If L is nilpotent then $K\cap L^n \not=0$

Let $K$ be a proper ideal of a nilpotent Lie algebra $L$. If the nilpotency class of $L$ is $n$ (i.e $L^n\not = 0, L^{n+1}=0$). Is it correct that $K\cap L^n \not=0$?

• No, you made it very clear.. thanks – Ronald Feb 17 '16 at 21:01

This is not true in general. Consider a direct Lie algebra sum $L=\mathfrak{h}_n\oplus \mathbb{C}$ of the $2n+1$-dimensional Heisenberg Lie algebra and the $1$-dimensional abelian Lie algebra. We have $L^1=L$, $L^2=[L,L]=\langle z\rangle =Z(\mathfrak{h}_n)$, and $L^3=0$, but the center of $L$ is $2$-dimensional. Hence $L^2\neq Z(L)$. Now we could take the ideal $K=\mathbb{C}$of $L$, which satisfies $K\cap L^2=0$.

• Ahh, nice and simple. And of course this works with any non-abelian nilpotent Lie algebra. – Tobias Kildetoft Feb 16 '16 at 20:44
• Of course one could ask if its true if we require more on $L$, e.g., being indecomposable. – Dietrich Burde Feb 16 '16 at 20:47
• I just posted a question asking about the indecomposable case. – Tobias Kildetoft Feb 17 '16 at 19:23
• @TobiasKildetoft I found an example of an indecomposable nilpotent Lie algebra, too (if I calculated correctly). – Dietrich Burde Feb 17 '16 at 19:52