# How do the infinitesimal generators generate the whole compact Lie group?

Suppose that $G$ is a connected compact Lie group, $g(a)\in G$ is its element, and $a=(a_1,a_2,...,a_r)$ is the parameter of the group element $g(a)$. Then the infinitesimal generator of the group can be defined as $$X_i=\frac{\partial{g(a)}}{\partial{a_i}}\bigg|_{a=0}$$ Then an infinitesimal element $g(\delta a)=E+\sum_i \delta a_i X_i$, in which $\delta a\in O({\bf0})$, $E=g({\bf0})$ is the identity element and $O({\bf0})$ is the neighborhood of the ${\bf 0}$ point in parameter space. We know that the parameter of the product of two infinitesimal elements is the sum of the parameters of the two infinitesimal elements (to the first order of the infinitesimal parameter), that is $$g(\delta a)g(\delta b)=(E+\sum_i \delta a_i X_i)(E+\sum_i \delta b_i X_i)=E+\sum_i (\delta a_i+\delta b_i) X_i=g(\delta a+\delta b). \tag{1}$$ Then, in every book I have read, it says that, any finite element $g(a)$ can be generated by $\{X_i\}$: let $\delta a=a/N$, \begin{align} g(a)=g(N\frac{a}{N})=g(N\delta a)&=g(\delta a+\delta a+...+\delta a)\\ &=*=g(\delta a)g(\delta a)...g(\delta a)=g(\delta a)^N\\ &=(E+\sum_i\delta a_iX_i)^N=(E+\frac{\sum_i a_iX_i}{N})^N\\ &=\exp(\sum_i a_iX_i)~~[\text{let } N\rightarrow \infty] \tag{2} \end{align} In the above proof, the key step is the $(=*=)$ step, which uses the relation in Eq.(1) $g(\delta a+\delta b)=g(\delta a)g(\delta b)$. But this relation is ONLY valid for infinitesimal elements and between two infinitesimal elements, it can not be used to derive the $(=*=)$ step directly because both $a$ is finite and there are $N$ elements here.

My questions are:

1. Since the ($=*=$) step is wrong, is the conclusion Eq.(2) also wrong? (not limited to this proof)
2. If the the conclusion Eq.(2) is right, how to prove it in a right way?

I think I have a counterexample. Using the same logic as the above proof, we consider a scalar function $f(x)$ with single parameter. It has the property $f(0)=1$. Then the "infinitesimal generator" of $f(x)$ is just its derivative at 0, let it be $$t=\frac{\partial f(x)}{\partial x}\bigg|_{x=0}=f'(0)$$ Then for any infinitesimal parameter $\delta x$ and $\delta y$, to the first order, we have $f(\delta x)=f(0)+f'(0)\delta x=1+\delta x\cdot t$ and $f(\delta y)=1+\delta y\cdot t$, and $$f(\delta x)f(\delta y)=(1+\delta x\cdot t)(1+\delta y\cdot t)=1+(\delta x+\delta y)t=f(\delta x+\delta y)$$ Then, if the logic in Eq.(2) is right, we can also derive the conclusion that: "For any finite $x$, there is the relation $f(x)=\exp(tx)$". But this conclusion is obviously wrong! Because here $f(x)$ can be any function satisfying $f(0)=1$, e.g. $f(x)=1+\sin(x)$. Obviously, $1+\sin(x)\ne\exp(tx)$ !

Only when $f(x)$ satisfies $f(x+y)=f(x)f(y)$ for any $x$ and $y$ (not limited to infinitesimal parameters), we can derive $f(x)=\exp(tx)$.

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It does sound like you should find a better book. Are your groups already matrix groups? Otherwise it is difficult to think of a universe, where we could raise the $X_i$s to arbitrary powers. – Jyrki Lahtonen Sep 14 '13 at 18:34
In fact, I major in physics, and nearly all books I read are something like "group theory in physics". The above proof generally appears with SO(n) or SU(n) groups. But the proof seems to be universal. – goodluck Sep 17 '13 at 0:57
So the fact you're talking about is the surjectivity of the exponential map [from Lie algebra to the (connected component of 1 of the) Lie group], right? I don't think there is an easy proof — for one thing, this map is (in general) not surjective for non-compact groups! – Grigory M Sep 19 '13 at 20:52
– Grigory M Sep 19 '13 at 20:58
What do you mean by "the parameter of a group element"? What do you mean by "finite element"? I recommend that you pick up Brian Hall's book "Lie groups, Lie algebras, and representation theory." It will help. – Matt Hogancamp Jan 16 '14 at 14:04