# Geometric representation of all 2x2 matrices

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:

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.

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

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.