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In every beginner’s class on differential geometry, we learn Euler’s formula, which tells us about the normal curvatures at a point on a surface.

I know how to prove this formula, and I have even taught it in classes, but it seems entirely unbelievable, to me. It says that at any point on any surface, the variation of normal curvature as a function of angle is given by an absurdly simple little formula. My intuition says that this variation should be different for different surfaces. If I choose a point on a highly convoluted surface, I would expect the variation of normal curvature to be complicated, but it’s not. I can’t get my head around this at all. It’s astonishing. Can anyone elucidate? I have seen the standard proofs, and those don’t seem to help very much. I’m looking for intuition, please, rather than manipulation of symbols.

bubba
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    Apparently your severe warnings has dissuaded answers or even comments. Look at when the theorem is false for clues: the surface has to be twice differentiable at the point. With that condition, all surfaces look locally similar to quadrics, even highly convoluted ones. – Chrystomath Feb 13 '21 at 08:45
  • Ok. That’s helpful. Thanks. Can you fill in the reasoning a bit, please. Why does twice differentiable imply that the surface looks like a quadric, locally? And I don’t really see why the theorem holds for quadrics, though it’s a lot more plausible in that case. – bubba Feb 13 '21 at 09:51
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    Locally a surface is the graph of a $C^2$ function which has a Taylor series expansion in $x,y$. Keeping only the order $\le2$ terms gives a quadric. As far as the normal and curvatures are concerned, only these terms matter. If you measure anything smooth (eg curvature) as you turn in a circle, you'll get maxima and minima. The reason why quadrics have two principal curvatures is because they are two-dimensional. – Chrystomath Feb 13 '21 at 10:53
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    Good question. You’re thinking of waves going around a point? Something like $z=r^2 \sin 4\theta$ in polar coordinates? If you convert this or anything similar back to rectangular coordinates, you’ll find that the surface is either not twice differentiable or the second derivatives all vanish at the origin. If you assume the surface to get twice differentiable and choose coordinates so that it touches the origin and is tangent to the $xy$-plane there, then Euler’s formula is easily verified. – Deane Feb 13 '21 at 18:33
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    @Deane. Yes, I was thinking about surfaces with radial waves, or even just a monkey saddle. What you’re saying is that smooth surfaces like this are locally “flat”. That’s surprising, to me. I’ll have to try to convince myself that it’s true. Thanks. – bubba Feb 14 '21 at 00:08
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    Well, not locally flat but flat at the point in question, yeah. The monkey saddle is a nice example, because we see three-fold symmetry (i.e., rotation through $2\pi/3$) but symmetry of the second fundamental form gives a two-fold symmetry (i.e., rotation through $\pi$). The only resolution is that the second fundamental form be $0$. And it is, because the monkey saddle is the graph of a cubic polynomial :) – Ted Shifrin Feb 14 '21 at 00:34
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    In other words, a quadratic can’t create more than two waves, so you have to use a cubic or highe order. But if there’s a quadratic term, it will dominate the higher order terms near the origin so you’ll see the multiple waves only if there is no quadratic term. – Deane Feb 14 '21 at 01:40

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