Prove that $SO_2(\mathbb R)$ is a maximal subgroup of $SL_2(\mathbb{R})$

Question

I am trying to prove that $SO_2(\mathbb R) $ is a maximal subgroup of $SL_2(\mathbb R)$. There were some hints in the exercise:

Say we have $SO_2(\mathbb R)\subsetneq H \subset SL_2(\mathbb R)$. Let $f=gh\in H-SO_2(\mathbb{R})$, where $g\in O_2(\mathbb R)$ and $h$ is a positive definite matrix, thus we are looking at the polar decomposition of $f$.

1. Prove that there is a $\lambda>1$ such that $h_{\lambda}=\begin{pmatrix}\lambda & 0 \\ 0 & \frac{1}{\lambda}\end{pmatrix}\in H$.

Solution: The $h$ matrix upto some change of basis is the matrix $h_\lambda$.

2. Let $u_x=\begin{pmatrix}\cos x & -\sin x\\ \sin x & \cos x \end{pmatrix}$. Then prove that for any $x\in \mathbb R$, $f(x)=u_x h_\lambda u_x^{-1} h_\lambda$ and $f(x)^tf(x)$ are in $H$.

This is straightforward.

3. Show that $F:=\{||f(x)||\mid x\in \mathbb R\}$ contains the interval $[1,\lambda^2]$, where $||\cdot||$ denotes the operator norm.

I proved this using connectedness of the set $F$, because it a continuous image of a connected set. Also $f(0)=\begin{pmatrix}\lambda^2 & 0\\ 0 & \frac{1}{\lambda^2}\end{pmatrix}$, thus $||f(0)||\ge \lambda^2$ and $f(\frac{\pi}{2})=I$, this $||f(\frac {\pi} {2})||=1$.

4. Conclude that $H=SL_2(\mathbb R)$.

I am stuck in this part. I know that it is sufficient to prove(by polar decomposition again) that any positive definite matrix with determinant $1$ is in $H$, but I don't know how to prove it. I would like to get some hints rather than a complete solution.

Answer 1

Up to change of basis, any symmetric positive definite matrix with determinant $1$ has form $$ h_\lambda:= \begin{bmatrix} \lambda & 0 \\ 0 & 1/\lambda \end{bmatrix}.$$ Note that matrices of this form are (essentially) determined by their norm, as $\|h_\lambda\| = \lambda$ assuming $\lambda\geq 1$.

What you've shown is that, for some fixed $\lambda_0>0$, you can construct $f(x)\in H$ so that $\|f(x)\|$ achieves every value between $1$ and $\lambda_0^2$.

I suggest doing two things:

1. Prove that, if you can construct $k\in H$ (where $k$ is not necessarily symmetric positive definite) with $\|k\|=\lambda$, then $h_\lambda\in H$. It will then follow from your work that $h_\lambda\in H$ for all $\lambda\in [1,\lambda_0^2]$.
2. Iterate your recipe (e.g., replacing $\lambda_0$ with $\lambda_0^2$ shows you can construct $k\in H$ such that $\|k\|=\lambda$ for all $\lambda\in[1,(\lambda_0^2)^2]$). Continue inductively. You will find that $h_\lambda\in H$ for all $\lambda\in[1,\infty)$.

There is a bit of weirdness where you need to show that $h_\lambda\in H$ for all $0<\lambda<1$ still, but I will leave it to you to patch that up.