Are there any surfaces that contain both positive and negative Gaussian curvature?

Question

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

This post imported from StackExchange Mathematics at 2014-06-16 11:19 (UCT), posted by SE-user user29372

Answer 1

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).

This post imported from StackExchange Mathematics at 2014-06-16 11:19 (UCT), posted by SE-user Michiel

Answer 2

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.

This post imported from StackExchange Mathematics at 2014-06-16 11:19 (UCT), posted by SE-user Dylan Yott

Answer 3

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.

This post imported from StackExchange Mathematics at 2014-06-16 11:19 (UCT), posted by SE-user lavinia

Answer 4

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.

This post imported from StackExchange Mathematics at 2014-06-16 11:19 (UCT), posted by SE-user George E. Smith