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I want some example of nilpotent lie algebras, here I also want to see how the set of all $n\times n$ matrices $(a_{ij})$ where $a_{ij} = 0\ \forall\ i\ge j$ forms a nilpotent lie algebra under the lie multiplication $[AB]=AB-BA$.

I can visualize that a matrix of that form when raised to power $n$ gives $0$ matrix. are they lie group? I guess not because as they are nilpotent matrix they are closed subset of $M_n(R)$ right? hence they are not manifold hence not a lie group?

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up vote 3 down vote accepted

They do indeed form a manifold, homeomorphic to ${\mathbb R}^{n(n-1)/2}$. Indeed any finite-dimensional vector space over $\mathbb R$ is a manifold. Whatever gave you the idea that a manifold can't be a closed subset?

But the reason this Lie algebra is nilpotent is not that $A^n = 0$, but rather that $[A,[A,\ldots,[A,B]\ldots]] = 0$.

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Actually, a Lie algebra consisting of nilpotent endomorphisms of some vector space is nilpotent, so using that the matrices in question are nilpotent is also a fine argument, though it does require one to use an extra result. – Tobias Kildetoft Sep 27 '12 at 8:32

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