$ is nilpotent. Since the kernel of the adjoint representation is abelian, it's usually easy to reduce questions to a *;inear" Lie algebra. As occurs elsewhere in Bourbaki, this chapter is augmented by a varied list of exercises; some are marked with the special symbol for "challenging", for instance exer. 21 here (and the earlier exer. 11 to which it refers). These exercises are often quite interesting, but also frustrating when they provide no context or source.

3) Exercise 21 concerns a given Lie algebra $\mathfrak{g}$ having an automorphism $\sigma$. In part (a) it's observed that the bracket of two generalized eigenvectors is again a generalized eigenvector (for the product of the two eigenvalues), as noted by Vladimir. For the proof it's harmless to extend the base field to an algebraic closure. Starting the inductive proof is a bit tricky, so Bourbaki makes a helpful observation:

$(\sigma - \lambda \mu) [xy] = [(\sigma -\lambda)x, \sigma y] + [\lambda x, (\sigma-\mu)y]$.

4) Then part (b) of the exercise, cited by Smale, assumes that no eigenvalue of $\sigma$ in an algebraic closure of the field is a root of unity. From this it follows that $\mathfrak{g}$ is nilpotent. Again it's harmless to pass to an algebraic closure. Vladimir outlines the method of proof, which in Bourbaki refers back to Exercise 11. Actually, the needed discussion is done in II.2 of Jacobson's 1962 book Lie Algebras. While Jacobson's style differs a lot from Bourbaki's, he tends to include all the details for these relatively elementary steps.

Concerning the remarks by Yves, it should be emphasized that his proposed approach to the question requires at least that the field have characteristic 0 (not a problem for Smale, but a further complication here). Arguing in terms of semisimple Lie algebras and nondegeneracy of the Killing form (to ensure that derivations are inner) doesn't generalize at all well to other fields.

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Now I fully understood the stories and the theorems. Thank you very much! –  Pengfei Mar 13 '12 at 0:50
    
Thanks Jim! I knew that you and a few other people on MO have much more to say about it than I, but somehow I got quite excited about the question, and spent some time on it, which then justified writing down what I understood... –  Vladimir Dotsenko Mar 14 '12 at 13:02

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