1:59 AM
@ACuriousMind Trying to prove "a simply connected abelian Lie group is isomorphic to $\Bbb R^k$"
I'm thinking that $\exp:\mathfrak g\to G$ is surjective
oh...maybe it's a homomorphism, too
Now, we know that $\Bbb R^k$ is its own Lie algebra
And certainly $\Bbb R^k\cong \mathfrak g$ for some $k$.
so $G=\exp \mathfrak g=\exp \Bbb R^k=\Bbb R^k$
I never used simple connectedness
actually, maybe you get $\exp \Bbb R^k=\Bbb R^n\times T^l$
I know you can get a torus somehow
yes, apparently $\exp:\mathfrak g\to G$ is a surjective Lie group homomorphism
we also know that $\mathrm{dim}\,\mathfrak g=\mathrm{dim}\, G=:n$.
we can write $\mathfrak g=\mathfrak g_1\oplus\cdots\oplus \mathfrak g_n$ for any 1-dim subspaces.
So it's clear that we have $$\exp\mathfrak g=\exp\mathfrak g_1\times\cdots\times\exp\mathfrak g_n$$
but the only abelian Lie groups obtained in this manner are $\Bbb R$ and $S^1$
simply connected then eliminates the $S^1$ factors
we invoke the classification of 1-manifolds to get the $\Bbb R$ and $S^1$ thing
@ACuriousMind That proof is probably wrong
I guess I should prove that $\exp \mathfrak g=G$.
Ok, $\mathfrak g$ is abelian iff $G$ is, so $\exp$ is a homomorphism by...BCH.
BCH is probably too strong there but whatever
clearly $\exp$ is a Lie group homomorphism
so we can use some theorems about that stuff
Given connected Lie groups $G,H$ and a homomorphism $F:G\to H$, if the induced map $F_*:\mathfrak g\to\mathfrak h$ is an isomorphism, $F$ is surjective
Clearly $\mathrm{Lie}\,\mathfrak g=\mathfrak g$ and by definition $\mathrm{Lie}\,G=\mathfrak g$
and we know that $\exp_*=\mathrm{id}$
the identity is an isomorphism, so $\exp$ is surjective