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.

  • $\begingroup$ You need some assumption on the group for the exponential to be necessarily surjective. See e.g. en.wikipedia.org/wiki/… $\endgroup$ – Logan M Feb 19 '19 at 23:52
  • $\begingroup$ I’m not sure exactly what is the question. Maybe you can explain why you think exponentiating would be problematic? $\endgroup$ – ZeroTheHero Feb 19 '19 at 23:58
  • $\begingroup$ @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? $\endgroup$ – Antoni Parellada Feb 20 '19 at 0:04
  • $\begingroup$ They would have of course different generators so the exponential would give elements in a 1-parameter subgroup in $GL(n)$.... $\endgroup$ – ZeroTheHero Feb 20 '19 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 ${\rm Sp}(2n,{\mathbb R})$.

  • $\begingroup$ I got it from the upcoming sentence at this point on this online lecture. What do you think is the misunderstanding? $\endgroup$ – Antoni Parellada Feb 20 '19 at 0:07
  • $\begingroup$ I can't read the lecturer's mind. I think he is just plain wrong. The Wiki article on the Exponential map gives the conditions for the Expontial map to be surjective. $\endgroup$ – mike stone Feb 20 '19 at 0:22
  • 1
    $\begingroup$ Yeah that lecture is misleading. @mikestone highlights some problems with the symplectic case. The claim is true (AFAIK) for the unitaires and the orthogonals, which are the most common in particle physics, but otherwise handle with care. $\endgroup$ – ZeroTheHero Feb 20 '19 at 0:30
  • $\begingroup$ see math.stackexchange.com/questions/1089636/… for a partial answer applicable to connected compact Lie groups. $\endgroup$ – ZeroTheHero Feb 20 '19 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.