Let $\mathcal{M}$ denote the space of all real $2\times 2$ matrices,
equipped with the norm $||A||=\sqrt{\mbox{tr}(A^TA)}$, for
$A\in\mathcal{M}$ (here $A^T$ denotes the transpose of the matrix $A$
and for any $2\times 2$ real matrix $B$, $\mbox{tr}(B)$ denotes its
trace). Consider the map $F$ from $\mathbb{R}^2$ to $\mathcal{M}$
given by the formula, for any $(s,t)\in\mathbb{R}^2$: $$F(s,t) =
\frac{1}{2}\cdot\begin{vmatrix}\cos(t)+\cos(s)&&\sin(t)+\sin(s)\\-\sin(t)+\sin(s)&&\cos(t)-\cos(s)\end{vmatrix}.$$
Denote by $\mathcal{N}\subset\mathcal{M}$ the space of all real
$2\times2$ matrices of rank one and norm one.Prove that the image of the map $F$ is the space $\mathcal{N}$ and
that the map $F$ is a local homeomorphism to its image (the latter
with the induced topology).
My attempt is as follows.
- Show that $\mbox{Im}(F)=\mathcal{N}$ means to show that $\mbox{Im}(F)\subseteq\mathcal{N}$ and $\mathcal{N}\subseteq\mbox{Im}(F)$.
For arbitrary tuple $(s,t)$ consider
$$A = \frac{1}{2}\cdot\begin{vmatrix}\cos(t)+\cos(s)&&\sin(t)+\sin(s)\\-\sin(t)+\sin(s)&&\cos(t)-\cos(s)\end{vmatrix}, ~A\in\mbox{Im}(F).$$
Note that $\det(A)=0$, therefore rows are linearly dependent and $\dim\mbox{Ker}(A)=1$ ($\dim\mbox{Ker}(A)\neq2$ since $A$ isn't identically zero), so $\mbox{rk}(A)=1$ follows from the rank-nullity theorem.
One can also check directly that $||A||=\sqrt{\mbox{tr}(A^TA)}=1$, so we have proved that $\mbox{Im}(F)\subseteq\mathcal{N}$. How can we show another inclusion — that any $2\times2$ real matrix with rank one and norm one is apparently of that form?
- Show that $F$ is a local homeomorphism to $\mbox{Im}(F)$.
By definition, homeomorphism is a continuous bijection such that the inverse is also continuous. Here the real issues in my understanding begin. We can show that $F$ is onto by proving that $\mathcal{N}\subseteq\mbox{Im}(F)$ which is essentially my question above. However, how can $F$ be one-to-one and has an inverse if $\det(F)=0$? This immediately implies that matrix isn't invertible and has a non-empty kernel.
I will very appreciate any help. Thanks in advance.
Best Answer
DISCLAIMER: I'm not terribly familiar with topology so I don't know whether I've been rigorous enough with this answer. I think it brings the idea across if nothing else.
The set $\mathcal{N}$ can be characterised as the set: $$ \mathcal{N}=\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}:a^2+b^2+c^2+d^2=1\quad\land\quad ad-bc=0\right\} $$ If we rewrite these conditions a bit, we see that they are: $$ (a+d)^2+(a-d)^2+(b+c)^2+(b-c)^2=2\\ (a+d)^2-(a-d)^2-(b+c)^2+(b-c)^2=0 $$ Introducing $$ x=a+d\quad y=a-d\quad z=b+c\quad w=b-c $$ we get: $$ x^2+y^2+z^2+w^2=2\\ x^2-y^2-z^2+w^2=0 $$ But this system is equivalent to $$ x^2+w^2=1\\ y^2+z^2=1 $$ These solutions can be parametrised as: $$ x=\cos t\quad w=\sin t\\ y=\cos s\quad z=\sin s $$ The map $(s,t)\rightarrow (x(s,t),y(s,t),z(s,t),w(s,t))$ is locally a homeomorphism as it essentially just represents a parametrisation of two independent circles with radius $1$ by the angles $s,t$. From here, we recover $a,b,c,d$ as: $$ a=\frac{\cos t+\cos s}{2}\quad b=\frac{\sin t+\sin s}{2}\quad c=\frac{-\sin t+\sin s}{2}\quad d=\frac{\cos t-\cos s}{2}\quad $$ But the map $(x,y,z,w)\rightarrow (a,b,c,d)$ is just a linear transformation and as such also a homeomorphism. The composition $(s,t)\rightarrow(x,y,z,w)\rightarrow(a,b,c,d)\rightarrow\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is precisely the map $F$, so $F$ is indeed a local homeomorphism from $\mathbb{R}^2$ to $\mathcal{N}$.