Timeline for are all simply connected nilpotent lie groups matrix groups over $\mathbb{R}$?

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May 7, 2017 at 10:18	comment	added	YCor		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).
May 7, 2017 at 10:17	comment	added	YCor		@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.
May 7, 2017 at 9:53	comment	added	Mariano Suárez-Álvarez		@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
May 7, 2017 at 9:39	comment	added	YCor		It's a quite trivial elaboration of Ado's theorem that the Lie algebra has a faithful nilpotent representation.
May 7, 2017 at 7:47	comment	added	Qiaochu Yuan		I don't see how the claim follows from Ado's theorem. Can you elaborate?
May 7, 2017 at 1:55	history	answered	Igor Rivin	CC BY-SA 3.0