Consider a (finite-dimensional) real connected Lie group $G$ with Lie algebra $\frak{g}$. Take a generating set $\mathcal{G} = \{ X_1, \cdots X_n \} $ of $\frak{g}$, i.e. such that any element of $\frak{g}$ can be written as a linear combination of their nested commutators.
Any element in $g\in G$ may be written as a, possibly infinite, product of elements in the one-parameter subgroups of $\mathcal{G}$, $ g = e^{t_1 X_{i_1}} \cdots e^{t_k X_{i_k} } \cdots$.
For $\mathrm{SL}(2,\mathbb{R})$ all possible minimal generating sets have been characterized and the order of generation computed (this is the minimal number $N$ such that $\prod_{i=1}^N e^{t_i X_{i_N}}$ cover all of $G$). Namely, this number is finite (and not smaller than 4) as long as the generators are not all "compact". If we treat this group/algebra as a matrix group/algebra, this condition is equivalent to not all generators having imaginary eigenvalues.
Is there a condition on a Lie group $G$ which determines whether there exists a finite set of $\{X_i\}$ such that the "minimal sequences" necessary to obtain any $g\in G$ are finite, in the case that $G$ is connected semisimple (non-compact, otherwise I think this is a well known result)? Can one compute the minimal number of generators and corresponding "length of generation" $N$ for any such Lie group?
Assuming $G$ satisfies the above condition, are there some known necessary and/or sufficient conditions which determine whether a given set $\{ X_i \}$ will finitely generate $G$ in the meaning above?
What are some examples beyond $\operatorname{SL}(2,\mathbb{R})$? I'm especially interested in the simple classical Lie groups $\operatorname{SL}(n,\mathbb{C}), \operatorname{SO}(n,\mathbb{C}), \operatorname{Sp}(n,\mathbb{C})$.
Best Answer
Let $G$ be a connected simple Lie group with finite center. Let $S$ be a generating subset of $G$ (symmetric with $1$) such that $S^n$ has nonempty interior for some $n$ (this is automatic if $S$ is $\sigma$-compact, by the Baire theorem). In the paper "On lengths on semisimple groups" (J. Topol. Anal. 2009) I proved the following: either $G=S^N$ for some $n$, or $S^k$ has compact closure for all $k$.
(In the paper it is stated equivalently in terms of lengths: every locally bounded length is either bounded or proper. For the word length relative to $S$, "locally bounded" means that $S^n$ has nonempty interior for some $n$, bounded means $S^N=G$ for some $n$, proper means each $S^k$ has compact closure for each $k$.)
Let $X=(X_1,\dots,X_p)$ be a generating family of the Lie algebra, generating 1-parameter subgroups $$P_i=\{\exp(tX_i):t\in\mathbf{R}\}; \qquad S_X=\bigcup_{1\le i\le p}P_i.$$ Apply the above to $S=S_X$. Then we see that either $S^N=G$ for some $N$, or $S^k$ has compact closure for all $k$. In the second case, all $P_i$ have to have compact closure. In the second case, assuming $G$ non-compact, we deduce that for some $i$, $P_i$ has a noncompact closure.
When $G$ is semisimple with finite center, I think one can show something similar: namely, if $T_j$ are the noncompact simple factors of $G$ (view them as quotients of $G$), then for each $j$ there exists $i$ such that the projection of $\exp(\mathbf{R}X_i)$ to $T_j$ has noncompact closure. (I guess this follows from Proposition 1.6 of the above reference.)
I'm not sure what happens beyond these (important) cases.