Little g as a Lie algebra

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

• if $X,Y\in 𝔤$ then $X+Y\in 𝔤$ , and

• if $\lambda \in 𝐑$ and $X\in 𝔤$ then $\lambda X\in 𝔤$ .

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

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

3.05 Since $X,Y\in 𝔤$ , 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.

5.05 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.

5.50 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: $$\exp (X\sqrt{s})\exp (Y\sqrt{s})=\exp \left((X+Y)\sqrt{s}+\frac{s}{2}[X,Y]+𝒪(s^{3/2})\right)$$ 7.25 Similarly, $$\exp (-X\sqrt{s})\exp (-Y\sqrt{s})=\exp \left(-(X+Y)\sqrt{s}+\frac{s}{2}[X,Y]+𝒪(s^{3/2})\right)$$ Multiplying these two together (using BCH again): $${\gamma }_3(s)=\exp \left((X+Y)\sqrt{s}-(X+Y)\sqrt{s}+s[X,Y]-s[X+Y,X+Y]+𝒪(s^{3/2})\right)$$ 9.30 Therefore $${\gamma }_3(s)=\exp \left(s[X,Y]+𝒪(s^{3/2})\right).$$ If we differentiate with respect to $s$ and set $s=0$ then we just get $$\frac{d{\gamma }_3}{ds}=([X,Y]+𝒪(\sqrt{s}))\exp \left(s[X,Y]+𝒪(s^{3/2})\right)$$ and so $\frac{d{\gamma }_3}{ds}(0)=[X,Y]$ as required.