Let $\mathfrak {g}$ be a finite dimensional complex Lie algebra. Recall that an element $g \in \mathfrak {g}$ is called nilpotent element if ad$g: \mathfrak {g} \to \mathfrak {g}$ is nilpotent endomorphism.

Give an example of a linear Lie Algebra with a element $g$ such that $g$ is nilpotent as a matrix but ad$(g)$ is not nilpotent?

I am not sure how can we construct this.Please help.


Such an example does not exist, because of the following Lemma (from Humphreys book)

Lemma: Let $V$ be a vector space and $x\in \mathfrak{gl}(V)$ be nilpotent, i.e., $x^n=0$ for some $n$. Then $ad(x)\in \mathfrak{gl}(\mathfrak{gl}(V))$ is nilpotent too.

The proof is easy, one considers the linear maps $L\colon \mathfrak{gl}(V)\rightarrow \mathfrak{gl}(V)$ given by $y\mapsto xy$ and $R\colon \mathfrak{gl}(V)\rightarrow \mathfrak{gl}(V)$ given by $y\mapsto yx$, and notes that $L$ and $R$ commute because of $(LR)(y)=xyx=(RL)(y)$. Since $x^n=0$ we have $L^n=R^n=0$, and thus $$ ad(x)^{2n}=(L-R)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}L^{2n-k}(-R)^k=0. $$

Remark: The converse statement is not true in general. There are ad-nilpotent elements $x$, i.e., with $ad(x)$ nilpotent, where $x\in \mathfrak{gl}(V)$ is not nilpotent. Take the Lie algebra $\mathfrak{d}_n$ of diagonal matrices. This Lie algebra is abelian, so that $ad(d)=0$ is nilpotent for all $d$, but not all diagonal matrices are nilpotent.

  • $\begingroup$ I see..Thank you sir! Is it true that every Lie algebra has nilpotent element? I cant see it. $\endgroup$ – Math Lover Sep 9 '16 at 9:41
  • $\begingroup$ No, it is not true. The Lie algebra $L=span \{id_V\}$ consists only of non-nilpotent elements, except for $x=0$ of course, which is always a nilpotent element in any Lie algebra. $\endgroup$ – Dietrich Burde Sep 9 '16 at 11:23
  • $\begingroup$ I think i am missing something. So $L$ is just the scalar matrices ?Isn't it ad$(d)=0$ for every element $d$ of $L$ because every element of $L$ commutes with every other element? $\endgroup$ – Math Lover Sep 9 '16 at 11:38
  • $\begingroup$ No, you are not missing anything. If $d$ is a multiple of the identity, then of course $ad(d)=0$. But $d$ is not nilpotent, and that was your question. $\endgroup$ – Dietrich Burde Sep 9 '16 at 15:00
  • $\begingroup$ I am really sorry i think i dint ask the question properly.Let me frame it properly: Does every Lie Algebra $\mathfrak {g}$ has a element $g$ such that ad$(g)$ is nilpotent? $\endgroup$ – Math Lover Sep 9 '16 at 17:46

To answer the question in the comment, namely: show that every Lie algebra over an algebraically closed field $K$ of characteristic zero, of positive finite dimension, has an ad-nilpotent element:

Let $V$ be such a Lie algebra. Choose any nonzero $x$. If $\mathrm{ad}(x)$ is nilpotent, we are done. Assume it's not the case. Decompose $V=\bigoplus_{t\in K}V_t$ into characteristic subspaces with respect to $\mathrm{ad}(x)$. Then this is a Lie algebra grading in the sense that $[V_t,V_u]\subset V_{t+u}$ for all $t,u\in K$.

If $t\neq 0$ and $y\in V_t$ then $\mathrm{ad}(y)$ is nilpotent: indeed $\mathrm{ad}(y)^n(V_u)\subset V_{u+nt}$, and for some $n$ (actually, for all large enough $n$), $W\cap (W+nt)$ is empty, where $W=\{u:V_u\neq 0\}$ (because $W$ is finite). So $\mathrm{ad}(y)^n=0$.

The assumption that $\mathrm{ad}(x)$ is not nilpotent means that $V_t$ is nonzero for some $t$. So nonzero elements in $V_t$, which exist, are non-zero ad-nilpotent elements.


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