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.


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.

  • $\begingroup$ 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
  • $\begingroup$ For example do they need to be constants or something? $\endgroup$ – mela May 15 '12 at 8:24
  • $\begingroup$ 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
  • $\begingroup$ $F$ 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
  • $\begingroup$ Thanks again Willie! Your explanation is very helpful! :) $\endgroup$ – mela May 15 '12 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.