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Mathworld just says that the lower central series terminates: $\mathfrak{g}_1= [ \mathfrak{g}, \mathfrak{g}]$, $\mathfrak{g}_2= [ \mathfrak{g}, \mathfrak{g}_1]$ and $\mathfrak{g}_n= [ \mathfrak{g}, \mathfrak{g}_{n-1}]$ the best example I could think of is something like:

$$ \left( \begin{array}{cccc} 1 & \mathbb{R} & \mathbb{R} & \mathbb{R} \\ 0 & 1 & 0 & \mathbb{R} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

Another question on here says all simply connected nilpotent lie groups are matrix groups. It was phrased in a difficult way, so I am asking it again.

For example, another arrangement of $\mathbb{R}$'s that I can think of (at least in a $4 \times 4$ matrix) could be:

$$ \left( \begin{array}{cccc} 1 & 0 & \mathbb{R} & \mathbb{R} \\ 0 & 1 & \mathbb{R} & \mathbb{R} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

and I think there is only one more in the $5 \times 5$ case:

$$ \left( \begin{array}{ccccc} 1 & \mathbb{R} & \mathbb{R} & \mathbb{R} & \mathbb{R} \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right)$$

So in other words, I am claiming (almost surely false) an identification with nilpotent lie algebras and Ferrers boards (integer partitions

RRRR

RRR
R

RR
RR

This is surely wrong. Is there a classification of nilpotent Lie groups? Are all nilpotent Lie Groups matix groups?

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    $\begingroup$ The classification of nilpotent Lie algebras is wild in dimension something like $7$ and higher. $\endgroup$
    – Qiaochu Yuan
    Commented May 7, 2017 at 7:47
  • $\begingroup$ Not sure "the classification is wild" is correctly defined in this case. $\endgroup$
    – YCor
    Commented May 7, 2017 at 9:36

2 Answers 2

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A connected Lie group admits a faithful finite dimensional representation (i.e. is a matrix group) just when it is a semidirect product $S \rtimes R$ of a connected and simply connected solvable Lie group $S$ and a connected linearly reductive Lie group $R$. See Hilgert and Neeb, Structure and Geometry of Lie Groups, p. 595, Theorem 16.2.7. (This book is very well written and detailed.) In particular, since nilpotent Lie groups are solvable, simply connected nilpotent Lie groups are matrix groups.There is no classification of nilpotent Lie groups, or even of nilpotent Lie algebras. As Igor Rivin points out, there are nonlinear nilpotent Lie groups.

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  • $\begingroup$ I have never heard of this reference, thanks! $\endgroup$
    – Igor Rivin
    Commented May 7, 2017 at 14:13
  • $\begingroup$ I am searching for examples of connected locally compact group $G = N \rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. I tried to find such examples in the above reference, but couldn't find any. Do you know any such group? Thanks $\endgroup$
    – Mambo
    Commented Dec 13, 2018 at 3:34
  • $\begingroup$ If I understand your question, perhaps you could take $H$ to be the group of diagonal invertible $3 \times 3$ matrices, modulo rescaling by multiples of the identity matrix, and $N$ to be the group of strictly upper triangular $3 \times 3$ matrices. Then $H$ acts on $N$ by conjugation (as automorphisms of a nonabelian nilpotent Lie algebra), and the only fixed point is the identity element. $\endgroup$
    – Ben McKay
    Commented Dec 13, 2018 at 8:02
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That simply connected nilpotent groups are matrix groups follows from Ado's Theorem. In general nilpotent Lie groups might not be linear, see the answer to this question.

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    $\begingroup$ I don't see how the claim follows from Ado's theorem. Can you elaborate? $\endgroup$
    – Qiaochu Yuan
    Commented May 7, 2017 at 7:47
  • $\begingroup$ It's a quite trivial elaboration of Ado's theorem that the Lie algebra has a faithful nilpotent representation. $\endgroup$
    – YCor
    Commented May 7, 2017 at 9:39
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    $\begingroup$ @Ycor, while it may be "a quite trivial elaboration", saying so in no way helps anyone. Qiaochu said that he does not see it — surely elaborating on how this is an elaboration of Ado's theorem would be immensely more useful for everyone $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 7, 2017 at 9:53
  • $\begingroup$ @MarianoSuárez-Álvarez Here's it. Consider a Lie algebra over a field of char. 0. First (1) there's a nilpotent rep. with kernel the derived subalgebra $W$. By Ado's theorem, there's a faithful rep. It splits into a direct sum of (scalar+nilpotent) reps. Vanishing the scalars, we get a modified rep. But since $W$ already acted nilpotently, its action is unchanged and remains faithful in the new action. Taking the sum with (1) yields a faithful nilpotent rep. $\endgroup$
    – YCor
    Commented May 7, 2017 at 10:17
  • $\begingroup$ But anyway I'd look into the original paper by Ado, because certainly he explicitly did this (and the proof itself certainly yields a nilpotent rep). $\endgroup$
    – YCor
    Commented May 7, 2017 at 10:18

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