Why for a lie algebra L is $\frac{L^{k}}{L^{k+1}}$ contained in $ Z(L/L^{k+1})$, where $L^{k}$ is the k-th term of the lower central series?

Asked Dec 08 '17 at 18:05

Active Dec 08 '17 at 18:05

Viewed 26 times

According to Erdmann and Wilson, pg.31, the lower central series of a Lie algebra $L$ is defined as:

$$L^{1} = [L,L] \quad L^{k} = [L,L^{k-1}]$$

They then claim that $\frac{L^{k}}{L^{k+1}}$ is contained in $Z\big(\frac{L}{L^{k+1}}\big)$. Why is this true?

As far as I can see:

$$ Z\bigg( \frac{L}{L^{k+1}}\bigg) = \bigg\{ x \in \frac{L}{L^{k+1}} : [x,y] = 0 \; \forall \, y \in \frac{L}{L^{k+1}} \bigg\} = \{\alpha \in L : [\alpha,\beta]+L^{k+1} = 0 \; \forall \, \beta \in L \} $$

$$ \frac{L^{k}}{L^{k+1}} = \bigg\{ \lambda + L^{k+1} : \lambda \in L^{k} \bigg\} = \big\{ [v,w] + L^{k+1} : v\in L , w \in L^{k-1} \big\} $$

How does this imply $ \frac{L^{k}}{L^{k+1}} \subseteq Z\big(\frac{L}{L^{k+1}}\big) $? Does it rely on the fact that the series is descending somehow?

asked Dec 08 '17 at 18:05

thesundayscientist

It's because $[L,L^k]\subseteq L^{k+1}$. – Angina Seng Dec 08 '17 at 18:38
Doesn't $L^{k+1} = [L,L^{k}]$ by definition- it's not possible for it to be a proper subset? – thesundayscientist Dec 08 '17 at 19:16
Indeed it does! But $A=B$ implies $A\subseteq B$, surely? – Angina Seng Dec 08 '17 at 19:17
Aha yes, sure - sorry I was confused because the series is descending, so in most cases it is true that $ L^{k+1} \subset L^{k}$ - i.e $ L^{n} \subseteq L^{n-1} \subseteq.... L^{1} \subseteq L$. I'm afraid I still don't understand why this helps us any however? – thesundayscientist Dec 09 '17 at 12:25
An element $a$ of a Lie algebra is in its centre if $[a,L]=0$. – Angina Seng Dec 09 '17 at 12:54

Why for a lie algebra L is $\frac{L^{k}}{L^{k+1}}$ contained in $ Z(L/L^{k+1})$, where $L^{k}$ is the k-th term of the lower central series?

0 Answers0