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.

  • 4
    $\begingroup$ This looks like applying the Baker–Campbell–Hausdorff formula backwards: $e^Xe^Y=e^Z$ for $Z=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+\ldots$. Now the problem reduces to finding for a given $Z$ such matrices $X$ and $Y$ with $Y$ diagonal and $X$ with zeroes on the diagonal. This has something to do with the fact that a lot of these repeated commutators are zero on the diagonal. Does not make the problem much simpler, though. $\endgroup$ Apr 13, 2021 at 16:30

2 Answers 2


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 $(m_1,m_2,m_3) = (1,1,0)$.

  • $\begingroup$ This is very useful input! I seems to me, though, that this example hinges on the fact that $x$ is skew diagonal with entries that commute. This gives some special structure to odd and even powers of $x$. I wonder if any of this generalises for other cases. $\endgroup$
    – Mads G
    Apr 15, 2021 at 9:13

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.

  • 3
    $\begingroup$ These formulas blow up if exactly one of $p$ and $s$ is nonzero. See my answer. $\endgroup$ Apr 14, 2021 at 21:52
  • $\begingroup$ @JoshuaMundinger's answer. $\endgroup$
    – LSpice
    Apr 14, 2021 at 22:40

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