# Poor Phrasing of a Lie Group Theorem

1. Mar 21, 2014

### Mandelbroth

I found what might be the worst written book on Lie Groups. Ever. Until I find one I like better, I'm going to see if I can persevere through the sludge. I'll write out the theorem word for word and then explain what I can. Hopefully someone can decipher it.

Typically, I use the term "chart" to mean the inverse of what the author uses here. Also, in case it wasn't clear from the title, $G$ is a Lie group.

What does this mean, explicitly using the chart (preferably using $\varphi_\alpha^{-1}$ for his $\varphi_\alpha$)?

Last edited: Mar 21, 2014
2. Mar 23, 2014

### Mandelbroth

For those of you who might be looking for this in the future, I think I've figured out what the author is trying to say.

Pretend $G$ is an abelian Lie group. Then, the multiplication map $\mu:G\times G\to G, (a,b)\mapsto ab$ is a homomorphism of Lie groups. Consider $\mu_1=\left.\mu\right|_{\{\mathrm{e}\}\times G}$. Identifying $\{\mathrm{e}\}\times G\cong G$, we note that $\mu_1$ is the identity. Thus, evaluating the pushforward of $0\oplus y$ by $\mu_1$, we get $\mathrm{d}(\mu_1)_{(\mathrm{e},\mathrm{e})}(0\oplus y)=y$. Similarly, for $\mu_2=\left.\mu\right|_{G\times\{\mathrm{e}\}}$, we get $\mathrm{d}(\mu_2)_{(\mathrm{e},\mathrm{e})}(x\oplus 0)=x$. By linearity, we get our result: $$\mathrm{d}\mu_{(\mathrm{e},\mathrm{e})}(x\oplus y)=\mathrm{d}\mu_{(\mathrm{e},\mathrm{e})}(x\oplus 0 + 0\oplus y)=x+y.$$

This is used, notably, to prove that $\exp$ is a homomorphism of groups (with an appropriate forgetful functor slapped onto the domain) if and only if the underlying Lie group is abelian.