The question is:

Show that the matrices \begin{pmatrix} \cosh \theta & \sinh \theta\\ \sinh \theta & \cosh \theta \end{pmatrix} leave invariant the real quadratic form $x_1^2-x_2^2$ and constitute the group $SO(1,1).$

I did the invariance part as follows : $$ x_1^\prime =x_1 \cosh \theta+ x_2\sinh \theta, \ \ \ x_2\prime =x_1\sinh \theta+x_2\cosh \theta $$ which gives $$(x_1^\prime) ^2- (x_2^\prime) ^2 = x_1^2-x_2^2.$$ But I don't know how to show that this transformation will give the group $SO(1,1).$

  • $\begingroup$ So it means that I will take an arbitrary matrix in $SO(1,1)$ and show that each element is in the form of either $\cosh \theta$ or $\sinh \theta.$ $\endgroup$ Feb 25, 2017 at 0:07
  • 1
    $\begingroup$ It doesn't look true to me: $-I$ is in $SO(1,1)$ but does not have this shape. $\endgroup$ Feb 25, 2017 at 0:10
  • $\begingroup$ I don't have any idea about this. This came in my exam, I was able to do only part1. $\endgroup$ Feb 25, 2017 at 0:24

1 Answer 1


This is 'true'.

Let $M$ be an matrix $2\times2$ with $\det(M)\neq 0$ of the form ($a,b,c,d \in \mathbb{R}$)

$$M = \left(\begin{matrix}a&b\\c&d\end{matrix}\right)$$

We have that $M \in SO(1,1)$ if, for $\eta := \left(\begin{matrix}1&0\\0&-1\end{matrix}\right)$ we have that

$$M^{-1}=\eta M^T \eta$$

and we have that $\det(M)=+1$. So, we have that

$$M^{-1} = \frac{1}{\det(M)}\left(\begin{matrix}d&-b\\-c&a\end{matrix}\right) =\left(\begin{matrix}d&-b\\-c&a\end{matrix}\right) $$


$$\eta M^T \eta = \left(\begin{matrix}a&-c\\-b&d\end{matrix}\right) $$

if set the equality quoted we get that $a=d$ and $b = c$ so the group have the form

$$SO(1,1) = \left\{M \text{ is a $2\times 2$ matrix}: M = \left(\begin{matrix}a&b\\b&a\end{matrix}\right),a,b \in \mathbb{R},\text{with } a^2-b^2=1 \right\} $$

Set $a = \cosh(\theta)$ for $\theta \in \mathbb{R}$ then you have your matrix. Note that you do not get all the group

As pointed out in the comments, we can see that $SO(1,1)$ have two connected components because this Lie-Group has a homeomorphism to $H_+ \cup H_-$ where

$$H_{\pm} = \left\{(x,y)\in\mathbb{R} : x = \pm \sqrt{y^2+1}, y\in \mathbb{R}\right\}$$

This implies that this group has two conex components, lets call them $L_+$ and $L_-$

$L_+$ is generated by matrix of the form, for $b\in \mathbb{R}$


and $L_-$ is generated by matrix of the form


The matrix pointed out in comments is in $L_-$. If there is some problem with my answer someone could explain in comments.

  • 1
    $\begingroup$ What is conex, is it not connected? That is there are two connected components. $\endgroup$ Feb 25, 2017 at 0:48
  • $\begingroup$ Very sorry I translate to English in a poor way, now i gaved an link $\endgroup$
    – R.W
    Feb 25, 2017 at 0:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .