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.
answered May 15 '12 at 7:51
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$\begingroup$ Thank you, Willie! Is there anything we can say about $E,F,G$ when the angles are only the same everywhere? $\endgroup$ – mela May 15 '12 at 8:20
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$\begingroup$ For example do they need to be constants or something? $\endgroup$ – mela May 15 '12 at 8:24
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$\begingroup$ Just about $F$: not really. However, by the definition of the angle using the law of cosine that $v\cdot w = |v|^2 + |w|^2 - |v||w|\cos \theta$, you have that cosine of the angle is equal to $F / (E + G - 2 \sqrt{EG})$. So this quantity must remain constant everywhere. $\endgroup$ – Willie Wong May 15 '12 at 8:24
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$\begingroup$ $F$ itself does not need to be constant. Start with the $(x,y)$ I wrote above. Let $y' = \exp y$. Since the change of variables $y' = \exp y$ does not change the level curves, you still have constant angles. But Now $\mathrm{d}y = \frac{1}{y'} \mathrm{d}y'$, and plugging this in you see that the new $F$ in this coordinates becomes $-1 / y'$ which is not constant. $\endgroup$ – Willie Wong May 15 '12 at 8:26
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$\begingroup$ Thanks again Willie! Your explanation is very helpful! :) $\endgroup$ – mela May 15 '12 at 8:44
