We now turn to the proof of the following theorem:
Let G be a topologically closed subgroup of big G L n R. Define little g to be the set of matrices X such that exp (t X) is in G for all t in R.
little g is a vector space.
X and Y in little g implies that X bracket Y is in little g.
little g is the tangent space to G at the identity.
exp from little g to big G is locally invertible.
We proved (3) in an earlier video; (4) is a bit technical so we will prove it in an optional video. We will now focus on (1) and (2). We will use the following theorem which we proved in the last video:
If gamma of s in G is a path such that gamma of zero equals the identity then d gamma by d s at zero is in little g.
To prove little g is a vector space, we need to show that:
if X and Y are in little g then X plus Y is in little g, and
if lambda is a real number and X is in little g then lambda X is in little g.
To prove (2), we need to show that:
if X and Y are in little g then X bracket Y is in little g.
In each case, we need to show that something lives in little g, so by our earlier theorem, it's enough to construct paths:
gamma 1 of s in G such that gamma 1 of zero equals the identity and d gamma 1 by d s at 0 equals X plus Y,
gamma 2 of s in G such that gamma 2 at zero equals the identity and d gamma 2 by d s at zero equals lambda X,
gamma 3 of s in G such that gamma 3 at zero equals the identity and d gamma 3 by d s at zero equals X bracket Y.
Since X and Y are in little g, we have paths exp of s X and exp of s Y in G and their product gamma 1 of s equals exp of s X times exp of s Y therefore also lives in G. This path gamma 1 satisfies the conditions we require because gamma 1 at zero equals exp 0 times exp 0 equals the identity and d gamma 1 by d s equals X (exp s X) times (exp s Y) plus (exp s X) times Y times (exp s Y) by the product rule, so d gamma 1 by d s at 0 equals X plus Y as required.
Next, we try gamma 2 of s equals exp of s lambda X. This has gamma 2 of zero equals exp 0 equals the identity and d gamma 2 by d s equals lambda X exp of s lambda X, so d gamma 2 by d s at zero equals lambda X as required.
For gamma 3 we try the following: gamma 3 of s equals (exp X root s) times (exp Y root s) times (exp of minus X root s) times (exp of minus Y root s) Certainly gamma 3 at zero equals the identity. We will evaluate gamma 3 bit-by-bit using the Baker-Campbell-Hausdorff formula: exp X root s times exp Y root s equals exp of ((X plus Y) times root s, plus (s over 2 times X bracket Y), plus terms of order s to the 3 halfths) Similarly, exp X root s times exp Y root s equals exp of (minus (X plus Y) times root s, plus (s over 2 times X bracket Y), plus terms of order s to the 3 halfths) Multiplying these two together (using BCH again): gamma 3 of s equals exp of ((X plus Y) times root s minus (X + Y) times root s, plus s times X bracket Y plus s times (X plus Y bracket X plus Y) plus terms of order s to the 3 halfths) Therefore gamma 3 of s equals exp of (s times X bracket Y plus higher order terms) If we differentiate with respect to s and set s = 0 then we just get d gamma 3 by d s equals (X bracket Y plus terms of order at least root s) times exp of (s X bracket Y plus terms of higher order) and so d gamma 3 by d s at 0 equals X bracket Y as required.
You might be worried that square root s is not differentiable at s = 0. However, because of the way we've multiplied these nondifferentiable functions together, the result gamma 3 is once-continuously differentiable (because s to the 3 halfths is differentiable) so our earlier theorem can be used (it only used first derivatives of \gamma). It's not smooth (infinitely differentiable) but that doesn't matter.