Grassmanian $(2, 4)$ homeomorphic to $S^2 \times S^2$

The only proof I know uses quaternions.

Identify $\bf R^4$ with quaternions, $\bf H$ a quaternion can be written as $q=t+xi+yj+zk$, with $i^2=j^2=k^2=-1, ij=-ji=k..$.

Let $H=R+ P$, with $P$ the 3 dimensional vector space of pure quaternions ($q^2\leq 0$, or $q=xi+yj+zk$).

Given an oriented plane $V$ the exists a unique pure quaternion of square $-1$, say $I$ such that $I^2=-1$ and $V$ is stable by $I$ (if $(u,v)$ is a direct orthonormal basis of $V$, then $I=v.\bar u$ ; note that orthonorml means that $v.\bar u=-u.\bar v$ so that $(v.\bar u)^2=-1$).

This maps $Gr(2,4)^+$ onto the unit sphere of $P$.

Now an element of this unit sphere is a complex structure $I$, and the plane $V$ is stable by $I$ iff $V$ is stable by $I$ i.e is complex line.

The set of $I$-complex lines is $CP^1$ homeomorphic to $S^2$.

This gives on the Grassmanian the structure of a $S^2$ bundle over $S^2$.

To prove that this bundle is trivial let us fix a pure quaternion $I$ of square $-1$, (for instnace $i=I)$ and $x$ any quaternion of norm $1$, the plane $P(x,I)=<x,Ix>$ (the plane generated by $x, Ix$ with this orientation) is $I$ stable. Note that $P(y,J)=P(x,I)$ iff $y\in P(x,I)$, and $I=J$ : indeed $P(y,J)=P(x,I)$ implies that $y\in P(x,Ix)$ as well as $Jy$, so that $J=I$ is the unique rotation of angle $\pi/2$ of this oriented plane.

This yields the trivialisation $CP^1\times S^2 \to Gr^+(2,4)$