# Proof or Counter example: Every non zero Lie algebra has an ad-nilpotent element?

I am trying to understand a definition in chapter $$1$$ of Humphreys Lie Algebra. I think I need the following result to understand the definition properly:

Let $$\mathcal L$$ be a non zero Lie Algebra over some (algebraically closed) field $$F$$. Recall that $$a \in \mathcal L$$ is called ad-nilpotent if $$ad(a): \mathcal L \to \mathcal L$$ is a nilpotent endomorphism. Is it true that every non zero Lie algebra has an ad-nilpotent element?

• Yes, it does but you need to know some structural results about Lie algebra in order to prove it. Commented Aug 15, 2016 at 9:31
• @studiosus Could you please sketch the argument?My intuition is that the proof wont be hard at least for semi simple case. Commented Aug 15, 2016 at 9:32
• Sorry, it works only in finite dimensional case, I suspect that in general the answer is negative. I will write down some details. Commented Aug 15, 2016 at 14:29
• In the free Lie algebra on 2 generators, no element is ad-nilpotent.
– YCor
Commented Aug 15, 2016 at 22:34
• It's perhaps worth noting that this is false if we don't assume algebraically closed. Any compact form of a semisimple Lie algebra over $\mathbb{R}$ (e.g. $\mathfrak{su}_n( \mathbb{R})$) will have no ad-nilpotent elements. Commented Aug 15, 2016 at 23:44

Here's an argument (with no use of the solvable radical and semisimple quotient).

First, if $\mathfrak{g}$ (finite-dimensional) admits a grading in $\mathbf{Z}$, then every element of $\mathfrak{g}_n$ for $n\neq 0$ is ad-nilpotent (clear).

So a Lie algebra with the property that no nonzero element is ad-nilpotent has no grading in $\mathbf{Z}$. In general this can happen (e.g., the real Lie algebra $\mathfrak{so}(n)$ for $n\ge 3$). However, if the field is algebraically closed of characteristic zero, this implies that the Lie algebra is nilpotent. But then every element is ad-nilpotent; since it was assumed that no nonzero element is ad-nilpotent, this implies that the Lie algebra is zero.

Now let us check that $\mathfrak{g}$ is nilpotent (assuming the field algebraically closed). Indeed, any 1-dimensional multiplicative group of automorphisms yields such a nontrivial grading. Hence if there's no nontrivial grading in $\mathbf{Z}$ (by trivial grading I mean the grading concentrated in degree 0), the automorphism group of $\mathfrak{g}$ is virtually unipotent. In particular, the Lie algebra of derivations is nilpotent. Hence the Lie algebra of inner derivations is nilpotent, which in turn implies that $\mathfrak{g}$ is nilpotent.

I'm curious about this argument in positive characteristic (although the conclusion is known with an elementary proof Benkaart and Isaacs, 1977). Namely, does there exist a non-nilpotent finite-dimensional Lie algebra, in positive characteristic (over an algebraically closed field), with no nontrivial grading in $\mathbf{Z}$ (or equivalently, with $\mathrm{Aut}(\mathfrak{g})_0$ unipotent)? Added: This is precisely Question (c) in Section 7 of D. Winter, On groups of automorphisms of Lie algebras, J. Algebra 8, 131-142 (1968).

• Dear @YCor: Thank you! Since I have just started to learn about Lie Algebras so i dont understand the maximum part of your answer.Could you please explain the "Now let us check..." paragraph in more detail.(Like how do we get grading from group of automorphisms etc.)Sorry for troubling you. Commented Aug 16, 2016 at 5:42
• this part is rather basics on algebraic groups: if we have a finite-dimensional representation $V$ of the 1-dimensional torus $GL_1$, then it decomposes as $V=\bigoplus_{n\in Z}V_n$ where $x\cdot v=x^nv$ for all $x\in GL_1$ and $v\in V_n$. If $V$ is a Lie algebra and this is an action by automorphisms, then this is a Lie algebra grading.
– YCor
Commented Aug 16, 2016 at 8:12

Here is a sketch of the proof in the finite-dimensional case. You start with the Levi-Malcev decomposition of your Lie algebra ${\mathfrak g}= {\mathfrak r} \oplus {\mathfrak s}$, where ${\mathfrak r}$ is the solvable radical and ${\mathfrak s}$ is the semisimple part. Note that the action of ${\mathfrak s}$ on ${\mathfrak r}$ is, in general, nontrivial. Next, you use the Gauss decomposition ${\mathfrak s}= {\mathfrak n}_+ \oplus {\mathfrak t} \oplus {\mathfrak n}_-$, where ${\mathfrak n}_\pm$ are nilpotent subalgebras of ${\mathfrak s}$. If ${\mathfrak s}\ne 0$, so are ${\mathfrak n}_\pm$. Furthermore the adjoint action of ${\mathfrak n}_\pm$ on ${\mathfrak r}$ is also nilpotent. (This is a general fact about semisimple Lie algebras: Nilpotent elements have nilpotent action under any representation.) Thus, if ${\mathfrak s}\ne 0$, then ${\mathfrak g}$ contains nonzero nilpotent elements (any nonzero element of ${\mathfrak n}_\pm$ would do). Consider now the case when ${\mathfrak s}=0$, ${\mathfrak g}={\mathfrak r}$. Then you use the further decomposition ${\mathfrak r}= {\mathfrak n}\oplus {\mathfrak t}$, where ${\mathfrak t}$ is abelian and ${\mathfrak n}$ is the nilponent radical of ${\mathfrak r}$. The subalgebra ${\mathfrak n}$ consists of nilponent elements. Moreover, ${\mathfrak r}$ admits a faithful linear representation where ${\mathfrak t}$ maps to the diagonal matrices and ${\mathfrak n}$ maps to strictly upper triangular ones. In any case, each nonzero element of ${\mathfrak n}$ gives a nonzero nilpotent element of ${\mathfrak r}$. Lastly, if ${\mathfrak n}=0$ then ${\mathfrak g}$ is abelian and every element is nilpotent (by the definition).

• Dear @studiosus,thank you for your answer.Unfortunately i dont understand the maximum part of your solution since i am new to Lie Algebras. Commented Aug 16, 2016 at 5:43

For more general fields see Varea, "EXISTENCE OF AD-NILPOTENT ELEMENTS AND SIMPLE LIE ALGEBRAS WITH SUBALGEBRAS OF CODIMENSION ONE", Proc. A.M.S. 104 (1988), 363-368.

• – YCor
Commented Aug 15, 2016 at 22:50
• But actually the relevant reference is rather: G. M. Benkart and I. M. Isaacs, On the existence of ad-nilpotent elements, Proc. Amer. Math. Soc. 63 (1977), 39-40 ams.org/journals/proc/1977-063-01/S0002-9939-1977-0432721-6
– YCor
Commented Aug 15, 2016 at 22:53