Little g as a Lie algebra

Recap of theorem

We now turn to the proof of the following theorem:

Theorem:

Let G be a topologically closed subgroup of GL(n,𝐑) . Define 𝔤={X∈𝔤𝔩(n,𝐑):exp(tX)∈G∀t∈𝐑}.

  1. 𝔤 is a vector space.

  2. X,Y∈𝔤 implies that [X,Y]∈𝔤 .

  3. 𝔤 is the tangent space to G at the identity.

  4. exp:𝔤→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:

Theorem:

If γ(s)∈G is a path such that γ(0)=I then dγds(0)∈𝔤 .

Proof of the main theorem

Proof:

To prove 𝔤 is a vector space, we need to show that:

  • if X,Y∈𝔤 then X+Y∈𝔤 , and

  • if λ∈𝐑 and X∈𝔤 then λX∈𝔤 .

To prove (2), we need to show that:

  • if X,Y∈𝔤 then [X,Y]∈𝔤 .

    • In each case, we need to show that something lives in 𝔤 , so by our earlier theorem, it's enough to construct paths:

  • γ1(s)∈G such that γ1(0)=I and dγ1ds(0)=X+Y ,

  • γ2(s)∈G such that γ2(0)=I and dγ2ds(0)=λX ,

  • γ3(s)∈G such that γ3(0)=I and dγ3ds(0)=[X,Y] .

    • Since X,Y∈𝔤 , we have paths exp(sX) and exp(sY) in G and their product γ1(s)=exp(sX)exp(sY) therefore also lives in G . This path γ1 satisfies the conditions we require because γ1(0)=exp(0)exp(0)=I and dγ1ds=Xexp(sX)exp(sY)+exp(sX)Yexp(sY),

    by the product rule, so dγ1ds(0)=X+Y as required.

    Next, we try γ2(s)=exp(sλX) . This has γ2(0)=exp(0)=I and dγ2ds=λXexp(sλX) , so dγ2ds(0)=λX as required.

    For γ3 we try the following: γ3(s)=exp(X√s)exp(Y√s)exp(-X√s)exp(-Y√s).

    Certainly γ3(0)=I . We will evaluate γ3 bit-by-bit using the Baker-Campbell-Hausdorff formula: exp(X√s)exp(Y√s)=exp((X+Y)√s+s2[X,Y]+𝒪(s3/2))
    Similarly, exp(-X√s)exp(-Y√s)=exp(-(X+Y)√s+s2[X,Y]+𝒪(s3/2))
    Multiplying these two together (using BCH again): γ3(s)=exp((X+Y)√s-(X+Y)√s+s[X,Y]-s[X+Y,X+Y]+𝒪(s3/2))
    Therefore γ3(s)=exp(s[X,Y]+𝒪(s3/2)).
    If we differentiate with respect to s and set s=0 then we just get dγ3ds=([X,Y]+𝒪(√s))exp(s[X,Y]+𝒪(s3/2))
    and so dγ3ds(0)=[X,Y] as required.

    Remark:

    You might be worried that √s is not differentiable at s=0 . However, because of the way we've multiplied these nondifferentiable functions together, the result γ3 is once-continuously differentiable (because s3/2 is differentiable) so our earlier theorem can be used (it only used first derivatives of γ ). It's not smooth (infinitely differentiable) but that doesn't matter.