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If $I=(du)^2+\sin^2u(dv)^2=II$ then show that the surface is given by $$\underline{x}=\sin u\cos v\:e_1+\sin u\sin v\:e_2+\cos u\:e_3+\underline{c}$$

This exercise question is really interesting like the fundamental form are enough to determine the surface, but my book didn't show anything similar. Like I know here,

$$E=1,F=0,G=\sin^2 u$$ $$e=1,f=0,g=\sin^2 u$$ Now I stuck, like what approach should I follow to get the surface equation? Some random thought suggest to find curvature ($K=\frac{eg-f^2}{EG-F^2}=\frac{\sin^2 u}{\sin^2 u}=1$) or normal of surface which might help to determine the surface but I didn't get any equation like the question.

It seems using Weingarten equation and Gauss equation I got $X_{uuu}+X_u=0$ which actually led me to the surface equation (up to rigid motion: translation and rotation). Does this approach is general enough? I got another M.SE thread where it suggest to verify compatibility equations (Gauss' equation and the Codazzi-Mainardi Equations). Does it need so?

N00BMaster
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  • Hint: Use Frenet-Serret for curves to conclude constant-$u$ and constant-$v$ curves are planar circles. – user10354138 Oct 10 '23 at 12:59
  • I afraid I couldn't get your hint, can you expand it little bit? @user10354138 And does you mean $u$ and $v$ parametric curves here? – N00BMaster Oct 10 '23 at 16:31
  • Yes, constant-$v$ curves are $u$-parametric curves parametrised by arclength (in this case it is also $u$). You will find $T=X_u$, $T_u=-N$ and now $N_u$ has no $X_v$ component so there is no torsion, hence a constant curvature 1 planar curve so you can write down the curve, etc. – user10354138 Oct 10 '23 at 16:46
  • I understand your hint now. But it seems using Weingarten equation and Gauss equation I got $X_{uuu}+X_u=0$ which actually lead me to the surface equation (up to rigid motion: translation and rotation). Does this approach is general enough? I got another M.SE thread where it suggest to verify compatibility equations (Gauss' equation and the Codazzi-Mainardi Equations). Does it need so? @user10354138 – N00BMaster Oct 11 '23 at 07:02
  • Yes, the Gauss and Codazzi-Mainardi equations are necessary and sufficient (locally) for there to exist a surface with the prescribed I and II. But integrating, in general, is very, very, very nontrivial. I think you should recognize that your I and II come from the unit sphere and then use the uniqueness part of the Fundamental Theorem of Surface Theory. – Ted Shifrin Oct 11 '23 at 20:03
  • Thank you @TedShifrin. I think you should recognize that your I and II come from the unit sphere, for this case I can. But I was wondering what if I couldn't at exam for unfamiliar $I,II$. – N00BMaster Oct 13 '23 at 19:27
  • Then, as I said, you end up with a messy system of overdetermined partial differential equations to solve. This is typically beyond the scope of an undergraduate course. – Ted Shifrin Oct 13 '23 at 19:39

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