Prove 2-dimensional Riemann manifold is conformally flat

[PhyPsy]

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
[itex]\lambda[/itex] = [itex]\lambda[/itex](x⁰, x¹), [itex]\nu[/itex] = [itex]\nu[/itex](x⁰, x¹)
satisfying
g^ab[itex]\lambda[/itex]_,a[itex]\lambda[/itex]_,b = g^ab[itex]\nu[/itex]_,a[itex]\nu[/itex]_,b = 0
and show that the line element reduces to the form
ds² = e^{2[itex]\mu[/itex]}d[itex]\lambda[/itex]d[itex]\nu[/itex]
and finally introduce new coordinates [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]) and [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])

Homework Equations

Conformally flat metric: g_ab = [itex]\Omega[/itex]²[itex]\eta[/itex]_ab ([itex]\eta[/itex]_ab is a flat metric)
ds² = g_abdx^adx^b

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), ([itex]\lambda[/itex], [itex]\nu[/itex]), and [[itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]), [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])]
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 ([itex]\lambda[/itex], [itex]\nu[/itex]) coordinate system is:
ds² = g_{[itex]\lambda\lambda[/itex]}d[itex]\lambda[/itex]² + g_{[itex]\lambda\nu[/itex]}d[itex]\lambda[/itex]d[itex]\nu[/itex] + g_{[itex]\nu\lambda[/itex]}d[itex]\nu[/itex]d[itex]\lambda[/itex] + g_{[itex]\nu\nu[/itex]}d[itex]\nu[/itex]²
and if I assume the metric is flat in this coordinate system, then that equation can be reduced to:
ds² = g_{[itex]\lambda\lambda[/itex]}(d[itex]\lambda[/itex]² - d[itex]\nu[/itex]²) + 2g_{[itex]\lambda\nu[/itex]}d[itex]\lambda[/itex]d[itex]\nu[/itex]
where I used the symmetry of the metric. If I am to reduce this equation to
ds² = e^{2[itex]\mu[/itex]}d[itex]\lambda[/itex]d[itex]\nu[/itex]
then I need to show that
1. g_{[itex]\lambda\lambda[/itex]}d[itex]\lambda[/itex]² + g_{[itex]\nu\nu[/itex]}d[itex]\nu[/itex]² = 0
2. 2g_{[itex]\lambda\nu[/itex]} = e^{2[itex]\mu[/itex]}.

I don't know how to do either. I have substituted d[itex]\lambda[/itex] with ([itex]\partial\lambda[/itex]/[itex]\partial[/itex]x⁰)dx⁰ + ([itex]\partial\lambda[/itex]/[itex]\partial[/itex]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_{[itex]\lambda\nu[/itex]} = e^{2[itex]\mu[/itex]}. Could I get some help, please?

 

[PhyPsy]

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

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

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