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By curvature I mean intrinsic curvature, if there even is such a thing as extrinsic curvature of a torus.

If a torus has negative curvature in some points and positive in others shouldn't there be points with zero curvature since it is a continuous surface?

Puppet master
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2 Answers2

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If by intrinsic curvature you mean Gaussian curvature, then a torus has points with zero Gaussian curvature. Namely the points that are "at the top" or "the bottom" of the torus when the revolution axis is vertical.

mathcounterexamples.net
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    The points at the top and bottom are probably of positive curvature. – Lee Mosher Feb 26 '19 at 18:29
  • The Gaussian curvature is $R = \frac{\cos v}{a(c+a \cos v)}$. This vanishes for $v =\pm \pi/2$, i.e. the « top » or « bottom » points. See here for torus metrics computation. The Gaussian curvature is continuous. Therefore it has to vanish as it takes positive and negative values as the OP noticed. – mathcounterexamples.net Feb 26 '19 at 20:29
  • Ah, I missed the significance of the requirement that it be a torus of revolution around a vertical axis, assuming that the OP had such a requirement in mind. – Lee Mosher Feb 26 '19 at 21:28
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Your argument using continuity is essentially correct, as long as you augment it with the Gauss-Bonnet argument mentioned in the comment of @MichaelAlbanese. That argument tells you that the integral of sectional curvature with respect to area is equal to zero (and "sectional curvature" is what you ought to take as "intrinsic curvature").

As a consequence, either the sectional curvature is a constant equal to zero, or it has a nonzero value. Furthermore, if the sectional curvature has a nonzero value then it must have a value of opposite sign --- if, say, the value is positive at some point then the set of points where it has positive value has nonzero area, and the integral of sectional curvature over that set is positive, so there must exist points with negative sectional curvature so that the total integral of sectional curvature comes out to be zero. And finally, once one knows that the sectional curvature has both positive and negative values, your continuity argument applies.

Lee Mosher
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  • Ok that all makes sense. I get that when for example you integrate gaussian curvature over a sphere you get 4π which equals 2π times the Euler characteristic. But what exactly is that gaussian curvature for example of a sphere or a torus. It is supposed to be intrinsic but people always say it's 1/r squared which is not intrinsic to the sphere , or they use a basis. – Puppet master Feb 27 '19 at 12:10
  • Perhaps your real question is: What is the intrinsic definition of curvature? For that, you should go pick up a differential geometry book and read it. But, don't confuse the intrinsic concept of curvature with the outcome of a computational formula for the curvature in specific examples. Sure, the sphere of radius $r=1$ is not isometric to the sphere of radius $r=2$, and (among other reasons) that is because of an intrinsic computation that one can perform, with the outcome that their curvatures are unequal real numbers, curvature $1$ for radius $1$, and curvature $1/4$ for radius $2$. – Lee Mosher Feb 27 '19 at 14:44
  • The point is if you're talking about intrinsic things you should be consistent with that and not talk about a radius of curvature. – Puppet master Feb 27 '19 at 19:30