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I'm trying to teach myself the classical differential geometry of 2D surfaces in 3D Euclidean space but I'm struggling to understand exactly how much information the Gaussian and mean curvature provide.

If I've understood the Bonnet theorem correctly, the Gauss–Codazzi equations are both necessary and sufficient conditions on the first and second fundamental forms to determine a surface (up to rigid motions). As elaborated on in this question Gaussian curvature and mean curvature sufficient to characterize a surface?, this implies that the Gaussian curvature and mean curvature alone should generally be insufficient to determine a surface.

Can someone give me an explicit example of two different surfaces, parameterised over the same patch, which have the same Gaussian and mean curvature everywhere?

towerglove
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1 Answers1

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First, you need simple connectivity to get a global result for sufficiency. The equality of Gaussian and mean curvatures tells you that the shape operators $\text{I}^{-1}\text{II}$ have the same eigenvalues. This is not telling you that the first and second fundamental forms of the two surfaces are equal.

Here's the easiest example I can think of. Take the catenoid $$x(u,v)=(u\cos v,u\sin v,v)$$ and the helicoid $$y(u,v)=\left(\sqrt{1+u^2}\cos v,\sqrt{1+u^2}\sin v, \sinh^{-1}u\right).$$ In fact, these two surfaces are locally isometric (and the first fundamental forms agree with these parametrizations), so they have the same Gaussian curvature. They are both minimal, so the mean curvatures are both $0$. But the surfaces are not even locally congruent: One is ruled and the other is not.

P.S. You may be more advanced at this point, but you might want to check out my undergraduate differential geometry text, (freely) linked in my profile.

Ted Shifrin
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    Plot of both surfaces: https://www.math3d.org/WsnNs862C – Al.G. Oct 17 '23 at 20:32
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    Thanks for that, @Al.G. I actually assume that every differential geometry student knows well what the two surfaces look like, but it never hurts :) – Ted Shifrin Oct 17 '23 at 20:34
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    Darn, I came to post exactly this is example. In Ivey and Landsberg's "Cartan for Beginners" they give an account of a generalization of this phenomenon called the Bonnet family. In minimal surfaces, every minimal surface comes in a $1$-parameter family by a sort of complex rotation. The part you see as the surface is the real part of an immersion into $\mathbb{C}^3$. The relevant keyword is "Weierstrass-Enneper representation." – A. Thomas Yerger Oct 18 '23 at 01:08
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    @A.ThomasYerger Yes, this is also in Spivak and in most every treatment of minimal surfaces. But I had never before thought of it from the perspective raised in this question. – Ted Shifrin Oct 18 '23 at 01:10
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    Oh, to be clear, the Weierstrass representation appears in every treatment under the sun. The part that was new to me, was that you can talk about Bonnet deformations for manifolds that are not minimal surfaces. Also, my comments are hopefully for the benefit of the OP. I am 100% sure you know all this, you've been answering my differential geometry questions here for a decade. – A. Thomas Yerger Oct 18 '23 at 01:13
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    Oh, @A.ThomasYerger, I missed that aspect of your comment. So you’re no longer integrating a holomorphic (or anti-holomorphic) Gauss map? Interesting. No, I don’t think I’ve thought about this generalization. – Ted Shifrin Oct 18 '23 at 01:18
  • Dear Prof. Ted Shifrin, I wonder if there is a pair of such surfaces (same H and K) when one of them is not a surface of revolution... with such a (u,v) parameterization. – Narasimham Nov 08 '23 at 14:15
  • @Narasimham Sure. The helicoid and catenoid are in a one-parameter family of isometric minimal surfaces (coming from the Weierstrass parametrization, and rotating the complex Gauss map by angle $\theta$). Take a general member of that family. – Ted Shifrin Nov 08 '23 at 16:57