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## 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](x

^{0}, x

^{1}), [itex]\nu[/itex] = [itex]\nu[/itex](x

^{0}, x

^{1})

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

ds

^{2}= 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)

ds

^{2}= g

_{ab}dx

^{a}dx

^{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:

(x

^{0}, x

^{1}), ([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:

ds

^{2}= 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:

ds

^{2}= 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

ds

^{2}= 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]x

^{0})dx

^{0}+ ([itex]\partial\lambda[/itex]/[itex]\partial[/itex]x

^{1})dx

^{1}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?