# Help with closing this novel proof of the equivalence of oriented Grassmannians $G^+(2, 4)$ and $S^2\times S^2$

I'm concluding my thesis on the Hopf fibration, in which I constructed it purely geometrically.

In a nutshell, you can treat $$S^3$$ as Quaternions, and then those Quaternions as rotations $$R_{(\cdot)}$$. So if you fix a special point $$q$$, the Hopf map is defined as $$h_q(r)=R_r(q)$$. The preimage $$h_q^{-1}(p)$$ of any $$p\in S^2$$ is a great circle in $$S^3$$, which contains all the Rotations which send $$q$$ to $$p$$. (You get the classic Hopf map by fixing $$q=(1, 0, 0)\equiv i$$.)

Now here is my proposed proof of the fact that the oriented Grassmannians $$G^+(2, 4)$$ are equivalent to $$S^2\times S^2$$.

Note that every oriented great circle $$C$$ in $$S^3$$ is equivalent to a plane $$H\in G^+(2,4)$$ since both are induced by a pair of orthonormal vectors. It is sufficient therefore to show that for any $$C$$ there is a unique pair $$(q, p)\in S^2\times S^2$$ such that $$h_q^{-1}(p) = C$$.

I have successfully proven, that for $$(q, p)\in S^2\times S^2$$ we have $$h_q^{-1}(p)=\frac{p+q}{\lVert p+q\rVert}\left(\cos t + p\sin t\right)$$ where we treat $$q, p$$ as pure Quaternions.

Now I'm stuck on the closing argument. I see two tantalizing choices:

1. Does someone see a direct argument, that for suitable choices of $$q, p$$ we get every possible oriented great circle in $$S^3$$?
2. If we write out the quaternions that are spanning the great circle as vectors \begin{align}h_q^{-1}(p) &= \frac{1}{\lVert p+q\rVert}\left( \begin{pmatrix} 0\\p_1+q_1\\p_2+q_2\\p_3+q_3 \end{pmatrix}\cos t + \begin{pmatrix} -\langle p, p+q\rangle\\p_3(p_2+q_2)-p_2(p_3+q_3)\\p_1(p_3+q_3)-p_3(p_1+q_1)\\p_2(p_1+q_1)-p_1(p_2+q_2) \end{pmatrix}\sin t \right)\\ &=\frac{1}{\lVert p+q\rVert}\left(\begin{pmatrix}0\\p+q\end{pmatrix}\cos t + \begin{pmatrix}\langle -p, p+q\rangle\\-p\times(p+q)\end{pmatrix}\right) \end{align} which gives two clearly independent vectors when taken as matrix rows are already fully reduced! This resembles a plane quite starcly, but how can we get from here to arbitrary planes?

My answer to (1) is quick but some brief bits of background first.

The orbit-stabilizer theorem says that when a nice topological group has a nice continuous action on a topological space, any choice of basepoint yields a bundle $$\mathrm{Stab}(x)\to G\to\mathrm{Orb}(x)$$. The fibration is $$\pi(g):=gx$$ with fibers $$\pi^{-1}(gx)=g\mathrm{Stab}(x)$$; that is, the fibers are cosets of stabilizers. Also helpful to keep in mind $$\mathrm{Stab}(gx)=g\mathrm{Stab}(x)g^{-1}$$, so points in the same orbit have conjugate stabilizers.

A quaternion is a formal sum of a scalar and a 3D vector, aka real and imaginary parts. The scalar and vector components of the product of two vectors is $$\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$$, so to multiply two arbitrary quaternions you would FOIL and then use this formula. This entails a lot: the sqrts of $$+1$$ are $$\pm1$$, the sqrts of $$-1$$ are the unit vectors, two quaternions commute iff their vector parts are parallel, two vectors anticommute (meaning $$\mathbf{vu}=-\mathbf{uv}$$) iff they are perpendicular. The conjugate of a quaternion has the vector part negated, and the natural inner product is $$\langle x,y\rangle=\mathrm{Re}(\overline{x}y)$$, which basically means $$\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$$ is an orthonormal basis. Nonreal quaternions have unique polar forms $$p=re^{\theta\mathbf{v}}$$ where $$r>0$$ is the magnitude, $$0<\theta<\pi$$ is a convex angle, and $$\mathbf{v}$$ is a unit vector. You can expand this with $$\cos$$ and $$\sin$$ with Euler's formula. The effect of conjugation $$p\mathbf{x}p^{-1}$$ on 3D vectors $$\mathbf{x}$$ is to rotate it around $$\mathbf{v}$$ by an angle of $$2\theta$$.

As a group action this gives a double covering $$S^3\to\mathrm{SO}(3)$$ with kernel $$S^0=\{\pm1\}$$, which means $$S^3$$ is a spin group $$\mathrm{Spin}(3)$$. By using both left and right multiplication simultaneously, every pair $$(p,q)$$ of quaternions yields a 4D rotation $$x\mapsto px\overline{q}$$ (treating quaternions themselves as a 4D Euclidean space). This gives a double covering $$S^3\times S^3\to\mathrm{SO}(4)$$, so we can say that $$S^3\times S^3\cong\mathrm{Spin}(4)$$.

Here's where your formula for the fibers comes from: given two nonparallel unit vectors $$\mathbf{u}$$ and $$\mathbf{v}$$, we can classify the set of unit quaternions ("versors") $$p$$ for which $$p\mathbf{u}p^{-1}=\mathbf{v}$$. Its just a coset of the stabilizer $$\mathrm{Stab}(\mathbf{u})$$, which is the unit circle subgroup of $$\mathbb{R}[\mathbf{u}]\cong\mathbb{C}$$. To know which stabilizer, it suffices to exhibit a single coset representative. We could use $$p=e^{\phi\mathbf{w}}$$ where $$\mathbf{w}$$ is perpendicular to $$\mathbf{u}$$ and $$\mathbf{v}$$ and $$\phi$$ is half the angle between $$\mathbf{u}$$ and $$\mathbf{v}$$; with the half-angle formula this would yield

$$p = \sqrt{\frac{1+\mathbf{u}\cdot\mathbf{v}}{2}} + \sqrt{\frac{1-\mathbf{u}\cdot\mathbf{v}}{2}}\frac{\mathbf{u}\times\mathbf{v}}{\|\mathbf{u}\times\mathbf{v}\|},$$

which you could write in other formats if you wanted. Or, you could use an axis $$\mathbf{z}$$ halfway between $$\mathbf{u}$$ and $$\mathbf{v}$$ and a $$180^\circ$$ rotation, corresponding to

$$p = \exp\left(\frac{\pi}{2}\mathbf{z}\right)=\mathbf{z}=\frac{\mathbf{x}+\mathbf{y}}{\|\mathbf{x}+\mathbf{y}\|}.$$

Any other choice of $$p$$ uses an axis on the great circle of $$S^2$$ containing $$\mathbf{w}$$ and $$\mathbf{z}$$. Note the axis $$\pm\mathbf{w}$$ is perpendicular to the axis $$\pm\mathbf{z}$$. Also note the left coset $$p\mathrm{Stab}(\mathbf{u})$$ is the right coset $$\mathrm{Stab}(\mathbf{v})p$$.

Proposition. Every great circle of $$S^3$$ is a coset of a circle subgroup.

[ Suppose $$C$$ is a great circle of $$S^3$$. Pick two perpendicular nonreal quaternions $$p$$ and $$q$$ from it. Then $$\mathbf{u}=p^{-1}q$$ is a unit vector, and $$C$$ is the coset $$p\{\exp(\theta\mathbf{u})\}$$. ]

Note you're talking about oriented 2D subspaces, hence oriented great circles. To accommodate for that in the above, replace circle subgroup with one-parameter circle subgroup.

It's a nice idea to use all Hopf fibrations (over all choices of basepoints) simultaneously. There's a pretty way to package this all together using the parallelizability of $$S^3$$. On the one hand, there's a kind of "currying" map $$S^3\times S^2\to S^2\times S^2$$ given by $$(p,\mathbf{v})\mapsto (p\mathbf{v}p^{-1},\mathbf{v})$$, where a Hopf map is applied to the first component and the second component determines which Hopf map is applied. On the other hand, $$S^3\times S^2 = \mathrm{UT}S^3$$ is the unit tangent bundle of $$S^3$$ via the correspondence $$(p,p^{-1}q)\leftrightarrow (p,q)$$, and each pair $$(p,q)$$ of perpendicular quaternions induces an oriented 2D subspaces. The projections $$S^3\times S^2\to S^2\times S^2$$ and $$\mathrm{UT}S^3\to\widetilde{\mathbb{G}}_2\mathbb{R}^4$$ both have circular fibers, which correspond to each other via $$S^3\times S^2\simeq\mathrm{UT}S^3$$.

In summary, we have a bundle isomorphism

$$\begin{array}{ccccc} S^1 & \longrightarrow & S^3\times S^2 & \longrightarrow & S^2\times S^2 \\ \updownarrow & & \updownarrow & & \updownarrow \\ S^1 & \longrightarrow & \mathrm{UT}S^3 & \longrightarrow & \widetilde{\mathbb{G}}_2\mathbb{R}^4 \end{array}$$

for which on the right half we can element-chase

$$\begin{array}{ccc} (p,p^{-1}q) & \mapsto & (qp^{-1},p^{-1}q) \\ \updownarrow & & \updownarrow \\ (p,q) & \mapsto & \mathrm{span}\{p,q\} \end{array}$$

In other words, $$\mathrm{span}\{p,q\}\in\widetilde{\mathbb{G}}_2\mathbb{R}^4$$ (with orientation interpreted so $$q$$ is a positive right angle from $$q$$) corresponds to the pair of unit vectors $$(qp^{-1},p^{-1}q)\in S^2\times S^2$$. Note for any 2D subspace $$\Pi$$ there is an $$S^1$$s worth of choices of ordered orthonormal bases $$\{p,q\}$$, but $$qp^{-1}$$ and $$p^{-1}q$$ do not depend on this choice so this is all well-defined.

Here's another way to see $$\widetilde{\mathbb{G}}_2\mathbb{R}^4\simeq S^2\times S^2$$.

In general, we can embed $$\widetilde{\mathbb{G}}_k\mathbb{R}^n\hookrightarrow\Lambda^k\mathbb{R}^n$$ via $$\mathrm{span}\{v_1,\cdots,v_k\}\mapsto v_1\wedge\cdots\wedge v_k$$ (using ordered orthonormal bases, this is well-defined). The Hodge-star operator $$\star:\Lambda^k\mathbb{R}^n\leftrightarrow\Lambda^{n-k}\mathbb{R}^n$$ linearly extends orthogonal-complmentation between $$\widetilde{\mathbb{G}}_k\mathbb{R}^n\leftrightarrow\widetilde{\mathbb{G}}_{n-k}\mathbb{R}^n$$. In the event of $$n=2k=4$$, the star operator as $$\pm1$$-eigenspaces, each 3D. The space $$\Lambda^2\mathbb{R}^4$$ inherits an inner product from $$\mathbb{R}^4$$, in which case $$\Lambda^2=\Lambda^2_+\oplus\Lambda^2_-$$ is an orthogonal direct sum, with $$a\wedge b\leftrightarrow a\wedge b\pm c\wedge d$$ for every ordered orthonormal basis $$\{a,b,c,d\}$$. We can verify $$\widetilde{\mathbb{G}}_2\mathbb{R}^4$$ corresponds to $$S^2\times S^2$$ wrt this.

It is a standard fact that $$\mathfrak{so}(n)\cong\Lambda^2\mathbb{R}^n$$, where $$a\wedge b$$ represents the right-angle rotation from $$a$$ to $$b$$ (assuming $$a,b$$ are perpendicular unit vectors) in the $$ab$$-plane and acts as the $$0$$ map in the complement of the $$ab$$-plane. The $$\star$$-eigendecomposition basically corresponds to $$\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$$, which is the infinitessimal version of $$\mathrm{Spin}(4)\cong\mathrm{Spin}(3)\times\mathrm{Spin}(3)$$.

In conclusion, $$\widetilde{\mathbb{G}}_2\mathbb{R}^4\simeq S^2\times S^2$$ is like the infinitessimal version of $$\mathrm{Spin}(4)\cong S^3\times S^3$$.