# Lie algebra of nilpotent Lie group

Let $G$ be a nilpotent closed subgroup of the group $UT(n,\mathbb{R})$ of upper triangular matrices with $1$-s on the diagonal.

I can construct a Lie algebra from $G$ in two ways:

1. The usual Lie algebra of a matrix Lie group consisting of matrices $A\in M_n(\mathbb{R})$ for which $\exp(tA) \in G$.

2. I can take the lower central series of $G$, take a direct sum of the factors and define Lie brackets in the natural (grading respecting) way.

Are the two Lie Algebras isomorphic? If so, canonically? (I'm new to this business)

• The second version will not immediately give you a Lie algebra over a field like the first one does (only a Lie algebra over $\mathbb{Z}$). And if you don't do some tensoring to correct this, then any finite subgroup will supply a counter example. – Tobias Kildetoft Aug 10 '17 at 12:00
• I don't understand your second construction. What's the “natural (grading respecting) way” of defining Lie brackets that you mention? And where are you defining them? – José Carlos Santos Aug 10 '17 at 12:00
• @JoséCarlosSantos This is a standard construction for nilpotent groups. The central series gives a filtration with abelian quotients and one can take the associated graded module corresponding to that, which becomes a Lie algebra over $\mathbb{Z}$ via the commutator by definition of the series. – Tobias Kildetoft Aug 10 '17 at 12:02
• @TobiasKildetoft Thanks. I didn't know it. – José Carlos Santos Aug 10 '17 at 12:05
• @TobiasKildetoft: Won't all factors be isomporphic to $\mathbb{R}^n$? How can $G$ be nontrivial finite if it lives in $UT(n,\mathbb{R})$? – Pablso Swanson Aug 10 '17 at 12:58

No. If the Lie algebra (i.e. the first construction) is denoted $\mathfrak{g}$, then the second one will be isomorphic to the associated Carnot-graded Lie algebra $\mathrm{Car}(\mathfrak{g})$.
Since there exist finite-dimensional nilpotent real Lie algebras (in dimension $\ge 5$) $\mathfrak{g}$ that are not isomorphic to their associated Carnot-graded Lie algebra, and since any finite-dimensional nilpotent real Lie algebra is isomorphic to a subalgebra of $\mathfrak{ut}(n,\mathbf{R})$ for some $n$, you get a negative answer to your question (at least for $n$ large enough, probably for all $n\ge 4$ actually).