4
$\begingroup$

I am trying to understand the content on pages 4-6 of this paper describing a geometric representation of all 2x2 matrices.

So far, I have confirmed that all orthogonal matrices lie on $S^3(\sqrt{2})$ since they can be represented as $\begin{bmatrix}\cos\theta&\sin\theta\\\mp \sin\theta&\pm \cos\theta\end{bmatrix}$ for some $\theta$. I have also confirmed that, when representing $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ in 4-space by changing coordinates from $x$, $w$, $y$, and $z$ with \begin{align} a&=(x-y)/\sqrt{2}\\ b&=(z-w)/\sqrt{2}\\ c&=(z+w)/\sqrt{2}\\ d&=(x+y)/\sqrt{2} \end{align} the set of rotations $SO(2)$ is the great circle on $S^3(\sqrt{2})$ in the $xw$-plane (with $a^2+b^2+c^2+d^2=2\implies x^2+w^2+y^2+z^2=2$ and $\det A = 1 \implies x^2+w^2-y^2-z^2=2$, implying $x^2+w^2=2$, $y=0$ and $z=0$) and the set of reflections $O^-(2)$ is the great circle in the $yz$-plane (with $\det A = -1$ so $y^2+z^2=2$, $x=0$ and $w=0$). Finally, I have confirmed that all singular matrices on $S^3(\sqrt{2})$ lie on the Clifford torus defined by $x^2+w^2=1$ and $y^2+z^2=1$ (since we have $\det A = ad-bc = x^2+w^2-y^2-z^2=0$ and know that $x^2+w^2+y^2+z^2=2$). So this drawing from the paper makes sense to me:

enter image description here

However, I am trying to confirm the following:

The complement of this Clifford torus consists of two open solid tori which are tubular neighborhoods of the great circles $SO(2)$ and $O^-(2)$. The cross-sectional disks in these tubular neighborhoods are round disks of radius $(\pi/4)\sqrt{2}$ lying on great 2-spheres which meet the great circles $SO(2)$ and $O^-(2)$ orthogonally.

I can't seem to wrap my mind around this. Any explanations / tips / hints would be greatly appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

So within $S^3(\sqrt{2})$, the singular matrices form the torus parametrized by

$$ \begin{cases} x=\cos\theta \\ w=\sin\theta \\ y=\cos\phi \\ z=\sin\phi \end{cases} $$

Based on your picture, let $\theta$ be toroidal coordinates, and $\phi$ poloidal coordinates. (Supposedly that's opposite of the usual convention but I'm already invested in it this way.)

The rotations in $SO(2)$ and reflections in $O(2)\setminus SO(2)$ are respectively given by

$$ \begin{cases} x=\sqrt{2}\cos\theta \\ w=\sqrt{2}\sin\theta \\ y=0 \\ z=0 \end{cases} \qquad \begin{cases} x=0 \\ w=0 \\ y=\sqrt{2}\cos\phi \\ z=\sqrt{2}\sin\phi \end{cases} $$

It ought to be obvious, topologically, how a torus is made up of a bunch of disks. But let's get specific.

For solid torus with $SO(2)$ running through it, the disk "centered" at the $\theta$-point on $SO(2)$ is

$$ \begin{cases} x=\sqrt{2-y^2-z^2}\cos\theta \\ w=\sqrt{2-y^2-z^2}\sin\theta \\ 1<y^2+z^2\le 2 \end{cases} $$

The "center" (on $SO(2)$) occurs when $y^2+z^2=2$. Set $y^2+z^2=2\cos^2\rho$ with $\rho\in[0,\frac{\pi}{4})$. Then we can parametrize the geodesic arc from the $\theta$-point on $SO(2)$ and the $(\theta,\phi)$-point on the torus via

$$ \begin{cases} x=\sqrt{2}\cos\rho \cos\theta \\ w=\sqrt{2}\cos\rho \sin\theta \\ y=\sqrt{2}\sin\rho \cos\phi \\ z=\sqrt{2}\sin\rho \sin\phi \end{cases} $$

For example when $\cos\theta=0$ these are spherical coordinates for a spherical cap of "diameter" $\frac{\pi}{2}$ (angle between antipodal points) in the $wyz$-plane. Letting $\rho$ run wild parametrizes a great $2$-sphere with poles

$$ \pm\begin{bmatrix} \sqrt{2}\cos\theta \\ \sqrt{2}\sin\theta \\ 0 \\ 0 \end{bmatrix}, \quad \pm\begin{bmatrix} 0 \\ 0 \\ \sqrt{2} \\ 0 \end{bmatrix}, \quad \pm\begin{bmatrix} 0 \\ 0 \\ 0 \\ \sqrt{2} \end{bmatrix}. $$

If we differentiate this parametrization at the $\theta$-point on $SO(2)$, where $\cos\rho=1$ the derivative is $\cos'\rho=0$ so the $xw$-components are $0$. On the other hand, leaving $\cos\rho=1$ and differentiating with respect to $\theta$ leaves the $yz$-components $0$. Thus, at the $\theta$-point on $SO(2)$, the tangent vectors of $SO(2)$ and of the $2$-spherical cap are orthogonal.

I will let you do the same thing for the other solid torus (which, under stereographic projection, is "inside out"). In your $\mathbb{R}^3$ visualization (again, under stereographic projection), if $O(2)\setminus SO(2)$ represents a straight line (which becomes a circle with the "point at infinity"), the great $2$-spheres containing these new caps will be all possible spheres through the circle representing $SO(2)$, plus the plane through said circle (which becomes a $2$-sphere with the "point at infinity").

Moreover, these $2$-spherical caps have orthogonally intersecting boundary circles. That is, in the original parametrization of the torus, let $\theta$ be fixed and differentiate with respect to $\phi$ for one tangent vector, and let $\phi$ be fixed and differentiate with respect to $\theta$ for another, and these tangent vectors will be orthogonal.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .