Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Would a torus shape be considered to have both positive and negative Gaussian curvature?

share|improve this question

migrated from physics.stackexchange.com Sep 16 '13 at 7:55

This question came from our site for active researchers, academics and students of physics.

add comment

4 Answers 4

Yes, there are many of those surfaces. In general saddle points will result in negative Gaussian curvature because the two principle radii of curvature are opposite in sign whereas peaks and holes will result in positive Gaussian curvature because their principle radii of curvature have the same sign (either both negative or both positive).

A few examples of surfaces with both positive and negative Gaussian curvature can be readily found in nature: a pear (usually), a peanut shell, a baseball bat (some negative Gaussian curvature at the handle).

A torus indeed has both positive and negative Gaussian curvature, because it has saddle points, a whole ring of them at the inside of the torus, and it has `peaks' at the outside. So it depends whether you are inside (in the hole) or outside the torus whether the Gaussian curvature is negative respectively positive (see e.g. this module on curvature).

share|improve this answer
add comment

Consider a torus $T$ in $\Bbb R^{3}$. By the following question, $T$ has a point of positive curvature.

Any compact embedded $2$-dimensional hypersurface in $\mathbb R^3$ has a point of positive Gaussian curvature

By the Gauss-Bonet theorem, $\int_T K \ dA = \chi(T)=2-2g=0$, where $g=1$ is the genus of the torus. Thus, since $T$ has a point (and therefore a neighborhood) of positive curvature, it must have a neighborhood of negative curvature in order to make the above integral vanish.

This might be overkill, but I like it.

share|improve this answer
add comment

It is a theorem of Hilbert that any closed smooth surface without boundary in 3 space must have a point of positive Gauss curvature. So any surface that also has a point of negative Gauss curvature will have both.

The integral of the Gauss curvature over a surface is 2pi times its Euler characteristic. The torus has Euler characteristic zero so it must always have regions of negative Gauss curvature to cancel the regions of positive curvature guaranteed in Hilbert's theorem. There is no way to warp it so that its Gauss curvature has only one sign. Similarly surfaces with more than one handle all have negative Euler characteristic and thus must always have both positive and negative Gauss curvature.

This is not true for the torus in 4 space. Here the torus can be given zero Gauss curvature everywhere.

share|improve this answer
add comment

Such surfaces are called "anticlastic" surfaces. A "synclastic" surface has the center of curvature of any plane intersection always on the same side of the surface.

It can be proved, that for any continuous anticlastic surface, at any point on the surface, there is at least one plane intersection direction where the curvature is zero.

Well it has to go to zero to get from positive curvature to negative curvature, if it is continuous.

As a result, anticlastic surfaces are susceptible to buckling, along a direction of near zero curvature. That's why egg shells, are built synclastic, and not anticlastic.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.