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

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.

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}