0

Let $\mathfrak{g}$ be a matrix Lie algebra of dimension $n$ containing elements $x,y,z$ such that: $z=[x,y], [x,z]=[y,z]=0$. Show that $z$ only has $0$ as an eigenvalue.

Immediately from $z=[x,y]=xy-yx$ we get that $tr(z)=0$ but I don't know how to proceed. I tried assuming $zu=\lambda u$ for some $\lambda\ne 0$ and $u\in\mathbb{F}^n$ but I don't see how to obtain something from this.

Dian_Horton92
  • 73
  • $z$ is ad-nilpotent so it is nilpotent as a matrix, too. This is the so-called Heisenberg algebra. $z$ is the center of it. – Marius S.L. Jul 29 '23 at 21:22
  • @MariusS.L. how is $z$ ad-nilpotent though? $x,y,z$ are not the full Lie algebra just 3 elements – Dian_Horton92 Jul 29 '23 at 21:36
  • Ooops, my fault, sorry. – Marius S.L. Jul 29 '23 at 21:39
  • But $\mathfrak{H}:=\langle x,y,z \rangle$ is a nilpotent Lie algebra, a subalgebra of $\mathfrak{g}$ that is still a subalgebra of $\mathfrak{gl}(V).$ Maybe this helps. – Marius S.L. Jul 29 '23 at 22:39
  • Any representation of $\mathfrak{g}$ descends to a representation of this subalgebra $\langle x,y,z \rangle$ by restriction. So you can assume $\mathfrak{g} = \langle x,y,z \rangle$. Ad-nilpotent is not enough regardless as that only works for semisimple Lie algebras. – Callum Jul 30 '23 at 05:21
  • @Callum ok even if you restrict it though you still have $n\times n$ matrices so how do I go from there? – Dian_Horton92 Jul 30 '23 at 11:53
  • In a matrix Lie algebra $[x,y]=xy-yx$. So $0=[z,x]=zx-xz$ says that $z$ commutes with $x$. The same with $0=[z,y]$. So $z$ is nilpotent. – Dietrich Burde Jul 30 '23 at 12:05

0 Answers0