A specific coset decomposition of $\mathrm{GL}_n(\mathbb{C})$ Disclaimer: I am a theoretical chemist (not a mathematician). I have tried asking this question at Math SE with no luck (https://math.stackexchange.com/questions/4080696/a-specific-coset-decomposition-of-mathrmgl-n-mathbbc).
I am reading an old paper [1] where they introduce (without proof) a specific decomposition of a unitary matrix. I would like (if possible) to generalize this decomposition to complex, invertible matrices. The claim I would like to prove goes as follows:
Let $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ be an element of the general linear group (i.e. an $n \times n$, complex, invertible matrix). Then $\mathbf{M}$ can be written as
\begin{equation} \tag{1}
   \mathbf{M} = \exp(\mathbf{m}) = \exp(\mathbf{m'}) \exp(\mathbf{m''})
\end{equation}
where $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ is an element of the general linear Lie algebra (i.e. an $n \times n$, complex matrix). The matrices $\mathbf{m}'$ and $\mathbf{m}''$ are $n \times n$, complex, block-matrices of the form
\begin{align}
   \mathbf{m}' &= 
   \begin{bmatrix}
   \mathbf{0}'_{00} & \mathbf{m}'_{01} & \mathbf{m}'_{02} \\
   \mathbf{m}'_{10} & \mathbf{0}'_{11} & \mathbf{m}'_{12} \\
   \mathbf{m}'_{20} & \mathbf{m}'_{21} & \mathbf{0}'_{22}
   \end{bmatrix}\tag{2} \\
   \mathbf{m}'' &=
   \begin{bmatrix}
   \mathbf{m}''_{00} & \mathbf{0}''_{01} & \mathbf{0}''_{02} \\
   \mathbf{0}''_{10} & \mathbf{m}''_{11} & \mathbf{0}''_{12} \\
   \mathbf{0}''_{20} & \mathbf{0}''_{21} & \mathbf{m}''_{22}
   \end{bmatrix} \tag{3}.
\end{align}
The diagonal blocks are square and of matching dimensions ($\mathbf{0}'_{00}$ has the same dimensions as $\mathbf{m}''_{00}$, say $n_0 \times n_0$, and so on). In essence, $\mathbf{m}'$ has zero blocks on the diagonal while $\mathbf{m}''$ is block-diagonal. I realize that the number of blocks is not essential for the problem; I'm using three by three blocks for illustration.
I know that the exponential map of the general linear group is surjective, meaning that for every $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ there exists some $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ such that $\mathbf{M} = \exp(\mathbf{m})$. This is standard group theory. I can also see that block-diagonal matrices of the form (3) form a Lie algebra, say $\mathfrak{b}_n(\mathbb{C})$, which generates a Lie group of $n \times n$, complex, invertible, block-diagonal matrices, say $\mathrm{B}_n(\mathbb{C})$, which is a subgroup of $\mathrm{GL}_n(\mathbb{C})$. The paper suggests that the factorisation in (1) should be viewed as a coset decomposition. As far as I understand, the group $\mathrm{GL}_n(\mathbb{C})$ is the union of the left cosets
\begin{equation}
   g \, \mathrm{B}_n(\mathbb{C}) = \{ g \, h \;|\; h \in \mathrm{B}_n(\mathbb{C}) \},
   \quad g \in \mathrm{GL}_n(\mathbb{C}).
\end{equation}
I also know that some cosets may be identical so one doesn't need all cosets in order get the whole group (so to speak). What I don't understand is how to prove or disprove the specific form of the matrix $\mathbf{m}'$. Following a comment on my original question on Math SE, we could ask more generally if the general linear Lie group is a product of exponentials of a Lie subalgebra and the complementary vector space of that subalgebra.
Any help or references would be much appreciated.
[1] J. Linderberg and Y. Öhrn, Int. J. Quantum Chem. 12(1), 161–191 (1977). State vectors and propagators in many-electron theory. A unified approach.
EDIT: Any ideas for the less general unitary case are also very welcome.
 A: Let
$$\mathfrak g = \left\{ \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} \mid a,b \in \mathbb C\right\}, \qquad \qquad \mathfrak p = \left\{ \begin{bmatrix} 0 & b \\c & 0\end{bmatrix} \mid b,c \in \mathbb C\right\}.$$
Then for $x = \begin{bmatrix} 0 & b \\ c & 0 \end{bmatrix} \in \mathfrak p$,
$$\exp(x) = 1 + \begin{bmatrix} 0 & b \\ c & 0 \end{bmatrix} + \frac{1}{2} \begin{bmatrix} bc & 0 \\ 0 & bc \end{bmatrix} + ...$$
Note that every odd power of $x$ is zero on the diagonal, while every even power of $x$ has equal diagonal elements. Hence, the diagonal entries of $exp(x)$ are equal.
This shows that if $y \in \mathfrak g$, then the two diagonal entries of $\exp(x)\exp(y)$ are either both zero or both nonzero. But there are $A \in GL_2(\mathbb C)$ not of this form, for instance
$$ A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}.$$
Thus $A$ (for example) is not of the form $\exp(x)\exp(y)$ for $(x,y) \in \mathfrak g \oplus \mathfrak p$. This is at least a counterexample for $(m_1,m_2,m_3) = (1,1,0)$.
A: For the simplest case, some exploration with Maple tells me that
$$ \begin{pmatrix} p & q \\ r & s \end{pmatrix} = 
   \exp \begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix}
   \exp \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}
$$
where
$$ \begin{array}{rl}
 2a &= \ln(ps-qr) + \ln(p) - \ln(s) \\
 2d &= \ln(ps-qr) - \ln(p) + \ln(s) \\
 2b &= \sqrt{\frac{pq}{rs}}\ln\left(\frac{\sqrt{ps}+\sqrt{qr}}{\sqrt{ps}-\sqrt{qr}}\right) \\
 2c &= \sqrt{\frac{rs}{pq}}\ln\left(\frac{\sqrt{ps}+\sqrt{qr}}{\sqrt{ps}-\sqrt{qr}}\right) \\
\end{array} $$
For this to work correctly you need to choose compatible branches of the various logs and square roots, and I have not worked out the details of that.
