Lie groups generated by finitely many Lie algebra elements Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A standard fact is that $G$ is generated by $\exp(\mathfrak{g})$, i.e. every $g \in G$ can be written as $g=\exp(x_1)\cdots\exp(x_n)$ for some $x_i\in\mathfrak{g}$. A natural follow-up question is whether there is a bound on the number $n$ of Lie algebra elements. I suppose this depends on the group, so let me formulate my question as follows.
Question. What are natural conditions on a Lie group $G$ which imply that there is a finite number $n$ such that
$$G = \underbrace{\exp(\mathfrak{g}) \cdots \exp(\mathfrak{g})}_{n \text{ times}}?$$
For instance, is this true for complex semisimple groups?
Of course, there are many cases in which $\exp$ is surjective (e.g. $G$ compact or nilpotent) so that $n = 1$.
I'm looking for a larger class of groups, preferably in the linear algebraic category, or just interesting specific examples where $n > 1$.
 A: $\def\fg{\mathfrak{g}}$The following answer is a paraphrase of material from
Philips, Christopher N., How many exponentials?, Am. J. Math. 116, No. 6, 1513-1543 (1994). ZBL0839.46054.
We'll say that a Lie group $G$ has exponential rank $\leq k$ if every element of $G$ is a product of $\leq k$ exponentials.
We'll say that $G$ has exponential rank $\leq k+\epsilon$ if the set of products of elements $\leq k$ exponents is dense in $G$. In other words, in a group of exponential rank $\leq k+\epsilon$, for any open neighborhood $\Omega$ of the identity, we have $G = \exp(g_1) \exp(g_2) \cdots \exp(g_k) u$ for $g_1$, $g_2$, ..., $g_k\in\fg$ and $u \in \Omega$. Since it is known that $\exp(\fg)$ contains an open neighborhood of the identity, any group of exponential rank $\leq k+\epsilon$ also has exponential rank $\leq k+1$. We define the exponential rank of $G$ to be the minimum $r$ in  $\{ 1 < 1+\epsilon < 2 < 2+\epsilon < 3 < \cdots \}$ such that $G$ has exponential rank $\leq r$.
Note that, if $G = AB$ (for Lie subgroups $A$ and $B$) and the exponential ranks of $A$ and $B$ are $(a,b)$, $(a+\epsilon, b)$, $(a, b+\epsilon)$ or $(a+\epsilon, b+\epsilon)$, then the exponential rank of $G$ is bounded above by $a+b$, $a+b+\epsilon$, $a+b+\epsilon$, $a+b+\epsilon$ respectively.
In this vocabulary, Phillips proves (Section 5) that

*

*Connected compact groups have exponential rank $\leq 1$.


*Unipotent groups have exponential rank $\leq 1$.


*Connected complex groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved, because $SL_2(\mathbb{C})$ has exponential rank $1+\epsilon$ (the matrix $\left[ \begin{smallmatrix} -1&1 \\ 0&-1 \end{smallmatrix} \right]$ is not an exponential)).


*Connected solvable groups have exponential rank $\leq 1+\epsilon$. This bound cannot be improved: Consider the Borel $\left[ \begin{smallmatrix} z & u \\ 0 & z^{-1} \\ \end{smallmatrix} \right]$ inside $SL_2(\mathbb{C})$ and the same matrix as before.


*All connected Lie groups have rank $\leq 2+\epsilon$.


*$SL_2(\mathbb{R})$ has exponential rank $2$.
The proof that all connected Lie groups have rank $\leq 2+\epsilon$ is credited to Djokovic: The point is that any connected $G$ has an Iwasawa decomposition $K(ANR)$ where $K$ is compact connected (so exponential rank $\leq 1$) and $ANR$ is connected solvable (so exponential rank $\leq 1+\epsilon$).
Phillips states it as an open problem whether there is a connected Lie group of exponential rank $2+\epsilon$, or whether, in fact, exponential rank $2$ is always enough.
