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On surface, for which $$ds^2=du^2+dv^2$$ find the angle between lines $v=u$ and $v=-u$.

This exercise is related to the first fundamental form. I think I need to find the angle between curves with parametrization: $$u(t)=t, v(t)=t$$ and $$u(s)=s, v(s)=-s$$ therefore $$du=dt, dv=dt$$ and $$du*=ds, dv*=-ds$$ Now I can make tangent vectors to both curves $$dr=\frac{dr}{du} dt + \frac{dr}{dv} dt$$ $$dr*=\frac{dr*}{du} ds - \frac{dr*}{dv} ds$$ Am I thinking right? Unfortunately at this point I don't have more ideas. For finding the angle I can use $$cos(a)=\frac{dr .dr*}{|dr| . |dr*|}$$ And then I could use the numbers "1", "0", "1" from the first fundamental form. But I'm absolutely unsure, so I would like to ask you about it. Thank you.

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  • $\begingroup$ Do you know that we have always $ds^2=dr\cdot dr$? $\endgroup$ – Mikasa Jun 11 '13 at 13:26
  • $\begingroup$ No, where did you get this from? $\endgroup$ – Andrew Jun 11 '13 at 17:47
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You can parametrize your curves as $c_1(t)=(t,t)$ and $c_2(t)=(t,-t)$ (first coordinate: $u$). You are searching for the angle between the tangent vectors $v=(1,1)$ and $w=(1,-1)$ (tangent vectors along $c_1$ and $c_2$: note that they do not depend on $t$) using the metric $g=du^2+dv^2$ with components $g_{ij}(u,v)=\delta_{ij}$ for $i,j=1,2$.

Then the angle $\theta$ is s.t. $\cos\theta=\frac{\langle v,w\rangle}{\|v\| \|w\|}$,

with $\langle v,w\rangle=g_{ij}v_iw_j$ and $\|v\|=\sqrt{g_{ij}v_iv_j}$ (and similarly for $\|w\|$). We sum over repeated indices.

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  • $\begingroup$ I tried one more time to do it alone. After several tries and operating on $dt$, $ds$ etc. I've finally got a result. And now I've read your answer, which is much more easier to understand (in some points of mine result I didn't know what I was doing in fact) and got the same result, but took just a few minutes. Thank you! $\endgroup$ – Andrew Jun 11 '13 at 19:29
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    $\begingroup$ I am happy to read that! Bye, Avitus $\endgroup$ – Avitus Jun 11 '13 at 19:44

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