Making sense of $\frac{\mathrm{GL}(2n,\mathbb{R})}{\mathrm{GL}(n,\mathbb{C})}$ I came across this post that says
$\frac{\mathrm{GL}(2n,\mathbb{R})}{\mathrm{GL}(n,\mathbb{C})}$ 
is the set of all almost complex structures. However, I do not understand
why. Can any point out an reference to this?
Also, any geometric/physical interpretation is warmly welcomed!
 A: An almost-complex structure is a matrix $J$ such that $J^2 = -I$ is the negative identity. As you said, one example of such a matrix $J$ is $$\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}.$$
Interpreting $GL(n,\mathbb{C})$ as a subgroup of $GL(2n;\mathbb{R})$ depends on having fixed such an almost-complex structure. Once we have a matrix $J$, we can call a matrix $A \in GL(2n;\mathbb{R})$ complex-linear if it commutes with $J$, i.e. $AJA^{-1} = J.$
(The idea is that $\mathbb{C}$-linear maps $T$ are just real linear maps with the additional property that $T(iv) = i T(v)$ for all vectors $v$)
Given any matrix $A \in GL(2n;\mathbb{R})$, we get another almost-complex structure $AJA^{-1}$. This is the same almost-complex structure $J$ if and only if $A \in GL(n;\mathbb{C}).$ On the other hand, all almost-complex structures are similar (although it may take some work to be convincing that they are similar over $\mathbb{R}$ and not only $\mathbb{C}$) since they are diagonalizable with the same eigenvalues $\pm i$. That gives you a bijection $$GL(2n;\mathbb{R}) / GL(n;\mathbb{C}) \longrightarrow \{\mathrm{almost}-\mathrm{complex}\; \mathrm{structures}\}$$ under which a class $A \cdot GL(n;\mathbb{C})$ corresponds to the almost-complex structure $AJA^{-1}.$
