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

Question

Prove that the Grassmanian manifold $G(2, 4)$ of all real two-dimensional planes in $\mathbb{R}^4$ that pass through the origin is homeomorphic to the product of two two-dimensional spheres $S^2 \times S^2$.

I need a relatively straightforward way of solving this if possible (there is another answer that goes into group actions, stabilisers etc. which I don't get: $Gr_2^+(\mathbb R^4) \cong S^2 \times S^2$)

I get the idea for $G(1,2)$. That is, that every line through the origin in $\mathbb{R^2}$ can be represented as a point on the unit circle (with antipodal points removed). I assume some sort of argument similar to this but in the relevant dimensions will do, but I haven't been able to come up with it. Any help would be greatly appreciated.

Answer 1

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 $CP^1\times S^2 \to Gr^+(2,4)$

Answer 2

Here may be one way to think about it. Let $Gr(1,3)$ be the one dimensional vector subspaces of $\mathbb{R}^{3}$, and consider $Gr(2,4)$ as adding an extra vector in $Gr(1,3)$ and the whole $\mathbb{R}^{3}$ in the same time. We know that $Gr(1,3) \cong \mathbb{RP}^{2}$. If the extra vector belongs to $\mathbb{R}^{3}$ already, then we get another copy of $\mathbb{RP}^2$ since $Gr(1,3)\cong Gr(2,3)$. Otherwise the extra vector falls into $\mathbb{R}^{4}$ minus a hyperplane, and the space of extra vectors we can choose is essentially $Gr(1,4)\cong \mathbb{RP}^{3}$. If we already chosen a vector in $Gr(1,3)$, then the space of extra vectors lying on the same plane spanned by it should fiber over a circle. One is thus lead to believe the desired manifold should be $\mathbb{S}^{2}\times \mathbb{S}^{2}$ as we are taking the oriented double cover. I think this is essentially the same as Qiaochu Yuan's argument except the group action is not as explicit.

Answer 3

I have just posted a complete answer in $Gr_2^+(\mathbb R^4) \cong S^2 \times S^2$ which uses the explicit covering $S^3 \times S^3 \to SO(4)$ to get an explicit diffeomorphism.