Standard representative for a coset space Motivation:
finding convenient real coordinates for the full complex flag manifold $ \mathrm{Flag} \left( 1, \ldots, n \right) $.
Details:
Consider the Lie group $ \mathrm{SU} \left( n \right) $ and its maximal torus $ \mathrm{T} := 
\left( \mathrm{U} \left( 1 \right) \times \ldots \times \mathrm{U} \left( 1 \right) \right) \cap \mathrm{SU} \left( n \right) $ - i.e. the subgroup of diagonal matrices. We know that the complete flag manifold may be described as a coset space:
\begin{equation}
\mathrm{Flag} \left( 1, \ldots, n \right) = \mathrm{SU} \left( n \right) / \mathrm{T}  .
\end{equation}
Now, using the physicists' convention where the exponential map is defined by $ A \mapsto e^{iA} $, the Lie algebra $ \mathfrak{su} \left( n \right) $ consists of traceless Hermitian matrices, and the torus is generated by the subalgebra $ \mathfrak{h} $ (a Cartan subalgebra of the complexified algebra), which consists of diagonal traceless Hermitian matrices. For example, for $ n = 2 $ we have that $ \mathfrak{su} \left( 2 \right) = \mathrm{Span}_{\mathbb{R}} \left\{ \sigma_x, \sigma_y , \sigma_z \right\} $ (the Pauli matrices) and $ \mathfrak{h} = \mathrm{Span}_{\mathbb{R}} \left\{ \sigma_z \right\} $.
Now, I read on lecture notes for some physics course that one can parametrize the coset space using what they call a standard representative. The groups involved were different, and no proof was given, but basically what it implies here is that one can parametrize the coset space using representatives of the form:
\begin{equation}
U \left(x, y \right) := e^{i \left( x \sigma_x + y\sigma_y \right)}
\end{equation}
That is, any $ V \in \mathrm{SU} \left( 2 \right) $ may be written in the form:
$$ V = e^{i \left( x \sigma_x + y\sigma_y \right)} e^{i z \sigma_z} \tag{1}\label{1} $$
for some $ x,y,z \in \mathbb{R} $. Thus, we obtain very convenient intrinsic real coordinates for $ \mathrm{Flag} \left( 1, 2 \right) $. Now, generally the RHS of \eqref{1} should be computed using the Baker-Campbell-Hausdorff formula. However, for the case of $ \mathrm{SU} \left( 2 \right) $ the exponentials may be computed explicitly, so I was able to convince myself that this indeed works. My question is regarding $ n > 2 $.
Question: Let us label the generators of $ \mathfrak{su} \left( n \right) $ as follows: $ \left\{ H_1, \ldots, H_{n-1}, A_1, \ldots A_{n \left(n-1\right)} \right\} $, such that $ \left\{ H_1, \ldots, H_{n-1} \right\} $ are diagonal. Is it true that any matrix $ V \in \mathrm{SU} \left( n \right) $ may be written as a product of the following form?
$$ V = \exp \left(i \sum_j x_j A_j \right) \exp \left(i \sum_k y_k H_k\right) , \quad x_j, y_k \in \mathbb{R} . $$
 A: This is actually much more general.
Let $G$ denote any compact Lie group and $H\subseteq G$ a closed connected subgroup.  Let's pick a bi-invariant Riemannian metric on $G$.  Then, we obtain an orthogonal splitting $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{p}$, where $\mathfrak{p}$ is merely a vector subspaces (i.e., it's not necessarily a sub-Lie algebra.)
The theorem is:

Any $g\in G$ is of the form $\exp(Y)\exp(X)$ for some $X\in \mathfrak{h}$ and $Y\in \mathfrak{p}$.

Here's a sketch of a proof.  Given $g\in G$, let $\pi(g)$ denote the image of $g$ in the homogeneous space $G/H$.  Let $\gamma:[0,1]\rightarrow G/H$ be a minimal geodesic from $eH$ to $\pi(g)$ (where the metric on $G/H$ is induced from the chosen one on $G$ and where $e\in G$ is the identity element).  Then $\gamma$ has a unique horizontal lift to a geodesic $\overline{\gamma}$ in $G$ which begins at $e\in G$.
Being horizontal means that $\overline{\gamma}'(0)\in \mathfrak{p}$.  Since with respect to a bi-invariant metric, geodesics are the same as one parameter subgroups, it follows that $\overline{\gamma}(t) = \exp(tY)$ for some $Y\in \mathfrak{p}$.
Now, since $\pi\circ \overline{\gamma} = \gamma$, we see that $\pi(g) = \pi(\overline{\gamma}(1)) = \pi(\exp(Y))$, so there is an $h\in H$ with $g = \exp(Y)h$.  Since $H$ is connected and compact, we may write $h = \exp(X)$.  $\square$
