# A direc sum of nilpotent Lie algebras such that $C^n(\mathfrak{g}) \subsetneq \mathfrak{z}(\mathfrak{g})$

It is known that the center of nilpotent Lie algebra is never trivial as it always contain $$C^n(\mathfrak{g})$$ if $$\mathfrak{g}$$ is nilpotent of class $$n+1$$

Let $$C^n$$ denote descending central series. I am looking for an example of direct sum of 2 nilpotent algebras $$\mathfrak{g}=\mathfrak{g_1}\oplus\mathfrak{g_2}$$ such that $$\mathfrak{g}$$ is nilpotent of class $$n+1$$ and $$C^n(\mathfrak{g}) \subsetneq \mathfrak{z}(\mathfrak{g})$$.

What I did:

I thought about the simplest nilpotent algebra I know: the strictly upper triangular matrices with entries in $$\Bbb R$$: $$\mathfrak{g}=\mathfrak{b_n}\oplus \mathfrak{b_n}$$ (notation $$\mathfrak{b_n}$$ here to denote strictly upper triangular matrices is an abuse).

I found for $$n=2$$ that $$\mathfrak{z}(\mathfrak{b_2}) = \mathfrak{b_2}$$ but I have a feeling it is true for every $$n>2$$. If this is the case, $$\mathfrak{z}(\mathfrak{g})=\mathfrak{z}(\mathfrak{b_n})\oplus \mathfrak{z}(\mathfrak{b_n})=\mathfrak{b_n}\oplus \mathfrak{b_n}$$ and $$C^n(\mathfrak{g}) \neq \mathfrak{g}$$ so $$C^n(\mathfrak{g}) \subsetneq \mathfrak{z}(\mathfrak{g})$$.

Is that correct? Are there other interesting examples?

Thank you for your help.

• Since $L$ is nilpotent, $C^n(L)=0$ for some $n\ge 1$. Hence $C^n(L)\subset Z(L)$ for this $n$? – Dietrich Burde May 10 at 9:25
• @DietrichBurde Sorry my question is unclear. Actually $C^{n+1}(\mathfrak{g})=0$ and $C^{n}(\mathfrak{g})\ne0$. I fixed that, thank you! – PerelMan May 10 at 10:06
• I'm not sure what really is the question. The easiest recipe to produce a nilpotent Lie algebra in which the last nonzero term of the lower central series is strictly contained in the center, is to take the direct product of two nonzero nilpotent Lie algebras of distinct nilpotency class; the smallest example being (3-dim Heisenberg)$\times$(1-dimensional). Then one can wonder about directly indecomposable examples, and Dietrich provides one. – YCor May 11 at 15:29

## 1 Answer

Here is an example of an indecomposable nilpotent Lie algebra $$L$$ of dimension $$7$$ with $$C^5(L)=0$$ but $$C^4(L)\subsetneq Z(L)$$. Here $$C^1(L)=L$$ and $$[L,C^k(L)]=C^{k+1}(L)$$. Consider the Lie brackets with respect to a basis $$(x_1,\ldots ,x_7)$$ given by $$[x_1,x_2]=x_4,\; [x_1,x_4]=x_5,\;[x_1,x_5]=x_7,\; [x_2,x_3]=x_7,\; [x_2,x_4]=x_6.$$ Then $$Z(L)=\langle x_6,x_7\rangle$$, but $$C^4(L)=\langle x_7\rangle$$ and $$C^5(L)=0$$.

Now take the direct sum $$L\oplus L$$.

• Interesting example thank you! I wonder if such algebra has a name? Did you construct it by hand? – PerelMan May 10 at 12:38
• Yes, it has the name $\mathfrak{g}_{7,2.8}$ of Magnin's classification. I had it ready at my hand for some other question, so I took it. – Dietrich Burde May 10 at 12:41
• Actually in the list of 6-dim. Lie algebras, four already work: $\mathfrak{g}_{6,i}$ in Magnin's notation for $i\in\{4,6,7,14\}$, which are $L_{6,19}(0)$, $L_{6,25}$, $L_{6,23}$, and $L_{6,21}(0)$ in de Graaf's notation. All have 2-dimensional center and last term of the central series of dimension 1. The first 3 have class 3, the last has class 4. For instance, $\mathfrak{g}_{6,4}$ has the basis $(x_i)_{1\le i\le 6}$ with nonzero brackets $[x_1,x_2]=x_4$, $[x_2,x_3]=x_5$, $[x_2,x_4]=x_6$; the center has basis $(x_5,x_6)$ while the third term in the LCS has basis $(x_6)$. Six is optimal. – YCor May 11 at 15:53