# How can the exponential generator apply to all Lie groups (not just rotation)?

How can it be shown that any element of a Lie group can be represented as $$A=e^{ig_A V^A}$$?

I think this results from the exponential map.

In the case of $$SO(3)$$ it can be shown through the Taylor expansion of the matrix components of, for instance, the rotation around the $$x$$ axis

\begin{align}& R_x(\theta)=\begin{bmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{bmatrix}= \small\begin{bmatrix}1&0&0\\0&\small{1-\frac {\theta ^2}{2!}+\frac {\theta^4 }{4!}-\cdots}&-\small{\theta+\frac {\theta ^3}{3!}-\frac {\theta^5 }{5!}+\cdots}\\0&\small{\theta-\frac {\theta ^3}{3!}+\frac {\theta^5 }{5!}-\cdots}&\small{1-\frac {\theta ^2}{2!}+\frac {\theta^4 }{4!}-\cdots}\end{bmatrix} \end{align}

compared to

$$R_x(\theta)=e^{ig_{Rx}\theta}=I +ig_{Rx}\theta-\frac{\theta^2}{2!}g^2_{Rx}+\frac{\theta^3}{3!}i^3g^3_{Rx}-\cdots$$

that $$g_{Rx}=\begin{bmatrix}0&0&0\\0&0&i\\0&-i&0\end{bmatrix}.$$

But the statement seems to apply to any Lie group, not just rotations, such as in the case of $$SO(3),$$ as stated, for instance, at this point on a lecture at at Colorado School of Mines.

• You need some assumption on the group for the exponential to be necessarily surjective. See e.g. en.wikipedia.org/wiki/… Commented Feb 19, 2019 at 23:52
• I’m not sure exactly what is the question. Maybe you can explain why you think exponentiating would be problematic? Commented Feb 19, 2019 at 23:58
• @ZeroTheHero If you take a look at the expansions above re: SO(3), it is clear why we are talking about rotations given the matrix $\tiny \begin{bmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{bmatrix},$ but what about other Lie groups, such as the general linear group, which I think encompasses any linear basis transformations? Commented Feb 20, 2019 at 0:04
• They would have of course different generators so the exponential would give elements in a 1-parameter subgroup in $GL(n)$.... Commented Feb 20, 2019 at 0:06

It's not even true. Many Lie groups are not of "exponential type," an example being the symplectic group $${\rm Sp}(2n,{\mathbb R})$$ that describes Gaussian optics. Even in the part of the group that is connected to the origin there are elements that do not lie in any one parameter subgroup of the form $${\rm Exp}(it V)$$. See the Wkipedia article on the symplectic group, in particular the section on $${\rm Sp}(2n,{\mathbb R})$$.