How to determine if this geometry is flat How do I determine whether the following geometry is flat?
$$\mathrm{d}s^2=\left(\frac{v}{u}+\frac{1}{uv}\right)\mathrm{d}u^2+\left(\frac{u}{v}+\frac{u}{v^3}\right)\mathrm{d}v^2+2\left(1-\frac{1}{v^2}\right)\mathrm{d}u\mathrm{d}v$$
Thanks.
Josh
 A: Just compute the Riemann tensor. The result is $R^\mu_{\nu\alpha\beta}=0$. This metric is flat.

A: If you haven't seen the Riemann tensor yet, then I guess the best way would be to guess functions x and y of u and v such that your statement there is true, and equals a flat geometry.
I would also think that, if this is the only method you have access to, then this will be a flat geometry, since it's less possible to guess that there isn't a solution.
First I note that, if there are coordinates x and y related to u and v such that:
$$\mathrm{d}s^2 = \mathrm{d}x(u,v)^2  +  \mathrm{d}y(u,v)^2$$
then by substituting the following:
$$\mathrm{d}x(u,v) = \frac{\partial x}{\partial u }\mathrm{d}u + \frac{\partial x}{\partial v }\mathrm{d}v  \\\ \mathrm{d}y(u,v) =\frac{\partial y }{\partial u }\mathrm{d}u + \frac{\partial y}{\partial v }\mathrm{d}v$$
We get:
$$\mathrm{d}s^2 = \left[ \left(\frac{\partial x}{\partial u }\right)^2 + \left(\frac{\partial y}{\partial u }\right)^2\right]\mathrm{d}u^2  +  \left[ \left(\frac{\partial x}{\partial v }\right)^2 + \left(\frac{\partial y}{\partial v }\right)^2\right] \mathrm{d}v^2 + 2  \left[\frac{\partial x}{\partial u }\frac{\partial x}{\partial v } + \frac{\partial y}{\partial u } \frac{\partial y}{\partial v }  \right]\mathrm{d}u \ \mathrm{d}v $$
Which gives me some clues to guess with when I compare with your original
$$\mathrm{d}s^2=\left(\frac{v}{u}+\frac{1}{uv}\right)\mathrm{d}u^2+\left(\frac{u}{v}+\frac{u}{v^3}\right)\mathrm{d}v^2+2\left(1-\frac{1}{v^2}\right)\mathrm{d}u\mathrm{d}v$$
Specifically:
$$\left(\frac{\partial x}{\partial u }\right)^2 + \left(\frac{\partial y}{\partial u }\right)^2 = \frac{v}{u}+\frac{1}{uv} $$
$$  \left(\frac{\partial x}{\partial v }\right)^2 + \left(\frac{\partial y}{\partial v }\right)^2 = \frac{u}{v}+\frac{u}{v^3}$$
and
$$\frac{\partial x}{\partial u }\frac{\partial x}{\partial v } + \frac{\partial y}{\partial u } \frac{\partial y}{\partial v }  = 1-\frac{1}{v^2} $$
My first guess was that the derivatives of x corresponded to v/u and u/v, but that was an assumption. It turned out be be correct however, as I found the consistent functions:
$$x = 2\sqrt{u \ v}$$
$$y = 2\sqrt{\frac{u}{v}}$$
for which the derivative equations hold, and hence there do exist coordinates for which the line element is flat, and so the geometry is flat.
A: Disregard the previous answer, I asked him and he said it was wrong. 
You want to find a way of expressing the result in the form ds^2 = dx^2 + dy^2 so find which terms come from squaring an expression of adu+bdv where a and b are the coefficients involving u and v.
OLD:
Are you in the UoN fourth year Gravity module? I believe u and v are light cone co-ordinates although I'm not 100% on this - check page 47 of the notes if you're on my course (he's capitalised them for some reason).
Just get them in terms of x and y and evaluate the metric components, then check if they meet the flatness condition in section  8.
