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Consider a compact, connected and simply connected Lie Group $G$, and two elements in the corresponding Lie algebra $X$ and $Y$. By successive action of exponential map you can get the following element in the Lie group $$e^{\alpha_1 X}e^{\alpha_2 Y}...e^{\alpha_{n-1} X}e^{\alpha_n Y} \in G. $$ Suppose $X$ and $Y$ can generate the whole Lie algebra, then the whole Lie group $G$ can be generated.

For example, consider $SU(2)$ and $X=i\sigma_x, Y=i\sigma_y$ (where $\sigma$ are Pauli matrices), then they can generate the whole $SU(2)$ by the famous parametrization of Euler angles.

The question is: what is the minimum number of $n$ for a general Lie group $G$? For example, for a $SU(2)$, the number is 3.

Thank you for your attention!

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    $\begingroup$ Wait, with your numbering Euler angles give the minimum $n$ for $\textrm{SO}(3)$ as two, not three: $A$ $=$ $e^{\psi J_3}e^{\theta J_1}e^{\phi J_3}$ $=$ $e^{\alpha_1 J_3}e^{\beta_1 J_1}e^{\alpha_2 J_3}$. $\endgroup$
    – Francois Ziegler
    Commented Apr 13, 2018 at 14:00
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    $\begingroup$ I understand the question as: what is the minimal $n$ such that $G$ is setwise product of $n$ 1-parameter subgroups? Clearly $\dim(G)$ is a lower bound. $\endgroup$
    – YCor
    Commented Apr 13, 2018 at 14:04
  • $\begingroup$ Thank you for your correction! Yes, I have changed the numbering so that it will cause no confusion, and YCor's comment is perfectly correct $\endgroup$
    – Weicheng Ye
    Commented Apr 13, 2018 at 14:05
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    $\begingroup$ Actually no, I don't think it's an equivalent restatement... $\endgroup$
    – YCor
    Commented Apr 13, 2018 at 14:06
  • $\begingroup$ Why? Do you mean that the order of $X$ and $Y$ will cause trouble $\endgroup$
    – Weicheng Ye
    Commented Apr 13, 2018 at 14:09

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