1
$\begingroup$

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?
$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\mathrm{Spin}(4)\cong S^3\times S^3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.