# 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 $GL(n,\mathbf{R})$ . Define $\mathfrak{g}=\{X\in\mathfrak{gl}(n,\mathbf{R})\ :\ \exp(tX)\in G\ \forall t\in% \mathbf{R}\}.$

1. $\mathfrak{g}$ is a vector space.

2. $X,Y\in\mathfrak{g}$ implies that $[X,Y]\in\mathfrak{g}$ .

3. $\mathfrak{g}$ is the tangent space to $G$ at the identity.

4. $\exp\colon\mathfrak{g}\to G$ 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 $\gamma(s)\in G$ is a path such that $\gamma(0)=I$ then $\frac{d\gamma}{ds}(0)\in\mathfrak{g}$ .

## Proof of the main theorem

Proof:

To prove $\mathfrak{g}$ is a vector space, we need to show that:

• if $X,Y\in\mathfrak{g}$ then $X+Y\in\mathfrak{g}$ , and

• if $\lambda\in\mathbf{R}$ and $X\in\mathfrak{g}$ then $\lambda X\in\mathfrak{g}$ .

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

• if $X,Y\in\mathfrak{g}$ then $[X,Y]\in\mathfrak{g}$ .

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

• $\gamma_{1}(s)\in G$ such that $\gamma_{1}(0)=I$ and $\frac{d\gamma_{1}}{ds}(0)=X+Y$ ,

• $\gamma_{2}(s)\in G$ such that $\gamma_{2}(0)=I$ and $\frac{d\gamma_{2}}{ds}(0)=\lambda X$ ,

• $\gamma_{3}(s)\in G$ such that $\gamma_{3}(0)=I$ and $\frac{d\gamma_{3}}{ds}(0)=[X,Y]$ .

• Since $X,Y\in\mathfrak{g}$ , we have paths $\exp(sX)$ and $\exp(sY)$ in $G$ and their product $\gamma_{1}(s)=\exp(sX)\exp(sY)$ therefore also lives in $G$ . This path $\gamma_{1}$ satisfies the conditions we require because $\gamma_{1}(0)=\exp(0)\exp(0)=I$ and $\frac{d\gamma_{1}}{ds}=X\exp(sX)\exp(sY)+\exp(sX)Y\exp(sY),$ by the product rule, so $\frac{d\gamma_{1}}{ds}(0)=X+Y$ as required.

Next, we try $\gamma_{2}(s)=\exp(s\lambda X)$ . This has $\gamma_{2}(0)=\exp(0)=I$ and $\frac{d\gamma_{2}}{ds}=\lambda X\exp(s\lambda X)$ , so $\frac{d\gamma_{2}}{ds}(0)=\lambda X$ as required.

For $\gamma_{3}$ we try the following: $\gamma_{3}(s)=\exp\left(X\sqrt{s}\right)\exp\left(Y\sqrt{s}\right)\exp\left(-X% \sqrt{s}\right)\exp\left(-Y\sqrt{s}\right).$ Certainly $\gamma_{3}(0)=I$ . We will evaluate $\gamma_{3}$ bit-by-bit using the Baker-Campbell-Hausdorff formula: $\exp(X\sqrt{s})\exp(Y\sqrt{s})=\exp\left((X+Y)\sqrt{s}+\frac{s}{2}[X,Y]+% \mathcal{O}(s^{3/2})\right)$ Similarly, $\exp(-X\sqrt{s})\exp(-Y\sqrt{s})=\exp\left(-(X+Y)\sqrt{s}+\frac{s}{2}[X,Y]+% \mathcal{O}(s^{3/2})\right)$ Multiplying these two together (using BCH again): $\gamma_{3}(s)=\exp\left((X+Y)\sqrt{s}-(X+Y)\sqrt{s}+s[X,Y]-s[X+Y,X+Y]+\mathcal% {O}(s^{3/2})\right)$ Therefore $\gamma_{3}(s)=\exp\left(s[X,Y]+\mathcal{O}(s^{3/2})\right).$ If we differentiate with respect to $s$ and set $s=0$ then we just get $\frac{d\gamma_{3}}{ds}=([X,Y]+\mathcal{O}(\sqrt{s}))\exp\left(s[X,Y]+\mathcal{% O}(s^{3/2})\right)$ and so $\frac{d\gamma_{3}}{ds}(0)=[X,Y]$ as required.

Remark:

You might be worried that $\sqrt{s}$ 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 $s^{3/2}$ 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.