# Proof of Lie's theorem, how to obtain a Lie group homomorphism from a Lie algebra homomorphism

Part of Lie's theorem (I assume we are working over the field of complex numbers) states that if you have a Lie algebra homomorphism $$f: \mathfrak{g} \rightarrow \mathfrak{h}$$ between the Lie algebras of the path-connected Lie groups $$G$$ and $$H$$, with $$G$$ simply connected, then there is a unique Lie group homomorphism $$F:G \rightarrow H$$ whose differential at 1 is $$f$$.

In a neighbourhood of the identity the exponential map is a diffeomorphism, and we can define, for $$X$$ in this neighbourhood $$F(e^X)= e^{f(X)}$$ (actually we are forced to define it like that, as there is a 1-1 correspondence between vectors in the Lie algebra and 1-parameter subgroups, which simply associates to the 1-parameter subgroup the speed to the curve at 1, and the curves $$F(e^{tX})$$, $$e^{tf(X)}$$ have both speed $$f(X)$$, once I show $$F$$ exists).

To define $$F$$ we use the fact that for a connected Lie group, a neighbourhood of the identity generates the whole group, and if $$g \in G$$ is written as $$g= u_1^{m_1} \cdots u_k^{m_k}$$ then I define $$F(g) = F(u_1)^{m_1} \cdots F(u_k)^{m_k}$$. To finish the proof we will need to show that the definition does not depend on the way we expressed $$g$$ i.e. that if $$g= v_1^{l_1}\cdots v_s^{l_s}$$ then $$F(v_1)^{l_1}\cdots F(v_s)^{l_s}$$ coincides with our previous definition. For this I guess we need to use the simply connecteness of $$G$$.

I don't know how if this would be a valid proof or how to show that the definition of $$F$$ is independent of the representation of $$g$$, can some one take it form here or point me out to a reference where this approach is used?

• Lie's theorem states that if $V$ is a finite dimensional vector space over an algebraically closed field of characteristic $0$, then for any solvable Lie algebra of endomorphisms of $V$ there is a vector that is an eigenvector for every element of the Lie algebra. – Dietrich Burde Oct 31 '18 at 10:03
• @DietrichBurde yes, that's another theorem of Lie, but I think there is no a priori relation with the proof I am asking. – inquisitor Oct 31 '18 at 12:04
• No, I was not referring to the proof, but only to the name. Sometimes the result you are interested in, has the name "Lie's three Theorems", whereas "Lie's Theorem" alone usually is about triangularizable Lie algebras. But, I guess, this is not so clear. – Dietrich Burde Oct 31 '18 at 12:35

Indeed, since you are only left to show the independence of how to represent $$g$$, the proof will carry over almost 1:1!
You can use path-conectness of the Lie Group (which holds since $$G$$ is a smooth Manifold) to join any element $$g \in G$$ by a path $$\gamma:[0,1] \to G$$ with $$\gamma(0)=1$$. Now, you use a partition of the path, say, $$t_0<... with $$\gamma(t_i)\gamma(t_{i+1})^{-1} \in U$$ where $$U$$ is a neighbourhood of $$1$$ such that we can define our homomorphism on $$U$$. The first step is to prove independence of this partition.
If we are done, the ''hard'' part begins: We have to show independence of the chosen path, which (naturally) uses the simple connectness of $$G$$. Taking any other path joining $$1$$ to $$g$$, we can find a homotopy with fixed endpoints that deformes the two path into each other. The next idea is to ''slice'' the homotopy, by which I mean that you use paths ''close enough'' to each other and then ''jump'' from one path to another until you reached your final path. Indeed we want them so close, that they are almost the same, and that the area in which they are not can be omitted. This will guarantee that we can proceed that way, showing that the value of our sought homomorphism is indeed independend of the choice of path.