Pushforward 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 try to 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.

I’ve tried to get a metric from this preserved quantity but have not been successful.

How can I push the Lorentz metric forward?

Thanks so much.

Answer 1

I would like to explain my comments to the OP's post here. As this is too long to be in the comments, I have decided to include them in an "answer format".

The metric on $2$-dimensional Minkowski spacetime is

$$ g = dt^2 - dx^2.$$

Of course, some people prefer to write it as $-g$. In the end, this is just a matter of convention. Now, let

$u = f_1(t,x) := t+x \quad \text{ and } \quad v = f_2(t,x) := t-x.$

Denote by $f$ the resulting map which maps $(t,x)$ to $(u,v)$, and thus has as components $f_1$ and $f_2$.

Therefore, if we let $h = dudv$, then

$$f^*(h) = dt^2 - dx^2 = g.$$

In other words, $f$ is an isometry from $\mathbb{R}^2$ with coordinates $(t,x)$ and Minkowski metric $g$ onto $\mathbb{R}^2$ with coordinates $(u,v)$ and metric $h = dudv$.

Finally, note that both $du$ and $dv$ are null, meaning that

$h^{-1}(du,du) = 0 \quad \text{ and } h^{-1}(dv,dv) = 0.$

This is what I meant by the statement that $u$ and $v$ are null coordinates.

Answer 2

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 $ds^2=dxdy.$

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