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Let $L$ be an indecomposable nilpotent Lie algebra (finite dimensional and over $\mathbb{C}$). Is it possible for the last non-zero term of the central series to be strictly smaller than the center?

For context, as shown in The center of a nilpotent Lie algebra intersects each ideal any non-trivial ideal will intersect non-trivially with the center. But in If L is nilpotent then $K\cap L^n \not=0$ it is shown that it is possible to have a non-trivial ideal which intersects the last non-zero term of the central series trivially. However, the example given is decomposable, and the answerer himself wonders in a comment whether it is also possible with an indecomposable Lie algebra.
Also note that a Lie algebra has the property that any non-trivial ideal has non-trivial intersection with the last non-zero term of the central series iff that term coincides with the center.

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  • $\begingroup$ Where can I find a proof for your last statement: "Also note that a Lie algebra has the property that any non-trivial ideal has non-trivial intersection with the last non-zero term of the central series iff that term coincides with the center."? $\endgroup$ – Ronald Jul 20 '18 at 8:22
  • $\begingroup$ @Ronald It is fairly straightforward to show once you know that this last non-zero term is contained in the center (consider what happens if it is smaller than the center and use this to construct an ideal which has trivial intersection with it). $\endgroup$ – Tobias Kildetoft Jul 20 '18 at 8:25
  • $\begingroup$ So if the algebra is nilpotent, then the center intersects the last non-zero term (actually the last non-zero term is central). Then the last non-zero term is exactly the center? $\endgroup$ – Ronald Jul 20 '18 at 8:33
  • $\begingroup$ @Ronald Actually, I think the way I stated it is not correct (any simple Lie algebra is a counter example). But once the algebra is nilpotent, then if the last non-zero term is strictly smaller than the center, we can just take a complement of it inside the center to get something non-trivial intersection it trivially). $\endgroup$ – Tobias Kildetoft Jul 20 '18 at 8:46
  • $\begingroup$ Any subspace of the center is an abelian ideal (this you really should be able to show directly). $\endgroup$ – Tobias Kildetoft Jul 20 '18 at 8:54
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Yes, it is possible. Here is an example of an indecomposable $7$-dimensional $5$-step nilpotent Lie algebra $L$, with basis $(x_1,\ldots ,x_7)$ and brackets $$ [x_1,x_2]=x_3,\,[x_1,x_3]=x_4,\,[x_1,x_4]=x_6,\,[x_1,x_6]=x_7,\,[x_2,x_3]=x_5, $$ where we have $L^1=L$, $L^2=\langle x_3,\ldots ,x_7\rangle $, $L^3=\langle x_4,\ldots ,x_7\rangle $, $L^4=\langle x_6,x_7\rangle $, $L^5=\langle x_7\rangle $, and $L^6=0$, but where the center has dimension $2$, namely $Z(L)=\langle x_5, x_7\rangle$.

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  • $\begingroup$ Thanks. I will need to think a bit to see this. Are the missing brackets assumed to be $0$, or does this in fact follow from the written ones? I also need to think a bit about why it is indecomposable (which seems mainly to need me to write things up). $\endgroup$ – Tobias Kildetoft Feb 18 '16 at 6:16
  • $\begingroup$ Right, the missing brackets are zero; the algebra is number $7.27$ in Magnin's classification of indecomposable complex nilpotent Lie algebras of dimension $7$. $\endgroup$ – Dietrich Burde Feb 18 '16 at 8:37

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