Lie algebra admitting some hyperbolic automorphism is nilpotent

Question

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!

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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 $b\ge n$.

Answer 1

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 $\phi$.

Answer 2

As already mentioned, we can enlarge the field, hence let us assume it to be algebraically closed and not algebraic over a finite field. Then in this context, for finite-dimensional algebras (Lie or not) $\mathfrak{g}$, the following are equivalent:

1. there exists an automorphism with no root of unity as eigenvalue
2. there exists a diagonalizable automorphism with no root of unity as eigenvalue
3. there exists an algebra grading in a torsion-free abelian group with $\mathfrak{g}_0=0$
4. there exists an algebra grading in $\mathbf{Z}$ with $\mathfrak{g}_0=0$

One goes from (1) to (2) by picking the diagonalizable part in the Jordan decomposition. (2) yields (3) since the eigenspace decomposition provides a grading in the quotient of $K^*$ by its subgroup of roots of unity. (3) yields (4) by picking the subgroup generated by the degrees, and picking a suitable homomorphism to $\mathbf{Z}$, and finally (4) yields (1) by picking a non-root of unity $t$ and define an automorphism to be multiplication by $t^n$ on $\mathfrak{g}_n$.

Note that all we use of $\mathfrak{g}$ is that its automorphism group is Zariski-closed in $\mathrm{GL}(\mathfrak{g})$; that $\mathfrak{g}$ is Lie plays no role.

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We next show that (4) implies nilpotent; this works in a considerably broader context.

If $\mathfrak{g}$ is an arbitrary Lie algebra (over an arbitrary commutative ring), graded in $\mathbf{Z}\smallsetminus\{0\}$ and with bounded degree (i.e., $\mathfrak{g}_n=0$ for $|n|$ large), then $\mathfrak{g}$ is nilpotent.

More precisely, if $\mathfrak{g}$ is graded in $\{-n,\dots,-1,1,\dots,n\}$, then it is $5n$-nilpotent (possibly this is not optimal; the filiform Lie algebras show that we cannot do better than $\sim 2n$).

Let us prove this in the broader realm of Leibniz algebras. An algebra is Leibniz if it satisfies the Leibniz identity $(xy)z-(xz)y=x(yz)$ for all $x,y,z$. Note that a Lie algebra is a Leibniz algebra satisfying in addition $xx=0$ for all $x$.

First immediate fact: in a Leibniz algebra, every product of $n$ elements is a linear combination of $n$-fold right products, i.e., of the form $((\cdots ((x_1x_2)x_3)\dots )x_n)$.

Write $R_x(y)=yx$. Then $R_xR_y=R_yR_x+R_{yx}$ by the Leibniz identity. Thus a Leibniz algebra $\mathfrak{g}$ is $k$-nilpotent if and only if $R_{x_1}\dots R_{x_k}=0$ for all $x_1,\dots,x_k\in\mathfrak{g}$.

Now suppose that $\mathfrak{g}$ is graded in $\mathbf{Z}\smallsetminus\{0\}$ and we have $x_1,\dots,x_k$, with $x_i\in\mathfrak{g}_{n_i}$, with all $|n_i|\le n$. Consider $v=R_{x_1}\dots R_{x_k}$. We call the set of $v$ obtained this way by $r(n,k)$.

Whenever $n_{i+1}>0>n_i$, we can write $R_{x_i}R_{x_{i+1}}=R_{x_{i+1}}R_{x_i}+R_{x_{i+1}x_i}$, noting that $x_{i+1}x_i$ has degree $n_i+n_{i+1}$ and $|n_i+n_{i+1}|\le n-1$. We reiterate the process with the new term involving $R_{x_{i+1}}R_{x_i}$, so as to eventually obtain $v=R_{x_{\sigma(1)}}\dots R_{x_{\sigma(k)}}+w$, where $w$ is a sum of elements in $r(n-1,k-1)$ and, for some $j$, all $n_{\sigma(i)}$ are $>0$ for $i\le j$ and $<0$ for $i>j$. Write $r^+(n,k)$ the set elements in $r(n,k)$ with this additional condition. Let $R(n,k)$ and $R^+(n,k)$ be the set of sums of elements in $r(n,k)$, resp $r^+(n,k)$.

Then we have proved $R(n,k)\subset R^+(n,k)+R(n-1,k-1)$. By induction, this yields $$R(n,k)\subset R^+(n,k)+R^+(n-1,k-1)+\dots,R^+(1,k-n+1)$$

thus $$R(n,5n)\subset R^+(n,5n)+R^+(n-1,5n-1)+\dots,R^+(1,4n+1)$$

Now assume that $\mathfrak{g}$ is graded in $\{-n,\dots,n\}\smallsetminus\{0\}$. Then the maximal difference between two degrees is $2n$. It follows that every element in $R^+(n,k)$ for $k\ge 4n+1$ is zero: indeed, such an element either involves $2n+1$ consecutive $R_{x_i}$ for $n_i>0$, or $2n+1$ consecutive $R_{x_i}$ for $n_i<0$, which shifts the degree by $\ge 2n+1$ and hence is zero. It follows that $R(n,5n)=0$.

By linearity, this implies that $R_{x_1}\dots R_{x_{5n}}=0$ for all $x_i$, and by the Leibniz identity this implies that $\mathfrak{g}$ is $5n$-nilpotent.

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Initial answer (for finite-dimensional Lie algebras in characteristic zero only):

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

Since every self-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 (*), it follows that $\mathfrak{g}$ is nilpotent.

(*) 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.

It remains the question whether more generally, in arbitrary characteristic, a finite-dimensional Lie algebra admitting an invertible self-derivation, is always nilpotent. [Edit: in every positive characteristic $p$, there exist finite-dimensional simple Lie algebras with invertible self-derivations.

See "G. Benkart - A.I. Kostrikin - M.I. Kuznetsov: Finite-Dimensional Simple Lie Algebras with a Nonsingular Derivation, J. Algebra 171 (1995), 894-916" Sciencedirect, and thanks to Salvatore Siciliano for the reference and link.]

Answer 3

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.

Answer 4

This is just for answering the last question posed by Yves. In positive characteristic, a finite-dimensional Lie algebra admitting an invertible derivation need not be nilpotent. For an example, look at the paper:

"G. Benkart - A.I. Kostrikin - M.I. Kuznetsov: Finite-Dimensional Simple Lie Algebras with a Nonsingular Derivation, J. Algebra 171 (1995), 894-916" (http://www.sciencedirect.com/science/article/pii/S0021869385710411)

However, under some assumption the conclusion is indeed true. For instance, Jacobson proved that this is just the case when $L$ is a finite-dimensional restricted Lie algebra admitting an invertible restricted derivation.