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Let $\mathfrak{g}$ be a Lie Algebra (finite dimensional, over $\mathbb{C}$). Engel's theorem tells us that if there exists a $m\in \mathbb{N}$ such that $\mathrm{ad}(x)^m = 0$, $\forall x\in \mathfrak{g}$, then $\mathfrak{g}$ is nilpotent. And if $\mathfrak{g}$ is $(k-1)$-step nilpotent (i.e. the $k$-th term of the lower central series of $\mathfrak{g}$ is the first one that is 0, or equivalently $\mathrm{ad}(x_1)\mathrm{ad}(x_2) \ldots \mathrm{ad}(x_k) = 0$ $\forall x_1, \ldots, x_k \in \mathfrak{g}$), it is clear that $$\min \big\{m\in \mathbb{N} : \mathrm{ad}(x)^m = 0 \forall x\in \mathfrak{g} \big\} \leq k.$$ Can we find an example where the previous inequality is not an equality?

If this is a very basic fact in the theory, I apologize.

Edit: I've been asked to share de Graaf's example, here it is. It is a 17-dimensional (nilpotent) Lie algebra over $\mathbb{Q}$ that is 3-Engel and of nilpotency class 4.

(The table has to be read as follows: $\langle 2,4,17|-3\rangle$ means that $[x_2,x_4] = -3 x_{17}$. If there are more tuples starting with $\langle 2,4$ then one has to take the sum, so if there also was $\langle 2,4,13|-2\rangle$ then $[x_2,x_4] = -2 x_{13} - 3 x_{17}$.)

$$[ \langle2, 4, 17| -3\rangle,\; \langle2, 6, 3| -3\rangle,\; \langle2, 7, 1| 1\rangle,\; \langle2, 9, 1| 2\rangle,\; \langle2, 12, 9| 1\rangle,\; \langle2, 13, 10| 1\rangle,\; \langle2, 14, 11| 1\rangle,\; \langle2, 15, 13| 1\rangle,\; \langle2, 16, 14| 1\rangle,\; \langle5, 16, 17| -3\rangle,\; \langle7, 15, 17| 1\rangle,\; \langle7, 16, 3| -2\rangle,\; \langle8, 15, 3| 3\rangle,\; \langle9, 15, 17| -1\rangle,\; \langle9, 16, 3| -1\rangle,\; \langle10, 16, 1| 3\rangle,\; \langle11, 15, 1| -3\rangle,\; \langle12, 13, 17| 4\rangle,\; \langle12, 14, 3| 4\rangle,\; \langle12, 15, 4| -1\rangle,\; \langle12, 16, 6| -1\rangle,\; \langle13, 14, 1| -4\rangle,\; \langle13, 15, 5| -1\rangle,\; \langle13, 16, 7| -1\rangle,\; \langle14, 15, 7| -1\rangle,\; \langle14, 15, 9| 1\rangle,\; \langle14, 16, 8| -1\rangle,\; \langle15, 16, 12| -1\rangle ]$$

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  • $\begingroup$ Note that Engel's Theorem is not really involved here, but rather its more elementary converse: given a nilpotent Lie algebra, each operator $ad(x)$ is nilpotent. (On the other hand, combining Engel's Theorem with Ado's Theorem allows one to regard an abstract nilpotent Lie algebra as a subalgebra of the upper triangular matrices.) It might be helpful to recall more explicitly what is meant by $k$-step nilpotent. $\endgroup$
    – Jim Humphreys
    Commented May 19, 2011 at 21:18
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    $\begingroup$ As we are in characteristic 0 we may by polarisation express the product of linear operators as a linear combination of powers (of linear combinations of the operators). This implies that we have an equality. $\endgroup$
    – Torsten Ekedahl
    Commented May 20, 2011 at 5:00
  • $\begingroup$ @Torsten: Your approach looks attractively straightforward, but if it works it should be written down in more detail as an answer (preferably with a source in the literature). As it is, there seem to be two contradictory answers to the question asked. I've never worked directly with nilpotent Lie algebras and don't have any intuitive feeling about what is actually true here. But the usual polarisation technique seems to require commuting operators. $\endgroup$
    – Jim Humphreys
    Commented May 23, 2011 at 12:40
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    $\begingroup$ Yesterday I was given an explicit counterexample by Willem de Graaf. It is a 17-dimensional Lie algebra over $\mathcal{Q}$ that is 3-Engel and of nilpotency class 4. I hope he posts it here. $\endgroup$
    – Ramiro Lafuente
    Commented May 24, 2011 at 17:50
  • $\begingroup$ Would you post de Graaf's example that you mentioned? You can mark it community wiki so as not to get reputation from someone else's answer. $\endgroup$
    – LSpice
    Commented Dec 19, 2019 at 18:10

2 Answers 2

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A Lie algebra $\mathfrak{g}$ satisfying the equation $ad(x)^m = 0$ for all $x \in \mathfrak{g}$ is called an Engel-$m$ Lie algebra. One of the key steps in Zelmanov's solution of the restricted Burnside problem is that any finitely generated Engel-$m$ Lie algebra is nilpotent. Zelmanov won the Fields medal for this work.

Let $E(t,m)$ be the free Lie algebra on $t$-generators subject to the Engel-$m$ identity. Then it follows that $E(t,m)$ is nilpotent and therefore finite dimensional, although its dimension and class depend on the characteristic of the ground field $k$.

Let us assume that $char(k) = 0$ for simplicity. Then it is not difficult to show that $E(t,2)$ is nilpotent of class $2$, and $E(2,3)$ is nilpotent of class $3$. The dimension of $E(t,2)$ is $t + \binom{t}{2}$, and the dimension of $E(2,3)$ is $5$.

But already $E(3,3)$ is actually nilpotent of class $4$, so this gives an example of a Lie algebra where the required inequality is strict.

More information can be found in the works of Michael Vaughan-Lee and Gunnar Traustason --- see, for example, http://people.bath.ac.uk/gt223/paper01.pdf. Traustason observes at the bottom of page 12 of this paper that "it is easy to construct an Engel-$3$ Lie algebra of class $4$".

I don't know the dimension of $E(3,3)$, but I'm sure that Willem de Graaf does. This paper contains more relevant information.

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    $\begingroup$ Presumably the original question was asked and answered many decades ago, so it would be helpful to see a reference earlier than the recent technical papers quoted here. Is there a smallest nilpotent Lie algebra for which the inequality in the question is strict, and if so where is the argument written down explicitly? The contrary answer proposed briefly by Torsten Ekedahl leaves me still feeling confused about the status of what seems at first sight to be an elementary question. $\endgroup$
    – Jim Humphreys
    Commented May 22, 2011 at 22:22
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    $\begingroup$ I was also under the impression that this question (and the example) should be in the literature. But apart from the remark of Traustason and the computations of de Graaf et al, I could not find the details of the example online. It's not clear to me there is a "smallest" nilpotent Lie algebra for which the inequality is strict, although of course there must be one of least possible dimension. If we look for an example with $m = 3$, it seems reasonable then to construct it as a quotient of $E(3,3)$ because $E(2,3)^4 = 0$: the example must have at least $3$ generators. $\endgroup$
    – user91132
    Commented May 22, 2011 at 23:07
  • $\begingroup$ P.S. I was wondering about what was known in the Mal'cev era, for instance. On the other hand, I don't see anything relevant in Bourbaki's foundational treatment of nilpotent Lie algebras (including many exercises): Chapter I, Section 4 of Groupes et algebres de Lie. $\endgroup$
    – Jim Humphreys
    Commented May 23, 2011 at 12:43
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I computed the example of Traustason (see his remark). We start with the free-nilpotent Lie algebra of class 4 with 3 generators $x_1$, $x_2$, $x_3$. It has dimension 32. Then we divide out the ideal generated by all brackets containing $x_1$ three times, or containing $x_2$, $x_3$ at least two times. The quotient is an Engel-3-Lie algebra of nilpotency class 4 and of dimension 11. It is easy to write down explicit Lie brackets. This should be an example of least possible dimension. However, the next case, to find an Engel-4-Lie algebra of nilpotency class 7, of least possible dimension is more complicated, and I am frightened to do the calculation.

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    $\begingroup$ I was already frightened with 3 generators! Hence the incomplete nature of my answer. I would be interested to see all 11 brackets that form a basis of your 11-dimensional algebra. I wonder how your example compares with de Graaf's 17-dimensional Lie algebra. $\endgroup$
    – user91132
    Commented May 27, 2011 at 21:17
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    $\begingroup$ The 11 elements are x1,x2,x3,[x1,x2],[x1,x3],[x2,x3],[x1,[x1,x2]], [x1,[x1,x3]], [x2,[x1,x3]], [x1,[x2,x3]] and [x1,[x2,[x1,x3]]]. I have written down all Lie brackets explicitly, in case someone wants to see it. We have ad(x)^3=0 for all x, and the nilpotency class is 4. The Lie algebra admits nonsingular derivations, its derivation algebra has dimension 27. $\endgroup$
    – Dietrich Burde
    Commented Jun 10, 2011 at 11:35
  • $\begingroup$ De Graaf's example is the free algebra on 3 generators in the variety of Lie algebras satisfying $[a,b,c,d,e]$ and $[a,a,a,b]$. Hence this 11-dimensional example is necessarily a quotient of it. Actually it just consists in removing (killing) the coefficients $4,5,6,8, 3,17$. $\endgroup$
    – YCor
    Commented Jun 2, 2020 at 9:54
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    $\begingroup$ Here's an explicit basis, if my computations are correct, for this 11-dimensional Lie algebra: $(x_i)_{0\le i\le 10}$ with, writing (when possible) $ij=\pm k$ for $[x_i,x_j]=\pm x_k$: $01=3,20=4,12=5$, $03=6,04=7$, $14=8,23=9,05=-8-9$, $[x_3,x_4]=2x_{10}$, $[x_0,x_8]=[x_0,x_9]=x_{10}$, $[x_2,x_6]=[x_1,x_7]=-x_{10}$. $\endgroup$
    – YCor
    Commented Jun 2, 2020 at 11:00

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