For the exam I'm trying to solve some problems. Today I found this exercise and need some help:
For the group S0(1,1) of the Lorentz transformation I have $\phi \in \mathbb{R}$ and $A_{\phi}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$
$$A_{\phi} = \begin{pmatrix} \cosh(\phi) & \sinh(\phi) \\ \sinh(\phi) & \cosh(\phi) \end{pmatrix} $$
Now there is a symetric bilinear form $$\langle \langle u_1, u_2 \rangle \rangle:= t_1t_2 - x_1x_2$$
with $u = \begin{pmatrix}t\\x\end{pmatrix}$ $\in \mathbb{R}^2$
Now I wanted to know if $\langle \langle u_1, u_2 \rangle \rangle$ is a dot product on the $\mathbb{R}^2$. So I did some calculations with the axioms for dot products but it does not work. How can I do it?
Furthermore I tried to show that $$\langle \langle A_{\phi}u_1, A_{\phi}u_2 \rangle \rangle = \langle \langle u_1, u_2 \rangle \rangle.$$ I tried to show it by calculation. But it also does not work. Do I need for this exercise some mathematical tricks?
I hope some of you can help me.
Thank you!
It's not a dot product since it's not positive definite, e.g. $(1,1)$ has pseudo-norm $0$. (If I'm understanding you correctly.)
To see that $\langle\langle A_{\phi}u_1,A_{\phi}u_2\rangle\rangle = \langle\langle u_1,u_2\rangle\rangle$, note that if $u_1 = (w,x)$ and $u_2 = (y,z)$, then
\begin{eqnarray} \langle\langle A_{\phi}u_1,A_{\phi}u_2\rangle\rangle &=& \left\langle\left\langle \left(\begin{array}{r}w\cosh(\phi)+x\sinh(\phi) \\ w\sinh(\phi) + x\cosh(\phi)\end{array}\right),\left(\begin{array}{r}y\cosh(\phi)+z\sinh(\phi) \\ y\sinh(\phi) + z\cosh(\phi)\end{array}\right)\right\rangle\right\rangle \\ &=& (w\cosh(\phi)+x\sinh(\phi))(y\cosh(\phi)+z\sinh(\phi)) \\ &-& (w\sinh(\phi)+x\cosh(\phi))(y\sinh(\phi)+z\cosh(\phi)) \\ &=& \color{purple}{wy\cosh^2(\phi)} \color{darkblue}{+wz\sinh(\phi)\cosh(\phi)} \color{green}{+xy\sinh(\phi)\cosh(\phi)} \color{darkred}{+xz\sinh^2(\phi)} \\ &\color{purple}{-}& \color{purple}{wy\sinh^2(\phi)}\, \color{darkblue}{-wz\sinh(\phi)\cosh(\phi)} \color{green}{-xy\sinh(\phi)\cosh(\phi)} \color{darkred}{-xz\cosh^2(\phi)} \\ &=& wy(\cosh^2(\phi)-\sinh^2(\phi)) - xz(\cosh^2(\phi)-\sinh^2(\phi)) \end{eqnarray}
From here, make use of some very convenient hyperbolic trig identities and you will have your result.