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Would I be right to think that if I have a coordinate system $(x,y)$ so that the lines/curves where one coordinate is fixed, so something like $x=a$ and $y=b$, always intersect at the same angle, then for the first fundamental form $F=0$?

Thank you.

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Only if the angle is a right angle.

Let $(u,v)$ be the standard Cartesian coordinate system on the plane, so the first fundamental form is (as a line element) $ \mathrm{d}s^2 = \mathrm{d}u^2 + \mathrm{d}v^2 $

Let $y = v$ and $x = u+v$. Clearly the level curves of $x$ and the level curves of $y$ always intersect at a 45 degree angle. But using that $u = x - y$ we get that $\mathrm{d}u = \mathrm{d}x -\mathrm{d}y$ while $\mathrm{d}v = \mathrm{d}y$. So $ \mathrm{d}s^2 = \left(\mathrm{d}x - \mathrm{d}y\right)^2 + \mathrm{d}y^2 = \mathrm{d}x^2 + 2\mathrm{d}y^2 - 2 \mathrm{d}x\mathrm{d}y$

Which means that in this $(x,y)$ coordinates, we have that $E = 1$, $F = -1$ and $G = 2$.


To put it in other words, $F = 0$ is equivalent to the coordinate system being such that the coordinate curves are everywhere mutually orthogonal. This condition is a bit more stringent then just having the same angle everywhere.

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    Thanks a$g$ain Willie! Your explanation is very helpful! :)2012-05-15