Is every simply-connected nilpotent Lie group algebraic

Question

Say a Lie group is a matrix group if it is a closed Lie subgroup of some general linear group. Say a Lie group is algebraic if it is the group of $\mathbb R$-points of a real linear algebraic group. Clearly every algebraic Lie group is a matrix group, but the converse is false: the simplest counter-example I can think of is the group $\text{GL}_n^+$ of positive-determinant $n\times n$ invertible matrices.

By Ado's theorem and the Lie algebra correspondence, every simply-connected nilpotent Lie group is a matrix group. But one could ask for a stronger property, that such groups are algebraic. Is this true? In other words, is every simply-connected nilpotent Lie group algebraic?

Answer 1

I believe the answer is positive, and that it can be deduced from Theorem 3.6.3 in Varadarajan's book "Lie Groups, Lie Algebras, and Their Representations" which I'll state in a more concise version:

If $\mathfrak{g}$ is a Lie subalgebra of $\mathfrak{gl}_n(\mathbb{R})$ consisting entriely of nilpotent endomorphisms, then the corresponding Lie subgroup of $GL_n(\mathbb{R})$ is algebraic.

Thus for your question, if $G$ is a simply connected nilpotent Lie group then as you mentioned it is a matrix group, but it is moreover a unipotent matrix group. Thus $\mathfrak{g}:=Lie(G)$ satisfies the condition of the theorem so that $G$ is algebraic.

I didn't fully look into the proof but I believe that this algebraicity is rooted in the fact that the Baker–Campbell–Hausdorff series is finite for nilpotent groups and thus polynomial.