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Suppose we have a Lie algebra $L$ over $\textbf{k}$ with basis $\{x,y,z\}$ and with

$$[x,y]=z, [y,z]=x, [z,x]=y.$$

How do I go about finding the lower central series and the derived series for $L$?

Edit: I have the definitions of the lower central series and the derived series, namely:

The Lower Central Series: Let $L$ be a Lie algebra over $\textbf{k}$ and set $L^1:=L$. Define recursively $L^{n+1}:=[L,L^n]$ for all $n \geq 1$. Then $L^{n+1}=[L,L^n] \subseteq L^n$ for all $n$. The lower central series of $L$ is the descending chain of ideals

$$L=L^1 \supseteq L^2\cdots\supseteq L^n \supseteq \cdots$$

The Derived Series: Let $L$ be a Lie algebra over $\textbf{k}$ and set $L^{(0)}:=L$. Define recursively $L^{(k+1)}:=[L^{(k)},L^{(k)}]$ for all $k \in \mathbb{Z}^\geq$. Then $L^{(k+1)} = [L^{(k)},L^{(k)}] \subseteq [L,L^{(k)}] \subseteq L^{(k)}$ for all $k$. The derived series of $L$ is the descending chain of ideals

$$L=L^{(0)} \supseteq L^{(1)} \supseteq \cdots \supseteq L^{(n)} \supseteq \cdots$$

My problem is that I really don't understand how to actually apply these definitions to a given Lie algebra, any advice would be greatly appreciated.

Edit 2: Suppose now we have a Lie algebra $L$ over a field $\mathbb{k}$ with basis $\{h,u,v,w,z\}$ such that

$$[h,u]=v, [h,v]=w, [h,w]=0, [u,v]=z, [w,x]=[z,x]=0 \,\,\,\,\,\,(\forall x \in L)$$

Am I right to begin finding the lower central series in the following manner?

Firstly set $L=L^1$, we then have that

\begin{align*} L^2&=[L,L] \\ &=\mathrm{span}\{[a,b] \, | \, a,b \in L\}\\ &=\mathrm{span}\{[h,u], [h,v], [h,w], [u,v], [w,x], [z,x]\}\\ &=\mathrm{span}\{v,w,0,z,0,0\} \\ &=\mathbb{k}v \oplus \mathbb{k}w \oplus \mathbb{k}z. \end{align*}

We would then have that

\begin{align} L^3 &= [L, L^2] \\ &=\mathrm{span}\{[a,b] \, | \, a \in L, b \in \mathbb{k}v \oplus \mathbb{k}w \oplus \mathbb{k}z\} \end{align}

I'm confused as to how to proceed from this point. Could anyone give me any pointers?

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  • $\begingroup$ What's the problem? Do you know definitions? $\endgroup$ – Adam Baranowski Apr 5 '15 at 19:06
  • $\begingroup$ Sorry, i've added the definitions I have. As I say in the edit, my problem is that I don't understand how to begin to apply the definitions to, for example, the Lie algebra I have given above. $\endgroup$ – Jamie3213 Apr 5 '15 at 20:26
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Since the given Lie algebra is perfect, i.e., satisfies $$ L=L^{(1)}=[L,L], $$ the derived series and the lower central series are constant to $L$. This is consistent with the definition of a solvable respectively nilpotent Lie algebra. A perfect Lie algebra is never solvable or nilpotent.

Indeed, $L^{(1)}$ contains $x=[y,z]$, $y=[z,x]$ and $z=[x,y]$, hence a basis $(x,y,z)$ of the three-dimensional vector space of $L$. Clearly then $L$ is perfect.

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  • $\begingroup$ Thank you, i've edited my post, would you be able to offer any advice? $\endgroup$ – Jamie3213 Apr 7 '15 at 18:20
  • $\begingroup$ Yes, your new Lie algebra $L_1$ satisfies $[L_1,L_1]=L$ so that $L_1^{(k)}=L^{(k-1)}=L$ for all $k\ge 2$. $\endgroup$ – Dietrich Burde Apr 7 '15 at 18:38
  • $\begingroup$ Say we're considering the second Lie algebra only, am I using the right approach to find $L^3$? $\endgroup$ – Jamie3213 Apr 7 '15 at 18:49

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