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I am a physicist doing some research and I come across with the following Lie algebra problem.

Consider the Lie Group $G$ (compact and connected, if you wish), and two generators in the corresponding Lie algebra $X$ and $Y$. By successive action of exponential map you can get the following element in the Lie group $$e^{\alpha_1 X}e^{\beta_1 Y}...e^{\alpha_n X}e^{\beta_n Y} \in G. $$

The question is: when will the whole Lie group be generated by the action above? And if some part of the Lie group cannot be generated, what is the subgroup that can be generated?

Extensions: What is the closure of the generated subgroup? Can you extend the above results to 3 or more generators?

Example: (1) Consider $SU(2)$ and $X=i\sigma_x, Y=i\sigma_y$ ($\sigma$ are Pauli matrices), then they can generate the whole $SU(2)$.

(2) Consider $S^1\times S^1$ as a Lie group and $X=Y=i(a,b)$ (in the naturally chosen coordinate system). Of course only a one dimensional subgroup can be generated. However, if $\frac{a}{b}$ is irrational, the closure is the whole group.

Thank you for your attention!

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Nice question. For $SU(2)$ or $SO(3)$ you get I think the Euler's angles, and there are analogous results for $SU(N)$. I don't know about the general case, it does look plausible.

Certainly the closure of the group generated by those exponential is the connected component of $e$. Either you use Cartan's theorem on closed subgroup, or use basic facts like $$\exp(X+Y) = \lim_{n\to\infty} (\exp\frac{1}{n}X\exp\frac{1}{n}Y)^n\\ \exp[X,Y]= \lim_{n\to\infty} (\exp \frac{1}{n}X\exp\frac{1}{n}Y\exp(-\frac{1}{n}X) \exp(-\frac{1}{n}Y) )^{n^2}$$

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  • $\begingroup$ Thank you for your answer! However, I do not think your second paragraph is correct. I think my second example illustrates that it has complicated dependence on the Lie algebra, which manifests itself in whether $a/b$ is rational or not (my intuition is that the first problem is a standard problem, but I have no clue). If you do not like two generators being the same, just take two copies of $S^1\times S^1$. $\endgroup$ – Weicheng Ye Apr 8 '18 at 19:57
  • $\begingroup$ @WeichengYe: I don't see how in your case $X$, $Y$ generate the Lie algebra of $S^1 \times S^1$. Their bracket is $0$. $\endgroup$ – orangeskid Apr 8 '18 at 20:09
  • $\begingroup$ Of course, but the closure of the generated subgroup is the whole group $\endgroup$ – Weicheng Ye Apr 9 '18 at 3:07
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About the main problem, I think I know the answer, which is just the sub-Lie group corresponding to the sub-Lie algebra generated by $X$ and $Y$.

The two-line argument is the following: It is obvious that these elements form a (path-connected) subgroup of the Lie group $G$, then according to the theorem by Yamabe,1950, the subgroup just has to be a Lie subgroup. The reference is

Yamabe, H. (1950). On an arcwise connected subgroup of a Lie group. Osaka Mathematical Journal, 2(1), 13-14.

(A comment: The paper uses the phrase "arcwise connected", which is supposed to refer to path-connectedness. However, I do think it uses local path-connectedness, which is fortunately not an issue in my situation since we have some elements in the Lie algebra in the first place.)

It is a two-step proof. First we need to show that the subalgebra generated by $X$ and $Y$ appear as those elements in the Lie algebra that are tangent to the subgroup. Then we need to show that the subgroup is homeomorphic to the Lie group corresponding to the generated sub-Lie algebra.

I think the question about the closure of the subgroup is still open.

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