2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.