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To give some context, perhaps irrelevant, define $s:SL(2,\mathbb{R}) \rightarrow O(1,2)$ where given $g\in SL(2,\mathbb{R})$, $s(g):x\mapsto gxg^T$, $x$ is a symmetric $\mathbb{R}^{2\times 2}$. I want to find the Kernel of s.

In other words, we want the set of $g$ s.t. $x \rightarrow gxg^T = x$, which should turn out to be ${\pm I}$.

In short, show that

$gxg^T=x, \forall x \in \mathbb{R}^{2\times 2} \implies g=\pm I$.

Is there a mistake in my logic, and if there isn't, to show this how should I start?

TwoStones
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Nugi
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    I think there are some dimensional errors, i.e. how can you mulitply a $2\times 2$ matrix and a $3\times 3$ matrix? – TwoStones Sep 01 '19 at 08:45

1 Answers1

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Using $x=I$ you obtain $g\cdot g^{T}=I$, so $g\in O(2,\mathbb{R})$. Thus $\exists \alpha \in [0,2\pi]: g=\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}(*)$ or $g=\begin{bmatrix} \cos\alpha & \sin\alpha\\ \sin\alpha & -\cos\alpha\end{bmatrix}(**)$.

Suppose $(*)$ then using $x=\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$ you have $gxg^{T}=\begin{bmatrix} \cos^2\alpha & -\sin\alpha\cos\alpha\\ -\sin\alpha\cos\alpha & \sin^2\alpha\end{bmatrix}$. From the imposition $gxg^{T}=x$ we obtain $\sin\alpha =0\iff \alpha =0$ or $\alpha = \pi$ from which the solutions $g=I$ and $g=-I$.

Supposing $(**)$ and using same $x$ we obtain the condition $\begin{bmatrix} \cos^2\alpha & \sin\alpha\cos\alpha\\ \sin\alpha\cos\alpha & \sin^2\alpha\end{bmatrix}=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$ so we have the possible values of $g=\pm\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$. Instead this two values of $g$ aren’t solution: you can demonstrate it by parametrising $x=\begin{bmatrix} a&b\\ c& d\end{bmatrix}$ and seeing that $gxg^{T}=\begin{bmatrix} a&-b\\ -c& d\end{bmatrix}$ that is in general different by $x$.

Alessandro Cigna
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  • I know that $(* )$ corresponds to rotation, but why can we write as $(**)$ when it is in $O(2)$? Also, can you see a intuitive illustration of why the second case is improper assumption as it does not produce solutions? – Nugi Sep 01 '19 at 14:50
  • About the first question: $$\begin{bmatrix} a&b\c&d\end{bmatrix}\in O(2,\mathbb{R})\iff \begin{bmatrix} a&b\c&d\end{bmatrix}\cdot\begin{bmatrix} a&c\b&d\end{bmatrix}=\begin{bmatrix} 1&0\0&1\end{bmatrix}\iff \begin{cases} a^2+b^2=c^2+d^2=1\ ac+bd=0\end{cases}$$from first condition $\exists \beta ,\gamma\in[0,2\pi]:, \begin{cases} a=\cos\beta \ b=\sin\beta\ c=\cos\gamma\ d=\sin\gamma\end{cases}$ then applying second condition $0=\cos\beta\cos\gamma +\sin\beta\sin\gamma=\cos(\beta -\gamma)\iff \beta=\gamma +\pi /2 $ or $\beta=\gamma -\pi /2$. Substitute this values nd have your answer – Alessandro Cigna Sep 01 '19 at 16:39
  • For second question you can use $x=\begin{bmatrix} 0&1\0&0\end{bmatrix}$: $$\begin{bmatrix} 1&0\0&-1\end{bmatrix}\cdot\begin{bmatrix} 0&1\0&0\end{bmatrix}\cdot\begin{bmatrix} 1&0\0&-1\end{bmatrix}=\begin{bmatrix} 0&1\0&0\end{bmatrix} \cdot\begin{bmatrix} 1&0\0&-1\end{bmatrix}=\begin{bmatrix} 0&-1\0&0\end{bmatrix}\ne \begin{bmatrix} 0&1\0&0\end{bmatrix}$$ – Alessandro Cigna Sep 01 '19 at 16:47
  • @YeonwookJung I forgot to tag you – Alessandro Cigna Sep 01 '19 at 21:47