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Let $\mathfrak{g}$ be a Lie algebra, if $\mathfrak{a},\mathfrak{b}$ are subspaces of $\mathfrak{g}$ we define: $$[\mathfrak{a},\mathfrak{b}]=\mathrm{span}\{[a,b]:a\in\mathfrak{a},b\in\mathfrak{b}\}. \tag{1}$$ The lower central series of $\mathfrak{g}$ is the sequence of ideals defined by: $$\mathfrak{g}^0=\mathfrak{g},$$ $$\mathfrak{g}^i=[\mathfrak{g},\mathfrak{g}^{i-1}], \forall i\in \mathbb{N}.$$ On this MathWorld page, I find that if $\mathfrak{g}$ if finite dimensional, the lower central series of $\mathfrak{g}$ is stable. I think that stable means that: $\exists k\in\mathbb{N}:\mathfrak{g}^n=\mathfrak{g}^k, \forall k\geq n$. Why $\mathrm{dim}(\mathfrak{g})<\infty$ implies that the lower central series is stable? I know that: $\mathfrak{g}^{i+1}\subset\mathfrak{g}^i, \forall i$.

Xander Henderson
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inoc
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  • Hint: Consider the dimensions of each subspace in the series. – Tobias Kildetoft Sep 29 '19 at 08:14
  • @TobiasKildetoft, so i have to find the dimension of the space in (1) in the question, can you show me how to proceed please? – inoc Sep 29 '19 at 08:20
  • The lower central series certainly stabilises, but the question is, does it end with the trivial Lie algebra $0$ or not? In case it does, $\mathfrak{g}$ is called nilpotent. Consider the Heisenberg Lie algebra. – Dietrich Burde Sep 29 '19 at 08:38
  • @DietrichBurde, why the lower central series certainly stabilizes? I have yet to study nilpotent Lie algebra and i never meet the Heisenberg Lie algebra. – inoc Sep 29 '19 at 08:46
  • It stabilises because of the arguments given already (have a look at the answer). It's a good idea to look up the Heisenberg Lie algebra. The more you do yourself before posting, the better! Then you can ask the really interesting questions for you. Examples are the best way to to test yourself (before posting). – Dietrich Burde Sep 29 '19 at 08:53

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Seems like if the dimension is not infinite, there is no way for there to be an infinite subset of nested subspaces that doesn't accumulate at some $\mathcal{g}^i$. At each iteration, either the dimension decreases, (in which case all future elements of the lower central series have lower dimensions), or it doesn't, in which case no change is made, since finite dimensional vector spaces of the same dimension are isomorphic.

Kevin Johnston
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  • If exists an index i such that g^{i+1} and g^i have the same dimension then the lower central series is stable. But if for all i the dimension of g^{i+1} is strictly less the dimension of g^i, why the lower central series is yet stable? – inoc Sep 29 '19 at 09:14