Optional: Local exponential charts

3.00 Let $U^{\prime }=U\cap 𝔤$ and $V^{\prime }=V\cap G$ . First, we note that $\exp$ does indeed go from $𝔤$ to $G$ by definition of $𝔤$ , and $\exp (U)=V$ , so $\exp (U^{\prime })\subset V^{\prime }$ .

4.30 So what is left to prove? The map $\exp :U\to V$ is invertible, hence injective, so its restriction ${\exp |}_{U^{\prime }}$ must also be injective. But is ${\exp |}_{U^{\prime }}$ surjective? In other words, is it clear that $\log (V^{\prime })\subset U^{\prime }$ ?

5.40 This figure shows what you might imagine going wrong (we will later show that this doesn't happen, at least if you shrink $U$ and $V$ ). It shows a cartoon of a subgroup $G\subset GL(n,𝐑)$ which "wraps back on itself" and gets very close to the identity but never quite gets there. You can now imagine that when you intersect with a very small $V$ (in red), $\exp (U^{\prime })$ (in blue) could end up missing this appendage of $G$ which wraps back towards the identity (in green), because to get to this appendage you have to exponentiate something very large. In other words, $\exp (U^{\prime })$ is the blue bit and $V^{\prime }$ is everything which is blue or green. In the end, we will show this doesn't happen, so this is a cartoon picture of something which doesn't happen. You therefore shouldn't be too annoyed if the picture doesn't make sense.

7.55 If surjectivity of ${\exp |}_{U^{\prime }}$ fails then there's an element $g\in V^{\prime }=G\cap V$ such that $g\notin \exp (U^{\prime })$ . We could try to fix this by shrinking $U$ and $V$ , but let's suppose that doesn't help us. This will mean there is a sequence $g_i\in V^{\prime }$ such that $g_i\to I$ and $g_i\notin \exp (U^{\prime })$ for all $i$ (you should imagine a sequence of matrices on the green appendage, tending to the origin in the picture).

9.30 Let's assume that there is such a sequence and aim to derive a contradiction.

1. Recall that $𝔤\subset 𝔤𝔩(n,𝐑)$ is a subspace. Pick a vector space complement $W$ for $𝔤$ , that is $𝔤\cap W=\{0\}$ and $𝔤+W=𝔤𝔩(n,𝐑)$ . 11.30 In the figure, $𝔤$ is supposed to be the horizontal axis (tangent to $G$ at the identity) and $W$ is supposed to be the vertical axis.

   12.10 I claim that the map $F:𝔤\oplus W\to GL(n,𝐑)$ defined by $F(v,w)=\exp (v)\exp (w)$ is locally invertible like $\exp$ , i.e. there exists a neighbourhood $N_1$ of $0\in 𝔤\oplus W$ and a neighbourhood $N_2$ of $I\in GL(n,𝐑)$ such that $F(N_1)=N_2$ and $F:N_1\to N_2$ is invertible. This is proved using the inverse function theorem, just like for $\exp$ (compute the derivative of $F$ at the zero map and show this derivative is invertible). I leave it as an exercise to fill in the details.

   14.35 Therefore, if $g_i$ is sufficiently close to the identity (which it is for large $i$ ), then $g_i=\exp (v_i)\exp (w_i)$ for some sequence $v_i\in 𝔤$ and $w_i\in W$ .

2. 15.20 Our sequence $g_i$ is not in $\exp (U^{\prime })$ , so $w_i\ne 0$ for all $i$ . In particular, we can divide $w_i$ by its matrix norm to get a matrix $w_i/\parallel w_i\parallel \in W$ with norm 1. Since the set of matrices with norm 1 is closed and bounded (compact), the sequence $w_i/\parallel w_i\parallel \in W$ converges to some matrix $w\in W$ with norm 1 (in particular, $w\ne 0$ ).

3. 17.55 We are going to prove that $w\in 𝔤$ ; this will give us a contradiction, as $w\in W$ and $W$ is a complement for $𝔤$ . For this, we need to show that $\exp (tw)\in G$ for all $t\in 𝐑$ .

   19.00 Fix $t$ . Consider $t/\parallel w_i\parallel$ and take its integer and fractional parts $t/\parallel w_i\parallel =n_i+{ϵ}_i$ , $n_i\in \mathbb{N}$ , $ϵ\in [0,1)$ . Since $g_i\to I$ as $i\to \mathrm{\infty }$ , we know that $w_i\to 0$ , so $\parallel w_i\parallel \to 0$ , hence $t/\parallel w_i\parallel \to \mathrm{\infty }$ (as $t$ is fixed) and hence $n_i\to \mathrm{\infty }$ .

4. 21.00 We want to show that $\exp (tw)\in G$ . First note that $\exp (w_i)\in G$ for all $i$ because $\exp (w_i)=\exp (-v_i)g_i$ and $\exp (-v_i)$ and $g_i$ are both in $G$ . Because $G$ is a group, ${(\exp (w_i))}^{n_i}\in G$ and ${(\exp (w_i))}^{n_i}=\exp (n_i{w}_i)$ because $w_i$ commutes with itself. The sequence $\exp (n_i{w}_i)=\exp (t{w}_i/\parallel w_i\parallel -{ϵ}_i{w}_i)$ converges to $\exp (tw)$ because $w_i/\parallel w_i\parallel \to w$ and ${ϵ}_i{w}_i\to 0$ .

   24.00 Because $G$ is topologically closed, this limit lives in $G$ , so $\exp (tw)\in G$ . This argument works for every $t$ , so we're done.