# Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group?

This question is closely related to this one.

Ado's theorem states that given a finite-dimensional Lie algebra $\mathfrak g$, there exists a faithful representation $\rho\colon\mathfrak g \to \mathfrak{gl}(V)$, with $V$ a finite-dimensional vector space. In the real or complex case one can take the exponent of the image and obtain a (virtual) Lie subgroup $\exp\rho(\mathfrak g)$ in $GL(V)$ having Lie algebra $\rho(\mathfrak g)$. But nothing guarantees that this subgroup will be closed in $GL(V)$.

So the question is: is every finite-dimensional Lie algebra the Lie algebra of some closed linear Lie group? I am primarily interested in the real and complex case, but it might be interesting to ask what happens in the ultrametric case as well.

I think that the answer is yes. It looks like you can prove it by relying on a convenient proof of Ado's theorem.

Procesi's book, "Lie groups: an approach through invariants and representations", has the following theorem preceding the proof of Ado's theorem:

Theorem 2. Given a Lie algebra $L$ with semismiple part $A$, we can embed it into a new Lie algebra $L'$ with the following properties:

1. $L'$ has the same semismiple part $A$ as $L$.
2. The solvable radical of $L'$ is decomposed as $B' \oplus N'$, where $N'$ is the nilpotent radical of $L'$, $B'$ is an abelian Lie algebra acting by semisimple derivations, and $[A, B'] = 0$.
3. $A \oplus B'$ is a subalgebra and $L' = (A \oplus B') \ltimes N'$.

With all of that, the idea is to first prove the refinement of Ado's theorem for $L'$. We need a particular refinement: Let $\tilde{A}$ be the maximal algebraic semisimple Lie group with Lie algebra $A$, and let $\tilde{B'}$ and $\tilde{N'}$ be the contractible Lie groups with Lie algebras $B'$ and $N'$. If we can find a closed embedding of $(\tilde{A} \times \tilde{B'}) \ltimes \tilde{N'}$ in a matrix group, then it will restrict to a closed embedding of the Lie subgroup of the original $L$.

In the proof of Ado's theorem that follows, the action of $N'$ is nilpotent, so the representation of $\tilde{N'}$ is closed and faithful. The Lie algebra $L'$ has a representation which is trivial on $B'$ and $N'$ and generates $\tilde{A}$. It has another representation which is trivial on $N'$ and $A$ and for which the action of $B'$ is nilpotent. If I have not made a mistake, the direct sum of these three representations is the desired representation of $L'$.

• Greg, thank you for your answer! I didn't understand few things: 1. Why "then it will restrict to a closed embedding of the Lie subgroup of the original L."? 2. If we take direct sum of the three rep's, it looks like ~N' will normalize ~A x ~B', am I wrong? 3. Do we work over C here? It looks like the proof of Theorem 2 is given for complex case only. Is it crucial? – mathreader Dec 3 '09 at 4:46
• 1. $L$ generates a closed subgroup $\tilde{L}$ of $\tilde{L'}$, and closed within closed is closed. 2. I don't understand what you're saying. The first summand is the most important one; in this summand $\tilde{N'}$ certainly does not normalize the other factors. 3. For the same reason that you can pass to $L'$, you can also pass to the complexification, so it should still work over the reals. – Greg Kuperberg Dec 3 '09 at 4:54

The answer is yes. According to Ado's theorem, every Lie algebra can be realized as subalgebra of $$\mathfrak{gl}(V)$$ for some $$V$$. According to the subalgebras-subgroups theorem, every Lie subalgebra of $$\mathfrak{gl}(V)$$ is the Lie algebra of an immersed Lie subgroup of $$\mathrm{GL}(V)$$. Call a Lie group linear if it admits a faithful representation. We have shown that every Lie algebra is the Lie algebra of a linear Lie group. It is known that every linear Lie group $$G$$ admits a faithful representation whose image is closed.

Here's a sketch of the proof of the last statement.

Let $$r\colon G\rightarrow GL(V)$$ be a faithful representation of $$G$$. The derived group $$r(G^{\prime})$$ of $$r(G)$$ is closed $$GL(V)$$ and hence in $$r(G)$$. Hence $$G/G^{\prime}$$ is an abelian Lie group. Therefore there exists a representation $$s\colon G\rightarrow GL(W)$$ such that $$Ker(s)=G^{\prime}$$ and $$s(G)$$ is closed in $$GL(W)$$. Consider the representation $$G\rightarrow GL(V\oplus W)$$ such $$a\in G$$ acts on $$V$$ as $$r(a)$$ and on $$W$$ as $$s(a)$$. This is faithful, and one can show that its image is closed --- Proposition 5 of

Djokovic, D. Ž. A closure theorem for analytic subgroups of real Lie groups. Canad. Math. Bull. 19 (1976), no. 4, 435--439. MR0442147

• Could you explain why $r(G')$ is closed in $GL(V)$? – Nicolas Tholozan Mar 1 at 22:59
• This is Proposition 2 of Djokovic's paper --- it is difficult for me to summarize the paper further (which is quite short and fairly elementary). – user166831 Mar 2 at 18:56
• Thanks for the reference ! – Nicolas Tholozan Mar 3 at 20:03