Set of 2-Dimensional Indefinite Real Orthogonal Matrices is not Compact in Normed Real Square Matrix Vector Space

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Theorem

Let $\struct {\R^{2 \times2}, \norm {\, \cdot \,}_\infty}$ be the normed real matrix vector space.

Let $\map O {1, 1} := \set {\mathbf R \in \R^{2 \times2} : \mathbf R^\intercal \mathbf J_{1,1} \mathbf R = \mathbf J_{1,1}}$ be the indefinite orthogonal group of degree $\paren {1, 1}$ over real numbers where:

$\ds \mathbf J_{1,1} := \begin{bmatrix} 1 & 0\\ 0 & -1 \\ \end{bmatrix}$

Then $\map O {1, 1}$ is not a compact set in $\struct {\R^{2 \times 2}, \norm {\, \cdot \,}_\infty}$.


Proof

Let:

$\begin{bmatrix} \map \cosh t& \map \sinh t\\ \map \sinh t & \map \cosh t \\ \end{bmatrix} := \map {\mathbf R} t$

We have that:

\(\ds \map {\mathbf R^\intercal} t \mathbf J_{1,1} \map {\mathbf R} t\) \(=\) \(\ds \begin{bmatrix} \map \cosh t & \map \sinh t \\ \map \sinh t & \map \cosh t \\ \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix} \map \cosh t & \map \sinh t \\ \map \sinh t & \map \cosh t \\ \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} \map {\cosh^2} t - \map {\sinh^2} t& 0 \\ 0 & \map {\sinh^2} t - \map {\cosh^2} t \\ \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\) Difference of Squares of Hyperbolic Cosine and Sine

Hence, $\map {\mathbf R} t \in \map O {1, 1}$.

From difference of squares of hyperbolic cosine and sine:

\(\ds \size {\map \cosh t}\) \(=\) \(\ds \sqrt {\map {\sinh^2} t + 1}\)
\(\ds \) \(>\) \(\ds \size {\map \sinh t}\)

The set of matrix elements constitutes a finite subset of the set of real numbers which is ordered.

By Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements, there is a maximal element.

Consider the supremum norm of $\map {\mathbf R} t$:

\(\ds \norm {\map {\mathbf R} t}_\infty\) \(=\) \(\ds \max_{ \begin {split} & 1 \mathop \le i \mathop \le 2\\ & 1 \mathop \le j \mathop \le 2 \end {split} } \size {\map {r_{i j} } t}\) Definition of Supremum Norm
\(\ds \) \(=\) \(\ds \size {\map \cosh t}\)

Furthermore:

$\ds \lim_{t \mathop \to \infty} \map \cosh t = \infty$

Hence, $\map O {1,1}$ is not bounded.

By Heine-Borel theorem, $\map O {1,1}$ is not compact in $\struct {\R^{2 \times 2}, \norm {\, \cdot \,}_\infty}$.

$\blacksquare$


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