One-Dimensional Projections in $M_{2}(\mathbb{C})$ Homeomorphic to $S^{2}$

Question

I am working on Exercise 3.8 in Rordam's book K-Theory for C$^{*}$-Algebras. The goal of the exercise is to prove that the space $G_{2,1}$ of all one-dimensional projections in $M_{2}(\mathbb{C})$ is homeomorphic the $2$-sphere $S^{2}$.

I have completed the first part of the problem, which is to prove that $$ G_{2,1}=\left\{\begin{pmatrix}t & \omega\sqrt{{t}(1-t)}\\\overline{\omega}\sqrt{t(1-t)} & 1-t\end{pmatrix} : \ t\in[0,1]\quad\text{and}\quad |\omega|=1 \right\}. $$

I am not sure how to use this result to exhibit an explicit homeomorphism between $G_{2,1}$ and $S^{2}$. This seems linked to stereographic projection somehow, but even upon reviewing that I am unable to construct the homeomorphism. Any help is appreciated.

Thanks.

Answer 1

Motivation: Notice that the parameters $t$ and $\omega$ determine a matrix in $G_{2,1}$ and conversely, given a matrix $\begin{pmatrix} a& b\\ c &d \end{pmatrix} \in G_{2,1}$ we can determine $t$ by $a$ and $\omega$ by $\frac{b}{\sqrt{a(1-a)}}$ except when $a=0,1$. So, we try to think of a map in terms of $t$ and $\omega$. Since we can think of $\omega$ in terms of $\theta:= \arg(\omega)$ intuitively this means these points should correspond to your North and South poles. Motivated by this, we get the geometric interpretation of $t$ as describing the height of the sphere and $\omega$ as describing the circle at the given height at time $t$. So we will use $t$ to describe the $z$ component on $S^2$ and the cross section at this height gives a circle in the $x,y$-plane which we describe with $\omega$. The following details arise from turning this picture into equations.

More precisely: Let's use $z$ to denote the height on $S^2$ so that $z\in [-1,1]$. We have $t\in [0,1]$ so at time $t$ we should be at height $z$ hence the relation $z=2t-1$.

Next, from $x^2+y^2+z^2=1$ we get $\frac{x^2}{1-z^2}+\frac{y^2}{1-z^2}=1$. Thus $\frac{x}{\sqrt{1-z^2}}$ and $\frac{y}{\sqrt{1-z^2}}$, represent the real and imaginary parts of a complex number on the unit circle. Namely, we get $\omega = \frac{x+iy}{\sqrt{1-z^2}}$.

From these, we can explicitly describe a map $(x,y,z) \mapsto \begin{pmatrix} t & \omega\sqrt{t(1-t)} \\ \bar{\omega}\sqrt{t(1-t)} & 1-t \end{pmatrix}$ and this will give the homeomorphism we want. Namely:

Let $\varphi: S^2\rightarrow G_{2,1}$ be the map given by $\varphi(x,y,z)= \begin{pmatrix} \frac{1+z}{2} &\frac{1}{2}(x+iy) \\ \frac{1}{2}(x-iy) & \frac{1-z}{2} \end{pmatrix}$

and $\psi:G_{2,1}\rightarrow S^2$ be given by $\psi \begin{pmatrix} a & b\\ c & d \end{pmatrix} = (b+c,\frac{1}{i}(b-c),2a-1)$.

We claim that $\varphi$ and $\psi$ are the desired homeomorphism. They are both clearly continuous and it is not hard to check they are inverses to each other (remembering for $\begin{pmatrix} a & b \\ c & d\end{pmatrix}\in G_{2,1}$ we have $d=1-a$ and assuming I didn't make a computation error).