Orthogonal complex structures on $\mathbb{R}^4$ How does one see that (the space of isomorphism classes of) orthogonal complex structures on $\mathbb{R}^4$ are a 2-sphere?
It seems that one has to take a quotient of $O(4,\mathbb{R})$ by $U(2,\mathbb{C})$, but how does one see that it is homeomorphic to $S^2$?

The (orientation preserving) orthogonal complex structure $J$ is actually controlled by its action on a fixed nonzero vector $v$: $v$, $Jv$ would spans a two plane $L$ in $\mathbb R^4$, and by fixing the orientation, the action of $J$ on $L^\perp$ is also fixed. 
Now 
$$g(v, Jv) = g(Jv, J^2v) = -g(Jv, v) \Rightarrow g(v, Jv) = 0 \Rightarrow Jv\in v^\perp.$$
Thus the (orientation preserving) orthogonal complex structure is parametrized by $\{w\in v^\perp : \|v\| = 1\}$, which is a $\mathbb S^2$.
It seems that if orientation is not fixed, there would be two copies of $\mathbb S^2$
