# Construction of the simply connected Lie group of a given Lie algebra

Given a finite dimensional real Lie algebra $\mathfrak{g}$, I am trying to obtain a concrete realization of its simply connected Lie group $G$, with $\mathrm{Lie}(G) \cong \mathfrak{g}$.

Let us assume that $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$, where $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are given by $\mathbb{R}^p$ and $\mathbb{R}^q$ respectively with certain commutator relations. I know in general how to obtain its simply connected Lie group $G$.

1. By a corollary to Ado's Theorem one can conclude that $$\mathfrak{g} \cong \mathfrak{g}_0 \subseteq \mathfrak{gl}(n,\mathbb{R})$$

2. One can find a connected Lie subgroup $G_0 \subseteq \mathrm{GL}(n,\mathbb{R})$ that has $\mathfrak{g}_0$ as its Lie algebra

3. The universal covering group of $G_0$, denoted by $G$, gives $\mathrm{Lie}(G) \cong \mathrm{Lie}(G_0) \cong \mathfrak{g}_0 \cong \mathfrak{g}$.

I would appreciate if anyone could provide me with an example by using this method, especially since all points include rather abstract realization of the regarding spaces (item 1 needs the universal enveloping algebra of a given Lie algebra, in item 3 one has to construct the simply connected covering group).

How is this procedure achieved in practice? Does one guess the right simply connected Lie group or does one apply the described procedure. If yes, where can I see in detail how this method works, for a rather a simple example. Some sort of enlightenment would definitely help. Many thanks in advance

• In the special case of a nilpotent Lie algebra; Does one know where I can look up the fact, that for a nilpotent Lie algebra the exponential map is a diffeomorphism onto its simply connected Lie group? Group multiplication is then given by the Baker Campell Hausdorff formula. Furthermore what is the exponential of a semidirect product of Lie algebras? – varsop Aug 21 '14 at 14:32
• This might seem helpful in the case of a nilpotent Lie algebra: (Local-Global Theorem for Nilpotent Lie Groups) books.google.de/… – varsop Aug 21 '14 at 15:09
• There is a direct construction which avoids Ado's theorem in Duistermaat and Kolk's Lie Groups; see section 1.14. It has the conceptual advantage of being manifestly functorial. – Qiaochu Yuan Aug 22 '14 at 18:55

## 1 Answer

The Lie functor from Lie groups to Lie algebras can be inverted on the subcategory of simply connected Lie groups. The essential surjectivity of the functor is called the third fundamental theorem of Lie, namely, every Lie algebra of finite dimension over the real or complex numbers is isomorphic to the Lie algebra of a Lie group. There are at least two ways to prove this. We could use either Ado's theorem, or the Levi-Malcev decomposition $\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{m}$, the semiidrect sum decomposition into a solvable ideal $\mathfrak{m}$ and a semisimple subalgebra $\mathfrak{s}$.

How are the steps 1.,2., 3. in practise ? If the Lie algebra has trivial center, then the adjoint representation is faithful, and we do not need Ado's theorem anymore. This fact, and other refinements of Ado's proof help to find a faithful linear representation. If the Lie algebra is nilpotent, then the adjoint representation is never faithful, so that we have to do more. This is often not difficult, and well known, e.g., for the Heisenberg Lie algebras of dimension $2n+1$ etc.