# Prove 2-dimensional Riemann manifold is conformally flat

## Homework Statement

Establish the theorem that any 2-dimensional Riemann manifold is conformally flat in the case of a metric of signature 0.
Hint: Use null curves as coordinate curves, that is, change to new coordinate curves
$\lambda$ = $\lambda$(x0, x1), $\nu$ = $\nu$(x0, x1)
satisfying
gab$\lambda$,a$\lambda$,b = gab$\nu$,a$\nu$,b = 0
and show that the line element reduces to the form
ds2 = e2$\mu$d$\lambda$d$\nu$
and finally introduce new coordinates $\frac{1}{2}$($\lambda$ + $\nu$) and $\frac{1}{2}$($\lambda$ - $\nu$)

## Homework Equations

Conformally flat metric: gab = $\Omega$2$\eta$ab ($\eta$ab is a flat metric)

## The Attempt at a Solution

It says in the hint that the metric has a signature of 0, so it must be flat in a specific set of coordinates, but there are 3 different coordinate systems the hint tells me to use in this problem:
(x0, x1), ($\lambda$, $\nu$), and [$\frac{1}{2}$($\lambda$ + $\nu$), $\frac{1}{2}$($\lambda$ - $\nu$)]
so I don't know in which coordinate system I should make the assumption that the metric is flat.

I know that the line element equation in the ($\lambda$, $\nu$) coordinate system is:
ds2 = g$\lambda\lambda$d$\lambda$2 + g$\lambda\nu$d$\lambda$d$\nu$ + g$\nu\lambda$d$\nu$d$\lambda$ + g$\nu\nu$d$\nu$2
and if I assume the metric is flat in this coordinate system, then that equation can be reduced to:
ds2 = g$\lambda\lambda$(d$\lambda$2 - d$\nu$2) + 2g$\lambda\nu$d$\lambda$d$\nu$
where I used the symmetry of the metric. If I am to reduce this equation to
ds2 = e2$\mu$d$\lambda$d$\nu$
then I need to show that
1. g$\lambda\lambda$d$\lambda$2 + g$\nu\nu$d$\nu$2 = 0
2. 2g$\lambda\nu$ = e2$\mu$.

I don't know how to do either. I have substituted d$\lambda$ with ($\partial\lambda$/$\partial$x0)dx0 + ($\partial\lambda$/$\partial$x1)dx1 and tried all sorts of algebraic manipulations, but I have not been able to cancel out those 2 terms or figure out how 2g$\lambda\nu$ = e2$\mu$. Could I get some help, please?

## Answers and Replies

I've managed to work through the steps in the hint, but it didn't lead me to the gab = $\Omega$2$\eta$ab equation that I should be getting. The only way I see to eliminate the d$\lambda$2 and d$\nu$2 terms in the line element equation is to assume that d$\lambda$2 = d$\nu$2. Since the ($\lambda$, $\nu$) coordinate system is arbitrary, I should be allowed to make this assumption.
Then, I set $\mu$ = $\frac{1}{2}$$\int$($\partial$$\lambda$/$\partial$x0)2[(dx0 + dx1)/(dx0 - dx1)], and this allows me to define g$\lambda\nu$ + g$\nu\lambda$ = e2$\mu$. I came up with the equation for $\mu$ from the identity $\frac{d}{dx}$ln x = $\frac{1}{x}$.

I did the coordinate transformation to [$\frac{1}{2}$($\lambda$ + $\nu$), $\frac{1}{2}$($\lambda$ - $\nu$)] and came up with this:
g$\lambda\lambda$ = gyz/2
g$\nu\nu$ = -gyz/2
g$\lambda\nu$ = gyy/2
where y = $\frac{1}{2}$($\lambda$ + $\nu$) and z = $\frac{1}{2}$($\lambda$ - $\nu$)

There are 2 problems with this. One, g$\lambda\lambda$ should be a function of gyy, g$\lambda\nu$ should be a function of gyz, etc. if I am to show that gab = $\Omega$2$\eta$ab.
Two, even if the above were true, the metric in the ($\lambda$, $\nu$) coordinate system is not flat, because if it was, then g$\lambda\nu$ would be 0, not e2$\mu$/2, and ds2 would be 0. So all I would have proven is that $\overline{g}$ab = $\Omega$2gab, not gab = $\Omega$2$\eta$ab. Is there someone who can help?