This is a simple terminology question. Let $S$ be a (let's say smooth) surface in $\mathbb{R}^3$, and $p$ a point on $S$. Suppose the principle curvatures $\kappa_1$ and $\kappa_2$ at $p$ are both negative. I am imagining $p$ sitting at the bottom of a dent in the surface. Is there an accepted term to describe such a point? The difficulty is that the Gaussian curvature $\kappa_1 \kappa_2$ is positive, so intrinsically $p$ is no different than if it were on a bump rather than a dent. I could make up my own term of course, e.g., valley point or cup point, but I'd rather follow convention.


  • $\begingroup$ I've never come across specific terminology for this situation. I think if $p$ is really at a "dent" in the surface, not only should the Gauss curvature be positive there, but it should decrease (roughly with the distance to $p$) and become negative near $p$. $\endgroup$ – Ryan Budney Sep 12 '10 at 1:10
  • $\begingroup$ The only term I could find is "elliptical" point (the signs of $\kappa_1$ and $\kappa_2$ are the same), so maybe you can just invent a new adjective to add to "elliptical". $\endgroup$ – J. M. is a poor mathematician Sep 12 '10 at 1:12
  • $\begingroup$ @Ryan: Yes, I see there are constraints nearby. @JM: Ah, elliptical! I didn't know that term. Your suggestion of modifying that term is perhaps my solution. Thanks! $\endgroup$ – Joseph O'Rourke Sep 12 '10 at 1:23

This question (does established terminology for a certain situation exist) has been answered in the comments (no). But for posterity--- ie, for students who may meet "principal curvature" terminology before they learn a lot of differential geometry, and who may run across this question on web searches for related topics--- I can't resist emphasizing a point hinted at in the question itself: having both principal curvatures negative at a point is not a well defined property of a point on a surface; the signs of the principal curvatures depend on the coordinate system you calculate them in.

For example, if you pick a point $p$ on a sphere in $\mathbb{R}^3$ and a coordinate patch around $p$ in which to compute principal curvatures, you find that both principal curvatures are positive at $p$ if the normal vector field associated to the coordinate patch points inward (ie, toward the center of the sphere), and both negative if the normal vector field associated to the patch points outward. The choice of normal is yours to make; it doesn't come with the sphere. (I suppose this point might be better made with a complicated surface that does not correspond to a well-known shape--- e.g. the surface formed by a tangle of ribbon after you have unwrapped a gift. If you punch a hole in this ribbon and replace the missing disc with a hemisphere, whether this is a "valley" or a "cup" is entirely up to you. In choosing the normal direction, you are deciding whether or not curves in the hemisphere are "curving" "toward" the normal or "away from" it.)

Now, this indeterminancy up to sign is the worst thing that can happen: if $S \subseteq \mathbb{R}^3$ is a smooth surface, and the principal curvatures at a point $p$ in $S$ are found in some coordinate system to be $\kappa_1$ and $\kappa_2$, then in any other coordinate system they will be either found to be exactly the same (that is, $\kappa_1$ and $\kappa_2$) or the same, but with opposite sign (that is, $-\kappa_1$ and $-\kappa_2$). How you prove this depends on how you define the principal curvatures, but arises e.g. from the fact that they are the maximum and minimum normal components of accelerations of unit-speed curves in the surface at $p$, and there are only two possible choices of unit normal at $p$, differing only by sign. So "both principal curvatures are negative" is a well-defined property of an oriented smooth surface in $\mathbb{R}^3$ (a property which, we learned in the comments, apparently does not have a well-established name; for what it's worth, I like "bump point").

This same discussion shows that the slightly weaker condition that "both principal curvatures at $p$ have the same sign" is well-defined independent of a choice of normal. (As pointed out in the comments, it does have a well-established name: such a $p$ is said to be an elliptic point.) What the above discussion does not show, but is nevertheless true, is that this weaker condition is even independent of how one "embeds" $S$ in any "ambient space" (what the asker meant when he referred to the Gaussian curvature at $p$ as being "intrinsic"). For more, open up any book on the differential geometry of surfaces and look for the sections around Gauss's Theorema Egregium.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.