Write every element of a nilpotent Lie subgroup as product of exponentials of simple generators I have a question about Lie groups. Let $G$ be a simply connected semi-simple complex Lie group and $\mathfrak{g}$ its Lie algebra. We consider a Cartan-Weyl basis of $\mathfrak{g}$, giving the usual decomposition $\mathfrak{g}=\mathfrak{h} \oplus \mathfrak{n}_+ \oplus \mathfrak{n}_-$. We denote $N$ the nilpotent subgroup corresponding to $\mathfrak{n}_+$ and $\alpha_1,...,\alpha_l$ the simple roots with the associated generators $E_1,...,E_n$.
Now, take any element $u \in N$, my question is the following : can we always write $u$ as
$$u = \exp\left(c_1 E_{i_1}\right) ... \exp\left(c_n E_{i_n}\right)$$
with $i_1,...,i_n \in \lbrace1,...,l\rbrace$ and $c_1,...,c_n \in \mathbb{C}^*$ ?
If it is possible, is there an algorithm to find such a form ? Do we have informations about the minimal number $n$ of exponentials needeed ?
It seems quite plausible to me but I can't find a demonstration or a reference about that. I've tried explicit computations for $\mathfrak{sl}_3$ (i.e. with two simple roots) : apparently, it works. Most of the time 3 exponentials are enough, for some particular cases we need 4 of them (and of course sometimes 1 or 2 suffice).
Does anyone have an idea, or a reference on the subject ?
Thank you,
Sylvain
 A: The following is meant as a comment. I just type it as an answer because it is so long.
There is something ambiguous here: if $G$ is a Lie group its actual Lie algebra  (often denoted $\mathfrak{g}_0$ to distinguish it from its complexification $\mathfrak{g}$) is a real vector space and need not have a Cartan-Weyl decomposition of the type you describe. We can pass to its complexification $\mathfrak{g}$ which contains the generators $E_i$ you talk about, but it is not obvious that we can choose $c_i$ such that the $c_i E_i$ do live in the real subspace $\mathfrak{g}_0$ where the exponential map is defined. In fact I believe we can choose $\mathfrak{h}$ in such a way that we always can find such $c_i$, but we have to be a bit careful in our choice of $\mathfrak{h}$. (Perhaps I'm mistaken and any choice of $\mathfrak{h}$ works.)
Anyway, having said that, let's turn to a concrete example: $G = SU(2)$. Now $\mathfrak{g}_0 = \mathfrak{su}_2$  (think of it as $\mathbb{R}^3$ with the vector product), $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$ and $\mathfrak{n}$ is one (complex) dimenional, so that it's intersection with $\mathfrak{g}_0$ is one (real) dimensional and its image under the exponential map is a circle. (A great circle on $S^3$ to which $SU_2$ is diffeomorphic because it can be viewed as the unit sphere in the quaternions. Actually now that I type this: if we identiy $SU(2)$ with a subset of the quaternions, the plane through 0 cutting out this great circle can be viewed as just the ordinary complex numbers and the exponential map sending $\mathfrak{n} \cap \mathfrak{g}_0$ to $\mathbb{C}$ is just the ordinary exponential sending $i \mathbb{R}$ to the unit circle.)
Clearly in this case the answer to your question is yes, but there is also something somewhat arbitrary in calling this circle $N$ since any great circle on $SU(2)$ could have been $N$ with the right choice of $\mathfrak{h}$. Of course this is not really a problem. Still it feels a a bit more naturall to me to work with the Iwasawa decomposition instead to define the group $N$.
A: Rather than exponentials, I prefer to talk about root subgroups (I usually work with algebraic reductive groups). "Linear algebraic groups" by Springer gives relevant information. Proposition 8.2.1 says that any $u\in N$ can be written as a product of one element from each positive root subgroup.
Now you want to know how to generate an arbitrary element in each root subgroup, or at least such an element times elements which belong to higher root subgroups (so you get a triangular recursion). For this you can use the fact that each positive root can be reached starting from a simple root, and then adding one simple root at a time, always staying in the root system (Bourbaki Lie Ch. VI, 1.6, Prop. 19), and Springer commutator relations (Springer 8.2.3), noting that in that formula you must have a nontrivial contribution in the root subgroup associated to $\alpha + \beta$, if it is a root.
So, in principle, it's possible. You can work out some bound on the number of exponentials from there (it's probably not optimal).
I'm not sure how practical this is, but let me make a remark. You ask if there is an algorithm but you did not specify in which form your element $u \in N$ in given. If you want an algorithm, it is important to be precise about the form of the input and of the output. You specified only the output. In any case, calculations are likely to be involved and you will certainly need to use a computer program like Magma to do it in practice (except in very small cases).
