# Why a nilpotent Lie group must be a matrix group?

The question may be a little naive (or even appear as a duplicate) as I guess the result is well known. I saw on the other thread that

" c) A solvable Lie group G is linear iff its commutator subgroup G′ is closed, and G′ has no non-trivial compact subgroup (Thm 3.2 in Chap. XVIII) "

But I cannot find the reference book anywhere. May I ask someone to give a hint why this is true?

Background:

Try to find a non-trivial example of infranil manifold which is not a matrix Lie group. Then the speaker visiting my university claiming every nilpotent Lie group must be a matrix Lie group. Since nilpotent is defined entirely algebraically, I feel there should be a proof based on algebraic techniques. But I do not know how to prove the above statement.

• Simply connected nilpotent Lie groups are linear; $\begin{pmatrix}1&\mathbf R&\mathbf R\\0&1&\mathbf R\\0&0&1\end{pmatrix}/\begin{pmatrix}1&0&\mathbf Z\\0&1&0\\0&0&1\end{pmatrix}$ isn't. See Ado's theorem. – Francois Ziegler Oct 3 '14 at 2:24
• @FrancoisZiegler: I see. I always thought Ado's theorem only tell us the Lie algebra is linear, and the Lie group could have covers which are not linear. I did not thought about this example. – Bombyx mori Oct 3 '14 at 2:27
• I still need to think carefully why your example would not be linear (is it a variation of Heisenberg group?) But thanks! – Bombyx mori Oct 3 '14 at 2:28
• Yes, a variation due to Garrett Birkhoff, Lie groups simply isomorphic with no linear group (1936). – Francois Ziegler Oct 3 '14 at 2:31
• @FrancoisZiegler: Thanks! I guess this is enough food for thought. I am glad I asked it at here and did not took the statement blindly. – Bombyx mori Oct 3 '14 at 2:34

Another reference which was not yet mentioned, I think, is the article of M. Moskowitz, "Faithful Representations and a local property of Lie groups", Math. Z. $143$, 1975. There the question is discussed when all analytic groups with a given Lie algebra $\mathfrak{g}$ have a faithful linear representation. He proves among other things the following result:
Theorem All connected Lie groups $G$ with Lie algebra $\mathfrak{g}$ have a faithful linear representation if and only if $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$ and $\tilde{S}$ has a faithful representation.
Here $\tilde{G}=rad(\tilde{G})\cdot \tilde{S}$ is the Levi decomposition for the simply connected group $\tilde{G}$ with Lie algebra $\mathfrak{g}$.
Corollary Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If $ad(\mathfrak{g})$ has nontrivial center then there is a locally isomorphic group without any faithful representation.