rt.representation theory - Nicest coset representatives of the symplectic group in the general linear group

I suppose that "nice" is very much application-dependent, but let me give it a try.

The first thing to notice is that, of course, you will not be able to find a global coset representative, since the principal bundle
$$mathrm{Sp}(2k,mathbb{C}) to mathrm{GL}(2k,mathbb{C}) to M = mathrm{GL}(2k,mathbb{C})/mathrm{Sp}(2k,mathbb{C})$$
is not trivial and hence it has no global (continous) section. So the best you can do is find a section over some $U subset M$.

The subgroup $mathrm{Sp}(2k,mathbb{C})$ is the stabilizer of a symplectic structure $Omega$ on $mathbb{C}^{2k}$. A convenient choice is
$$Omega = pmatrix{ 0 & -mathbf{1} cr mathbf{1} & 0}$$
where $mathbf{1}$ is the $ktimes k$ identity matrix.

The Lie algebra $mathfrak{sp}(2k,mathbb{C})$ consists of those $2k times 2k$ complex matrices $X$ such that $Omega X$ is symmetric. A complementary vector subspace of $mathfrak{sp}(2k,mathbb{C})$ in $mathfrak{gl}(2k,mathbb{C})$ is given by
$$mathfrak{sp}(2k,mathbb{C})^perp := biglbrace Omega X mid X in mathfrak{so}(2k,mathbb{C})bigrbrace$$

Explicitly and for our choice of $Omega$ above, this subspace consists of the matrices of the form
$$pmatrix{B^t & C cr A & B}$$
for $ktimes k$ matrices $A,B,C$ with $A$ and $C$ skewsymmetric.

You can now exponentiate these matrices to find a coset representative. Depending on the calculation, though, you might it easier to write the coset representative as a product of exponentials,...