Pushforward of Lorentz metric: $ds^2=dxdy$

Question

Consider an open embedding $\varphi:\Bbb R^{1,1}_- \hookrightarrow (0,1)^2$ with $\varphi(x_1,x_2)=(e^{x_1},e^{x_2})$

I need to put a metric on $(0,1)^2$.

Here’s what I know so far. I need to transport the Lorentz metric: $ds^2=dxdy. $

Since my map is a diffeomorphism I can push forward the Lorentz metric to $(0,1)^2$.

I know that $n=\log(x)\log(y)$ is a preserved metric of a transformation on $(0,1)^2$ that is related to the classical Lorentz transformation.

However I don't think I can obtain a metric purely from this fact.

How can I push the Lorentz metric forward?

My wild guess is: $ds^2=\frac{1}{x}dx\frac{1}{y}dy.$

Also how can one generalize this metric to dimension 3?

I think it has to do with adding in a positive definite matrix of dimension $n-2$ somewhere into the metric. This makes sense because that matrix will do the job of rotating the coordinate system.

Answer 1

I figured out the first question. The new metric in $(0,1)^2$ is $ds^2=\frac{1}{u}du\frac{1}{v}dv.$

This is done by pulling back the Lorentz metric $ds^2=dxdy.$

$\begin{align} (\varphi^{-1})^* (dxdy) &=( d((\varphi^{-1})^*x))( d((\varphi^{-1})^*y)) \\ &= (d (x\circ \varphi^{-1}))(d (y\circ \varphi^{-1})) \\ &= (d \log u)(d\log v) \\ &= \left( \frac{1}{u} du\right)\left( \frac{1}{v} dv\right) \\ &= \frac{1}{u} du\frac{1}{v} dv. \end{align}$

Still not sure about the dimension 3 case.