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How should I see that the set of almost complex structures on $R^4$ preserving the positive orientation, namely $\{J\in GL^{+}(4,R), J^2=-I\}$ is homotopy equialent to $S^2$.

There is a similar question asked before, but I need more details as why this set is $GL^{+}(4,R)/GL(2,C)$, etc.

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Fix a nonzero vector $v$. Then $Jv$ will be perpendicular to $v$, which lies in $S^2\subset \mathbb{R}^3=\langle v\rangle^\perp$. The orientation of $\mathbb{R}^4$ defines the latter two directions $u_1,u_2=Ju_1$ after fixing $u_1$. Note that picking $v$ and $u_1$ was just for presenting a basis of $\mathbb{R}^4$, so that to build $J$ we only have an $S^2$ worth of choices (the $\mathbb{R}v\oplus\mathbb{R}u_1$ is contractible).

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    $\begingroup$ In general $Jv$ is not perpendicular to $v$, unless $J$ preserve the inner product. $\endgroup$
    – user99914
    May 11 '15 at 21:06

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